Endogenous Quantal Response Equilibrium

Evan Friedmandagger

January 29 2019

Abstract

We endogenize the precision parameter of logit quantal response equilibrium (LQRE)

(McKelvey and Palfrey 1995 ) In the first stage of an endogenous quantal response equilibrium

(EQRE) each player chooses precision optimally subject to costs given correct beliefs over

other playersrsquo (second-stage) actions In the second stage playersrsquo actions form a heterogeneous

LQRE given the first-stage choices of precision We establish existence and basic results for

normal form games In the case where each player has only two actions we show that EQRE

satisfies a modified version of the regularity axioms (Goeree et al 2005 ) and provide analogues

to classic results for LQRE including those for equilibrium selection For generalized matching

pennies we show that the sets of EQRE and LQRE (ie indexed by their respective parameters)

are curves in the unit square that cross at finite points that we give explicitly and hence the

modelsrsquo predictions are generically distinct We apply EQRE to experimental data

Keywords quantal response equilibrium logit endogenous experimental economics

JEL Classification C44 C72 C92

I thank Alessandra Casella Mark Dean Navin Kartik RC Lim Felix Mauersberger Jeffrey Mensch Suanna Oh Tom Palfrey Michael Woodford and participants at the PSE Summer School on Bounded Rationality (2018) Stony Brook Game Theory Conference (2016) EconCon (Princeton 2016) The Columbia-NYU-Wharton Graduate Students Conference in Experimental Economics (2016) and seminars at Columbia (Microtheory Colloquium Applied Theory Colloquium Cognition and Decisions Lab and Experimental Lunch) for helpful comments All errors are my own

daggerEmail ekf2119columbiaedu

1 Introduction

In games played in the laboratory the empirical frequency of actions systematically deviates from

that predicted by Nash equilibrium Quantal response equilibrium (QRE) introduced by McKelvey

and Palfrey [1995] was developed as an alternative equilibrium theory that better explains what we

do observe In essence it assumes that each player makes mistakes in executing strategies though

tends to play better actions (by expected utility) with greater probability according to a ldquoquantal response functionrdquo That is QRE replaces the ldquobest responserdquo of Nash with ldquobetter responserdquo but otherwise maintains the equilibrium structure In a QRE players better respond to their beliefs over other playersrsquo actions and the probabilistic mistakes of all players are consistent with those

beliefs In general player irsquos quantal response function maps his vector of expected utilities (each element

corresponding to one of his actions) to a distribution over his actions In applications it is most common for the quantal response function to take the logit form (1) where parameter 2 [01)

is exogenous and controls the sensitivity to payoff differences1 The resulting model is called logit QRE (LQRE) and each value of gives specific equilibrium predictions Typically the best-fit is estimated in-sample from data and the resulting prediction is compared to that of other models The fit of LQRE is often very good for individual games but the parameter estimates are unstable

estimated in one game often makes poor predictions for another and it is usually easy to reject the restriction that is constant across games For instance McKelvey et al [2000] document this using 2 2 matching pennies-type games

Since LQRE performs well when is allowed to adjust several authors (see Related Literature

below) have suggested to endogenize such that its value is determined as part of the equilibrium In such a theory exogenous is replaced with endogenous

i ( p i

) for each player i where

captures the relevant features of the game p i is the opponentsrsquo (equilibrium) distribution of actions

and is a parameter of the process that determines i At this level of generality the theory nests

LQRE and it may or may not be that i is the same for all players The important feature is

that parameter is the same for all players and any heterogeneity across players is determined

endogenously In such a model a single value of predicts a different = ( 1

n

) for each

game so it may be that such a theory preserves LQRErsquos ability to explain data from individual games while making better out-of-sample predictions and better explaining the data pooled together from several games

Following this motivation we introduce endogenous quantal response equilibrium (EQRE) in i (vi ) optimally subject to costs where v

i

utilities (determined in equilibrium) and is a parameter of the cost function One interpretation

is that is the precision with which a player calculates or perceives the expected utility to each

= 0 implies uniformly mixing over all actions independently of payoffs and when = 1 the player simply best responds

which each player i chooses is irsquos vector of expected

1

1

action Errors in evaluating expected utility lead to costly mistakes but reducing errors is itself costly so optimal precision balances these tradeoffs EQRE applies this process to games via two

stages In the first stage each player i chooses i

In the second stage players take actions In

equilibrium the first-stage choice of i

is a best response to othersrsquo (second-stage) actions and

second-stage actions form an LQRE given the endogenous precision profile ( 1

n

) EQRE is defined for a given cost function c and parameter is a positive scaling factor These

are common to all players but since the benefit to precision depends on onersquos role in the game each player i will generically have a different

i

Hence EQRE provides a theory not only of how

depends on the game but also on the heterogeneity of i

s within a game Since LQRE assumes homogeneity the modelsrsquo predictions are generically distinct

For general normal form games we prove existence of equilibria provide conditions for the

equilibrium to be unique and show other basic properties such as that the limit points (as cost parameter 0) are Nash equilibria

In the case where each player has only two actions we have a much tighter characterization of each agentrsquos stochastic choice In particular we establish that endogenous quantal response satisfies a modified version of the regularity axioms (Goeree et al [2005]) This suggests that EQRE might behave similarly to LQRE in individual games and yet we show that there are are systematic

deviations from the LQRE predictions related to the endogenous heterogeneity of the i

s We also

establish that just like for LQRE the EQRE graph2 has a manifold structure that can be used to

select a unique Nash equilibrium Using the example of generalized matching pennies we provide nearly a full characterization of

EQRE We show that for any given cost function the EQRE set (ie indexed by ) is a curve in

the unit square that crosses the LQRE set (ie indexed by ) at finite points that we give explicitly

as a function of the gamersquos parameters Importantly the crossings and qualitative shape of the

EQRE set does not depend on the cost function Furthermore we establish bounds on the set of all possible EQRE (ie for any cost function) and this typically excludes a large measure of possible

datasets As the example of generalized matching pennies illustrates we dispel two false intuitions that

the reader might have First EQRE is not simply a reparametrization of LQRE far from it the

predictions overlap on a set of measure zero The reason is that for any given cost function the

asymmetry in player roles implies endogenous heterogeneity in i

s for almost every Second EQRE cannot ldquoexplain everythingrdquondasheven if we have the freedom to choose any cost function3

There are several difficulties in the analysis of EQRE First there is no closed form for optimal precision Second the optimal precision may not be unique and hence equilibria may require that players mix over different levels of precision We overcome these obstacles by establishing sufficient

2The EQRE graph is the graph of the correspondence which associates an equilibrium to every value of 2 (0 1) 3The fact that EQRE belongs to the class of (modified) regular QRE implies testable restrictions (Goeree et al

[2005]) but we find a much tighter bound than that implied by regularity alone

2

conditions for optimal precision to be unique and using implicit methods to analyze the behavior of i We compare EQRE and LQRE in experimental data from existing studies on generalized match-

ing pennies For most of the games individually EQRE outperforms LQRE for any cost function In terms of our original motivation however we conclude that there is little evidence that estimates of EQRE- are more stable across games than those of LQRE- and EQRE does not make better out-of-sample predictions than LQRE Though these are empirical findings we show that they may

not be too surprising on theoretical grounds since EQRE and LQRE both belong to the class of translation invariant4 regular QRE results from our related paper Friedman [2018] suggest that there are similarities in their across-game restrictions (ie for fixed parameter values)

This paper is organized as follows In the remainder of this section we discuss related literature Section 2 reviews QRE Section 3 gives the EQRE model and compares endogenous to exogenous quantal response Section 4 gives the equilibrium selection and related results Section 5 studies the

EQRE of generalized matching pennies Section 6 specializes results for EQRE using the family of power cost functions Section 7 applies EQRE to experimental datasets and Section 8 concludes

Related Literature The literature on QRE is vast and in several papers there is a discussion

about where comes fromndashincluding the original 1995 paper which discusses its potential relation-ship to models of learning To our knowledge the first explicit discussion of endogenizing comes from McKelvey and Palfrey [1996] in a section called ldquoEndogenizing rdquo McKelvey et al [2000] suggest endogenizing to account for the fact that estimates are unstable in their data which

they relate in their discussion to learning and heterogeneity Selten and Chmura [2008] who test the empirical performance of a number of ldquostationary conceptsrdquo including QRE remark ldquoIt would be

desirable to complement [QRE] by a theory that permits the computation of the noise parameter as a function of the payoffs of the gamerdquo5 A recent book on QRE Goeree et al [2016] has a one-page chapter called ldquoEndogenizing the Quantal Response Parameterrdquo within the section ldquoEpilogue Some Thoughts about Future Researchrdquo To our knowledge the only formal theory of endogenous

comes from McKelvey et al [1997] Our EQRE is the same as their endogenous rationality equi-librium (ERE) but they do not provide analysis so this paper can be regarded as an extension and

deepening of theirs In particular our approach based on implicit methods is novel This paper is most related to the literature on control costs (eg Stahl [1990]) which models

errors as resulting from trembles that can be reduced through costly effort Both Stahl [1990] and

Mattsson and Weibull [2002] derive the multinomial logit from such an optimization The approach

can be contrasted with the literature on rational inattention in games such as Yang [2015] and

4That is unaffected by adding the same constant to each payoff See for example Goeree et al [2005]5They continue ldquoExtended in this way [QRE] could become a much more powerful stationary concept Since

at the moment no theory of is available we have applied quantal response theory with interpreted as a natural constant which is the same for all gamesrdquo

3

Denti [2015] in which stochastic choice is driven by learning about some unknown state6 This paper is also related to other critical analyses of QRE such as Haile et al [2008] which

shows that structural QRE7 can rationalize any data without restrictions on errors Other ap-proaches for studying bounded rationality in games include Nagel [1995] Jehiel [2005] and Camerer et al [2004] There is also a relationship to Alaoui and Penta [2015] which endogenizes a parameter of a well known class of non-equilibrium models Finally Weibull et al [2007] provide some results for the stochastic choice of an agent who chooses logit precision optimally

2 Quantal response equilibrium

In this section we introduce the notation for normal form games that will be used throughout the

paper and review QRE

21 Normal form games

A finite normal form game = NA u is defined by a set of players N = 1 n action space

A = A1 An with A

i = ai1 a

iJ(i) such that each player i has J(i) possible pure actions P

(J = J(i) actions total) and a vector of payoff functions u = (u1 un) with ui A R

i

Let Ai be the set of probability measures on A

i

Elements of Ai are of the form p

i Ai R P

J(i)where aij ) = 1 and p

i

(aij ) 0 For simplicity set p

ij pi

(aij ) Define A = A1

j=1 pi(

An and A

i = k 6 A

k with typical elements p 2 A and p i 2 A

i

As standard extend =i P

payoff functions u = (u1 un) to be defined over A via ui

(p) = a2A p(a)ui(a)

As is standard in the literature on QRE we use additional notation for expected utilities Given

p i 2 A

i

player irsquos vector of expected utilities is given by ui

(p i

) = (ui1(p

i

) uiJ(i)(p

i

)) 2

RJ(i) where uij (p

i

) = ui

(aij p

i

) is the expected utility to action aij given behavior of the oppo-

nents We use vi = (v

i1 viJ(i)) 2 RJ(i) as shorthand for an arbitrary vector of expected utilities

That is vi is understood to satisfy v

i = ui

(p 0 i

) for some p 0 i

22 Quantal response equilibrium

Player irsquos quantal response function Qi = (Q

i1 QiJ(i)) RJ(i) A

i maps his vector of expected

utilities to a distribution over actions For any vi 2 RJ(i) component Q

ij (vi) gives the probability

assigned to action j Intuitively Qi allows for arbitrary probabilistic mistakes in taking actions

given the payoffs to each action 6It is well-known from Matejka and McKay [2015] that if the state is a vector of payoffs (ie the payoff to

each action) then the solution to the rational inattention problem with mutual information costs is a generalized multinomial logit that depends on the prior Such an interpretation does not readily extend to QRE since the vector of expected utilities is deterministic in equilibrium

7In a structural QRE quantal response results from choosing the action that maximizes expected utility plus a random error term If the errors are iid across actions then there are testable restrictions Regularity also imposes testable restrictions (Goeree et al [2005])

4

A quantal response equilibrium (QRE) is obtained when the distribution over all playersrsquo actions is consistent with their quantal response functions Letting Q = (Q1 Qn

) and u = (u1 un) QRE is a fixed point of the composite function Q u A A

Definition 1 Fix Q A QRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i) pij = Q

ij (ui(p i

))

QRE can explain any behavior without some basic restrictions on the quantal response function In practice the weakest restrictions imposed are given by the regularity axioms (Goeree et al [2005])

(A1) Interiority Qij (vi) 2 (0 1) for all j = 1 J(i) for all v

i 2 RJ(i)

(A2) Continuity Qij (vi) is a continuous and differentiable function for all v

i 2 RJ(i)

(A3) Monotonicity vij gt v

ik =) Qij (vi) gt Q

ik

(vi

) for all j k = 1 J(i) Q

ij (vi)(A4) Responsiveness v

ij gt 0 for all j = 1 J(i) and for all v

i 2 RJ(i)

These capture a sensitivity to payoff differences players take actions that give higher expected

utility with higher probability and the greater the payoff differences the greater the effect Goeree

et al [2005] show that these axioms are sufficient to impose testable restrictions on the data Often quantal response is derived through the ldquostructuralrdquo approach given v

i 2 RJ(i) the

probability player i takes action j is given by

Qij (vi) = Pv

ij + ij v

ik + ik 8k = 1 J(i)

where ik is an action-specific mean-zero random error Errors are assumed independent across

players but not necessarily across actions In the structural approach players maximize the real-ization of noisy expected utility where the exogenous noise is in perceiving or calculating expected

utility If the errors are iid across actions (and satisfy other mild conditions) structural QRE models

are regular (Goeree et al [2005]) and thus impose testable restrictions on the data This is an

important observation given the results of Haile et al [2008] who show that structural QRE can

rationalize the data from any one game without such restrictions on the errors For applications it is common to assume that the errors

ik are distributed according to an

extreme value ldquologitrdquo distribution with precision 2 [01)8 In this case the quantal response

function takes the logit form

v

ije Q

ij (vi ) (1)P

J(i) k=1 e vik

e ik8Specifically ik is distributed iid according to a mean-zero Gumbel distribution with CDF F (

ik

) = e

where 05772 is the Euler-Mascheroni constant The variance of ik is given by

6

2

2 which is strictly decreasing in and hence the interpretation of as precision

5

In this paper we take as our starting point the structure of logit quantal response (1) We do

this as a convenient parametrization for the task we outlined in the introduction to endogenize

the precision of errors given here by Logit is also extremely common in applications and fairly

representative of QRE models in that it is structural and satisfies regularity A logit QRE or LQRE

is defined in the obvious way

Definition 2 Fix An LQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i) pij = Q

ij (ui(p i

) )

3 Endogenous quantal response equilibrium

In an endogenous quantal response equilibrium (EQRE) is no longer exogenous Each player i chooses precision (or a distribution over different s) optimally subject to costs in a first stage i i

given beliefs over the opponentsrsquo second-stage actions In equilibrium the second-stage actions form a heterogeneous LQRE given the endogenously determined vector = ( 1 )

n

We begin by analyzing the first-stage optimal precision choice problem of such an agent holding

fixed beliefs Beliefs only enter the problem via the expected utilities they induce Hence given

vi 2 RJ(i) and cost parameter 2 (01) player i solves

J(i)X

i (vi ) argmax Q

ik

(vi

)vik c( ) (2)

2[01) k=1| z

b( vi

)

where Qik is given by (1) and cost function c [01) [01) satisfies c 0 (0) = 0 lim c

0 ( ) = 1

1 kand for all gt 0 c 0 ( ) gt 0 and c

00 ( ) gt 09 For example c could be a power function c( ) =

for k gt 1 For most of the paper we think of the cost function c as held fixed though for some

later results we consider the predictions obtained in the limit by a sequence of cost functions or the

set of predictions obtainable for some cost function Note that we define b(middot vi

) [01) [01)

as the benefit function for convenience This is merely player irsquos overall expected utility given the

expected utility to each of his actions and his quantal response Intuitively greater precision leads to better decisions (as we will show) but precision is itself

costly so optimal precision balances these tradeoffs Since the value of making better decisions depends on payoffs the benefit function depends on v

i

We favor an as if interpretation to the first-stage problem because the benefit function depends on a noiseless perception of v

i

even though the

agent will (and knows that he will) perceive vi with error in the second stage Such an interpretation

is essentially the same as those offered for models of rational inattention10 in which the benefit to 00 009It is fine if c (0) = 0 or lim c ( ) = 1

0+ 10Some classic references include Sims [1998] Sims [2003] Woodford [2009] and Caplin and Dean [2015]

6

an information structure depends on the stochastic choice it induces and the value to each action When v

ij = vik for all j k then b( v

i

) = the constant function that assigns utility for all levels of precision When v

ij 6= vik for some j k the following properties of b are easily verified

(see Appendix 91) P11 b(0 v

i

) = j vij and lim b( v

i

) = max vij

J(i) 1 j

b( vi

) vi

)2 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

b( vi

)

2b( v

i

) vi

)3 lim = 0 lim = 0 and 2b( lt 0 for sufficiently high

2

2 1 1

4 b( vi + e

J(i)) = b( vi

) + for all 2 R where eJ(i) = (1 1)

These are intuitive Property 1 when = 0 the agent uniformly randomizes so expects to

receive the unweighted average utility As 1 the agent takes (one of) the best action(s) with

probability approaching one Property 2 increasing will always lead to better decisions even on

the margin when = 0 so the first derivative is positive Unsurprisingly it is also bounded We also

establish that the second derivative is bounded but it cannot be signed in general which we will return to as it complicates the analysis Property 3 We establish some limiting behaviors for the

first two derivatives which are unsurprising given that b is strictly increasing and bounded Property

4 b shifts up by constant when the utility to each alternative increases by This is a simple

vi

)11consequence of the translation invariance of logit (1) which states that Qi

(vi + e

J(i)) = Qi

( Hence it is without loss to assume that v

ij gt 0 for all i and j which will simplify a number of arguments

It is easy to show that a solution to the first-stage problem exists If vij = v

ik for all j k then

the solution is trivial i = 0 which results in a uniform distribution over all actions Otherwise

any solution is strictly greater than 0 and satisfies a first-order condition This is summarized in

Theorem 1 along with basic properties of

i (

i

that follow from properties of b and c

i (Theorem 1 A solution to (2) exists ( 6= ) If vi

) is a solution it satisfies vi

) 2

2P

J(i) P

J(i)2 v

il v

il l=1 vile

l=1 vile c

0 ( ) = 0 (3)

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

1 inf (vi

) gt 0i

2 inf (vi

) and supi

i (vi ) are strictly decreasing in

3 lim 0

(inf vi

)) = 1 and lim 1

(sup vi

)) = 0 i ( i (

11See Goeree et al [2005] and Friedman [2018] for discussions of translation invariance and its implications

7

Proof See Appendix 92

One may expect equilibrium to be defined exactly as in Definition 2 but with Q ui

ij (ui ) Q

ij (macri (ui ))

replacing Qij (ui ) This is almost right except for the fact that the set of maximizers

i (vi )

is not a singleton in general (hence the use of inf and sup) This is because b(middot vi

) is not concave

for some vi

In other words the marginal benefit to may actually increase over portions of the

domain We provide an example of this in Appendix 95 which is due to Weibull et al [2007] Fortunately

i (vi ) is still well-behaved in the following sense

Lemma 1 Correspondence i RJ(i) (01) [01) is upper hemi-continuous

Proof See Appendix 92

i is upper hemi-continuous and not simply continuous because

i (vi ) may not be a singleton The next lemma establishes that

i (vi ) is a singleton however for both sufficiently low and

sufficiently high The result is proved by establishing that for extreme the solution to the

first-stage problem is in a region where the objective is locally concave

i (vi ) is a singleton for sufficiently low and sufficiently high Lemma 2 For all vi 2 RJ(i)

Proof See Appendix 92

In general equilibria may involve mixing over 2 i (middot ) Hence we define optimal mixed

precision by

i (vi )

to be any distribution over optimal selections which is also clearly optimal Finally the endogenous

i (vi )

( )vi

i i 2 [01) supp(

i

)

quantal response correspondence Qi (middot ) RJ(i)

Z 1

Ai is defined by

Qi

( )| i 2 (4)

i (vi ) Qi

(vi

)d 0

= ConvTi

(vi

)

where i (vi )

is the set of action frequencies from any optimal pure precision and Conv is the convex hull For all

Ti

(vi

) = Qi

(vi

)| 2

(vi

) an element p 0 i 2 Q

i (vi ) represents an action frequency induced by some optimal (possibly 0 P

J(i) 0 0 and

k=1 pmixed) precision and satisfies p = 1ij

ik 1 Q

to Definition 2 as a fixed point of the composite correspondence Q u A A Defining Q = (Q ) (suppressing in the notation) equilibrium is defined analogously

n

An EQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i)

Definition 3 Fix

pi 2 Q

i (ui(p i

) )

8

We establish several properties of Q Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Q i (middot ) is convex-valued and upper hemi-continuous on RJ(i)

3 For all i and j k = 1 J(i)

00 0for any pv

ij gt vik =) p

i 2 Q ij (vi )ij gt p

ik

Q

(vi

)4 If vij = max

k

vik

gt vil for some l ij gt 0 for sufficiently low and high

v

ij

Properties 1-3 are trivial12 and Property 4 is established in Appendix 93 Properties 1 and 2 are

technical Property 3 is generalized monotonicity (A3) that allows for possible non-uniqueness in

optimal quantal response Property 4 is a weaker form of responsiveness (A4) Note that it is weaker in two ways First it only holds for extreme which is required to ensure that Q is single-valued

Q

(vi

)ij 13(and differentiable) Second we are only able to show that gt 0 if v

ij = maxk

vik

gt vil

v

ij

The first two properties of Q imply existence

Theorem 2 Fix There exists an EQRE

Proof An EQRE is a fixed point of Q u Since Qi (middot ) is convex-valued and upper hemi-

continuous on RJ(i) and u is continuous on A Q u A A is convex-valued and upper hemi-continuous Because A is a compact subset of RJ Q u is also closed graph Summarizing Q u A A is a non-empty and convex-valued correspondence with a closed graph mapping

from a non-empty compact convex set to itself By Kakutanirsquos fixed point theorem Q u has a

fixed point

31 Binary actions

We now specialize the theory to the class of binary action games in which each player has two

actions Games in this class are widely used in experiments covering 2 2 games14 various voting

type games and more The theory could also be extended to an endogenous agent QRE (AQRE) (McKelvey and Palfrey [1998]) and used to analyze extensive form games in which there are two

12Since i Q

Since Qi RJ(i) [0 1)

i (vi ) = ConvT

i

(vi

) Q RJ(i) (0 1) [0 1) is upper hemi-continuous

is convex-valued 6= 6= i

also Because of the representation as Q

A

i is a continuous function andi

T

i RJ(i) (0 1) A

i is also upper hemi-continuous It is easy to check that Q i as the convexification of T

i

is also upper hemi-continuous Finally since v

ij gt v

ik =) Q

ij (vi ) gt Q

ik

(vi

) for any gt 0 it also holds when integrating over any s

13An increase in any vij will generically change

i (in either direction) Our result relies on showing that an increase in v

ij increases the product i vij if vij

implies an increase in Q ij if vij

= max

k

vik

(we conjecture that this is true even if vij 6= max

k

vik

) which = max

k

vik

(which is not necessarily true if vij 6= max

k

vik

)14Common ones used in experiments include generalized matching pennies the prisonerrsquos dilemma battle of the

sexes and at least several more

9

actions at every node such as the centipede game Importantly for our purposes with binary

actions it becomes much easier to analyze the solution to the first-stage problem (2) from which

we derive a much tighter characterization of stochastic choice e v

ij 6v

ijk eIf J(i) = 2 Q

ij (vi ) = = where vijk = v

ij vik is the signed payoff

e v

ij +e v

ik 1+e 6v

ijk P2 6v

i

edifference It is also easy to show that k=1 Qik

(vi

)vik = v

i 6v

i +const where v

i | vijk

|1+e

is the absolute payoff difference and the constant depends on the levels of vi = (v

ij vik

) Note that

vi 2 [0 1) by construction and that this is without loss We can now rewrite (2) as

fv

ie i ( v

i

) argmax vi c( ) (5)

fv

i1 + e2[01) | z

b( fv

i

)

Note that we have defined the net benefit function b(middot vi

) for this case Inspection reveals that it is simply the probability of taking the better action times the net benefit of doing so When

vi = 0 b( v

i

) = 0 for all When vi gt 0 it is easy to verify that

1 b(0 vi

) = 1 vi and lim b( v

i

) = vi

2 1

b( fv

i

)

2b( fv

i

)2 lim = 0 and lim = 0

2 1 1

b( fv

i

) fv

i

)3 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 0) for some 1 2 2 R++

2

For the most part these properties are intuitive specializations of the more general properties from

the previous section Unlike in the general case however property 3 states that b(middot vi

) and thus the first-stage objective (5) is strictly concave15 Due to this fact i ( v

i

) is a singleton and

satisfies

1 i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0

2 i ( v

i

) is strictly decreasing in if vi gt 0

3 lim i ( v

i

) = 1 and lim i ( v

i

) = 0 if vi gt 0

0 1

Since single-valued upper hemi-continuous correspondences are continuous functions we have the

following lemma as a direct corollary of Lemma 1

Lemma 3 [0 1) (0 1) [0 1) is continuous i

From Lemma 3 it follows that the binary action analogue of the endogenous quantal response

correspondence (4) is a continuous function Q (middot ) R2 (0 1) defined by ij

e (fv

i

)fv

ijk

Q vi

) = i

(6)ij ( (fv

i

)fv

ijk 1 + e i

3 vi

2b( 6vi) 6vi e

vi (e 1)15Taking derivatives 2 = lt 0 for v

i gt 0 (e vi +1)3

10

For reference we give the equilibrium definition and properties of Q for the binary action case

where J(i) = 2 for all i An EQRE is any p 2 A such that for all Definition 4 Fix i 2 1 n p

i1 = Q (ui

(p i

) ) i1

Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Qi ( middot ) is continuous and differentiable on R2

3 For all i and j k = 1 J(i)

=) Q vi

) gt Qij (

(ik vi )v

ij gt vik

Q

(vi

) Q

(vi

)ij ij4 = 0 and|

v

i1=v

i2 |v

i1v

ij v

ij gt 0=v

i2

and Lemma 3 Property 3 is monotonicity (A3) verbatim Property 4 is nearly responsiveness (A4) Q

(vi

)

6

Properties 1-3 are nearly immediate given the properties of general from the previous section Q

except that when (as opposed to being strictly positive) Property 4 follows ij = 0 = 0vi

v

ij

from properties of i presented later in this section

We note that the single-valuedness of i in the binary action case implies that all equilibria are

i s Also the single-valuedness of

is what makes Q a function as opposed to a correspondence Hence while existence is still ensured

by Theorem 2 Brouwerrsquos fixed point theorem would have been enough For the binary action case we seek to compare exogenous and endogenous quantal response

For this it is sufficient to track accuracy which we define as the probability of taking the better action under either model16

e fv

i

ai

( vi

) 1 + e fv

i

(fv

i

)fv

i

a i ( v

i

) i ()) = a

i

( vi

e i

(7)

pure in the sense that it is never optimal to mix over differenti

(fv

i

)fv

i1 + e i

We note that ai only depends on and is strictly increasing in the product v

i

In particular ai

is strictly increasing in vi for any gt 0 and strictly increasing in for any v

i gt 0 Thus to i While there is no closed form for

i understand quantal response we analyze properties of

much can be learned via implicit methods

Theorem 3

i (0 ) = 0 and i ((i) lim v

i

) = 0 fv

i

1

16Recalling that v

i is the absolute payoff difference between the two actions these are simply Qij and Q

ij if

v

ij gt v

ik

11

(ii) (middot ) is strictly single-peaked with macr() max ( vi

) and v() argmax ( vi

)i i i

fv

i

2(01) fv

i

2(01) i (fv

i

) i (fv

i

)(iii) fv

i fv

i

i (fv

i

) |0ltfv

i

ltfv() gt 0 and fv

i

|fv

i

=0 = 0 (iv) The product

|fv

i

gtfv() lt 0 ( v

i

)( vi

) vi is strictly increasing in v

i

i v

i = 0 and lim ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof See Appendix 92

Parts (i)-(iii) of the theorem state that i (middot ) is hump-shaped starting at 0 when v

i = 0 strictly increasing to its peak and then strictly decreasing to 0 as v

i 1 This is intuitive The good action is worth v

i more than the bad action so vi represents the cost of making a

mistake or equivalently the benefit of being accurate But for any given the agent is better able

to distinguish the good action from the bad and is thus more accurate when vi is high So when

vi is very low the benefit of accuracy is low and the agent optimally chooses low precision and

thus low accuracy and when vi is very high it is so easy to distinguish the good action from the

bad that the decision maker does not need high precision for high accuracy Parts (iv)-(vi) have implications for accuracy Part (iv) implies that a increases in v

i even i

though is non-monotonic in vi

Parts (v) and (vi) are more technical in nature with no obvious i

implication for decision problems though we use them extensively in the analysis of games in

subsequent sections Part (v) implies that the accuracy associated with peak precision macr() which macr()6v() e eobtains at v() is independent of a ( v() ) = macr()6macr = 092 Part (vi) states

i v()1+e 1+e

that the location of the peak v() increases in In Corollary 1 we compare accuracy under exogenous and endogenous quantal response This

follows immediately from properties of and ai

i

Corollary 1 (i) a

i (fv

i

) fv

i

(ii) a

|fv

i

=0 = 017

i (0 ) = ai

(0 ) = 1

a

i (fv

i

) )|fv

i

gt0 gt 0 and ai(fv

i

fv

i fv

i a

i

(lim lim

gt 0 and

i ( vi

) =fv

i

1 fv

i

1 vi

) = 1a 2

gt macr (iii) If () a ( vi

) lt ai

( vi

) for all vi

i

macr (iv) If 2 (0 ()) a ( vi

) gt ai

( vi

) for intermediate vi and a ( v

i

) lt ai

( vi

) i i

for extreme vi

Figure 1 graphically summarizes Theorem 3 and Corollary 1 The left panel of Figure 1 plots i (middot ) and

i (middot 0 ) as functions of v

i for lt 0 0 as well as the constant function and macr gt macr() gt

0 ( ) Since

i (are chosen such that peak precisions are ordered middot ) is hump-shaped

17Using the quotient rule take the derivative of (7) with respect to v

i and evaluate the expression at v

i = 0

i (0 ) = 0 and i (6vi )using that6vi

|6vi=0 = 0

12

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

action Errors in evaluating expected utility lead to costly mistakes but reducing errors is itself costly so optimal precision balances these tradeoffs EQRE applies this process to games via two

stages In the first stage each player i chooses i

In the second stage players take actions In

equilibrium the first-stage choice of i

is a best response to othersrsquo (second-stage) actions and

second-stage actions form an LQRE given the endogenous precision profile ( 1

n

) EQRE is defined for a given cost function c and parameter is a positive scaling factor These

are common to all players but since the benefit to precision depends on onersquos role in the game each player i will generically have a different

i

Hence EQRE provides a theory not only of how

depends on the game but also on the heterogeneity of i

s within a game Since LQRE assumes homogeneity the modelsrsquo predictions are generically distinct

For general normal form games we prove existence of equilibria provide conditions for the

equilibrium to be unique and show other basic properties such as that the limit points (as cost parameter 0) are Nash equilibria

In the case where each player has only two actions we have a much tighter characterization of each agentrsquos stochastic choice In particular we establish that endogenous quantal response satisfies a modified version of the regularity axioms (Goeree et al [2005]) This suggests that EQRE might behave similarly to LQRE in individual games and yet we show that there are are systematic

deviations from the LQRE predictions related to the endogenous heterogeneity of the i

s We also

establish that just like for LQRE the EQRE graph2 has a manifold structure that can be used to

select a unique Nash equilibrium Using the example of generalized matching pennies we provide nearly a full characterization of

EQRE We show that for any given cost function the EQRE set (ie indexed by ) is a curve in

the unit square that crosses the LQRE set (ie indexed by ) at finite points that we give explicitly

as a function of the gamersquos parameters Importantly the crossings and qualitative shape of the

EQRE set does not depend on the cost function Furthermore we establish bounds on the set of all possible EQRE (ie for any cost function) and this typically excludes a large measure of possible

datasets As the example of generalized matching pennies illustrates we dispel two false intuitions that

the reader might have First EQRE is not simply a reparametrization of LQRE far from it the

predictions overlap on a set of measure zero The reason is that for any given cost function the

asymmetry in player roles implies endogenous heterogeneity in i

s for almost every Second EQRE cannot ldquoexplain everythingrdquondasheven if we have the freedom to choose any cost function3

There are several difficulties in the analysis of EQRE First there is no closed form for optimal precision Second the optimal precision may not be unique and hence equilibria may require that players mix over different levels of precision We overcome these obstacles by establishing sufficient

2The EQRE graph is the graph of the correspondence which associates an equilibrium to every value of 2 (0 1) 3The fact that EQRE belongs to the class of (modified) regular QRE implies testable restrictions (Goeree et al

[2005]) but we find a much tighter bound than that implied by regularity alone

2

conditions for optimal precision to be unique and using implicit methods to analyze the behavior of i We compare EQRE and LQRE in experimental data from existing studies on generalized match-

ing pennies For most of the games individually EQRE outperforms LQRE for any cost function In terms of our original motivation however we conclude that there is little evidence that estimates of EQRE- are more stable across games than those of LQRE- and EQRE does not make better out-of-sample predictions than LQRE Though these are empirical findings we show that they may

not be too surprising on theoretical grounds since EQRE and LQRE both belong to the class of translation invariant4 regular QRE results from our related paper Friedman [2018] suggest that there are similarities in their across-game restrictions (ie for fixed parameter values)

This paper is organized as follows In the remainder of this section we discuss related literature Section 2 reviews QRE Section 3 gives the EQRE model and compares endogenous to exogenous quantal response Section 4 gives the equilibrium selection and related results Section 5 studies the

EQRE of generalized matching pennies Section 6 specializes results for EQRE using the family of power cost functions Section 7 applies EQRE to experimental datasets and Section 8 concludes

Related Literature The literature on QRE is vast and in several papers there is a discussion

about where comes fromndashincluding the original 1995 paper which discusses its potential relation-ship to models of learning To our knowledge the first explicit discussion of endogenizing comes from McKelvey and Palfrey [1996] in a section called ldquoEndogenizing rdquo McKelvey et al [2000] suggest endogenizing to account for the fact that estimates are unstable in their data which

they relate in their discussion to learning and heterogeneity Selten and Chmura [2008] who test the empirical performance of a number of ldquostationary conceptsrdquo including QRE remark ldquoIt would be

desirable to complement [QRE] by a theory that permits the computation of the noise parameter as a function of the payoffs of the gamerdquo5 A recent book on QRE Goeree et al [2016] has a one-page chapter called ldquoEndogenizing the Quantal Response Parameterrdquo within the section ldquoEpilogue Some Thoughts about Future Researchrdquo To our knowledge the only formal theory of endogenous

comes from McKelvey et al [1997] Our EQRE is the same as their endogenous rationality equi-librium (ERE) but they do not provide analysis so this paper can be regarded as an extension and

deepening of theirs In particular our approach based on implicit methods is novel This paper is most related to the literature on control costs (eg Stahl [1990]) which models

errors as resulting from trembles that can be reduced through costly effort Both Stahl [1990] and

Mattsson and Weibull [2002] derive the multinomial logit from such an optimization The approach

can be contrasted with the literature on rational inattention in games such as Yang [2015] and

4That is unaffected by adding the same constant to each payoff See for example Goeree et al [2005]5They continue ldquoExtended in this way [QRE] could become a much more powerful stationary concept Since

at the moment no theory of is available we have applied quantal response theory with interpreted as a natural constant which is the same for all gamesrdquo

3

Denti [2015] in which stochastic choice is driven by learning about some unknown state6 This paper is also related to other critical analyses of QRE such as Haile et al [2008] which

shows that structural QRE7 can rationalize any data without restrictions on errors Other ap-proaches for studying bounded rationality in games include Nagel [1995] Jehiel [2005] and Camerer et al [2004] There is also a relationship to Alaoui and Penta [2015] which endogenizes a parameter of a well known class of non-equilibrium models Finally Weibull et al [2007] provide some results for the stochastic choice of an agent who chooses logit precision optimally

2 Quantal response equilibrium

In this section we introduce the notation for normal form games that will be used throughout the

paper and review QRE

21 Normal form games

A finite normal form game = NA u is defined by a set of players N = 1 n action space

A = A1 An with A

i = ai1 a

iJ(i) such that each player i has J(i) possible pure actions P

(J = J(i) actions total) and a vector of payoff functions u = (u1 un) with ui A R

i

Let Ai be the set of probability measures on A

i

Elements of Ai are of the form p

i Ai R P

J(i)where aij ) = 1 and p

i

(aij ) 0 For simplicity set p

ij pi

(aij ) Define A = A1

j=1 pi(

An and A

i = k 6 A

k with typical elements p 2 A and p i 2 A

i

As standard extend =i P

payoff functions u = (u1 un) to be defined over A via ui

(p) = a2A p(a)ui(a)

As is standard in the literature on QRE we use additional notation for expected utilities Given

p i 2 A

i

player irsquos vector of expected utilities is given by ui

(p i

) = (ui1(p

i

) uiJ(i)(p

i

)) 2

RJ(i) where uij (p

i

) = ui

(aij p

i

) is the expected utility to action aij given behavior of the oppo-

nents We use vi = (v

i1 viJ(i)) 2 RJ(i) as shorthand for an arbitrary vector of expected utilities

That is vi is understood to satisfy v

i = ui

(p 0 i

) for some p 0 i

22 Quantal response equilibrium

Player irsquos quantal response function Qi = (Q

i1 QiJ(i)) RJ(i) A

i maps his vector of expected

utilities to a distribution over actions For any vi 2 RJ(i) component Q

ij (vi) gives the probability

assigned to action j Intuitively Qi allows for arbitrary probabilistic mistakes in taking actions

given the payoffs to each action 6It is well-known from Matejka and McKay [2015] that if the state is a vector of payoffs (ie the payoff to

each action) then the solution to the rational inattention problem with mutual information costs is a generalized multinomial logit that depends on the prior Such an interpretation does not readily extend to QRE since the vector of expected utilities is deterministic in equilibrium

7In a structural QRE quantal response results from choosing the action that maximizes expected utility plus a random error term If the errors are iid across actions then there are testable restrictions Regularity also imposes testable restrictions (Goeree et al [2005])

4

A quantal response equilibrium (QRE) is obtained when the distribution over all playersrsquo actions is consistent with their quantal response functions Letting Q = (Q1 Qn

) and u = (u1 un) QRE is a fixed point of the composite function Q u A A

Definition 1 Fix Q A QRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i) pij = Q

ij (ui(p i

))

QRE can explain any behavior without some basic restrictions on the quantal response function In practice the weakest restrictions imposed are given by the regularity axioms (Goeree et al [2005])

(A1) Interiority Qij (vi) 2 (0 1) for all j = 1 J(i) for all v

i 2 RJ(i)

(A2) Continuity Qij (vi) is a continuous and differentiable function for all v

i 2 RJ(i)

(A3) Monotonicity vij gt v

ik =) Qij (vi) gt Q

ik

(vi

) for all j k = 1 J(i) Q

ij (vi)(A4) Responsiveness v

ij gt 0 for all j = 1 J(i) and for all v

i 2 RJ(i)

These capture a sensitivity to payoff differences players take actions that give higher expected

utility with higher probability and the greater the payoff differences the greater the effect Goeree

et al [2005] show that these axioms are sufficient to impose testable restrictions on the data Often quantal response is derived through the ldquostructuralrdquo approach given v

i 2 RJ(i) the

probability player i takes action j is given by

Qij (vi) = Pv

ij + ij v

ik + ik 8k = 1 J(i)

where ik is an action-specific mean-zero random error Errors are assumed independent across

players but not necessarily across actions In the structural approach players maximize the real-ization of noisy expected utility where the exogenous noise is in perceiving or calculating expected

utility If the errors are iid across actions (and satisfy other mild conditions) structural QRE models

are regular (Goeree et al [2005]) and thus impose testable restrictions on the data This is an

important observation given the results of Haile et al [2008] who show that structural QRE can

rationalize the data from any one game without such restrictions on the errors For applications it is common to assume that the errors

ik are distributed according to an

extreme value ldquologitrdquo distribution with precision 2 [01)8 In this case the quantal response

function takes the logit form

v

ije Q

ij (vi ) (1)P

J(i) k=1 e vik

e ik8Specifically ik is distributed iid according to a mean-zero Gumbel distribution with CDF F (

ik

) = e

where 05772 is the Euler-Mascheroni constant The variance of ik is given by

6

2

2 which is strictly decreasing in and hence the interpretation of as precision

5

In this paper we take as our starting point the structure of logit quantal response (1) We do

this as a convenient parametrization for the task we outlined in the introduction to endogenize

the precision of errors given here by Logit is also extremely common in applications and fairly

representative of QRE models in that it is structural and satisfies regularity A logit QRE or LQRE

is defined in the obvious way

Definition 2 Fix An LQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i) pij = Q

ij (ui(p i

) )

3 Endogenous quantal response equilibrium

In an endogenous quantal response equilibrium (EQRE) is no longer exogenous Each player i chooses precision (or a distribution over different s) optimally subject to costs in a first stage i i

given beliefs over the opponentsrsquo second-stage actions In equilibrium the second-stage actions form a heterogeneous LQRE given the endogenously determined vector = ( 1 )

n

We begin by analyzing the first-stage optimal precision choice problem of such an agent holding

fixed beliefs Beliefs only enter the problem via the expected utilities they induce Hence given

vi 2 RJ(i) and cost parameter 2 (01) player i solves

J(i)X

i (vi ) argmax Q

ik

(vi

)vik c( ) (2)

2[01) k=1| z

b( vi

)

where Qik is given by (1) and cost function c [01) [01) satisfies c 0 (0) = 0 lim c

0 ( ) = 1

1 kand for all gt 0 c 0 ( ) gt 0 and c

00 ( ) gt 09 For example c could be a power function c( ) =

for k gt 1 For most of the paper we think of the cost function c as held fixed though for some

later results we consider the predictions obtained in the limit by a sequence of cost functions or the

set of predictions obtainable for some cost function Note that we define b(middot vi

) [01) [01)

as the benefit function for convenience This is merely player irsquos overall expected utility given the

expected utility to each of his actions and his quantal response Intuitively greater precision leads to better decisions (as we will show) but precision is itself

costly so optimal precision balances these tradeoffs Since the value of making better decisions depends on payoffs the benefit function depends on v

i

We favor an as if interpretation to the first-stage problem because the benefit function depends on a noiseless perception of v

i

even though the

agent will (and knows that he will) perceive vi with error in the second stage Such an interpretation

is essentially the same as those offered for models of rational inattention10 in which the benefit to 00 009It is fine if c (0) = 0 or lim c ( ) = 1

0+ 10Some classic references include Sims [1998] Sims [2003] Woodford [2009] and Caplin and Dean [2015]

6

an information structure depends on the stochastic choice it induces and the value to each action When v

ij = vik for all j k then b( v

i

) = the constant function that assigns utility for all levels of precision When v

ij 6= vik for some j k the following properties of b are easily verified

(see Appendix 91) P11 b(0 v

i

) = j vij and lim b( v

i

) = max vij

J(i) 1 j

b( vi

) vi

)2 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

b( vi

)

2b( v

i

) vi

)3 lim = 0 lim = 0 and 2b( lt 0 for sufficiently high

2

2 1 1

4 b( vi + e

J(i)) = b( vi

) + for all 2 R where eJ(i) = (1 1)

These are intuitive Property 1 when = 0 the agent uniformly randomizes so expects to

receive the unweighted average utility As 1 the agent takes (one of) the best action(s) with

probability approaching one Property 2 increasing will always lead to better decisions even on

the margin when = 0 so the first derivative is positive Unsurprisingly it is also bounded We also

establish that the second derivative is bounded but it cannot be signed in general which we will return to as it complicates the analysis Property 3 We establish some limiting behaviors for the

first two derivatives which are unsurprising given that b is strictly increasing and bounded Property

4 b shifts up by constant when the utility to each alternative increases by This is a simple

vi

)11consequence of the translation invariance of logit (1) which states that Qi

(vi + e

J(i)) = Qi

( Hence it is without loss to assume that v

ij gt 0 for all i and j which will simplify a number of arguments

It is easy to show that a solution to the first-stage problem exists If vij = v

ik for all j k then

the solution is trivial i = 0 which results in a uniform distribution over all actions Otherwise

any solution is strictly greater than 0 and satisfies a first-order condition This is summarized in

Theorem 1 along with basic properties of

i (

i

that follow from properties of b and c

i (Theorem 1 A solution to (2) exists ( 6= ) If vi

) is a solution it satisfies vi

) 2

2P

J(i) P

J(i)2 v

il v

il l=1 vile

l=1 vile c

0 ( ) = 0 (3)

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

1 inf (vi

) gt 0i

2 inf (vi

) and supi

i (vi ) are strictly decreasing in

3 lim 0

(inf vi

)) = 1 and lim 1

(sup vi

)) = 0 i ( i (

11See Goeree et al [2005] and Friedman [2018] for discussions of translation invariance and its implications

7

Proof See Appendix 92

One may expect equilibrium to be defined exactly as in Definition 2 but with Q ui

ij (ui ) Q

ij (macri (ui ))

replacing Qij (ui ) This is almost right except for the fact that the set of maximizers

i (vi )

is not a singleton in general (hence the use of inf and sup) This is because b(middot vi

) is not concave

for some vi

In other words the marginal benefit to may actually increase over portions of the

domain We provide an example of this in Appendix 95 which is due to Weibull et al [2007] Fortunately

i (vi ) is still well-behaved in the following sense

Lemma 1 Correspondence i RJ(i) (01) [01) is upper hemi-continuous

Proof See Appendix 92

i is upper hemi-continuous and not simply continuous because

i (vi ) may not be a singleton The next lemma establishes that

i (vi ) is a singleton however for both sufficiently low and

sufficiently high The result is proved by establishing that for extreme the solution to the

first-stage problem is in a region where the objective is locally concave

i (vi ) is a singleton for sufficiently low and sufficiently high Lemma 2 For all vi 2 RJ(i)

Proof See Appendix 92

In general equilibria may involve mixing over 2 i (middot ) Hence we define optimal mixed

precision by

i (vi )

to be any distribution over optimal selections which is also clearly optimal Finally the endogenous

i (vi )

( )vi

i i 2 [01) supp(

i

)

quantal response correspondence Qi (middot ) RJ(i)

Z 1

Ai is defined by

Qi

( )| i 2 (4)

i (vi ) Qi

(vi

)d 0

= ConvTi

(vi

)

where i (vi )

is the set of action frequencies from any optimal pure precision and Conv is the convex hull For all

Ti

(vi

) = Qi

(vi

)| 2

(vi

) an element p 0 i 2 Q

i (vi ) represents an action frequency induced by some optimal (possibly 0 P

J(i) 0 0 and

k=1 pmixed) precision and satisfies p = 1ij

ik 1 Q

to Definition 2 as a fixed point of the composite correspondence Q u A A Defining Q = (Q ) (suppressing in the notation) equilibrium is defined analogously

n

An EQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i)

Definition 3 Fix

pi 2 Q

i (ui(p i

) )

8

We establish several properties of Q Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Q i (middot ) is convex-valued and upper hemi-continuous on RJ(i)

3 For all i and j k = 1 J(i)

00 0for any pv

ij gt vik =) p

i 2 Q ij (vi )ij gt p

ik

Q

(vi

)4 If vij = max

k

vik

gt vil for some l ij gt 0 for sufficiently low and high

v

ij

Properties 1-3 are trivial12 and Property 4 is established in Appendix 93 Properties 1 and 2 are

technical Property 3 is generalized monotonicity (A3) that allows for possible non-uniqueness in

optimal quantal response Property 4 is a weaker form of responsiveness (A4) Note that it is weaker in two ways First it only holds for extreme which is required to ensure that Q is single-valued

Q

(vi

)ij 13(and differentiable) Second we are only able to show that gt 0 if v

ij = maxk

vik

gt vil

v

ij

The first two properties of Q imply existence

Theorem 2 Fix There exists an EQRE

Proof An EQRE is a fixed point of Q u Since Qi (middot ) is convex-valued and upper hemi-

continuous on RJ(i) and u is continuous on A Q u A A is convex-valued and upper hemi-continuous Because A is a compact subset of RJ Q u is also closed graph Summarizing Q u A A is a non-empty and convex-valued correspondence with a closed graph mapping

from a non-empty compact convex set to itself By Kakutanirsquos fixed point theorem Q u has a

fixed point

31 Binary actions

We now specialize the theory to the class of binary action games in which each player has two

actions Games in this class are widely used in experiments covering 2 2 games14 various voting

type games and more The theory could also be extended to an endogenous agent QRE (AQRE) (McKelvey and Palfrey [1998]) and used to analyze extensive form games in which there are two

12Since i Q

Since Qi RJ(i) [0 1)

i (vi ) = ConvT

i

(vi

) Q RJ(i) (0 1) [0 1) is upper hemi-continuous

is convex-valued 6= 6= i

also Because of the representation as Q

A

i is a continuous function andi

T

i RJ(i) (0 1) A

i is also upper hemi-continuous It is easy to check that Q i as the convexification of T

i

is also upper hemi-continuous Finally since v

ij gt v

ik =) Q

ij (vi ) gt Q

ik

(vi

) for any gt 0 it also holds when integrating over any s

13An increase in any vij will generically change

i (in either direction) Our result relies on showing that an increase in v

ij increases the product i vij if vij

implies an increase in Q ij if vij

= max

k

vik

(we conjecture that this is true even if vij 6= max

k

vik

) which = max

k

vik

(which is not necessarily true if vij 6= max

k

vik

)14Common ones used in experiments include generalized matching pennies the prisonerrsquos dilemma battle of the

sexes and at least several more

9

actions at every node such as the centipede game Importantly for our purposes with binary

actions it becomes much easier to analyze the solution to the first-stage problem (2) from which

we derive a much tighter characterization of stochastic choice e v

ij 6v

ijk eIf J(i) = 2 Q

ij (vi ) = = where vijk = v

ij vik is the signed payoff

e v

ij +e v

ik 1+e 6v

ijk P2 6v

i

edifference It is also easy to show that k=1 Qik

(vi

)vik = v

i 6v

i +const where v

i | vijk

|1+e

is the absolute payoff difference and the constant depends on the levels of vi = (v

ij vik

) Note that

vi 2 [0 1) by construction and that this is without loss We can now rewrite (2) as

fv

ie i ( v

i

) argmax vi c( ) (5)

fv

i1 + e2[01) | z

b( fv

i

)

Note that we have defined the net benefit function b(middot vi

) for this case Inspection reveals that it is simply the probability of taking the better action times the net benefit of doing so When

vi = 0 b( v

i

) = 0 for all When vi gt 0 it is easy to verify that

1 b(0 vi

) = 1 vi and lim b( v

i

) = vi

2 1

b( fv

i

)

2b( fv

i

)2 lim = 0 and lim = 0

2 1 1

b( fv

i

) fv

i

)3 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 0) for some 1 2 2 R++

2

For the most part these properties are intuitive specializations of the more general properties from

the previous section Unlike in the general case however property 3 states that b(middot vi

) and thus the first-stage objective (5) is strictly concave15 Due to this fact i ( v

i

) is a singleton and

satisfies

1 i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0

2 i ( v

i

) is strictly decreasing in if vi gt 0

3 lim i ( v

i

) = 1 and lim i ( v

i

) = 0 if vi gt 0

0 1

Since single-valued upper hemi-continuous correspondences are continuous functions we have the

following lemma as a direct corollary of Lemma 1

Lemma 3 [0 1) (0 1) [0 1) is continuous i

From Lemma 3 it follows that the binary action analogue of the endogenous quantal response

correspondence (4) is a continuous function Q (middot ) R2 (0 1) defined by ij

e (fv

i

)fv

ijk

Q vi

) = i

(6)ij ( (fv

i

)fv

ijk 1 + e i

3 vi

2b( 6vi) 6vi e

vi (e 1)15Taking derivatives 2 = lt 0 for v

i gt 0 (e vi +1)3

10

For reference we give the equilibrium definition and properties of Q for the binary action case

where J(i) = 2 for all i An EQRE is any p 2 A such that for all Definition 4 Fix i 2 1 n p

i1 = Q (ui

(p i

) ) i1

Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Qi ( middot ) is continuous and differentiable on R2

3 For all i and j k = 1 J(i)

=) Q vi

) gt Qij (

(ik vi )v

ij gt vik

Q

(vi

) Q

(vi

)ij ij4 = 0 and|

v

i1=v

i2 |v

i1v

ij v

ij gt 0=v

i2

and Lemma 3 Property 3 is monotonicity (A3) verbatim Property 4 is nearly responsiveness (A4) Q

(vi

)

6

Properties 1-3 are nearly immediate given the properties of general from the previous section Q

except that when (as opposed to being strictly positive) Property 4 follows ij = 0 = 0vi

v

ij

from properties of i presented later in this section

We note that the single-valuedness of i in the binary action case implies that all equilibria are

i s Also the single-valuedness of

is what makes Q a function as opposed to a correspondence Hence while existence is still ensured

by Theorem 2 Brouwerrsquos fixed point theorem would have been enough For the binary action case we seek to compare exogenous and endogenous quantal response

For this it is sufficient to track accuracy which we define as the probability of taking the better action under either model16

e fv

i

ai

( vi

) 1 + e fv

i

(fv

i

)fv

i

a i ( v

i

) i ()) = a

i

( vi

e i

(7)

pure in the sense that it is never optimal to mix over differenti

(fv

i

)fv

i1 + e i

We note that ai only depends on and is strictly increasing in the product v

i

In particular ai

is strictly increasing in vi for any gt 0 and strictly increasing in for any v

i gt 0 Thus to i While there is no closed form for

i understand quantal response we analyze properties of

much can be learned via implicit methods

Theorem 3

i (0 ) = 0 and i ((i) lim v

i

) = 0 fv

i

1

16Recalling that v

i is the absolute payoff difference between the two actions these are simply Qij and Q

ij if

v

ij gt v

ik

11

(ii) (middot ) is strictly single-peaked with macr() max ( vi

) and v() argmax ( vi

)i i i

fv

i

2(01) fv

i

2(01) i (fv

i

) i (fv

i

)(iii) fv

i fv

i

i (fv

i

) |0ltfv

i

ltfv() gt 0 and fv

i

|fv

i

=0 = 0 (iv) The product

|fv

i

gtfv() lt 0 ( v

i

)( vi

) vi is strictly increasing in v

i

i v

i = 0 and lim ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof See Appendix 92

Parts (i)-(iii) of the theorem state that i (middot ) is hump-shaped starting at 0 when v

i = 0 strictly increasing to its peak and then strictly decreasing to 0 as v

i 1 This is intuitive The good action is worth v

i more than the bad action so vi represents the cost of making a

mistake or equivalently the benefit of being accurate But for any given the agent is better able

to distinguish the good action from the bad and is thus more accurate when vi is high So when

vi is very low the benefit of accuracy is low and the agent optimally chooses low precision and

thus low accuracy and when vi is very high it is so easy to distinguish the good action from the

bad that the decision maker does not need high precision for high accuracy Parts (iv)-(vi) have implications for accuracy Part (iv) implies that a increases in v

i even i

though is non-monotonic in vi

Parts (v) and (vi) are more technical in nature with no obvious i

implication for decision problems though we use them extensively in the analysis of games in

subsequent sections Part (v) implies that the accuracy associated with peak precision macr() which macr()6v() e eobtains at v() is independent of a ( v() ) = macr()6macr = 092 Part (vi) states

i v()1+e 1+e

that the location of the peak v() increases in In Corollary 1 we compare accuracy under exogenous and endogenous quantal response This

follows immediately from properties of and ai

i

Corollary 1 (i) a

i (fv

i

) fv

i

(ii) a

|fv

i

=0 = 017

i (0 ) = ai

(0 ) = 1

a

i (fv

i

) )|fv

i

gt0 gt 0 and ai(fv

i

fv

i fv

i a

i

(lim lim

gt 0 and

i ( vi

) =fv

i

1 fv

i

1 vi

) = 1a 2

gt macr (iii) If () a ( vi

) lt ai

( vi

) for all vi

i

macr (iv) If 2 (0 ()) a ( vi

) gt ai

( vi

) for intermediate vi and a ( v

i

) lt ai

( vi

) i i

for extreme vi

Figure 1 graphically summarizes Theorem 3 and Corollary 1 The left panel of Figure 1 plots i (middot ) and

i (middot 0 ) as functions of v

i for lt 0 0 as well as the constant function and macr gt macr() gt

0 ( ) Since

i (are chosen such that peak precisions are ordered middot ) is hump-shaped

17Using the quotient rule take the derivative of (7) with respect to v

i and evaluate the expression at v

i = 0

i (0 ) = 0 and i (6vi )using that6vi

|6vi=0 = 0

12

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

conditions for optimal precision to be unique and using implicit methods to analyze the behavior of i We compare EQRE and LQRE in experimental data from existing studies on generalized match-

ing pennies For most of the games individually EQRE outperforms LQRE for any cost function In terms of our original motivation however we conclude that there is little evidence that estimates of EQRE- are more stable across games than those of LQRE- and EQRE does not make better out-of-sample predictions than LQRE Though these are empirical findings we show that they may

not be too surprising on theoretical grounds since EQRE and LQRE both belong to the class of translation invariant4 regular QRE results from our related paper Friedman [2018] suggest that there are similarities in their across-game restrictions (ie for fixed parameter values)

This paper is organized as follows In the remainder of this section we discuss related literature Section 2 reviews QRE Section 3 gives the EQRE model and compares endogenous to exogenous quantal response Section 4 gives the equilibrium selection and related results Section 5 studies the

EQRE of generalized matching pennies Section 6 specializes results for EQRE using the family of power cost functions Section 7 applies EQRE to experimental datasets and Section 8 concludes

Related Literature The literature on QRE is vast and in several papers there is a discussion

about where comes fromndashincluding the original 1995 paper which discusses its potential relation-ship to models of learning To our knowledge the first explicit discussion of endogenizing comes from McKelvey and Palfrey [1996] in a section called ldquoEndogenizing rdquo McKelvey et al [2000] suggest endogenizing to account for the fact that estimates are unstable in their data which

they relate in their discussion to learning and heterogeneity Selten and Chmura [2008] who test the empirical performance of a number of ldquostationary conceptsrdquo including QRE remark ldquoIt would be

desirable to complement [QRE] by a theory that permits the computation of the noise parameter as a function of the payoffs of the gamerdquo5 A recent book on QRE Goeree et al [2016] has a one-page chapter called ldquoEndogenizing the Quantal Response Parameterrdquo within the section ldquoEpilogue Some Thoughts about Future Researchrdquo To our knowledge the only formal theory of endogenous

comes from McKelvey et al [1997] Our EQRE is the same as their endogenous rationality equi-librium (ERE) but they do not provide analysis so this paper can be regarded as an extension and

deepening of theirs In particular our approach based on implicit methods is novel This paper is most related to the literature on control costs (eg Stahl [1990]) which models

errors as resulting from trembles that can be reduced through costly effort Both Stahl [1990] and

Mattsson and Weibull [2002] derive the multinomial logit from such an optimization The approach

can be contrasted with the literature on rational inattention in games such as Yang [2015] and

4That is unaffected by adding the same constant to each payoff See for example Goeree et al [2005]5They continue ldquoExtended in this way [QRE] could become a much more powerful stationary concept Since

at the moment no theory of is available we have applied quantal response theory with interpreted as a natural constant which is the same for all gamesrdquo

3

Denti [2015] in which stochastic choice is driven by learning about some unknown state6 This paper is also related to other critical analyses of QRE such as Haile et al [2008] which

shows that structural QRE7 can rationalize any data without restrictions on errors Other ap-proaches for studying bounded rationality in games include Nagel [1995] Jehiel [2005] and Camerer et al [2004] There is also a relationship to Alaoui and Penta [2015] which endogenizes a parameter of a well known class of non-equilibrium models Finally Weibull et al [2007] provide some results for the stochastic choice of an agent who chooses logit precision optimally

2 Quantal response equilibrium

In this section we introduce the notation for normal form games that will be used throughout the

paper and review QRE

21 Normal form games

A finite normal form game = NA u is defined by a set of players N = 1 n action space

A = A1 An with A

i = ai1 a

iJ(i) such that each player i has J(i) possible pure actions P

(J = J(i) actions total) and a vector of payoff functions u = (u1 un) with ui A R

i

Let Ai be the set of probability measures on A

i

Elements of Ai are of the form p

i Ai R P

J(i)where aij ) = 1 and p

i

(aij ) 0 For simplicity set p

ij pi

(aij ) Define A = A1

j=1 pi(

An and A

i = k 6 A

k with typical elements p 2 A and p i 2 A

i

As standard extend =i P

payoff functions u = (u1 un) to be defined over A via ui

(p) = a2A p(a)ui(a)

As is standard in the literature on QRE we use additional notation for expected utilities Given

p i 2 A

i

player irsquos vector of expected utilities is given by ui

(p i

) = (ui1(p

i

) uiJ(i)(p

i

)) 2

RJ(i) where uij (p

i

) = ui

(aij p

i

) is the expected utility to action aij given behavior of the oppo-

nents We use vi = (v

i1 viJ(i)) 2 RJ(i) as shorthand for an arbitrary vector of expected utilities

That is vi is understood to satisfy v

i = ui

(p 0 i

) for some p 0 i

22 Quantal response equilibrium

Player irsquos quantal response function Qi = (Q

i1 QiJ(i)) RJ(i) A

i maps his vector of expected

utilities to a distribution over actions For any vi 2 RJ(i) component Q

ij (vi) gives the probability

assigned to action j Intuitively Qi allows for arbitrary probabilistic mistakes in taking actions

given the payoffs to each action 6It is well-known from Matejka and McKay [2015] that if the state is a vector of payoffs (ie the payoff to

each action) then the solution to the rational inattention problem with mutual information costs is a generalized multinomial logit that depends on the prior Such an interpretation does not readily extend to QRE since the vector of expected utilities is deterministic in equilibrium

7In a structural QRE quantal response results from choosing the action that maximizes expected utility plus a random error term If the errors are iid across actions then there are testable restrictions Regularity also imposes testable restrictions (Goeree et al [2005])

4

A quantal response equilibrium (QRE) is obtained when the distribution over all playersrsquo actions is consistent with their quantal response functions Letting Q = (Q1 Qn

) and u = (u1 un) QRE is a fixed point of the composite function Q u A A

Definition 1 Fix Q A QRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i) pij = Q

ij (ui(p i

))

QRE can explain any behavior without some basic restrictions on the quantal response function In practice the weakest restrictions imposed are given by the regularity axioms (Goeree et al [2005])

(A1) Interiority Qij (vi) 2 (0 1) for all j = 1 J(i) for all v

i 2 RJ(i)

(A2) Continuity Qij (vi) is a continuous and differentiable function for all v

i 2 RJ(i)

(A3) Monotonicity vij gt v

ik =) Qij (vi) gt Q

ik

(vi

) for all j k = 1 J(i) Q

ij (vi)(A4) Responsiveness v

ij gt 0 for all j = 1 J(i) and for all v

i 2 RJ(i)

These capture a sensitivity to payoff differences players take actions that give higher expected

utility with higher probability and the greater the payoff differences the greater the effect Goeree

et al [2005] show that these axioms are sufficient to impose testable restrictions on the data Often quantal response is derived through the ldquostructuralrdquo approach given v

i 2 RJ(i) the

probability player i takes action j is given by

Qij (vi) = Pv

ij + ij v

ik + ik 8k = 1 J(i)

where ik is an action-specific mean-zero random error Errors are assumed independent across

players but not necessarily across actions In the structural approach players maximize the real-ization of noisy expected utility where the exogenous noise is in perceiving or calculating expected

utility If the errors are iid across actions (and satisfy other mild conditions) structural QRE models

are regular (Goeree et al [2005]) and thus impose testable restrictions on the data This is an

important observation given the results of Haile et al [2008] who show that structural QRE can

rationalize the data from any one game without such restrictions on the errors For applications it is common to assume that the errors

ik are distributed according to an

extreme value ldquologitrdquo distribution with precision 2 [01)8 In this case the quantal response

function takes the logit form

v

ije Q

ij (vi ) (1)P

J(i) k=1 e vik

e ik8Specifically ik is distributed iid according to a mean-zero Gumbel distribution with CDF F (

ik

) = e

where 05772 is the Euler-Mascheroni constant The variance of ik is given by

6

2

2 which is strictly decreasing in and hence the interpretation of as precision

5

In this paper we take as our starting point the structure of logit quantal response (1) We do

this as a convenient parametrization for the task we outlined in the introduction to endogenize

the precision of errors given here by Logit is also extremely common in applications and fairly

representative of QRE models in that it is structural and satisfies regularity A logit QRE or LQRE

is defined in the obvious way

Definition 2 Fix An LQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i) pij = Q

ij (ui(p i

) )

3 Endogenous quantal response equilibrium

In an endogenous quantal response equilibrium (EQRE) is no longer exogenous Each player i chooses precision (or a distribution over different s) optimally subject to costs in a first stage i i

given beliefs over the opponentsrsquo second-stage actions In equilibrium the second-stage actions form a heterogeneous LQRE given the endogenously determined vector = ( 1 )

n

We begin by analyzing the first-stage optimal precision choice problem of such an agent holding

fixed beliefs Beliefs only enter the problem via the expected utilities they induce Hence given

vi 2 RJ(i) and cost parameter 2 (01) player i solves

J(i)X

i (vi ) argmax Q

ik

(vi

)vik c( ) (2)

2[01) k=1| z

b( vi

)

where Qik is given by (1) and cost function c [01) [01) satisfies c 0 (0) = 0 lim c

0 ( ) = 1

1 kand for all gt 0 c 0 ( ) gt 0 and c

00 ( ) gt 09 For example c could be a power function c( ) =

for k gt 1 For most of the paper we think of the cost function c as held fixed though for some

later results we consider the predictions obtained in the limit by a sequence of cost functions or the

set of predictions obtainable for some cost function Note that we define b(middot vi

) [01) [01)

as the benefit function for convenience This is merely player irsquos overall expected utility given the

expected utility to each of his actions and his quantal response Intuitively greater precision leads to better decisions (as we will show) but precision is itself

costly so optimal precision balances these tradeoffs Since the value of making better decisions depends on payoffs the benefit function depends on v

i

We favor an as if interpretation to the first-stage problem because the benefit function depends on a noiseless perception of v

i

even though the

agent will (and knows that he will) perceive vi with error in the second stage Such an interpretation

is essentially the same as those offered for models of rational inattention10 in which the benefit to 00 009It is fine if c (0) = 0 or lim c ( ) = 1

0+ 10Some classic references include Sims [1998] Sims [2003] Woodford [2009] and Caplin and Dean [2015]

6

an information structure depends on the stochastic choice it induces and the value to each action When v

ij = vik for all j k then b( v

i

) = the constant function that assigns utility for all levels of precision When v

ij 6= vik for some j k the following properties of b are easily verified

(see Appendix 91) P11 b(0 v

i

) = j vij and lim b( v

i

) = max vij

J(i) 1 j

b( vi

) vi

)2 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

b( vi

)

2b( v

i

) vi

)3 lim = 0 lim = 0 and 2b( lt 0 for sufficiently high

2

2 1 1

4 b( vi + e

J(i)) = b( vi

) + for all 2 R where eJ(i) = (1 1)

These are intuitive Property 1 when = 0 the agent uniformly randomizes so expects to

receive the unweighted average utility As 1 the agent takes (one of) the best action(s) with

probability approaching one Property 2 increasing will always lead to better decisions even on

the margin when = 0 so the first derivative is positive Unsurprisingly it is also bounded We also

establish that the second derivative is bounded but it cannot be signed in general which we will return to as it complicates the analysis Property 3 We establish some limiting behaviors for the

first two derivatives which are unsurprising given that b is strictly increasing and bounded Property

4 b shifts up by constant when the utility to each alternative increases by This is a simple

vi

)11consequence of the translation invariance of logit (1) which states that Qi

(vi + e

J(i)) = Qi

( Hence it is without loss to assume that v

ij gt 0 for all i and j which will simplify a number of arguments

It is easy to show that a solution to the first-stage problem exists If vij = v

ik for all j k then

the solution is trivial i = 0 which results in a uniform distribution over all actions Otherwise

any solution is strictly greater than 0 and satisfies a first-order condition This is summarized in

Theorem 1 along with basic properties of

i (

i

that follow from properties of b and c

i (Theorem 1 A solution to (2) exists ( 6= ) If vi

) is a solution it satisfies vi

) 2

2P

J(i) P

J(i)2 v

il v

il l=1 vile

l=1 vile c

0 ( ) = 0 (3)

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

1 inf (vi

) gt 0i

2 inf (vi

) and supi

i (vi ) are strictly decreasing in

3 lim 0

(inf vi

)) = 1 and lim 1

(sup vi

)) = 0 i ( i (

11See Goeree et al [2005] and Friedman [2018] for discussions of translation invariance and its implications

7

Proof See Appendix 92

One may expect equilibrium to be defined exactly as in Definition 2 but with Q ui

ij (ui ) Q

ij (macri (ui ))

replacing Qij (ui ) This is almost right except for the fact that the set of maximizers

i (vi )

is not a singleton in general (hence the use of inf and sup) This is because b(middot vi

) is not concave

for some vi

In other words the marginal benefit to may actually increase over portions of the

domain We provide an example of this in Appendix 95 which is due to Weibull et al [2007] Fortunately

i (vi ) is still well-behaved in the following sense

Lemma 1 Correspondence i RJ(i) (01) [01) is upper hemi-continuous

Proof See Appendix 92

i is upper hemi-continuous and not simply continuous because

i (vi ) may not be a singleton The next lemma establishes that

i (vi ) is a singleton however for both sufficiently low and

sufficiently high The result is proved by establishing that for extreme the solution to the

first-stage problem is in a region where the objective is locally concave

i (vi ) is a singleton for sufficiently low and sufficiently high Lemma 2 For all vi 2 RJ(i)

Proof See Appendix 92

In general equilibria may involve mixing over 2 i (middot ) Hence we define optimal mixed

precision by

i (vi )

to be any distribution over optimal selections which is also clearly optimal Finally the endogenous

i (vi )

( )vi

i i 2 [01) supp(

i

)

quantal response correspondence Qi (middot ) RJ(i)

Z 1

Ai is defined by

Qi

( )| i 2 (4)

i (vi ) Qi

(vi

)d 0

= ConvTi

(vi

)

where i (vi )

is the set of action frequencies from any optimal pure precision and Conv is the convex hull For all

Ti

(vi

) = Qi

(vi

)| 2

(vi

) an element p 0 i 2 Q

i (vi ) represents an action frequency induced by some optimal (possibly 0 P

J(i) 0 0 and

k=1 pmixed) precision and satisfies p = 1ij

ik 1 Q

to Definition 2 as a fixed point of the composite correspondence Q u A A Defining Q = (Q ) (suppressing in the notation) equilibrium is defined analogously

n

An EQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i)

Definition 3 Fix

pi 2 Q

i (ui(p i

) )

8

We establish several properties of Q Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Q i (middot ) is convex-valued and upper hemi-continuous on RJ(i)

3 For all i and j k = 1 J(i)

00 0for any pv

ij gt vik =) p

i 2 Q ij (vi )ij gt p

ik

Q

(vi

)4 If vij = max

k

vik

gt vil for some l ij gt 0 for sufficiently low and high

v

ij

Properties 1-3 are trivial12 and Property 4 is established in Appendix 93 Properties 1 and 2 are

technical Property 3 is generalized monotonicity (A3) that allows for possible non-uniqueness in

optimal quantal response Property 4 is a weaker form of responsiveness (A4) Note that it is weaker in two ways First it only holds for extreme which is required to ensure that Q is single-valued

Q

(vi

)ij 13(and differentiable) Second we are only able to show that gt 0 if v

ij = maxk

vik

gt vil

v

ij

The first two properties of Q imply existence

Theorem 2 Fix There exists an EQRE

Proof An EQRE is a fixed point of Q u Since Qi (middot ) is convex-valued and upper hemi-

continuous on RJ(i) and u is continuous on A Q u A A is convex-valued and upper hemi-continuous Because A is a compact subset of RJ Q u is also closed graph Summarizing Q u A A is a non-empty and convex-valued correspondence with a closed graph mapping

from a non-empty compact convex set to itself By Kakutanirsquos fixed point theorem Q u has a

fixed point

31 Binary actions

We now specialize the theory to the class of binary action games in which each player has two

actions Games in this class are widely used in experiments covering 2 2 games14 various voting

type games and more The theory could also be extended to an endogenous agent QRE (AQRE) (McKelvey and Palfrey [1998]) and used to analyze extensive form games in which there are two

12Since i Q

Since Qi RJ(i) [0 1)

i (vi ) = ConvT

i

(vi

) Q RJ(i) (0 1) [0 1) is upper hemi-continuous

is convex-valued 6= 6= i

also Because of the representation as Q

A

i is a continuous function andi

T

i RJ(i) (0 1) A

i is also upper hemi-continuous It is easy to check that Q i as the convexification of T

i

is also upper hemi-continuous Finally since v

ij gt v

ik =) Q

ij (vi ) gt Q

ik

(vi

) for any gt 0 it also holds when integrating over any s

13An increase in any vij will generically change

i (in either direction) Our result relies on showing that an increase in v

ij increases the product i vij if vij

implies an increase in Q ij if vij

= max

k

vik

(we conjecture that this is true even if vij 6= max

k

vik

) which = max

k

vik

(which is not necessarily true if vij 6= max

k

vik

)14Common ones used in experiments include generalized matching pennies the prisonerrsquos dilemma battle of the

sexes and at least several more

9

actions at every node such as the centipede game Importantly for our purposes with binary

actions it becomes much easier to analyze the solution to the first-stage problem (2) from which

we derive a much tighter characterization of stochastic choice e v

ij 6v

ijk eIf J(i) = 2 Q

ij (vi ) = = where vijk = v

ij vik is the signed payoff

e v

ij +e v

ik 1+e 6v

ijk P2 6v

i

edifference It is also easy to show that k=1 Qik

(vi

)vik = v

i 6v

i +const where v

i | vijk

|1+e

is the absolute payoff difference and the constant depends on the levels of vi = (v

ij vik

) Note that

vi 2 [0 1) by construction and that this is without loss We can now rewrite (2) as

fv

ie i ( v

i

) argmax vi c( ) (5)

fv

i1 + e2[01) | z

b( fv

i

)

Note that we have defined the net benefit function b(middot vi

) for this case Inspection reveals that it is simply the probability of taking the better action times the net benefit of doing so When

vi = 0 b( v

i

) = 0 for all When vi gt 0 it is easy to verify that

1 b(0 vi

) = 1 vi and lim b( v

i

) = vi

2 1

b( fv

i

)

2b( fv

i

)2 lim = 0 and lim = 0

2 1 1

b( fv

i

) fv

i

)3 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 0) for some 1 2 2 R++

2

For the most part these properties are intuitive specializations of the more general properties from

the previous section Unlike in the general case however property 3 states that b(middot vi

) and thus the first-stage objective (5) is strictly concave15 Due to this fact i ( v

i

) is a singleton and

satisfies

1 i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0

2 i ( v

i

) is strictly decreasing in if vi gt 0

3 lim i ( v

i

) = 1 and lim i ( v

i

) = 0 if vi gt 0

0 1

Since single-valued upper hemi-continuous correspondences are continuous functions we have the

following lemma as a direct corollary of Lemma 1

Lemma 3 [0 1) (0 1) [0 1) is continuous i

From Lemma 3 it follows that the binary action analogue of the endogenous quantal response

correspondence (4) is a continuous function Q (middot ) R2 (0 1) defined by ij

e (fv

i

)fv

ijk

Q vi

) = i

(6)ij ( (fv

i

)fv

ijk 1 + e i

3 vi

2b( 6vi) 6vi e

vi (e 1)15Taking derivatives 2 = lt 0 for v

i gt 0 (e vi +1)3

10

For reference we give the equilibrium definition and properties of Q for the binary action case

where J(i) = 2 for all i An EQRE is any p 2 A such that for all Definition 4 Fix i 2 1 n p

i1 = Q (ui

(p i

) ) i1

Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Qi ( middot ) is continuous and differentiable on R2

3 For all i and j k = 1 J(i)

=) Q vi

) gt Qij (

(ik vi )v

ij gt vik

Q

(vi

) Q

(vi

)ij ij4 = 0 and|

v

i1=v

i2 |v

i1v

ij v

ij gt 0=v

i2

and Lemma 3 Property 3 is monotonicity (A3) verbatim Property 4 is nearly responsiveness (A4) Q

(vi

)

6

Properties 1-3 are nearly immediate given the properties of general from the previous section Q

except that when (as opposed to being strictly positive) Property 4 follows ij = 0 = 0vi

v

ij

from properties of i presented later in this section

We note that the single-valuedness of i in the binary action case implies that all equilibria are

i s Also the single-valuedness of

is what makes Q a function as opposed to a correspondence Hence while existence is still ensured

by Theorem 2 Brouwerrsquos fixed point theorem would have been enough For the binary action case we seek to compare exogenous and endogenous quantal response

For this it is sufficient to track accuracy which we define as the probability of taking the better action under either model16

e fv

i

ai

( vi

) 1 + e fv

i

(fv

i

)fv

i

a i ( v

i

) i ()) = a

i

( vi

e i

(7)

pure in the sense that it is never optimal to mix over differenti

(fv

i

)fv

i1 + e i

We note that ai only depends on and is strictly increasing in the product v

i

In particular ai

is strictly increasing in vi for any gt 0 and strictly increasing in for any v

i gt 0 Thus to i While there is no closed form for

i understand quantal response we analyze properties of

much can be learned via implicit methods

Theorem 3

i (0 ) = 0 and i ((i) lim v

i

) = 0 fv

i

1

16Recalling that v

i is the absolute payoff difference between the two actions these are simply Qij and Q

ij if

v

ij gt v

ik

11

(ii) (middot ) is strictly single-peaked with macr() max ( vi

) and v() argmax ( vi

)i i i

fv

i

2(01) fv

i

2(01) i (fv

i

) i (fv

i

)(iii) fv

i fv

i

i (fv

i

) |0ltfv

i

ltfv() gt 0 and fv

i

|fv

i

=0 = 0 (iv) The product

|fv

i

gtfv() lt 0 ( v

i

)( vi

) vi is strictly increasing in v

i

i v

i = 0 and lim ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof See Appendix 92

Parts (i)-(iii) of the theorem state that i (middot ) is hump-shaped starting at 0 when v

i = 0 strictly increasing to its peak and then strictly decreasing to 0 as v

i 1 This is intuitive The good action is worth v

i more than the bad action so vi represents the cost of making a

mistake or equivalently the benefit of being accurate But for any given the agent is better able

to distinguish the good action from the bad and is thus more accurate when vi is high So when

vi is very low the benefit of accuracy is low and the agent optimally chooses low precision and

thus low accuracy and when vi is very high it is so easy to distinguish the good action from the

bad that the decision maker does not need high precision for high accuracy Parts (iv)-(vi) have implications for accuracy Part (iv) implies that a increases in v

i even i

though is non-monotonic in vi

Parts (v) and (vi) are more technical in nature with no obvious i

implication for decision problems though we use them extensively in the analysis of games in

subsequent sections Part (v) implies that the accuracy associated with peak precision macr() which macr()6v() e eobtains at v() is independent of a ( v() ) = macr()6macr = 092 Part (vi) states

i v()1+e 1+e

that the location of the peak v() increases in In Corollary 1 we compare accuracy under exogenous and endogenous quantal response This

follows immediately from properties of and ai

i

Corollary 1 (i) a

i (fv

i

) fv

i

(ii) a

|fv

i

=0 = 017

i (0 ) = ai

(0 ) = 1

a

i (fv

i

) )|fv

i

gt0 gt 0 and ai(fv

i

fv

i fv

i a

i

(lim lim

gt 0 and

i ( vi

) =fv

i

1 fv

i

1 vi

) = 1a 2

gt macr (iii) If () a ( vi

) lt ai

( vi

) for all vi

i

macr (iv) If 2 (0 ()) a ( vi

) gt ai

( vi

) for intermediate vi and a ( v

i

) lt ai

( vi

) i i

for extreme vi

Figure 1 graphically summarizes Theorem 3 and Corollary 1 The left panel of Figure 1 plots i (middot ) and

i (middot 0 ) as functions of v

i for lt 0 0 as well as the constant function and macr gt macr() gt

0 ( ) Since

i (are chosen such that peak precisions are ordered middot ) is hump-shaped

17Using the quotient rule take the derivative of (7) with respect to v

i and evaluate the expression at v

i = 0

i (0 ) = 0 and i (6vi )using that6vi

|6vi=0 = 0

12

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

Denti [2015] in which stochastic choice is driven by learning about some unknown state6 This paper is also related to other critical analyses of QRE such as Haile et al [2008] which

shows that structural QRE7 can rationalize any data without restrictions on errors Other ap-proaches for studying bounded rationality in games include Nagel [1995] Jehiel [2005] and Camerer et al [2004] There is also a relationship to Alaoui and Penta [2015] which endogenizes a parameter of a well known class of non-equilibrium models Finally Weibull et al [2007] provide some results for the stochastic choice of an agent who chooses logit precision optimally

2 Quantal response equilibrium

In this section we introduce the notation for normal form games that will be used throughout the

paper and review QRE

21 Normal form games

A finite normal form game = NA u is defined by a set of players N = 1 n action space

A = A1 An with A

i = ai1 a

iJ(i) such that each player i has J(i) possible pure actions P

(J = J(i) actions total) and a vector of payoff functions u = (u1 un) with ui A R

i

Let Ai be the set of probability measures on A

i

Elements of Ai are of the form p

i Ai R P

J(i)where aij ) = 1 and p

i

(aij ) 0 For simplicity set p

ij pi

(aij ) Define A = A1

j=1 pi(

An and A

i = k 6 A

k with typical elements p 2 A and p i 2 A

i

As standard extend =i P

payoff functions u = (u1 un) to be defined over A via ui

(p) = a2A p(a)ui(a)

As is standard in the literature on QRE we use additional notation for expected utilities Given

p i 2 A

i

player irsquos vector of expected utilities is given by ui

(p i

) = (ui1(p

i

) uiJ(i)(p

i

)) 2

RJ(i) where uij (p

i

) = ui

(aij p

i

) is the expected utility to action aij given behavior of the oppo-

nents We use vi = (v

i1 viJ(i)) 2 RJ(i) as shorthand for an arbitrary vector of expected utilities

That is vi is understood to satisfy v

i = ui

(p 0 i

) for some p 0 i

22 Quantal response equilibrium

Player irsquos quantal response function Qi = (Q

i1 QiJ(i)) RJ(i) A

i maps his vector of expected

utilities to a distribution over actions For any vi 2 RJ(i) component Q

ij (vi) gives the probability

assigned to action j Intuitively Qi allows for arbitrary probabilistic mistakes in taking actions

given the payoffs to each action 6It is well-known from Matejka and McKay [2015] that if the state is a vector of payoffs (ie the payoff to

each action) then the solution to the rational inattention problem with mutual information costs is a generalized multinomial logit that depends on the prior Such an interpretation does not readily extend to QRE since the vector of expected utilities is deterministic in equilibrium

7In a structural QRE quantal response results from choosing the action that maximizes expected utility plus a random error term If the errors are iid across actions then there are testable restrictions Regularity also imposes testable restrictions (Goeree et al [2005])

4

A quantal response equilibrium (QRE) is obtained when the distribution over all playersrsquo actions is consistent with their quantal response functions Letting Q = (Q1 Qn

) and u = (u1 un) QRE is a fixed point of the composite function Q u A A

Definition 1 Fix Q A QRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i) pij = Q

ij (ui(p i

))

QRE can explain any behavior without some basic restrictions on the quantal response function In practice the weakest restrictions imposed are given by the regularity axioms (Goeree et al [2005])

(A1) Interiority Qij (vi) 2 (0 1) for all j = 1 J(i) for all v

i 2 RJ(i)

(A2) Continuity Qij (vi) is a continuous and differentiable function for all v

i 2 RJ(i)

(A3) Monotonicity vij gt v

ik =) Qij (vi) gt Q

ik

(vi

) for all j k = 1 J(i) Q

ij (vi)(A4) Responsiveness v

ij gt 0 for all j = 1 J(i) and for all v

i 2 RJ(i)

These capture a sensitivity to payoff differences players take actions that give higher expected

utility with higher probability and the greater the payoff differences the greater the effect Goeree

et al [2005] show that these axioms are sufficient to impose testable restrictions on the data Often quantal response is derived through the ldquostructuralrdquo approach given v

i 2 RJ(i) the

probability player i takes action j is given by

Qij (vi) = Pv

ij + ij v

ik + ik 8k = 1 J(i)

where ik is an action-specific mean-zero random error Errors are assumed independent across

players but not necessarily across actions In the structural approach players maximize the real-ization of noisy expected utility where the exogenous noise is in perceiving or calculating expected

utility If the errors are iid across actions (and satisfy other mild conditions) structural QRE models

are regular (Goeree et al [2005]) and thus impose testable restrictions on the data This is an

important observation given the results of Haile et al [2008] who show that structural QRE can

rationalize the data from any one game without such restrictions on the errors For applications it is common to assume that the errors

ik are distributed according to an

extreme value ldquologitrdquo distribution with precision 2 [01)8 In this case the quantal response

function takes the logit form

v

ije Q

ij (vi ) (1)P

J(i) k=1 e vik

e ik8Specifically ik is distributed iid according to a mean-zero Gumbel distribution with CDF F (

ik

) = e

where 05772 is the Euler-Mascheroni constant The variance of ik is given by

6

2

2 which is strictly decreasing in and hence the interpretation of as precision

5

In this paper we take as our starting point the structure of logit quantal response (1) We do

this as a convenient parametrization for the task we outlined in the introduction to endogenize

the precision of errors given here by Logit is also extremely common in applications and fairly

representative of QRE models in that it is structural and satisfies regularity A logit QRE or LQRE

is defined in the obvious way

Definition 2 Fix An LQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i) pij = Q

ij (ui(p i

) )

3 Endogenous quantal response equilibrium

In an endogenous quantal response equilibrium (EQRE) is no longer exogenous Each player i chooses precision (or a distribution over different s) optimally subject to costs in a first stage i i

given beliefs over the opponentsrsquo second-stage actions In equilibrium the second-stage actions form a heterogeneous LQRE given the endogenously determined vector = ( 1 )

n

We begin by analyzing the first-stage optimal precision choice problem of such an agent holding

fixed beliefs Beliefs only enter the problem via the expected utilities they induce Hence given

vi 2 RJ(i) and cost parameter 2 (01) player i solves

J(i)X

i (vi ) argmax Q

ik

(vi

)vik c( ) (2)

2[01) k=1| z

b( vi

)

where Qik is given by (1) and cost function c [01) [01) satisfies c 0 (0) = 0 lim c

0 ( ) = 1

1 kand for all gt 0 c 0 ( ) gt 0 and c

00 ( ) gt 09 For example c could be a power function c( ) =

for k gt 1 For most of the paper we think of the cost function c as held fixed though for some

later results we consider the predictions obtained in the limit by a sequence of cost functions or the

set of predictions obtainable for some cost function Note that we define b(middot vi

) [01) [01)

as the benefit function for convenience This is merely player irsquos overall expected utility given the

expected utility to each of his actions and his quantal response Intuitively greater precision leads to better decisions (as we will show) but precision is itself

costly so optimal precision balances these tradeoffs Since the value of making better decisions depends on payoffs the benefit function depends on v

i

We favor an as if interpretation to the first-stage problem because the benefit function depends on a noiseless perception of v

i

even though the

agent will (and knows that he will) perceive vi with error in the second stage Such an interpretation

is essentially the same as those offered for models of rational inattention10 in which the benefit to 00 009It is fine if c (0) = 0 or lim c ( ) = 1

0+ 10Some classic references include Sims [1998] Sims [2003] Woodford [2009] and Caplin and Dean [2015]

6

an information structure depends on the stochastic choice it induces and the value to each action When v

ij = vik for all j k then b( v

i

) = the constant function that assigns utility for all levels of precision When v

ij 6= vik for some j k the following properties of b are easily verified

(see Appendix 91) P11 b(0 v

i

) = j vij and lim b( v

i

) = max vij

J(i) 1 j

b( vi

) vi

)2 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

b( vi

)

2b( v

i

) vi

)3 lim = 0 lim = 0 and 2b( lt 0 for sufficiently high

2

2 1 1

4 b( vi + e

J(i)) = b( vi

) + for all 2 R where eJ(i) = (1 1)

These are intuitive Property 1 when = 0 the agent uniformly randomizes so expects to

receive the unweighted average utility As 1 the agent takes (one of) the best action(s) with

probability approaching one Property 2 increasing will always lead to better decisions even on

the margin when = 0 so the first derivative is positive Unsurprisingly it is also bounded We also

establish that the second derivative is bounded but it cannot be signed in general which we will return to as it complicates the analysis Property 3 We establish some limiting behaviors for the

first two derivatives which are unsurprising given that b is strictly increasing and bounded Property

4 b shifts up by constant when the utility to each alternative increases by This is a simple

vi

)11consequence of the translation invariance of logit (1) which states that Qi

(vi + e

J(i)) = Qi

( Hence it is without loss to assume that v

ij gt 0 for all i and j which will simplify a number of arguments

It is easy to show that a solution to the first-stage problem exists If vij = v

ik for all j k then

the solution is trivial i = 0 which results in a uniform distribution over all actions Otherwise

any solution is strictly greater than 0 and satisfies a first-order condition This is summarized in

Theorem 1 along with basic properties of

i (

i

that follow from properties of b and c

i (Theorem 1 A solution to (2) exists ( 6= ) If vi

) is a solution it satisfies vi

) 2

2P

J(i) P

J(i)2 v

il v

il l=1 vile

l=1 vile c

0 ( ) = 0 (3)

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

1 inf (vi

) gt 0i

2 inf (vi

) and supi

i (vi ) are strictly decreasing in

3 lim 0

(inf vi

)) = 1 and lim 1

(sup vi

)) = 0 i ( i (

11See Goeree et al [2005] and Friedman [2018] for discussions of translation invariance and its implications

7

Proof See Appendix 92

One may expect equilibrium to be defined exactly as in Definition 2 but with Q ui

ij (ui ) Q

ij (macri (ui ))

replacing Qij (ui ) This is almost right except for the fact that the set of maximizers

i (vi )

is not a singleton in general (hence the use of inf and sup) This is because b(middot vi

) is not concave

for some vi

In other words the marginal benefit to may actually increase over portions of the

domain We provide an example of this in Appendix 95 which is due to Weibull et al [2007] Fortunately

i (vi ) is still well-behaved in the following sense

Lemma 1 Correspondence i RJ(i) (01) [01) is upper hemi-continuous

Proof See Appendix 92

i is upper hemi-continuous and not simply continuous because

i (vi ) may not be a singleton The next lemma establishes that

i (vi ) is a singleton however for both sufficiently low and

sufficiently high The result is proved by establishing that for extreme the solution to the

first-stage problem is in a region where the objective is locally concave

i (vi ) is a singleton for sufficiently low and sufficiently high Lemma 2 For all vi 2 RJ(i)

Proof See Appendix 92

In general equilibria may involve mixing over 2 i (middot ) Hence we define optimal mixed

precision by

i (vi )

to be any distribution over optimal selections which is also clearly optimal Finally the endogenous

i (vi )

( )vi

i i 2 [01) supp(

i

)

quantal response correspondence Qi (middot ) RJ(i)

Z 1

Ai is defined by

Qi

( )| i 2 (4)

i (vi ) Qi

(vi

)d 0

= ConvTi

(vi

)

where i (vi )

is the set of action frequencies from any optimal pure precision and Conv is the convex hull For all

Ti

(vi

) = Qi

(vi

)| 2

(vi

) an element p 0 i 2 Q

i (vi ) represents an action frequency induced by some optimal (possibly 0 P

J(i) 0 0 and

k=1 pmixed) precision and satisfies p = 1ij

ik 1 Q

to Definition 2 as a fixed point of the composite correspondence Q u A A Defining Q = (Q ) (suppressing in the notation) equilibrium is defined analogously

n

An EQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i)

Definition 3 Fix

pi 2 Q

i (ui(p i

) )

8

We establish several properties of Q Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Q i (middot ) is convex-valued and upper hemi-continuous on RJ(i)

3 For all i and j k = 1 J(i)

00 0for any pv

ij gt vik =) p

i 2 Q ij (vi )ij gt p

ik

Q

(vi

)4 If vij = max

k

vik

gt vil for some l ij gt 0 for sufficiently low and high

v

ij

Properties 1-3 are trivial12 and Property 4 is established in Appendix 93 Properties 1 and 2 are

technical Property 3 is generalized monotonicity (A3) that allows for possible non-uniqueness in

optimal quantal response Property 4 is a weaker form of responsiveness (A4) Note that it is weaker in two ways First it only holds for extreme which is required to ensure that Q is single-valued

Q

(vi

)ij 13(and differentiable) Second we are only able to show that gt 0 if v

ij = maxk

vik

gt vil

v

ij

The first two properties of Q imply existence

Theorem 2 Fix There exists an EQRE

Proof An EQRE is a fixed point of Q u Since Qi (middot ) is convex-valued and upper hemi-

continuous on RJ(i) and u is continuous on A Q u A A is convex-valued and upper hemi-continuous Because A is a compact subset of RJ Q u is also closed graph Summarizing Q u A A is a non-empty and convex-valued correspondence with a closed graph mapping

from a non-empty compact convex set to itself By Kakutanirsquos fixed point theorem Q u has a

fixed point

31 Binary actions

We now specialize the theory to the class of binary action games in which each player has two

actions Games in this class are widely used in experiments covering 2 2 games14 various voting

type games and more The theory could also be extended to an endogenous agent QRE (AQRE) (McKelvey and Palfrey [1998]) and used to analyze extensive form games in which there are two

12Since i Q

Since Qi RJ(i) [0 1)

i (vi ) = ConvT

i

(vi

) Q RJ(i) (0 1) [0 1) is upper hemi-continuous

is convex-valued 6= 6= i

also Because of the representation as Q

A

i is a continuous function andi

T

i RJ(i) (0 1) A

i is also upper hemi-continuous It is easy to check that Q i as the convexification of T

i

is also upper hemi-continuous Finally since v

ij gt v

ik =) Q

ij (vi ) gt Q

ik

(vi

) for any gt 0 it also holds when integrating over any s

13An increase in any vij will generically change

i (in either direction) Our result relies on showing that an increase in v

ij increases the product i vij if vij

implies an increase in Q ij if vij

= max

k

vik

(we conjecture that this is true even if vij 6= max

k

vik

) which = max

k

vik

(which is not necessarily true if vij 6= max

k

vik

)14Common ones used in experiments include generalized matching pennies the prisonerrsquos dilemma battle of the

sexes and at least several more

9

actions at every node such as the centipede game Importantly for our purposes with binary

actions it becomes much easier to analyze the solution to the first-stage problem (2) from which

we derive a much tighter characterization of stochastic choice e v

ij 6v

ijk eIf J(i) = 2 Q

ij (vi ) = = where vijk = v

ij vik is the signed payoff

e v

ij +e v

ik 1+e 6v

ijk P2 6v

i

edifference It is also easy to show that k=1 Qik

(vi

)vik = v

i 6v

i +const where v

i | vijk

|1+e

is the absolute payoff difference and the constant depends on the levels of vi = (v

ij vik

) Note that

vi 2 [0 1) by construction and that this is without loss We can now rewrite (2) as

fv

ie i ( v

i

) argmax vi c( ) (5)

fv

i1 + e2[01) | z

b( fv

i

)

Note that we have defined the net benefit function b(middot vi

) for this case Inspection reveals that it is simply the probability of taking the better action times the net benefit of doing so When

vi = 0 b( v

i

) = 0 for all When vi gt 0 it is easy to verify that

1 b(0 vi

) = 1 vi and lim b( v

i

) = vi

2 1

b( fv

i

)

2b( fv

i

)2 lim = 0 and lim = 0

2 1 1

b( fv

i

) fv

i

)3 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 0) for some 1 2 2 R++

2

For the most part these properties are intuitive specializations of the more general properties from

the previous section Unlike in the general case however property 3 states that b(middot vi

) and thus the first-stage objective (5) is strictly concave15 Due to this fact i ( v

i

) is a singleton and

satisfies

1 i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0

2 i ( v

i

) is strictly decreasing in if vi gt 0

3 lim i ( v

i

) = 1 and lim i ( v

i

) = 0 if vi gt 0

0 1

Since single-valued upper hemi-continuous correspondences are continuous functions we have the

following lemma as a direct corollary of Lemma 1

Lemma 3 [0 1) (0 1) [0 1) is continuous i

From Lemma 3 it follows that the binary action analogue of the endogenous quantal response

correspondence (4) is a continuous function Q (middot ) R2 (0 1) defined by ij

e (fv

i

)fv

ijk

Q vi

) = i

(6)ij ( (fv

i

)fv

ijk 1 + e i

3 vi

2b( 6vi) 6vi e

vi (e 1)15Taking derivatives 2 = lt 0 for v

i gt 0 (e vi +1)3

10

For reference we give the equilibrium definition and properties of Q for the binary action case

where J(i) = 2 for all i An EQRE is any p 2 A such that for all Definition 4 Fix i 2 1 n p

i1 = Q (ui

(p i

) ) i1

Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Qi ( middot ) is continuous and differentiable on R2

3 For all i and j k = 1 J(i)

=) Q vi

) gt Qij (

(ik vi )v

ij gt vik

Q

(vi

) Q

(vi

)ij ij4 = 0 and|

v

i1=v

i2 |v

i1v

ij v

ij gt 0=v

i2

and Lemma 3 Property 3 is monotonicity (A3) verbatim Property 4 is nearly responsiveness (A4) Q

(vi

)

6

Properties 1-3 are nearly immediate given the properties of general from the previous section Q

except that when (as opposed to being strictly positive) Property 4 follows ij = 0 = 0vi

v

ij

from properties of i presented later in this section

We note that the single-valuedness of i in the binary action case implies that all equilibria are

i s Also the single-valuedness of

is what makes Q a function as opposed to a correspondence Hence while existence is still ensured

by Theorem 2 Brouwerrsquos fixed point theorem would have been enough For the binary action case we seek to compare exogenous and endogenous quantal response

For this it is sufficient to track accuracy which we define as the probability of taking the better action under either model16

e fv

i

ai

( vi

) 1 + e fv

i

(fv

i

)fv

i

a i ( v

i

) i ()) = a

i

( vi

e i

(7)

pure in the sense that it is never optimal to mix over differenti

(fv

i

)fv

i1 + e i

We note that ai only depends on and is strictly increasing in the product v

i

In particular ai

is strictly increasing in vi for any gt 0 and strictly increasing in for any v

i gt 0 Thus to i While there is no closed form for

i understand quantal response we analyze properties of

much can be learned via implicit methods

Theorem 3

i (0 ) = 0 and i ((i) lim v

i

) = 0 fv

i

1

16Recalling that v

i is the absolute payoff difference between the two actions these are simply Qij and Q

ij if

v

ij gt v

ik

11

(ii) (middot ) is strictly single-peaked with macr() max ( vi

) and v() argmax ( vi

)i i i

fv

i

2(01) fv

i

2(01) i (fv

i

) i (fv

i

)(iii) fv

i fv

i

i (fv

i

) |0ltfv

i

ltfv() gt 0 and fv

i

|fv

i

=0 = 0 (iv) The product

|fv

i

gtfv() lt 0 ( v

i

)( vi

) vi is strictly increasing in v

i

i v

i = 0 and lim ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof See Appendix 92

Parts (i)-(iii) of the theorem state that i (middot ) is hump-shaped starting at 0 when v

i = 0 strictly increasing to its peak and then strictly decreasing to 0 as v

i 1 This is intuitive The good action is worth v

i more than the bad action so vi represents the cost of making a

mistake or equivalently the benefit of being accurate But for any given the agent is better able

to distinguish the good action from the bad and is thus more accurate when vi is high So when

vi is very low the benefit of accuracy is low and the agent optimally chooses low precision and

thus low accuracy and when vi is very high it is so easy to distinguish the good action from the

bad that the decision maker does not need high precision for high accuracy Parts (iv)-(vi) have implications for accuracy Part (iv) implies that a increases in v

i even i

though is non-monotonic in vi

Parts (v) and (vi) are more technical in nature with no obvious i

implication for decision problems though we use them extensively in the analysis of games in

subsequent sections Part (v) implies that the accuracy associated with peak precision macr() which macr()6v() e eobtains at v() is independent of a ( v() ) = macr()6macr = 092 Part (vi) states

i v()1+e 1+e

that the location of the peak v() increases in In Corollary 1 we compare accuracy under exogenous and endogenous quantal response This

follows immediately from properties of and ai

i

Corollary 1 (i) a

i (fv

i

) fv

i

(ii) a

|fv

i

=0 = 017

i (0 ) = ai

(0 ) = 1

a

i (fv

i

) )|fv

i

gt0 gt 0 and ai(fv

i

fv

i fv

i a

i

(lim lim

gt 0 and

i ( vi

) =fv

i

1 fv

i

1 vi

) = 1a 2

gt macr (iii) If () a ( vi

) lt ai

( vi

) for all vi

i

macr (iv) If 2 (0 ()) a ( vi

) gt ai

( vi

) for intermediate vi and a ( v

i

) lt ai

( vi

) i i

for extreme vi

Figure 1 graphically summarizes Theorem 3 and Corollary 1 The left panel of Figure 1 plots i (middot ) and

i (middot 0 ) as functions of v

i for lt 0 0 as well as the constant function and macr gt macr() gt

0 ( ) Since

i (are chosen such that peak precisions are ordered middot ) is hump-shaped

17Using the quotient rule take the derivative of (7) with respect to v

i and evaluate the expression at v

i = 0

i (0 ) = 0 and i (6vi )using that6vi

|6vi=0 = 0

12

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

A quantal response equilibrium (QRE) is obtained when the distribution over all playersrsquo actions is consistent with their quantal response functions Letting Q = (Q1 Qn

) and u = (u1 un) QRE is a fixed point of the composite function Q u A A

Definition 1 Fix Q A QRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i) pij = Q

ij (ui(p i

))

QRE can explain any behavior without some basic restrictions on the quantal response function In practice the weakest restrictions imposed are given by the regularity axioms (Goeree et al [2005])

(A1) Interiority Qij (vi) 2 (0 1) for all j = 1 J(i) for all v

i 2 RJ(i)

(A2) Continuity Qij (vi) is a continuous and differentiable function for all v

i 2 RJ(i)

(A3) Monotonicity vij gt v

ik =) Qij (vi) gt Q

ik

(vi

) for all j k = 1 J(i) Q

ij (vi)(A4) Responsiveness v

ij gt 0 for all j = 1 J(i) and for all v

i 2 RJ(i)

These capture a sensitivity to payoff differences players take actions that give higher expected

utility with higher probability and the greater the payoff differences the greater the effect Goeree

et al [2005] show that these axioms are sufficient to impose testable restrictions on the data Often quantal response is derived through the ldquostructuralrdquo approach given v

i 2 RJ(i) the

probability player i takes action j is given by

Qij (vi) = Pv

ij + ij v

ik + ik 8k = 1 J(i)

where ik is an action-specific mean-zero random error Errors are assumed independent across

players but not necessarily across actions In the structural approach players maximize the real-ization of noisy expected utility where the exogenous noise is in perceiving or calculating expected

utility If the errors are iid across actions (and satisfy other mild conditions) structural QRE models

are regular (Goeree et al [2005]) and thus impose testable restrictions on the data This is an

important observation given the results of Haile et al [2008] who show that structural QRE can

rationalize the data from any one game without such restrictions on the errors For applications it is common to assume that the errors

ik are distributed according to an

extreme value ldquologitrdquo distribution with precision 2 [01)8 In this case the quantal response

function takes the logit form

v

ije Q

ij (vi ) (1)P

J(i) k=1 e vik

e ik8Specifically ik is distributed iid according to a mean-zero Gumbel distribution with CDF F (

ik

) = e

where 05772 is the Euler-Mascheroni constant The variance of ik is given by

6

2

2 which is strictly decreasing in and hence the interpretation of as precision

5

In this paper we take as our starting point the structure of logit quantal response (1) We do

this as a convenient parametrization for the task we outlined in the introduction to endogenize

the precision of errors given here by Logit is also extremely common in applications and fairly

representative of QRE models in that it is structural and satisfies regularity A logit QRE or LQRE

is defined in the obvious way

Definition 2 Fix An LQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i) pij = Q

ij (ui(p i

) )

3 Endogenous quantal response equilibrium

In an endogenous quantal response equilibrium (EQRE) is no longer exogenous Each player i chooses precision (or a distribution over different s) optimally subject to costs in a first stage i i

given beliefs over the opponentsrsquo second-stage actions In equilibrium the second-stage actions form a heterogeneous LQRE given the endogenously determined vector = ( 1 )

n

We begin by analyzing the first-stage optimal precision choice problem of such an agent holding

fixed beliefs Beliefs only enter the problem via the expected utilities they induce Hence given

vi 2 RJ(i) and cost parameter 2 (01) player i solves

J(i)X

i (vi ) argmax Q

ik

(vi

)vik c( ) (2)

2[01) k=1| z

b( vi

)

where Qik is given by (1) and cost function c [01) [01) satisfies c 0 (0) = 0 lim c

0 ( ) = 1

1 kand for all gt 0 c 0 ( ) gt 0 and c

00 ( ) gt 09 For example c could be a power function c( ) =

for k gt 1 For most of the paper we think of the cost function c as held fixed though for some

later results we consider the predictions obtained in the limit by a sequence of cost functions or the

set of predictions obtainable for some cost function Note that we define b(middot vi

) [01) [01)

as the benefit function for convenience This is merely player irsquos overall expected utility given the

expected utility to each of his actions and his quantal response Intuitively greater precision leads to better decisions (as we will show) but precision is itself

costly so optimal precision balances these tradeoffs Since the value of making better decisions depends on payoffs the benefit function depends on v

i

We favor an as if interpretation to the first-stage problem because the benefit function depends on a noiseless perception of v

i

even though the

agent will (and knows that he will) perceive vi with error in the second stage Such an interpretation

is essentially the same as those offered for models of rational inattention10 in which the benefit to 00 009It is fine if c (0) = 0 or lim c ( ) = 1

0+ 10Some classic references include Sims [1998] Sims [2003] Woodford [2009] and Caplin and Dean [2015]

6

an information structure depends on the stochastic choice it induces and the value to each action When v

ij = vik for all j k then b( v

i

) = the constant function that assigns utility for all levels of precision When v

ij 6= vik for some j k the following properties of b are easily verified

(see Appendix 91) P11 b(0 v

i

) = j vij and lim b( v

i

) = max vij

J(i) 1 j

b( vi

) vi

)2 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

b( vi

)

2b( v

i

) vi

)3 lim = 0 lim = 0 and 2b( lt 0 for sufficiently high

2

2 1 1

4 b( vi + e

J(i)) = b( vi

) + for all 2 R where eJ(i) = (1 1)

These are intuitive Property 1 when = 0 the agent uniformly randomizes so expects to

receive the unweighted average utility As 1 the agent takes (one of) the best action(s) with

probability approaching one Property 2 increasing will always lead to better decisions even on

the margin when = 0 so the first derivative is positive Unsurprisingly it is also bounded We also

establish that the second derivative is bounded but it cannot be signed in general which we will return to as it complicates the analysis Property 3 We establish some limiting behaviors for the

first two derivatives which are unsurprising given that b is strictly increasing and bounded Property

4 b shifts up by constant when the utility to each alternative increases by This is a simple

vi

)11consequence of the translation invariance of logit (1) which states that Qi

(vi + e

J(i)) = Qi

( Hence it is without loss to assume that v

ij gt 0 for all i and j which will simplify a number of arguments

It is easy to show that a solution to the first-stage problem exists If vij = v

ik for all j k then

the solution is trivial i = 0 which results in a uniform distribution over all actions Otherwise

any solution is strictly greater than 0 and satisfies a first-order condition This is summarized in

Theorem 1 along with basic properties of

i (

i

that follow from properties of b and c

i (Theorem 1 A solution to (2) exists ( 6= ) If vi

) is a solution it satisfies vi

) 2

2P

J(i) P

J(i)2 v

il v

il l=1 vile

l=1 vile c

0 ( ) = 0 (3)

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

1 inf (vi

) gt 0i

2 inf (vi

) and supi

i (vi ) are strictly decreasing in

3 lim 0

(inf vi

)) = 1 and lim 1

(sup vi

)) = 0 i ( i (

11See Goeree et al [2005] and Friedman [2018] for discussions of translation invariance and its implications

7

Proof See Appendix 92

One may expect equilibrium to be defined exactly as in Definition 2 but with Q ui

ij (ui ) Q

ij (macri (ui ))

replacing Qij (ui ) This is almost right except for the fact that the set of maximizers

i (vi )

is not a singleton in general (hence the use of inf and sup) This is because b(middot vi

) is not concave

for some vi

In other words the marginal benefit to may actually increase over portions of the

domain We provide an example of this in Appendix 95 which is due to Weibull et al [2007] Fortunately

i (vi ) is still well-behaved in the following sense

Lemma 1 Correspondence i RJ(i) (01) [01) is upper hemi-continuous

Proof See Appendix 92

i is upper hemi-continuous and not simply continuous because

i (vi ) may not be a singleton The next lemma establishes that

i (vi ) is a singleton however for both sufficiently low and

sufficiently high The result is proved by establishing that for extreme the solution to the

first-stage problem is in a region where the objective is locally concave

i (vi ) is a singleton for sufficiently low and sufficiently high Lemma 2 For all vi 2 RJ(i)

Proof See Appendix 92

In general equilibria may involve mixing over 2 i (middot ) Hence we define optimal mixed

precision by

i (vi )

to be any distribution over optimal selections which is also clearly optimal Finally the endogenous

i (vi )

( )vi

i i 2 [01) supp(

i

)

quantal response correspondence Qi (middot ) RJ(i)

Z 1

Ai is defined by

Qi

( )| i 2 (4)

i (vi ) Qi

(vi

)d 0

= ConvTi

(vi

)

where i (vi )

is the set of action frequencies from any optimal pure precision and Conv is the convex hull For all

Ti

(vi

) = Qi

(vi

)| 2

(vi

) an element p 0 i 2 Q

i (vi ) represents an action frequency induced by some optimal (possibly 0 P

J(i) 0 0 and

k=1 pmixed) precision and satisfies p = 1ij

ik 1 Q

to Definition 2 as a fixed point of the composite correspondence Q u A A Defining Q = (Q ) (suppressing in the notation) equilibrium is defined analogously

n

An EQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i)

Definition 3 Fix

pi 2 Q

i (ui(p i

) )

8

We establish several properties of Q Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Q i (middot ) is convex-valued and upper hemi-continuous on RJ(i)

3 For all i and j k = 1 J(i)

00 0for any pv

ij gt vik =) p

i 2 Q ij (vi )ij gt p

ik

Q

(vi

)4 If vij = max

k

vik

gt vil for some l ij gt 0 for sufficiently low and high

v

ij

Properties 1-3 are trivial12 and Property 4 is established in Appendix 93 Properties 1 and 2 are

technical Property 3 is generalized monotonicity (A3) that allows for possible non-uniqueness in

optimal quantal response Property 4 is a weaker form of responsiveness (A4) Note that it is weaker in two ways First it only holds for extreme which is required to ensure that Q is single-valued

Q

(vi

)ij 13(and differentiable) Second we are only able to show that gt 0 if v

ij = maxk

vik

gt vil

v

ij

The first two properties of Q imply existence

Theorem 2 Fix There exists an EQRE

Proof An EQRE is a fixed point of Q u Since Qi (middot ) is convex-valued and upper hemi-

continuous on RJ(i) and u is continuous on A Q u A A is convex-valued and upper hemi-continuous Because A is a compact subset of RJ Q u is also closed graph Summarizing Q u A A is a non-empty and convex-valued correspondence with a closed graph mapping

from a non-empty compact convex set to itself By Kakutanirsquos fixed point theorem Q u has a

fixed point

31 Binary actions

We now specialize the theory to the class of binary action games in which each player has two

actions Games in this class are widely used in experiments covering 2 2 games14 various voting

type games and more The theory could also be extended to an endogenous agent QRE (AQRE) (McKelvey and Palfrey [1998]) and used to analyze extensive form games in which there are two

12Since i Q

Since Qi RJ(i) [0 1)

i (vi ) = ConvT

i

(vi

) Q RJ(i) (0 1) [0 1) is upper hemi-continuous

is convex-valued 6= 6= i

also Because of the representation as Q

A

i is a continuous function andi

T

i RJ(i) (0 1) A

i is also upper hemi-continuous It is easy to check that Q i as the convexification of T

i

is also upper hemi-continuous Finally since v

ij gt v

ik =) Q

ij (vi ) gt Q

ik

(vi

) for any gt 0 it also holds when integrating over any s

13An increase in any vij will generically change

i (in either direction) Our result relies on showing that an increase in v

ij increases the product i vij if vij

implies an increase in Q ij if vij

= max

k

vik

(we conjecture that this is true even if vij 6= max

k

vik

) which = max

k

vik

(which is not necessarily true if vij 6= max

k

vik

)14Common ones used in experiments include generalized matching pennies the prisonerrsquos dilemma battle of the

sexes and at least several more

9

actions at every node such as the centipede game Importantly for our purposes with binary

actions it becomes much easier to analyze the solution to the first-stage problem (2) from which

we derive a much tighter characterization of stochastic choice e v

ij 6v

ijk eIf J(i) = 2 Q

ij (vi ) = = where vijk = v

ij vik is the signed payoff

e v

ij +e v

ik 1+e 6v

ijk P2 6v

i

edifference It is also easy to show that k=1 Qik

(vi

)vik = v

i 6v

i +const where v

i | vijk

|1+e

is the absolute payoff difference and the constant depends on the levels of vi = (v

ij vik

) Note that

vi 2 [0 1) by construction and that this is without loss We can now rewrite (2) as

fv

ie i ( v

i

) argmax vi c( ) (5)

fv

i1 + e2[01) | z

b( fv

i

)

Note that we have defined the net benefit function b(middot vi

) for this case Inspection reveals that it is simply the probability of taking the better action times the net benefit of doing so When

vi = 0 b( v

i

) = 0 for all When vi gt 0 it is easy to verify that

1 b(0 vi

) = 1 vi and lim b( v

i

) = vi

2 1

b( fv

i

)

2b( fv

i

)2 lim = 0 and lim = 0

2 1 1

b( fv

i

) fv

i

)3 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 0) for some 1 2 2 R++

2

For the most part these properties are intuitive specializations of the more general properties from

the previous section Unlike in the general case however property 3 states that b(middot vi

) and thus the first-stage objective (5) is strictly concave15 Due to this fact i ( v

i

) is a singleton and

satisfies

1 i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0

2 i ( v

i

) is strictly decreasing in if vi gt 0

3 lim i ( v

i

) = 1 and lim i ( v

i

) = 0 if vi gt 0

0 1

Since single-valued upper hemi-continuous correspondences are continuous functions we have the

following lemma as a direct corollary of Lemma 1

Lemma 3 [0 1) (0 1) [0 1) is continuous i

From Lemma 3 it follows that the binary action analogue of the endogenous quantal response

correspondence (4) is a continuous function Q (middot ) R2 (0 1) defined by ij

e (fv

i

)fv

ijk

Q vi

) = i

(6)ij ( (fv

i

)fv

ijk 1 + e i

3 vi

2b( 6vi) 6vi e

vi (e 1)15Taking derivatives 2 = lt 0 for v

i gt 0 (e vi +1)3

10

For reference we give the equilibrium definition and properties of Q for the binary action case

where J(i) = 2 for all i An EQRE is any p 2 A such that for all Definition 4 Fix i 2 1 n p

i1 = Q (ui

(p i

) ) i1

Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Qi ( middot ) is continuous and differentiable on R2

3 For all i and j k = 1 J(i)

=) Q vi

) gt Qij (

(ik vi )v

ij gt vik

Q

(vi

) Q

(vi

)ij ij4 = 0 and|

v

i1=v

i2 |v

i1v

ij v

ij gt 0=v

i2

and Lemma 3 Property 3 is monotonicity (A3) verbatim Property 4 is nearly responsiveness (A4) Q

(vi

)

6

Properties 1-3 are nearly immediate given the properties of general from the previous section Q

except that when (as opposed to being strictly positive) Property 4 follows ij = 0 = 0vi

v

ij

from properties of i presented later in this section

We note that the single-valuedness of i in the binary action case implies that all equilibria are

i s Also the single-valuedness of

is what makes Q a function as opposed to a correspondence Hence while existence is still ensured

by Theorem 2 Brouwerrsquos fixed point theorem would have been enough For the binary action case we seek to compare exogenous and endogenous quantal response

For this it is sufficient to track accuracy which we define as the probability of taking the better action under either model16

e fv

i

ai

( vi

) 1 + e fv

i

(fv

i

)fv

i

a i ( v

i

) i ()) = a

i

( vi

e i

(7)

pure in the sense that it is never optimal to mix over differenti

(fv

i

)fv

i1 + e i

We note that ai only depends on and is strictly increasing in the product v

i

In particular ai

is strictly increasing in vi for any gt 0 and strictly increasing in for any v

i gt 0 Thus to i While there is no closed form for

i understand quantal response we analyze properties of

much can be learned via implicit methods

Theorem 3

i (0 ) = 0 and i ((i) lim v

i

) = 0 fv

i

1

16Recalling that v

i is the absolute payoff difference between the two actions these are simply Qij and Q

ij if

v

ij gt v

ik

11

(ii) (middot ) is strictly single-peaked with macr() max ( vi

) and v() argmax ( vi

)i i i

fv

i

2(01) fv

i

2(01) i (fv

i

) i (fv

i

)(iii) fv

i fv

i

i (fv

i

) |0ltfv

i

ltfv() gt 0 and fv

i

|fv

i

=0 = 0 (iv) The product

|fv

i

gtfv() lt 0 ( v

i

)( vi

) vi is strictly increasing in v

i

i v

i = 0 and lim ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof See Appendix 92

Parts (i)-(iii) of the theorem state that i (middot ) is hump-shaped starting at 0 when v

i = 0 strictly increasing to its peak and then strictly decreasing to 0 as v

i 1 This is intuitive The good action is worth v

i more than the bad action so vi represents the cost of making a

mistake or equivalently the benefit of being accurate But for any given the agent is better able

to distinguish the good action from the bad and is thus more accurate when vi is high So when

vi is very low the benefit of accuracy is low and the agent optimally chooses low precision and

thus low accuracy and when vi is very high it is so easy to distinguish the good action from the

bad that the decision maker does not need high precision for high accuracy Parts (iv)-(vi) have implications for accuracy Part (iv) implies that a increases in v

i even i

though is non-monotonic in vi

Parts (v) and (vi) are more technical in nature with no obvious i

implication for decision problems though we use them extensively in the analysis of games in

subsequent sections Part (v) implies that the accuracy associated with peak precision macr() which macr()6v() e eobtains at v() is independent of a ( v() ) = macr()6macr = 092 Part (vi) states

i v()1+e 1+e

that the location of the peak v() increases in In Corollary 1 we compare accuracy under exogenous and endogenous quantal response This

follows immediately from properties of and ai

i

Corollary 1 (i) a

i (fv

i

) fv

i

(ii) a

|fv

i

=0 = 017

i (0 ) = ai

(0 ) = 1

a

i (fv

i

) )|fv

i

gt0 gt 0 and ai(fv

i

fv

i fv

i a

i

(lim lim

gt 0 and

i ( vi

) =fv

i

1 fv

i

1 vi

) = 1a 2

gt macr (iii) If () a ( vi

) lt ai

( vi

) for all vi

i

macr (iv) If 2 (0 ()) a ( vi

) gt ai

( vi

) for intermediate vi and a ( v

i

) lt ai

( vi

) i i

for extreme vi

Figure 1 graphically summarizes Theorem 3 and Corollary 1 The left panel of Figure 1 plots i (middot ) and

i (middot 0 ) as functions of v

i for lt 0 0 as well as the constant function and macr gt macr() gt

0 ( ) Since

i (are chosen such that peak precisions are ordered middot ) is hump-shaped

17Using the quotient rule take the derivative of (7) with respect to v

i and evaluate the expression at v

i = 0

i (0 ) = 0 and i (6vi )using that6vi

|6vi=0 = 0

12

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

In this paper we take as our starting point the structure of logit quantal response (1) We do

this as a convenient parametrization for the task we outlined in the introduction to endogenize

the precision of errors given here by Logit is also extremely common in applications and fairly

representative of QRE models in that it is structural and satisfies regularity A logit QRE or LQRE

is defined in the obvious way

Definition 2 Fix An LQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i) pij = Q

ij (ui(p i

) )

3 Endogenous quantal response equilibrium

In an endogenous quantal response equilibrium (EQRE) is no longer exogenous Each player i chooses precision (or a distribution over different s) optimally subject to costs in a first stage i i

given beliefs over the opponentsrsquo second-stage actions In equilibrium the second-stage actions form a heterogeneous LQRE given the endogenously determined vector = ( 1 )

n

We begin by analyzing the first-stage optimal precision choice problem of such an agent holding

fixed beliefs Beliefs only enter the problem via the expected utilities they induce Hence given

vi 2 RJ(i) and cost parameter 2 (01) player i solves

J(i)X

i (vi ) argmax Q

ik

(vi

)vik c( ) (2)

2[01) k=1| z

b( vi

)

where Qik is given by (1) and cost function c [01) [01) satisfies c 0 (0) = 0 lim c

0 ( ) = 1

1 kand for all gt 0 c 0 ( ) gt 0 and c

00 ( ) gt 09 For example c could be a power function c( ) =

for k gt 1 For most of the paper we think of the cost function c as held fixed though for some

later results we consider the predictions obtained in the limit by a sequence of cost functions or the

set of predictions obtainable for some cost function Note that we define b(middot vi

) [01) [01)

as the benefit function for convenience This is merely player irsquos overall expected utility given the

expected utility to each of his actions and his quantal response Intuitively greater precision leads to better decisions (as we will show) but precision is itself

costly so optimal precision balances these tradeoffs Since the value of making better decisions depends on payoffs the benefit function depends on v

i

We favor an as if interpretation to the first-stage problem because the benefit function depends on a noiseless perception of v

i

even though the

agent will (and knows that he will) perceive vi with error in the second stage Such an interpretation

is essentially the same as those offered for models of rational inattention10 in which the benefit to 00 009It is fine if c (0) = 0 or lim c ( ) = 1

0+ 10Some classic references include Sims [1998] Sims [2003] Woodford [2009] and Caplin and Dean [2015]

6

an information structure depends on the stochastic choice it induces and the value to each action When v

ij = vik for all j k then b( v

i

) = the constant function that assigns utility for all levels of precision When v

ij 6= vik for some j k the following properties of b are easily verified

(see Appendix 91) P11 b(0 v

i

) = j vij and lim b( v

i

) = max vij

J(i) 1 j

b( vi

) vi

)2 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

b( vi

)

2b( v

i

) vi

)3 lim = 0 lim = 0 and 2b( lt 0 for sufficiently high

2

2 1 1

4 b( vi + e

J(i)) = b( vi

) + for all 2 R where eJ(i) = (1 1)

These are intuitive Property 1 when = 0 the agent uniformly randomizes so expects to

receive the unweighted average utility As 1 the agent takes (one of) the best action(s) with

probability approaching one Property 2 increasing will always lead to better decisions even on

the margin when = 0 so the first derivative is positive Unsurprisingly it is also bounded We also

establish that the second derivative is bounded but it cannot be signed in general which we will return to as it complicates the analysis Property 3 We establish some limiting behaviors for the

first two derivatives which are unsurprising given that b is strictly increasing and bounded Property

4 b shifts up by constant when the utility to each alternative increases by This is a simple

vi

)11consequence of the translation invariance of logit (1) which states that Qi

(vi + e

J(i)) = Qi

( Hence it is without loss to assume that v

ij gt 0 for all i and j which will simplify a number of arguments

It is easy to show that a solution to the first-stage problem exists If vij = v

ik for all j k then

the solution is trivial i = 0 which results in a uniform distribution over all actions Otherwise

any solution is strictly greater than 0 and satisfies a first-order condition This is summarized in

Theorem 1 along with basic properties of

i (

i

that follow from properties of b and c

i (Theorem 1 A solution to (2) exists ( 6= ) If vi

) is a solution it satisfies vi

) 2

2P

J(i) P

J(i)2 v

il v

il l=1 vile

l=1 vile c

0 ( ) = 0 (3)

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

1 inf (vi

) gt 0i

2 inf (vi

) and supi

i (vi ) are strictly decreasing in

3 lim 0

(inf vi

)) = 1 and lim 1

(sup vi

)) = 0 i ( i (

11See Goeree et al [2005] and Friedman [2018] for discussions of translation invariance and its implications

7

Proof See Appendix 92

One may expect equilibrium to be defined exactly as in Definition 2 but with Q ui

ij (ui ) Q

ij (macri (ui ))

replacing Qij (ui ) This is almost right except for the fact that the set of maximizers

i (vi )

is not a singleton in general (hence the use of inf and sup) This is because b(middot vi

) is not concave

for some vi

In other words the marginal benefit to may actually increase over portions of the

domain We provide an example of this in Appendix 95 which is due to Weibull et al [2007] Fortunately

i (vi ) is still well-behaved in the following sense

Lemma 1 Correspondence i RJ(i) (01) [01) is upper hemi-continuous

Proof See Appendix 92

i is upper hemi-continuous and not simply continuous because

i (vi ) may not be a singleton The next lemma establishes that

i (vi ) is a singleton however for both sufficiently low and

sufficiently high The result is proved by establishing that for extreme the solution to the

first-stage problem is in a region where the objective is locally concave

i (vi ) is a singleton for sufficiently low and sufficiently high Lemma 2 For all vi 2 RJ(i)

Proof See Appendix 92

In general equilibria may involve mixing over 2 i (middot ) Hence we define optimal mixed

precision by

i (vi )

to be any distribution over optimal selections which is also clearly optimal Finally the endogenous

i (vi )

( )vi

i i 2 [01) supp(

i

)

quantal response correspondence Qi (middot ) RJ(i)

Z 1

Ai is defined by

Qi

( )| i 2 (4)

i (vi ) Qi

(vi

)d 0

= ConvTi

(vi

)

where i (vi )

is the set of action frequencies from any optimal pure precision and Conv is the convex hull For all

Ti

(vi

) = Qi

(vi

)| 2

(vi

) an element p 0 i 2 Q

i (vi ) represents an action frequency induced by some optimal (possibly 0 P

J(i) 0 0 and

k=1 pmixed) precision and satisfies p = 1ij

ik 1 Q

to Definition 2 as a fixed point of the composite correspondence Q u A A Defining Q = (Q ) (suppressing in the notation) equilibrium is defined analogously

n

An EQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i)

Definition 3 Fix

pi 2 Q

i (ui(p i

) )

8

We establish several properties of Q Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Q i (middot ) is convex-valued and upper hemi-continuous on RJ(i)

3 For all i and j k = 1 J(i)

00 0for any pv

ij gt vik =) p

i 2 Q ij (vi )ij gt p

ik

Q

(vi

)4 If vij = max

k

vik

gt vil for some l ij gt 0 for sufficiently low and high

v

ij

Properties 1-3 are trivial12 and Property 4 is established in Appendix 93 Properties 1 and 2 are

technical Property 3 is generalized monotonicity (A3) that allows for possible non-uniqueness in

optimal quantal response Property 4 is a weaker form of responsiveness (A4) Note that it is weaker in two ways First it only holds for extreme which is required to ensure that Q is single-valued

Q

(vi

)ij 13(and differentiable) Second we are only able to show that gt 0 if v

ij = maxk

vik

gt vil

v

ij

The first two properties of Q imply existence

Theorem 2 Fix There exists an EQRE

Proof An EQRE is a fixed point of Q u Since Qi (middot ) is convex-valued and upper hemi-

continuous on RJ(i) and u is continuous on A Q u A A is convex-valued and upper hemi-continuous Because A is a compact subset of RJ Q u is also closed graph Summarizing Q u A A is a non-empty and convex-valued correspondence with a closed graph mapping

from a non-empty compact convex set to itself By Kakutanirsquos fixed point theorem Q u has a

fixed point

31 Binary actions

We now specialize the theory to the class of binary action games in which each player has two

actions Games in this class are widely used in experiments covering 2 2 games14 various voting

type games and more The theory could also be extended to an endogenous agent QRE (AQRE) (McKelvey and Palfrey [1998]) and used to analyze extensive form games in which there are two

12Since i Q

Since Qi RJ(i) [0 1)

i (vi ) = ConvT

i

(vi

) Q RJ(i) (0 1) [0 1) is upper hemi-continuous

is convex-valued 6= 6= i

also Because of the representation as Q

A

i is a continuous function andi

T

i RJ(i) (0 1) A

i is also upper hemi-continuous It is easy to check that Q i as the convexification of T

i

is also upper hemi-continuous Finally since v

ij gt v

ik =) Q

ij (vi ) gt Q

ik

(vi

) for any gt 0 it also holds when integrating over any s

13An increase in any vij will generically change

i (in either direction) Our result relies on showing that an increase in v

ij increases the product i vij if vij

implies an increase in Q ij if vij

= max

k

vik

(we conjecture that this is true even if vij 6= max

k

vik

) which = max

k

vik

(which is not necessarily true if vij 6= max

k

vik

)14Common ones used in experiments include generalized matching pennies the prisonerrsquos dilemma battle of the

sexes and at least several more

9

actions at every node such as the centipede game Importantly for our purposes with binary

actions it becomes much easier to analyze the solution to the first-stage problem (2) from which

we derive a much tighter characterization of stochastic choice e v

ij 6v

ijk eIf J(i) = 2 Q

ij (vi ) = = where vijk = v

ij vik is the signed payoff

e v

ij +e v

ik 1+e 6v

ijk P2 6v

i

edifference It is also easy to show that k=1 Qik

(vi

)vik = v

i 6v

i +const where v

i | vijk

|1+e

is the absolute payoff difference and the constant depends on the levels of vi = (v

ij vik

) Note that

vi 2 [0 1) by construction and that this is without loss We can now rewrite (2) as

fv

ie i ( v

i

) argmax vi c( ) (5)

fv

i1 + e2[01) | z

b( fv

i

)

Note that we have defined the net benefit function b(middot vi

) for this case Inspection reveals that it is simply the probability of taking the better action times the net benefit of doing so When

vi = 0 b( v

i

) = 0 for all When vi gt 0 it is easy to verify that

1 b(0 vi

) = 1 vi and lim b( v

i

) = vi

2 1

b( fv

i

)

2b( fv

i

)2 lim = 0 and lim = 0

2 1 1

b( fv

i

) fv

i

)3 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 0) for some 1 2 2 R++

2

For the most part these properties are intuitive specializations of the more general properties from

the previous section Unlike in the general case however property 3 states that b(middot vi

) and thus the first-stage objective (5) is strictly concave15 Due to this fact i ( v

i

) is a singleton and

satisfies

1 i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0

2 i ( v

i

) is strictly decreasing in if vi gt 0

3 lim i ( v

i

) = 1 and lim i ( v

i

) = 0 if vi gt 0

0 1

Since single-valued upper hemi-continuous correspondences are continuous functions we have the

following lemma as a direct corollary of Lemma 1

Lemma 3 [0 1) (0 1) [0 1) is continuous i

From Lemma 3 it follows that the binary action analogue of the endogenous quantal response

correspondence (4) is a continuous function Q (middot ) R2 (0 1) defined by ij

e (fv

i

)fv

ijk

Q vi

) = i

(6)ij ( (fv

i

)fv

ijk 1 + e i

3 vi

2b( 6vi) 6vi e

vi (e 1)15Taking derivatives 2 = lt 0 for v

i gt 0 (e vi +1)3

10

For reference we give the equilibrium definition and properties of Q for the binary action case

where J(i) = 2 for all i An EQRE is any p 2 A such that for all Definition 4 Fix i 2 1 n p

i1 = Q (ui

(p i

) ) i1

Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Qi ( middot ) is continuous and differentiable on R2

3 For all i and j k = 1 J(i)

=) Q vi

) gt Qij (

(ik vi )v

ij gt vik

Q

(vi

) Q

(vi

)ij ij4 = 0 and|

v

i1=v

i2 |v

i1v

ij v

ij gt 0=v

i2

and Lemma 3 Property 3 is monotonicity (A3) verbatim Property 4 is nearly responsiveness (A4) Q

(vi

)

6

Properties 1-3 are nearly immediate given the properties of general from the previous section Q

except that when (as opposed to being strictly positive) Property 4 follows ij = 0 = 0vi

v

ij

from properties of i presented later in this section

We note that the single-valuedness of i in the binary action case implies that all equilibria are

i s Also the single-valuedness of

is what makes Q a function as opposed to a correspondence Hence while existence is still ensured

by Theorem 2 Brouwerrsquos fixed point theorem would have been enough For the binary action case we seek to compare exogenous and endogenous quantal response

For this it is sufficient to track accuracy which we define as the probability of taking the better action under either model16

e fv

i

ai

( vi

) 1 + e fv

i

(fv

i

)fv

i

a i ( v

i

) i ()) = a

i

( vi

e i

(7)

pure in the sense that it is never optimal to mix over differenti

(fv

i

)fv

i1 + e i

We note that ai only depends on and is strictly increasing in the product v

i

In particular ai

is strictly increasing in vi for any gt 0 and strictly increasing in for any v

i gt 0 Thus to i While there is no closed form for

i understand quantal response we analyze properties of

much can be learned via implicit methods

Theorem 3

i (0 ) = 0 and i ((i) lim v

i

) = 0 fv

i

1

16Recalling that v

i is the absolute payoff difference between the two actions these are simply Qij and Q

ij if

v

ij gt v

ik

11

(ii) (middot ) is strictly single-peaked with macr() max ( vi

) and v() argmax ( vi

)i i i

fv

i

2(01) fv

i

2(01) i (fv

i

) i (fv

i

)(iii) fv

i fv

i

i (fv

i

) |0ltfv

i

ltfv() gt 0 and fv

i

|fv

i

=0 = 0 (iv) The product

|fv

i

gtfv() lt 0 ( v

i

)( vi

) vi is strictly increasing in v

i

i v

i = 0 and lim ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof See Appendix 92

Parts (i)-(iii) of the theorem state that i (middot ) is hump-shaped starting at 0 when v

i = 0 strictly increasing to its peak and then strictly decreasing to 0 as v

i 1 This is intuitive The good action is worth v

i more than the bad action so vi represents the cost of making a

mistake or equivalently the benefit of being accurate But for any given the agent is better able

to distinguish the good action from the bad and is thus more accurate when vi is high So when

vi is very low the benefit of accuracy is low and the agent optimally chooses low precision and

thus low accuracy and when vi is very high it is so easy to distinguish the good action from the

bad that the decision maker does not need high precision for high accuracy Parts (iv)-(vi) have implications for accuracy Part (iv) implies that a increases in v

i even i

though is non-monotonic in vi

Parts (v) and (vi) are more technical in nature with no obvious i

implication for decision problems though we use them extensively in the analysis of games in

subsequent sections Part (v) implies that the accuracy associated with peak precision macr() which macr()6v() e eobtains at v() is independent of a ( v() ) = macr()6macr = 092 Part (vi) states

i v()1+e 1+e

that the location of the peak v() increases in In Corollary 1 we compare accuracy under exogenous and endogenous quantal response This

follows immediately from properties of and ai

i

Corollary 1 (i) a

i (fv

i

) fv

i

(ii) a

|fv

i

=0 = 017

i (0 ) = ai

(0 ) = 1

a

i (fv

i

) )|fv

i

gt0 gt 0 and ai(fv

i

fv

i fv

i a

i

(lim lim

gt 0 and

i ( vi

) =fv

i

1 fv

i

1 vi

) = 1a 2

gt macr (iii) If () a ( vi

) lt ai

( vi

) for all vi

i

macr (iv) If 2 (0 ()) a ( vi

) gt ai

( vi

) for intermediate vi and a ( v

i

) lt ai

( vi

) i i

for extreme vi

Figure 1 graphically summarizes Theorem 3 and Corollary 1 The left panel of Figure 1 plots i (middot ) and

i (middot 0 ) as functions of v

i for lt 0 0 as well as the constant function and macr gt macr() gt

0 ( ) Since

i (are chosen such that peak precisions are ordered middot ) is hump-shaped

17Using the quotient rule take the derivative of (7) with respect to v

i and evaluate the expression at v

i = 0

i (0 ) = 0 and i (6vi )using that6vi

|6vi=0 = 0

12

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

an information structure depends on the stochastic choice it induces and the value to each action When v

ij = vik for all j k then b( v

i

) = the constant function that assigns utility for all levels of precision When v

ij 6= vik for some j k the following properties of b are easily verified

(see Appendix 91) P11 b(0 v

i

) = j vij and lim b( v

i

) = max vij

J(i) 1 j

b( vi

) vi

)2 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

b( vi

)

2b( v

i

) vi

)3 lim = 0 lim = 0 and 2b( lt 0 for sufficiently high

2

2 1 1

4 b( vi + e

J(i)) = b( vi

) + for all 2 R where eJ(i) = (1 1)

These are intuitive Property 1 when = 0 the agent uniformly randomizes so expects to

receive the unweighted average utility As 1 the agent takes (one of) the best action(s) with

probability approaching one Property 2 increasing will always lead to better decisions even on

the margin when = 0 so the first derivative is positive Unsurprisingly it is also bounded We also

establish that the second derivative is bounded but it cannot be signed in general which we will return to as it complicates the analysis Property 3 We establish some limiting behaviors for the

first two derivatives which are unsurprising given that b is strictly increasing and bounded Property

4 b shifts up by constant when the utility to each alternative increases by This is a simple

vi

)11consequence of the translation invariance of logit (1) which states that Qi

(vi + e

J(i)) = Qi

( Hence it is without loss to assume that v

ij gt 0 for all i and j which will simplify a number of arguments

It is easy to show that a solution to the first-stage problem exists If vij = v

ik for all j k then

the solution is trivial i = 0 which results in a uniform distribution over all actions Otherwise

any solution is strictly greater than 0 and satisfies a first-order condition This is summarized in

Theorem 1 along with basic properties of

i (

i

that follow from properties of b and c

i (Theorem 1 A solution to (2) exists ( 6= ) If vi

) is a solution it satisfies vi

) 2

2P

J(i) P

J(i)2 v

il v

il l=1 vile

l=1 vile c

0 ( ) = 0 (3)

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

1 inf (vi

) gt 0i

2 inf (vi

) and supi

i (vi ) are strictly decreasing in

3 lim 0

(inf vi

)) = 1 and lim 1

(sup vi

)) = 0 i ( i (

11See Goeree et al [2005] and Friedman [2018] for discussions of translation invariance and its implications

7

Proof See Appendix 92

One may expect equilibrium to be defined exactly as in Definition 2 but with Q ui

ij (ui ) Q

ij (macri (ui ))

replacing Qij (ui ) This is almost right except for the fact that the set of maximizers

i (vi )

is not a singleton in general (hence the use of inf and sup) This is because b(middot vi

) is not concave

for some vi

In other words the marginal benefit to may actually increase over portions of the

domain We provide an example of this in Appendix 95 which is due to Weibull et al [2007] Fortunately

i (vi ) is still well-behaved in the following sense

Lemma 1 Correspondence i RJ(i) (01) [01) is upper hemi-continuous

Proof See Appendix 92

i is upper hemi-continuous and not simply continuous because

i (vi ) may not be a singleton The next lemma establishes that

i (vi ) is a singleton however for both sufficiently low and

sufficiently high The result is proved by establishing that for extreme the solution to the

first-stage problem is in a region where the objective is locally concave

i (vi ) is a singleton for sufficiently low and sufficiently high Lemma 2 For all vi 2 RJ(i)

Proof See Appendix 92

In general equilibria may involve mixing over 2 i (middot ) Hence we define optimal mixed

precision by

i (vi )

to be any distribution over optimal selections which is also clearly optimal Finally the endogenous

i (vi )

( )vi

i i 2 [01) supp(

i

)

quantal response correspondence Qi (middot ) RJ(i)

Z 1

Ai is defined by

Qi

( )| i 2 (4)

i (vi ) Qi

(vi

)d 0

= ConvTi

(vi

)

where i (vi )

is the set of action frequencies from any optimal pure precision and Conv is the convex hull For all

Ti

(vi

) = Qi

(vi

)| 2

(vi

) an element p 0 i 2 Q

i (vi ) represents an action frequency induced by some optimal (possibly 0 P

J(i) 0 0 and

k=1 pmixed) precision and satisfies p = 1ij

ik 1 Q

to Definition 2 as a fixed point of the composite correspondence Q u A A Defining Q = (Q ) (suppressing in the notation) equilibrium is defined analogously

n

An EQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i)

Definition 3 Fix

pi 2 Q

i (ui(p i

) )

8

We establish several properties of Q Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Q i (middot ) is convex-valued and upper hemi-continuous on RJ(i)

3 For all i and j k = 1 J(i)

00 0for any pv

ij gt vik =) p

i 2 Q ij (vi )ij gt p

ik

Q

(vi

)4 If vij = max

k

vik

gt vil for some l ij gt 0 for sufficiently low and high

v

ij

Properties 1-3 are trivial12 and Property 4 is established in Appendix 93 Properties 1 and 2 are

technical Property 3 is generalized monotonicity (A3) that allows for possible non-uniqueness in

optimal quantal response Property 4 is a weaker form of responsiveness (A4) Note that it is weaker in two ways First it only holds for extreme which is required to ensure that Q is single-valued

Q

(vi

)ij 13(and differentiable) Second we are only able to show that gt 0 if v

ij = maxk

vik

gt vil

v

ij

The first two properties of Q imply existence

Theorem 2 Fix There exists an EQRE

Proof An EQRE is a fixed point of Q u Since Qi (middot ) is convex-valued and upper hemi-

continuous on RJ(i) and u is continuous on A Q u A A is convex-valued and upper hemi-continuous Because A is a compact subset of RJ Q u is also closed graph Summarizing Q u A A is a non-empty and convex-valued correspondence with a closed graph mapping

from a non-empty compact convex set to itself By Kakutanirsquos fixed point theorem Q u has a

fixed point

31 Binary actions

We now specialize the theory to the class of binary action games in which each player has two

actions Games in this class are widely used in experiments covering 2 2 games14 various voting

type games and more The theory could also be extended to an endogenous agent QRE (AQRE) (McKelvey and Palfrey [1998]) and used to analyze extensive form games in which there are two

12Since i Q

Since Qi RJ(i) [0 1)

i (vi ) = ConvT

i

(vi

) Q RJ(i) (0 1) [0 1) is upper hemi-continuous

is convex-valued 6= 6= i

also Because of the representation as Q

A

i is a continuous function andi

T

i RJ(i) (0 1) A

i is also upper hemi-continuous It is easy to check that Q i as the convexification of T

i

is also upper hemi-continuous Finally since v

ij gt v

ik =) Q

ij (vi ) gt Q

ik

(vi

) for any gt 0 it also holds when integrating over any s

13An increase in any vij will generically change

i (in either direction) Our result relies on showing that an increase in v

ij increases the product i vij if vij

implies an increase in Q ij if vij

= max

k

vik

(we conjecture that this is true even if vij 6= max

k

vik

) which = max

k

vik

(which is not necessarily true if vij 6= max

k

vik

)14Common ones used in experiments include generalized matching pennies the prisonerrsquos dilemma battle of the

sexes and at least several more

9

actions at every node such as the centipede game Importantly for our purposes with binary

actions it becomes much easier to analyze the solution to the first-stage problem (2) from which

we derive a much tighter characterization of stochastic choice e v

ij 6v

ijk eIf J(i) = 2 Q

ij (vi ) = = where vijk = v

ij vik is the signed payoff

e v

ij +e v

ik 1+e 6v

ijk P2 6v

i

edifference It is also easy to show that k=1 Qik

(vi

)vik = v

i 6v

i +const where v

i | vijk

|1+e

is the absolute payoff difference and the constant depends on the levels of vi = (v

ij vik

) Note that

vi 2 [0 1) by construction and that this is without loss We can now rewrite (2) as

fv

ie i ( v

i

) argmax vi c( ) (5)

fv

i1 + e2[01) | z

b( fv

i

)

Note that we have defined the net benefit function b(middot vi

) for this case Inspection reveals that it is simply the probability of taking the better action times the net benefit of doing so When

vi = 0 b( v

i

) = 0 for all When vi gt 0 it is easy to verify that

1 b(0 vi

) = 1 vi and lim b( v

i

) = vi

2 1

b( fv

i

)

2b( fv

i

)2 lim = 0 and lim = 0

2 1 1

b( fv

i

) fv

i

)3 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 0) for some 1 2 2 R++

2

For the most part these properties are intuitive specializations of the more general properties from

the previous section Unlike in the general case however property 3 states that b(middot vi

) and thus the first-stage objective (5) is strictly concave15 Due to this fact i ( v

i

) is a singleton and

satisfies

1 i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0

2 i ( v

i

) is strictly decreasing in if vi gt 0

3 lim i ( v

i

) = 1 and lim i ( v

i

) = 0 if vi gt 0

0 1

Since single-valued upper hemi-continuous correspondences are continuous functions we have the

following lemma as a direct corollary of Lemma 1

Lemma 3 [0 1) (0 1) [0 1) is continuous i

From Lemma 3 it follows that the binary action analogue of the endogenous quantal response

correspondence (4) is a continuous function Q (middot ) R2 (0 1) defined by ij

e (fv

i

)fv

ijk

Q vi

) = i

(6)ij ( (fv

i

)fv

ijk 1 + e i

3 vi

2b( 6vi) 6vi e

vi (e 1)15Taking derivatives 2 = lt 0 for v

i gt 0 (e vi +1)3

10

For reference we give the equilibrium definition and properties of Q for the binary action case

where J(i) = 2 for all i An EQRE is any p 2 A such that for all Definition 4 Fix i 2 1 n p

i1 = Q (ui

(p i

) ) i1

Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Qi ( middot ) is continuous and differentiable on R2

3 For all i and j k = 1 J(i)

=) Q vi

) gt Qij (

(ik vi )v

ij gt vik

Q

(vi

) Q

(vi

)ij ij4 = 0 and|

v

i1=v

i2 |v

i1v

ij v

ij gt 0=v

i2

and Lemma 3 Property 3 is monotonicity (A3) verbatim Property 4 is nearly responsiveness (A4) Q

(vi

)

6

Properties 1-3 are nearly immediate given the properties of general from the previous section Q

except that when (as opposed to being strictly positive) Property 4 follows ij = 0 = 0vi

v

ij

from properties of i presented later in this section

We note that the single-valuedness of i in the binary action case implies that all equilibria are

i s Also the single-valuedness of

is what makes Q a function as opposed to a correspondence Hence while existence is still ensured

by Theorem 2 Brouwerrsquos fixed point theorem would have been enough For the binary action case we seek to compare exogenous and endogenous quantal response

For this it is sufficient to track accuracy which we define as the probability of taking the better action under either model16

e fv

i

ai

( vi

) 1 + e fv

i

(fv

i

)fv

i

a i ( v

i

) i ()) = a

i

( vi

e i

(7)

pure in the sense that it is never optimal to mix over differenti

(fv

i

)fv

i1 + e i

We note that ai only depends on and is strictly increasing in the product v

i

In particular ai

is strictly increasing in vi for any gt 0 and strictly increasing in for any v

i gt 0 Thus to i While there is no closed form for

i understand quantal response we analyze properties of

much can be learned via implicit methods

Theorem 3

i (0 ) = 0 and i ((i) lim v

i

) = 0 fv

i

1

16Recalling that v

i is the absolute payoff difference between the two actions these are simply Qij and Q

ij if

v

ij gt v

ik

11

(ii) (middot ) is strictly single-peaked with macr() max ( vi

) and v() argmax ( vi

)i i i

fv

i

2(01) fv

i

2(01) i (fv

i

) i (fv

i

)(iii) fv

i fv

i

i (fv

i

) |0ltfv

i

ltfv() gt 0 and fv

i

|fv

i

=0 = 0 (iv) The product

|fv

i

gtfv() lt 0 ( v

i

)( vi

) vi is strictly increasing in v

i

i v

i = 0 and lim ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof See Appendix 92

Parts (i)-(iii) of the theorem state that i (middot ) is hump-shaped starting at 0 when v

i = 0 strictly increasing to its peak and then strictly decreasing to 0 as v

i 1 This is intuitive The good action is worth v

i more than the bad action so vi represents the cost of making a

mistake or equivalently the benefit of being accurate But for any given the agent is better able

to distinguish the good action from the bad and is thus more accurate when vi is high So when

vi is very low the benefit of accuracy is low and the agent optimally chooses low precision and

thus low accuracy and when vi is very high it is so easy to distinguish the good action from the

bad that the decision maker does not need high precision for high accuracy Parts (iv)-(vi) have implications for accuracy Part (iv) implies that a increases in v

i even i

though is non-monotonic in vi

Parts (v) and (vi) are more technical in nature with no obvious i

implication for decision problems though we use them extensively in the analysis of games in

subsequent sections Part (v) implies that the accuracy associated with peak precision macr() which macr()6v() e eobtains at v() is independent of a ( v() ) = macr()6macr = 092 Part (vi) states

i v()1+e 1+e

that the location of the peak v() increases in In Corollary 1 we compare accuracy under exogenous and endogenous quantal response This

follows immediately from properties of and ai

i

Corollary 1 (i) a

i (fv

i

) fv

i

(ii) a

|fv

i

=0 = 017

i (0 ) = ai

(0 ) = 1

a

i (fv

i

) )|fv

i

gt0 gt 0 and ai(fv

i

fv

i fv

i a

i

(lim lim

gt 0 and

i ( vi

) =fv

i

1 fv

i

1 vi

) = 1a 2

gt macr (iii) If () a ( vi

) lt ai

( vi

) for all vi

i

macr (iv) If 2 (0 ()) a ( vi

) gt ai

( vi

) for intermediate vi and a ( v

i

) lt ai

( vi

) i i

for extreme vi

Figure 1 graphically summarizes Theorem 3 and Corollary 1 The left panel of Figure 1 plots i (middot ) and

i (middot 0 ) as functions of v

i for lt 0 0 as well as the constant function and macr gt macr() gt

0 ( ) Since

i (are chosen such that peak precisions are ordered middot ) is hump-shaped

17Using the quotient rule take the derivative of (7) with respect to v

i and evaluate the expression at v

i = 0

i (0 ) = 0 and i (6vi )using that6vi

|6vi=0 = 0

12

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

Proof See Appendix 92

One may expect equilibrium to be defined exactly as in Definition 2 but with Q ui

ij (ui ) Q

ij (macri (ui ))

replacing Qij (ui ) This is almost right except for the fact that the set of maximizers

i (vi )

is not a singleton in general (hence the use of inf and sup) This is because b(middot vi

) is not concave

for some vi

In other words the marginal benefit to may actually increase over portions of the

domain We provide an example of this in Appendix 95 which is due to Weibull et al [2007] Fortunately

i (vi ) is still well-behaved in the following sense

Lemma 1 Correspondence i RJ(i) (01) [01) is upper hemi-continuous

Proof See Appendix 92

i is upper hemi-continuous and not simply continuous because

i (vi ) may not be a singleton The next lemma establishes that

i (vi ) is a singleton however for both sufficiently low and

sufficiently high The result is proved by establishing that for extreme the solution to the

first-stage problem is in a region where the objective is locally concave

i (vi ) is a singleton for sufficiently low and sufficiently high Lemma 2 For all vi 2 RJ(i)

Proof See Appendix 92

In general equilibria may involve mixing over 2 i (middot ) Hence we define optimal mixed

precision by

i (vi )

to be any distribution over optimal selections which is also clearly optimal Finally the endogenous

i (vi )

( )vi

i i 2 [01) supp(

i

)

quantal response correspondence Qi (middot ) RJ(i)

Z 1

Ai is defined by

Qi

( )| i 2 (4)

i (vi ) Qi

(vi

)d 0

= ConvTi

(vi

)

where i (vi )

is the set of action frequencies from any optimal pure precision and Conv is the convex hull For all

Ti

(vi

) = Qi

(vi

)| 2

(vi

) an element p 0 i 2 Q

i (vi ) represents an action frequency induced by some optimal (possibly 0 P

J(i) 0 0 and

k=1 pmixed) precision and satisfies p = 1ij

ik 1 Q

to Definition 2 as a fixed point of the composite correspondence Q u A A Defining Q = (Q ) (suppressing in the notation) equilibrium is defined analogously

n

An EQRE is any p 2 A such that for all i 2 1 n and all j 2 1 J(i)

Definition 3 Fix

pi 2 Q

i (ui(p i

) )

8

We establish several properties of Q Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Q i (middot ) is convex-valued and upper hemi-continuous on RJ(i)

3 For all i and j k = 1 J(i)

00 0for any pv

ij gt vik =) p

i 2 Q ij (vi )ij gt p

ik

Q

(vi

)4 If vij = max

k

vik

gt vil for some l ij gt 0 for sufficiently low and high

v

ij

Properties 1-3 are trivial12 and Property 4 is established in Appendix 93 Properties 1 and 2 are

technical Property 3 is generalized monotonicity (A3) that allows for possible non-uniqueness in

optimal quantal response Property 4 is a weaker form of responsiveness (A4) Note that it is weaker in two ways First it only holds for extreme which is required to ensure that Q is single-valued

Q

(vi

)ij 13(and differentiable) Second we are only able to show that gt 0 if v

ij = maxk

vik

gt vil

v

ij

The first two properties of Q imply existence

Theorem 2 Fix There exists an EQRE

Proof An EQRE is a fixed point of Q u Since Qi (middot ) is convex-valued and upper hemi-

continuous on RJ(i) and u is continuous on A Q u A A is convex-valued and upper hemi-continuous Because A is a compact subset of RJ Q u is also closed graph Summarizing Q u A A is a non-empty and convex-valued correspondence with a closed graph mapping

from a non-empty compact convex set to itself By Kakutanirsquos fixed point theorem Q u has a

fixed point

31 Binary actions

We now specialize the theory to the class of binary action games in which each player has two

actions Games in this class are widely used in experiments covering 2 2 games14 various voting

type games and more The theory could also be extended to an endogenous agent QRE (AQRE) (McKelvey and Palfrey [1998]) and used to analyze extensive form games in which there are two

12Since i Q

Since Qi RJ(i) [0 1)

i (vi ) = ConvT

i

(vi

) Q RJ(i) (0 1) [0 1) is upper hemi-continuous

is convex-valued 6= 6= i

also Because of the representation as Q

A

i is a continuous function andi

T

i RJ(i) (0 1) A

i is also upper hemi-continuous It is easy to check that Q i as the convexification of T

i

is also upper hemi-continuous Finally since v

ij gt v

ik =) Q

ij (vi ) gt Q

ik

(vi

) for any gt 0 it also holds when integrating over any s

13An increase in any vij will generically change

i (in either direction) Our result relies on showing that an increase in v

ij increases the product i vij if vij

implies an increase in Q ij if vij

= max

k

vik

(we conjecture that this is true even if vij 6= max

k

vik

) which = max

k

vik

(which is not necessarily true if vij 6= max

k

vik

)14Common ones used in experiments include generalized matching pennies the prisonerrsquos dilemma battle of the

sexes and at least several more

9

actions at every node such as the centipede game Importantly for our purposes with binary

actions it becomes much easier to analyze the solution to the first-stage problem (2) from which

we derive a much tighter characterization of stochastic choice e v

ij 6v

ijk eIf J(i) = 2 Q

ij (vi ) = = where vijk = v

ij vik is the signed payoff

e v

ij +e v

ik 1+e 6v

ijk P2 6v

i

edifference It is also easy to show that k=1 Qik

(vi

)vik = v

i 6v

i +const where v

i | vijk

|1+e

is the absolute payoff difference and the constant depends on the levels of vi = (v

ij vik

) Note that

vi 2 [0 1) by construction and that this is without loss We can now rewrite (2) as

fv

ie i ( v

i

) argmax vi c( ) (5)

fv

i1 + e2[01) | z

b( fv

i

)

Note that we have defined the net benefit function b(middot vi

) for this case Inspection reveals that it is simply the probability of taking the better action times the net benefit of doing so When

vi = 0 b( v

i

) = 0 for all When vi gt 0 it is easy to verify that

1 b(0 vi

) = 1 vi and lim b( v

i

) = vi

2 1

b( fv

i

)

2b( fv

i

)2 lim = 0 and lim = 0

2 1 1

b( fv

i

) fv

i

)3 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 0) for some 1 2 2 R++

2

For the most part these properties are intuitive specializations of the more general properties from

the previous section Unlike in the general case however property 3 states that b(middot vi

) and thus the first-stage objective (5) is strictly concave15 Due to this fact i ( v

i

) is a singleton and

satisfies

1 i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0

2 i ( v

i

) is strictly decreasing in if vi gt 0

3 lim i ( v

i

) = 1 and lim i ( v

i

) = 0 if vi gt 0

0 1

Since single-valued upper hemi-continuous correspondences are continuous functions we have the

following lemma as a direct corollary of Lemma 1

Lemma 3 [0 1) (0 1) [0 1) is continuous i

From Lemma 3 it follows that the binary action analogue of the endogenous quantal response

correspondence (4) is a continuous function Q (middot ) R2 (0 1) defined by ij

e (fv

i

)fv

ijk

Q vi

) = i

(6)ij ( (fv

i

)fv

ijk 1 + e i

3 vi

2b( 6vi) 6vi e

vi (e 1)15Taking derivatives 2 = lt 0 for v

i gt 0 (e vi +1)3

10

For reference we give the equilibrium definition and properties of Q for the binary action case

where J(i) = 2 for all i An EQRE is any p 2 A such that for all Definition 4 Fix i 2 1 n p

i1 = Q (ui

(p i

) ) i1

Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Qi ( middot ) is continuous and differentiable on R2

3 For all i and j k = 1 J(i)

=) Q vi

) gt Qij (

(ik vi )v

ij gt vik

Q

(vi

) Q

(vi

)ij ij4 = 0 and|

v

i1=v

i2 |v

i1v

ij v

ij gt 0=v

i2

and Lemma 3 Property 3 is monotonicity (A3) verbatim Property 4 is nearly responsiveness (A4) Q

(vi

)

6

Properties 1-3 are nearly immediate given the properties of general from the previous section Q

except that when (as opposed to being strictly positive) Property 4 follows ij = 0 = 0vi

v

ij

from properties of i presented later in this section

We note that the single-valuedness of i in the binary action case implies that all equilibria are

i s Also the single-valuedness of

is what makes Q a function as opposed to a correspondence Hence while existence is still ensured

by Theorem 2 Brouwerrsquos fixed point theorem would have been enough For the binary action case we seek to compare exogenous and endogenous quantal response

For this it is sufficient to track accuracy which we define as the probability of taking the better action under either model16

e fv

i

ai

( vi

) 1 + e fv

i

(fv

i

)fv

i

a i ( v

i

) i ()) = a

i

( vi

e i

(7)

pure in the sense that it is never optimal to mix over differenti

(fv

i

)fv

i1 + e i

We note that ai only depends on and is strictly increasing in the product v

i

In particular ai

is strictly increasing in vi for any gt 0 and strictly increasing in for any v

i gt 0 Thus to i While there is no closed form for

i understand quantal response we analyze properties of

much can be learned via implicit methods

Theorem 3

i (0 ) = 0 and i ((i) lim v

i

) = 0 fv

i

1

16Recalling that v

i is the absolute payoff difference between the two actions these are simply Qij and Q

ij if

v

ij gt v

ik

11

(ii) (middot ) is strictly single-peaked with macr() max ( vi

) and v() argmax ( vi

)i i i

fv

i

2(01) fv

i

2(01) i (fv

i

) i (fv

i

)(iii) fv

i fv

i

i (fv

i

) |0ltfv

i

ltfv() gt 0 and fv

i

|fv

i

=0 = 0 (iv) The product

|fv

i

gtfv() lt 0 ( v

i

)( vi

) vi is strictly increasing in v

i

i v

i = 0 and lim ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof See Appendix 92

Parts (i)-(iii) of the theorem state that i (middot ) is hump-shaped starting at 0 when v

i = 0 strictly increasing to its peak and then strictly decreasing to 0 as v

i 1 This is intuitive The good action is worth v

i more than the bad action so vi represents the cost of making a

mistake or equivalently the benefit of being accurate But for any given the agent is better able

to distinguish the good action from the bad and is thus more accurate when vi is high So when

vi is very low the benefit of accuracy is low and the agent optimally chooses low precision and

thus low accuracy and when vi is very high it is so easy to distinguish the good action from the

bad that the decision maker does not need high precision for high accuracy Parts (iv)-(vi) have implications for accuracy Part (iv) implies that a increases in v

i even i

though is non-monotonic in vi

Parts (v) and (vi) are more technical in nature with no obvious i

implication for decision problems though we use them extensively in the analysis of games in

subsequent sections Part (v) implies that the accuracy associated with peak precision macr() which macr()6v() e eobtains at v() is independent of a ( v() ) = macr()6macr = 092 Part (vi) states

i v()1+e 1+e

that the location of the peak v() increases in In Corollary 1 we compare accuracy under exogenous and endogenous quantal response This

follows immediately from properties of and ai

i

Corollary 1 (i) a

i (fv

i

) fv

i

(ii) a

|fv

i

=0 = 017

i (0 ) = ai

(0 ) = 1

a

i (fv

i

) )|fv

i

gt0 gt 0 and ai(fv

i

fv

i fv

i a

i

(lim lim

gt 0 and

i ( vi

) =fv

i

1 fv

i

1 vi

) = 1a 2

gt macr (iii) If () a ( vi

) lt ai

( vi

) for all vi

i

macr (iv) If 2 (0 ()) a ( vi

) gt ai

( vi

) for intermediate vi and a ( v

i

) lt ai

( vi

) i i

for extreme vi

Figure 1 graphically summarizes Theorem 3 and Corollary 1 The left panel of Figure 1 plots i (middot ) and

i (middot 0 ) as functions of v

i for lt 0 0 as well as the constant function and macr gt macr() gt

0 ( ) Since

i (are chosen such that peak precisions are ordered middot ) is hump-shaped

17Using the quotient rule take the derivative of (7) with respect to v

i and evaluate the expression at v

i = 0

i (0 ) = 0 and i (6vi )using that6vi

|6vi=0 = 0

12

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

We establish several properties of Q Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Q i (middot ) is convex-valued and upper hemi-continuous on RJ(i)

3 For all i and j k = 1 J(i)

00 0for any pv

ij gt vik =) p

i 2 Q ij (vi )ij gt p

ik

Q

(vi

)4 If vij = max

k

vik

gt vil for some l ij gt 0 for sufficiently low and high

v

ij

Properties 1-3 are trivial12 and Property 4 is established in Appendix 93 Properties 1 and 2 are

technical Property 3 is generalized monotonicity (A3) that allows for possible non-uniqueness in

optimal quantal response Property 4 is a weaker form of responsiveness (A4) Note that it is weaker in two ways First it only holds for extreme which is required to ensure that Q is single-valued

Q

(vi

)ij 13(and differentiable) Second we are only able to show that gt 0 if v

ij = maxk

vik

gt vil

v

ij

The first two properties of Q imply existence

Theorem 2 Fix There exists an EQRE

Proof An EQRE is a fixed point of Q u Since Qi (middot ) is convex-valued and upper hemi-

continuous on RJ(i) and u is continuous on A Q u A A is convex-valued and upper hemi-continuous Because A is a compact subset of RJ Q u is also closed graph Summarizing Q u A A is a non-empty and convex-valued correspondence with a closed graph mapping

from a non-empty compact convex set to itself By Kakutanirsquos fixed point theorem Q u has a

fixed point

31 Binary actions

We now specialize the theory to the class of binary action games in which each player has two

actions Games in this class are widely used in experiments covering 2 2 games14 various voting

type games and more The theory could also be extended to an endogenous agent QRE (AQRE) (McKelvey and Palfrey [1998]) and used to analyze extensive form games in which there are two

12Since i Q

Since Qi RJ(i) [0 1)

i (vi ) = ConvT

i

(vi

) Q RJ(i) (0 1) [0 1) is upper hemi-continuous

is convex-valued 6= 6= i

also Because of the representation as Q

A

i is a continuous function andi

T

i RJ(i) (0 1) A

i is also upper hemi-continuous It is easy to check that Q i as the convexification of T

i

is also upper hemi-continuous Finally since v

ij gt v

ik =) Q

ij (vi ) gt Q

ik

(vi

) for any gt 0 it also holds when integrating over any s

13An increase in any vij will generically change

i (in either direction) Our result relies on showing that an increase in v

ij increases the product i vij if vij

implies an increase in Q ij if vij

= max

k

vik

(we conjecture that this is true even if vij 6= max

k

vik

) which = max

k

vik

(which is not necessarily true if vij 6= max

k

vik

)14Common ones used in experiments include generalized matching pennies the prisonerrsquos dilemma battle of the

sexes and at least several more

9

actions at every node such as the centipede game Importantly for our purposes with binary

actions it becomes much easier to analyze the solution to the first-stage problem (2) from which

we derive a much tighter characterization of stochastic choice e v

ij 6v

ijk eIf J(i) = 2 Q

ij (vi ) = = where vijk = v

ij vik is the signed payoff

e v

ij +e v

ik 1+e 6v

ijk P2 6v

i

edifference It is also easy to show that k=1 Qik

(vi

)vik = v

i 6v

i +const where v

i | vijk

|1+e

is the absolute payoff difference and the constant depends on the levels of vi = (v

ij vik

) Note that

vi 2 [0 1) by construction and that this is without loss We can now rewrite (2) as

fv

ie i ( v

i

) argmax vi c( ) (5)

fv

i1 + e2[01) | z

b( fv

i

)

Note that we have defined the net benefit function b(middot vi

) for this case Inspection reveals that it is simply the probability of taking the better action times the net benefit of doing so When

vi = 0 b( v

i

) = 0 for all When vi gt 0 it is easy to verify that

1 b(0 vi

) = 1 vi and lim b( v

i

) = vi

2 1

b( fv

i

)

2b( fv

i

)2 lim = 0 and lim = 0

2 1 1

b( fv

i

) fv

i

)3 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 0) for some 1 2 2 R++

2

For the most part these properties are intuitive specializations of the more general properties from

the previous section Unlike in the general case however property 3 states that b(middot vi

) and thus the first-stage objective (5) is strictly concave15 Due to this fact i ( v

i

) is a singleton and

satisfies

1 i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0

2 i ( v

i

) is strictly decreasing in if vi gt 0

3 lim i ( v

i

) = 1 and lim i ( v

i

) = 0 if vi gt 0

0 1

Since single-valued upper hemi-continuous correspondences are continuous functions we have the

following lemma as a direct corollary of Lemma 1

Lemma 3 [0 1) (0 1) [0 1) is continuous i

From Lemma 3 it follows that the binary action analogue of the endogenous quantal response

correspondence (4) is a continuous function Q (middot ) R2 (0 1) defined by ij

e (fv

i

)fv

ijk

Q vi

) = i

(6)ij ( (fv

i

)fv

ijk 1 + e i

3 vi

2b( 6vi) 6vi e

vi (e 1)15Taking derivatives 2 = lt 0 for v

i gt 0 (e vi +1)3

10

For reference we give the equilibrium definition and properties of Q for the binary action case

where J(i) = 2 for all i An EQRE is any p 2 A such that for all Definition 4 Fix i 2 1 n p

i1 = Q (ui

(p i

) ) i1

Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Qi ( middot ) is continuous and differentiable on R2

3 For all i and j k = 1 J(i)

=) Q vi

) gt Qij (

(ik vi )v

ij gt vik

Q

(vi

) Q

(vi

)ij ij4 = 0 and|

v

i1=v

i2 |v

i1v

ij v

ij gt 0=v

i2

and Lemma 3 Property 3 is monotonicity (A3) verbatim Property 4 is nearly responsiveness (A4) Q

(vi

)

6

Properties 1-3 are nearly immediate given the properties of general from the previous section Q

except that when (as opposed to being strictly positive) Property 4 follows ij = 0 = 0vi

v

ij

from properties of i presented later in this section

We note that the single-valuedness of i in the binary action case implies that all equilibria are

i s Also the single-valuedness of

is what makes Q a function as opposed to a correspondence Hence while existence is still ensured

by Theorem 2 Brouwerrsquos fixed point theorem would have been enough For the binary action case we seek to compare exogenous and endogenous quantal response

For this it is sufficient to track accuracy which we define as the probability of taking the better action under either model16

e fv

i

ai

( vi

) 1 + e fv

i

(fv

i

)fv

i

a i ( v

i

) i ()) = a

i

( vi

e i

(7)

pure in the sense that it is never optimal to mix over differenti

(fv

i

)fv

i1 + e i

We note that ai only depends on and is strictly increasing in the product v

i

In particular ai

is strictly increasing in vi for any gt 0 and strictly increasing in for any v

i gt 0 Thus to i While there is no closed form for

i understand quantal response we analyze properties of

much can be learned via implicit methods

Theorem 3

i (0 ) = 0 and i ((i) lim v

i

) = 0 fv

i

1

16Recalling that v

i is the absolute payoff difference between the two actions these are simply Qij and Q

ij if

v

ij gt v

ik

11

(ii) (middot ) is strictly single-peaked with macr() max ( vi

) and v() argmax ( vi

)i i i

fv

i

2(01) fv

i

2(01) i (fv

i

) i (fv

i

)(iii) fv

i fv

i

i (fv

i

) |0ltfv

i

ltfv() gt 0 and fv

i

|fv

i

=0 = 0 (iv) The product

|fv

i

gtfv() lt 0 ( v

i

)( vi

) vi is strictly increasing in v

i

i v

i = 0 and lim ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof See Appendix 92

Parts (i)-(iii) of the theorem state that i (middot ) is hump-shaped starting at 0 when v

i = 0 strictly increasing to its peak and then strictly decreasing to 0 as v

i 1 This is intuitive The good action is worth v

i more than the bad action so vi represents the cost of making a

mistake or equivalently the benefit of being accurate But for any given the agent is better able

to distinguish the good action from the bad and is thus more accurate when vi is high So when

vi is very low the benefit of accuracy is low and the agent optimally chooses low precision and

thus low accuracy and when vi is very high it is so easy to distinguish the good action from the

bad that the decision maker does not need high precision for high accuracy Parts (iv)-(vi) have implications for accuracy Part (iv) implies that a increases in v

i even i

though is non-monotonic in vi

Parts (v) and (vi) are more technical in nature with no obvious i

implication for decision problems though we use them extensively in the analysis of games in

subsequent sections Part (v) implies that the accuracy associated with peak precision macr() which macr()6v() e eobtains at v() is independent of a ( v() ) = macr()6macr = 092 Part (vi) states

i v()1+e 1+e

that the location of the peak v() increases in In Corollary 1 we compare accuracy under exogenous and endogenous quantal response This

follows immediately from properties of and ai

i

Corollary 1 (i) a

i (fv

i

) fv

i

(ii) a

|fv

i

=0 = 017

i (0 ) = ai

(0 ) = 1

a

i (fv

i

) )|fv

i

gt0 gt 0 and ai(fv

i

fv

i fv

i a

i

(lim lim

gt 0 and

i ( vi

) =fv

i

1 fv

i

1 vi

) = 1a 2

gt macr (iii) If () a ( vi

) lt ai

( vi

) for all vi

i

macr (iv) If 2 (0 ()) a ( vi

) gt ai

( vi

) for intermediate vi and a ( v

i

) lt ai

( vi

) i i

for extreme vi

Figure 1 graphically summarizes Theorem 3 and Corollary 1 The left panel of Figure 1 plots i (middot ) and

i (middot 0 ) as functions of v

i for lt 0 0 as well as the constant function and macr gt macr() gt

0 ( ) Since

i (are chosen such that peak precisions are ordered middot ) is hump-shaped

17Using the quotient rule take the derivative of (7) with respect to v

i and evaluate the expression at v

i = 0

i (0 ) = 0 and i (6vi )using that6vi

|6vi=0 = 0

12

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

actions at every node such as the centipede game Importantly for our purposes with binary

actions it becomes much easier to analyze the solution to the first-stage problem (2) from which

we derive a much tighter characterization of stochastic choice e v

ij 6v

ijk eIf J(i) = 2 Q

ij (vi ) = = where vijk = v

ij vik is the signed payoff

e v

ij +e v

ik 1+e 6v

ijk P2 6v

i

edifference It is also easy to show that k=1 Qik

(vi

)vik = v

i 6v

i +const where v

i | vijk

|1+e

is the absolute payoff difference and the constant depends on the levels of vi = (v

ij vik

) Note that

vi 2 [0 1) by construction and that this is without loss We can now rewrite (2) as

fv

ie i ( v

i

) argmax vi c( ) (5)

fv

i1 + e2[01) | z

b( fv

i

)

Note that we have defined the net benefit function b(middot vi

) for this case Inspection reveals that it is simply the probability of taking the better action times the net benefit of doing so When

vi = 0 b( v

i

) = 0 for all When vi gt 0 it is easy to verify that

1 b(0 vi

) = 1 vi and lim b( v

i

) = vi

2 1

b( fv

i

)

2b( fv

i

)2 lim = 0 and lim = 0

2 1 1

b( fv

i

) fv

i

)3 For all 2 [0 1) 2 (0 1) and 2b( 2 ( 2 0) for some 1 2 2 R++

2

For the most part these properties are intuitive specializations of the more general properties from

the previous section Unlike in the general case however property 3 states that b(middot vi

) and thus the first-stage objective (5) is strictly concave15 Due to this fact i ( v

i

) is a singleton and

satisfies

1 i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0

2 i ( v

i

) is strictly decreasing in if vi gt 0

3 lim i ( v

i

) = 1 and lim i ( v

i

) = 0 if vi gt 0

0 1

Since single-valued upper hemi-continuous correspondences are continuous functions we have the

following lemma as a direct corollary of Lemma 1

Lemma 3 [0 1) (0 1) [0 1) is continuous i

From Lemma 3 it follows that the binary action analogue of the endogenous quantal response

correspondence (4) is a continuous function Q (middot ) R2 (0 1) defined by ij

e (fv

i

)fv

ijk

Q vi

) = i

(6)ij ( (fv

i

)fv

ijk 1 + e i

3 vi

2b( 6vi) 6vi e

vi (e 1)15Taking derivatives 2 = lt 0 for v

i gt 0 (e vi +1)3

10

For reference we give the equilibrium definition and properties of Q for the binary action case

where J(i) = 2 for all i An EQRE is any p 2 A such that for all Definition 4 Fix i 2 1 n p

i1 = Q (ui

(p i

) ) i1

Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Qi ( middot ) is continuous and differentiable on R2

3 For all i and j k = 1 J(i)

=) Q vi

) gt Qij (

(ik vi )v

ij gt vik

Q

(vi

) Q

(vi

)ij ij4 = 0 and|

v

i1=v

i2 |v

i1v

ij v

ij gt 0=v

i2

and Lemma 3 Property 3 is monotonicity (A3) verbatim Property 4 is nearly responsiveness (A4) Q

(vi

)

6

Properties 1-3 are nearly immediate given the properties of general from the previous section Q

except that when (as opposed to being strictly positive) Property 4 follows ij = 0 = 0vi

v

ij

from properties of i presented later in this section

We note that the single-valuedness of i in the binary action case implies that all equilibria are

i s Also the single-valuedness of

is what makes Q a function as opposed to a correspondence Hence while existence is still ensured

by Theorem 2 Brouwerrsquos fixed point theorem would have been enough For the binary action case we seek to compare exogenous and endogenous quantal response

For this it is sufficient to track accuracy which we define as the probability of taking the better action under either model16

e fv

i

ai

( vi

) 1 + e fv

i

(fv

i

)fv

i

a i ( v

i

) i ()) = a

i

( vi

e i

(7)

pure in the sense that it is never optimal to mix over differenti

(fv

i

)fv

i1 + e i

We note that ai only depends on and is strictly increasing in the product v

i

In particular ai

is strictly increasing in vi for any gt 0 and strictly increasing in for any v

i gt 0 Thus to i While there is no closed form for

i understand quantal response we analyze properties of

much can be learned via implicit methods

Theorem 3

i (0 ) = 0 and i ((i) lim v

i

) = 0 fv

i

1

16Recalling that v

i is the absolute payoff difference between the two actions these are simply Qij and Q

ij if

v

ij gt v

ik

11

(ii) (middot ) is strictly single-peaked with macr() max ( vi

) and v() argmax ( vi

)i i i

fv

i

2(01) fv

i

2(01) i (fv

i

) i (fv

i

)(iii) fv

i fv

i

i (fv

i

) |0ltfv

i

ltfv() gt 0 and fv

i

|fv

i

=0 = 0 (iv) The product

|fv

i

gtfv() lt 0 ( v

i

)( vi

) vi is strictly increasing in v

i

i v

i = 0 and lim ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof See Appendix 92

Parts (i)-(iii) of the theorem state that i (middot ) is hump-shaped starting at 0 when v

i = 0 strictly increasing to its peak and then strictly decreasing to 0 as v

i 1 This is intuitive The good action is worth v

i more than the bad action so vi represents the cost of making a

mistake or equivalently the benefit of being accurate But for any given the agent is better able

to distinguish the good action from the bad and is thus more accurate when vi is high So when

vi is very low the benefit of accuracy is low and the agent optimally chooses low precision and

thus low accuracy and when vi is very high it is so easy to distinguish the good action from the

bad that the decision maker does not need high precision for high accuracy Parts (iv)-(vi) have implications for accuracy Part (iv) implies that a increases in v

i even i

though is non-monotonic in vi

Parts (v) and (vi) are more technical in nature with no obvious i

implication for decision problems though we use them extensively in the analysis of games in

subsequent sections Part (v) implies that the accuracy associated with peak precision macr() which macr()6v() e eobtains at v() is independent of a ( v() ) = macr()6macr = 092 Part (vi) states

i v()1+e 1+e

that the location of the peak v() increases in In Corollary 1 we compare accuracy under exogenous and endogenous quantal response This

follows immediately from properties of and ai

i

Corollary 1 (i) a

i (fv

i

) fv

i

(ii) a

|fv

i

=0 = 017

i (0 ) = ai

(0 ) = 1

a

i (fv

i

) )|fv

i

gt0 gt 0 and ai(fv

i

fv

i fv

i a

i

(lim lim

gt 0 and

i ( vi

) =fv

i

1 fv

i

1 vi

) = 1a 2

gt macr (iii) If () a ( vi

) lt ai

( vi

) for all vi

i

macr (iv) If 2 (0 ()) a ( vi

) gt ai

( vi

) for intermediate vi and a ( v

i

) lt ai

( vi

) i i

for extreme vi

Figure 1 graphically summarizes Theorem 3 and Corollary 1 The left panel of Figure 1 plots i (middot ) and

i (middot 0 ) as functions of v

i for lt 0 0 as well as the constant function and macr gt macr() gt

0 ( ) Since

i (are chosen such that peak precisions are ordered middot ) is hump-shaped

17Using the quotient rule take the derivative of (7) with respect to v

i and evaluate the expression at v

i = 0

i (0 ) = 0 and i (6vi )using that6vi

|6vi=0 = 0

12

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

For reference we give the equilibrium definition and properties of Q for the binary action case

where J(i) = 2 for all i An EQRE is any p 2 A such that for all Definition 4 Fix i 2 1 n p

i1 = Q (ui

(p i

) ) i1

Fix 2 (01)

1 Q(middot ) 2 A is non-empty

2 Qi ( middot ) is continuous and differentiable on R2

3 For all i and j k = 1 J(i)

=) Q vi

) gt Qij (

(ik vi )v

ij gt vik

Q

(vi

) Q

(vi

)ij ij4 = 0 and|

v

i1=v

i2 |v

i1v

ij v

ij gt 0=v

i2

and Lemma 3 Property 3 is monotonicity (A3) verbatim Property 4 is nearly responsiveness (A4) Q

(vi

)

6

Properties 1-3 are nearly immediate given the properties of general from the previous section Q

except that when (as opposed to being strictly positive) Property 4 follows ij = 0 = 0vi

v

ij

from properties of i presented later in this section

We note that the single-valuedness of i in the binary action case implies that all equilibria are

i s Also the single-valuedness of

is what makes Q a function as opposed to a correspondence Hence while existence is still ensured

by Theorem 2 Brouwerrsquos fixed point theorem would have been enough For the binary action case we seek to compare exogenous and endogenous quantal response

For this it is sufficient to track accuracy which we define as the probability of taking the better action under either model16

e fv

i

ai

( vi

) 1 + e fv

i

(fv

i

)fv

i

a i ( v

i

) i ()) = a

i

( vi

e i

(7)

pure in the sense that it is never optimal to mix over differenti

(fv

i

)fv

i1 + e i

We note that ai only depends on and is strictly increasing in the product v

i

In particular ai

is strictly increasing in vi for any gt 0 and strictly increasing in for any v

i gt 0 Thus to i While there is no closed form for

i understand quantal response we analyze properties of

much can be learned via implicit methods

Theorem 3

i (0 ) = 0 and i ((i) lim v

i

) = 0 fv

i

1

16Recalling that v

i is the absolute payoff difference between the two actions these are simply Qij and Q

ij if

v

ij gt v

ik

11

(ii) (middot ) is strictly single-peaked with macr() max ( vi

) and v() argmax ( vi

)i i i

fv

i

2(01) fv

i

2(01) i (fv

i

) i (fv

i

)(iii) fv

i fv

i

i (fv

i

) |0ltfv

i

ltfv() gt 0 and fv

i

|fv

i

=0 = 0 (iv) The product

|fv

i

gtfv() lt 0 ( v

i

)( vi

) vi is strictly increasing in v

i

i v

i = 0 and lim ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof See Appendix 92

Parts (i)-(iii) of the theorem state that i (middot ) is hump-shaped starting at 0 when v

i = 0 strictly increasing to its peak and then strictly decreasing to 0 as v

i 1 This is intuitive The good action is worth v

i more than the bad action so vi represents the cost of making a

mistake or equivalently the benefit of being accurate But for any given the agent is better able

to distinguish the good action from the bad and is thus more accurate when vi is high So when

vi is very low the benefit of accuracy is low and the agent optimally chooses low precision and

thus low accuracy and when vi is very high it is so easy to distinguish the good action from the

bad that the decision maker does not need high precision for high accuracy Parts (iv)-(vi) have implications for accuracy Part (iv) implies that a increases in v

i even i

though is non-monotonic in vi

Parts (v) and (vi) are more technical in nature with no obvious i

implication for decision problems though we use them extensively in the analysis of games in

subsequent sections Part (v) implies that the accuracy associated with peak precision macr() which macr()6v() e eobtains at v() is independent of a ( v() ) = macr()6macr = 092 Part (vi) states

i v()1+e 1+e

that the location of the peak v() increases in In Corollary 1 we compare accuracy under exogenous and endogenous quantal response This

follows immediately from properties of and ai

i

Corollary 1 (i) a

i (fv

i

) fv

i

(ii) a

|fv

i

=0 = 017

i (0 ) = ai

(0 ) = 1

a

i (fv

i

) )|fv

i

gt0 gt 0 and ai(fv

i

fv

i fv

i a

i

(lim lim

gt 0 and

i ( vi

) =fv

i

1 fv

i

1 vi

) = 1a 2

gt macr (iii) If () a ( vi

) lt ai

( vi

) for all vi

i

macr (iv) If 2 (0 ()) a ( vi

) gt ai

( vi

) for intermediate vi and a ( v

i

) lt ai

( vi

) i i

for extreme vi

Figure 1 graphically summarizes Theorem 3 and Corollary 1 The left panel of Figure 1 plots i (middot ) and

i (middot 0 ) as functions of v

i for lt 0 0 as well as the constant function and macr gt macr() gt

0 ( ) Since

i (are chosen such that peak precisions are ordered middot ) is hump-shaped

17Using the quotient rule take the derivative of (7) with respect to v

i and evaluate the expression at v

i = 0

i (0 ) = 0 and i (6vi )using that6vi

|6vi=0 = 0

12

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

(ii) (middot ) is strictly single-peaked with macr() max ( vi

) and v() argmax ( vi

)i i i

fv

i

2(01) fv

i

2(01) i (fv

i

) i (fv

i

)(iii) fv

i fv

i

i (fv

i

) |0ltfv

i

ltfv() gt 0 and fv

i

|fv

i

=0 = 0 (iv) The product

|fv

i

gtfv() lt 0 ( v

i

)( vi

) vi is strictly increasing in v

i

i v

i = 0 and lim ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof See Appendix 92

Parts (i)-(iii) of the theorem state that i (middot ) is hump-shaped starting at 0 when v

i = 0 strictly increasing to its peak and then strictly decreasing to 0 as v

i 1 This is intuitive The good action is worth v

i more than the bad action so vi represents the cost of making a

mistake or equivalently the benefit of being accurate But for any given the agent is better able

to distinguish the good action from the bad and is thus more accurate when vi is high So when

vi is very low the benefit of accuracy is low and the agent optimally chooses low precision and

thus low accuracy and when vi is very high it is so easy to distinguish the good action from the

bad that the decision maker does not need high precision for high accuracy Parts (iv)-(vi) have implications for accuracy Part (iv) implies that a increases in v

i even i

though is non-monotonic in vi

Parts (v) and (vi) are more technical in nature with no obvious i

implication for decision problems though we use them extensively in the analysis of games in

subsequent sections Part (v) implies that the accuracy associated with peak precision macr() which macr()6v() e eobtains at v() is independent of a ( v() ) = macr()6macr = 092 Part (vi) states

i v()1+e 1+e

that the location of the peak v() increases in In Corollary 1 we compare accuracy under exogenous and endogenous quantal response This

follows immediately from properties of and ai

i

Corollary 1 (i) a

i (fv

i

) fv

i

(ii) a

|fv

i

=0 = 017

i (0 ) = ai

(0 ) = 1

a

i (fv

i

) )|fv

i

gt0 gt 0 and ai(fv

i

fv

i fv

i a

i

(lim lim

gt 0 and

i ( vi

) =fv

i

1 fv

i

1 vi

) = 1a 2

gt macr (iii) If () a ( vi

) lt ai

( vi

) for all vi

i

macr (iv) If 2 (0 ()) a ( vi

) gt ai

( vi

) for intermediate vi and a ( v

i

) lt ai

( vi

) i i

for extreme vi

Figure 1 graphically summarizes Theorem 3 and Corollary 1 The left panel of Figure 1 plots i (middot ) and

i (middot 0 ) as functions of v

i for lt 0 0 as well as the constant function and macr gt macr() gt

0 ( ) Since

i (are chosen such that peak precisions are ordered middot ) is hump-shaped

17Using the quotient rule take the derivative of (7) with respect to v

i and evaluate the expression at v

i = 0

i (0 ) = 0 and i (6vi )using that6vi

|6vi=0 = 0

12

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

i

this implies that i ( v

i

) gt for intermediate vi and that

i for extreme vi

( vi

) lt 0 0 0

( vi

) lt(middot is chosen so that i

) is also hump-shaped but for all vi

0(middot ) (middot ) and as functions The right panel of Figure 1 plots the accuracy implied by

i i

of vi

Since accuracy is increasing in for any given vi

it is immediate from the left panel that a

i ( vi

) gt a i ( v

i

0 ) for all v

i and that a i ( v

i

) gt ai

( i

) for intermediate vi (and v

( vi

) lt ai

( vi

) for extreme vi

) Note how just like ai 1 as v

i 1 despite ai

ai

) a (fv

i

)iFinally note that while ai(fv

i

|fv

i

=0 gt 0

fv

i

When precision is fixed at the fact that

i 0 |fv

i

=0

gt 0 an infinitesimal = 0 This is

fv

i also intuitive from a comparison of andi

increase in vi from 0 to must increase accuracy However when precision is endogenous since

i (fv

i

) fv

i |fv

i

=0 = 0 an infinitesimal increase in vi from 0 to will have no effect on accuracy

since is still essentiallyi (0 ) = 0

i

Precision Accuracy

0 262 697 1418 2616 4603 0

01

02

03

λlowast i (∆vi θ)

λlowast i (∆vi θ prime)

λ

0 262 697 1418 2616 4603 05

075

1

a lowast i (∆vi θ)

a lowast i (∆vi θ prime)

ai(∆vi λ)

∆vi ∆vi

Figure 1 Endogenous vs exogenous quantal response

Noting from (6) and (7) that Q ij

( vi

)) part (i) (vi

) = 1vj

and hence Q

( vi

) + 1vj ltv

k (1 also satisfies the regularity axioms (A1)-(A3)

v

k

a ai i

of Corollary 1 implies property 4 of Qand a (very slightly) modified version of responsiveness (A4) It is well-known that the axioms impose testable restrictions on equilibria (Goeree et al [2005]) and easy to see that replacing (A4) with the modified version does not effect regular QRErsquos empirical content Since regularity imposes testable restrictions on data the fact that both LQRE and EQRE are regular implies that their performance in explaining data from individual games may be very similar a priori depending on

the game Nevertheless as we show in Section 5 the LQRE and EQRE sets (ie the sets of equilibria indexed by and respectively) may be very different

Regularity in of itself does not imply much in terms of across-game restrictions However regularity together with translation invariance which holds for both LQRE and EQRE does imply

considerable structure as shown in Friedman [2018] They give examples of QRErsquos comparative static

13

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

predictions that hold for any fixed quantal response function satisfying regularity and translation

invariance Hence across certain games LQRE and EQRE make similar qualitative predictions (ie for any fixed parameter values) Nevertheless Corollary 1 suggests that the models may make very

different quantitative predictions across games

4 The structure of equilibria and equilibrium selection

Analogous to the LQRE correspondence18 we define the EQRE correspondences p (0 1) A

and (0 1) [0 1)n by

p () = p 2 A = Q i

) ) 8i j pij

ij (ui(p

and () = 1(u1(p 1) ) (u

n

(p n

) ) [0 1)n p 2 p () n

1 2 1 2Theorem 4 Let 1 2 be a sequence such that lim t = 0 Let p p and be t1

t t t tcorresponding sequences with pt 2 p(t) and t = ( 1 ) = ( 1(u1(p 1) t) (u

n

(p ) t)) 2n n n

t t(t) for all t such that lim p = p Without loss assume that is a singleton for all i andi

t1 1t19 Then (i) p is a Nash equilibrium and (ii) for all players i such that p 6= for some j

ij J(i)

lim t

i = 1 t1

Proof (i) Suppose p is not a Nash equilibrium Then there is some player i and a pair of actions a

ij and aik such that p(a

ik

) gt 0 and ui

(aij p

i

) gt ui

(aik

pi

) or equivalently uij (p

i

) gt uik

(pi

) tSince u is continuous it follows that for sufficiently small gt 0 there is a T such that u

ij (pi

) gt tu

ik

(pi

) + for all t T But then since t 0 any finite i is such that there exists T

0 T such

tthat macr i cannot be a solution to (2) for any t T

0 Thus i 1 which implies that pt(a

ik

) 0 contradicting that p(a

ik

) gt 0 (ii) It is the case that either maxjk

uij (p

i

) uik

(pi

) gt 0 or tthat max

jk

uij (p

i

) uik

(pi

) = 0 In the former case it is clear that i 1 (using an argument

tsimilar to that given in the proof of part (i) above) In the latter case suppose that lim i lt 1

t1 t t t 1But then since max

jk

uij (p

i

) uik

(pi

) 0 this would imply that limt1p = for all j

ij J(i)

a contradiction

Part (i) of Theorem 4 is analogous to a result for LQRE Part (ii) is potentially interesting

because it gives a non-obvious relationship between a Nash equilibrium of a game and the limiting

behavior of the optimal precision required to approach it It is unsurprising that whenever the limiting Nash equilibrium p is such that max

jk

uij (p

i

) uik

(pi

) gt 0 (as is the case for many tpure strategy equilibria) 1 as 0 (the marginal benefit to precision is positive but the i

18See McKelvey and Palfrey [1995] ( ) = p 2 A pij = Q

ij (ui

(p i

) ) 8i j 19Since this result is only concerned with the limit as 0 take a strictly decreasing sequence t with 1

t t tsufficiently small such that in all equilibria p = i (ui

(p i

) t) is a singleton by Lemma 2i

14

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

marginal costs go to 0) Surprisingly the result also holds even for fully mixed equilibria where 1max

jk

uij (p

i

) uik

(pi

) = 0 as long as pij 6=

J(i) for all j ie that i is not uniformly mixing

over all of his pure actions The result is non-obvious because in this case both the marginal cost to precision and the marginal benefit go to 0 so it seems there may be an indeterminacy but the

condition that resolves it is related to the Nash strategy of player i The next theorem establishes several properties of the equilibrium correspondence in the binary

action case and is analogous to the well-known equilibrium selection result for LQRE Like the

classic theorem it is proved using results from differential topology though the proof differs at several steps

Theorem 5 For almost all games with J(i) = 2 for all i (i) p() is upper hemi-continuous20 (ii) p() is odd for almost all

(iii) The graph of p contains a unique branch which starts at the centroid21 for = 1 and

converges to a unique Nash equilibrium as 0

Proof See Appendix 94

Some steps of the proof go through for general normal form games such as Lemma 7 of Appendix

94 which states that the equilibrium is unique (and pure) for sufficiently high There are two

issues in generalizing the result fully First since the first-stage objective is not generally concave () and p() may be discontinuous in 22 Second without a better understanding of properties of Q (such as we have in the binary action case) we are unable to show that the equilibrium

graph is a manifold (even if we assume that () is well-behaved) Nevertheless we conjecture that the result does hold for almost all normal form games in which RJ(i) (0 1) [0 1) isi

single-valued for all i (part (i) is true in that case)

player 2 L R

player 1 U D

0 0 6 1 1 14 2 2

Table 1 Asymmetric Chicken

To illustrate the theorem we use the example of the asymmetric chicken game23 whose payoffs are given in Table 1 Letting p and q denote the probabilities that player 1 plays U and player

20This is always true not just generically 121That is where p

ij = J(i) for all i and j

22The example in Appendix 95 of an agent with a non-concave objective can be extended to games trivially by adding opponents with one action each It is immediate that () and p () in the resulting game are discontinuous

23This particular variant is from McKelvey and Palfrey [1996]

15

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

1 1

08 08

06 06

p q

04 04

02 02

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 10 8

8

6

6

λlowast λlowast 1 2

4

4

2

2

0 0 0004 0058 0873 13214 200 0004 0058 0873 13214 200

θ θ 2

18

λlowast 116

14

003 004 005 006 007 θ

Figure 2 EQRE of asymmetric chicken These figures plot equilibrium objects as a function of parameter 2 (0 1) In all cases the dashed line gives the main branch that can be ldquotracedrdquo from very high to very low values of The fifth figure ldquozooms inrdquo to get a better picture of the -graph 1

16

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

2 plays L respectively Figure 2 gives the EQRE correspondences Asymmetric chicken has three 4Nash Equilibria p q 2 0 1 12 1 0 all of which are limit points of p() as 013 5

and p q = 0 1 is the unique Nash equilibrium selected by EQRE the same equilibrium

selected by LQRE as 1 Also note that even though there are only three Nash equilibria for some values of there are five EQRE We do not have a specific intuition for this result but regard

it as interesting as standard LQRE gives no more than three equilibria for this game What is not clear from the figure (but follows from Theorem 4) is that along any branch 1 for all i asi

0 The results of this section indicate that EQRE and LQRE have similar limiting behavior as

functions of their respective parameters However to differentiate the models one must compare

the entire sets of obtainable equilibria without reference to parameters Accordingly we define the

EQRE set E( c) = p () 2 A 2 (01) (8)

which gives the collection of equilibria obtainable for some and the LQRE set is defined anal-ogously24 We emphasize in the notation that an EQRE set is defined for a given cost function c

which while arbitrary is held fixed as varies In the next section we compare the two equilibrium

sets in our leading example

5 Generalized matching pennies

In this section we study 22 games with the generalized ldquomatching penniesrdquo structure as in Figure

3 These are the simplest non-trivial games with unique fully mixed Nash equilibria and hence

have been played extensively in the lab We show that (1) the EQRE and LQRE sets are curves in

the unit square that cross at finite points (that we give explicitly as a function of the game) (2) the

EQRE set deviates systematically from the LQRE set at all other points and (3) these deviations are closely related to the endogenous heterogeneity of s Importantly these results do not depend i

at all on the EQRE cost function which we assume is arbitrary but fixed as we vary

The parameters aL

aR

bU and b

D in Figure 3 give the base payoffs These are often greater than or equal to zero in experiments due to behavioral considerations which is the case for all games considered in later sections The parameters c

L

cR

dU and d

D are the payoff differences which we

assume are strictly positive to maintain the relevant features25 As is well-known such games have

a unique fully mixed Nash equilibrium p q = d

D c

R d

U +d

D c

L

+c

R

The fixed points that define the equilibria of these games are easily plotted in the unit-square

of p-q space as in Figure 4 In this example the Nash equilibrium is given as the intersection 24If the LQRE correspondence is defined by ( ) the LQRE set is L( ) = ( ) 2 A 2 (0 1) 25Games in which the payoff differences are all strictly negative are equivalent up to the labelling of actions

17

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

L R

U

D

aL + c

L

bU

aR

bU + d

U

aL

bD + d

D

aR + c

R

bD

U up D down L left R right

player 1rsquos payoff in upper-left corner player 2rsquos payoff in lower-right corner

aL

aR

bU bD 0

cL

cR

dU dD gt 0

Figure 3 Structure of 2 2 Games The restrictions imposed ensure that the Nash equilibrium is unique and fully mixed

of the best response correspondences Similarly the unique LQRE and EQRE are given as the

intersections of their respective reaction functions which map the opponentrsquos mixed action into

quantal response That is for any fixed and the curves can be drawn and their intersections give the corresponding equilibria The LQRE and EQRE sets (not drawn here) are given by varying

and respectively from 0 to 1 and collecting the equilibria

1

p 05

0

NE

EQRELQRE

p lowastlowast 1 2

p q lowastlowast lowastlowast

1 lowastlowast 2 q 1

2 1 2

0 05 1 q

Figure 4 Equilibria as the intersection of reaction functions

It is easy to show that player 1rsquos LQRE and EQRE reactions are strictly increasing in q (re-sponsiveness (A4)) and pass through the point 1 q when player 2 is mixing according to his 2

Nash strategy player 1 is indifferent and must uniformly randomize (monotonicity (A3)) Similarly

18

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

2

player 2rsquos reactions are strictly decreasing in p and pass through pof all reaction functions induced by quantal response satisfying the regularity axioms (A1)-(A4) It

1 2 These restrictions are true

1 and q lt 1is easy to show that when p 2 2 (as in the example of Figure 4) the set of regular 1) (q 1

2) and

R2 = (2 p) (2

QRE26 which we call the regular set is given by R1 [ R2 [0 1]2 where R1 = (p

1) are two rectangular components2711 We draw the components of the regular set as gray rectangles

Since we only consider QRE-type models that are translation invariant the base payoffsndashaL

aR

bU

and bD

ndashcan be ignored for equilibrium analysis For what follows and for reasons that will be clear we reparameterize the game by specifying p q r and s defined as

dD

p = dU + d

D cR q =

cL + c

R cL + c

R r =

dU + d

D

s = cL + c

R

These are the Nash equilibrium the ratio of payoff differences and the sum of player 1rsquos payoff differences It is easy to see that from p q and r all the payoff differences can be recovered up

28to a positive scaling which is pinned down by s In this section none of the results depend on the scale of the game so s can be ignored for now while p q and r are essential and easier to

use in the analysis than the payoff differences directly are uninteresting29 and

For parsimony we detail the case in

11 = q 2 2301

Games with completely symmetric Nash equilibria pfor all other games it is without loss to consider p 2

which q lt 1 2 and Appendix 96 presents the case in which q gt 1

2 which can be analyzed using

the methods introduced here but looks qualitatively different We compare the EQRE and LQRE sets by relating the endogenous heterogeneity of i s to

features of the EQRE set Since the s depend on the absolute expected utility differences u1(q) =i

|(cL + c

R

)q cR

| and u2(p) = |(dU + d

D

)p dD

| that obtain in EQRE p q we first characterize

26ie all QRE that can be supported by quantal response functions satisfying the regularity axioms See Goeree et al [2005] for the regular QRE analysis of similar games

27A few notes about the regular set First if p 1 the second component is degenerate R2 = Second

2 is also included in = 2

if p gt

1 2 the point p

the regular set Third that R

1 2

and R1 and R

21

which is the unique intersection of the closures of Rare open sets results from the fact that the reaction functions are strictly

monotonic and hence cannot cross at the closure of these sets (excepting the point p 28Actually any sum of one or more of the payoff differences would do the job

1 2 if p

gt

1 2 )

29All regular QRE coincide with Nash in this case 30By switching columns and then switching rows we obtain from game a transformed game

0 where actions

0 0 0 0 0 0 0are relabelled U = D D = U L = R and R = L and payoffs are relabelled c = c

R

c = cL

d = dD and

L R U0 0 0 0 )1 1 U Player labels remain unchanged pd

D = d lt 2 () d

D lt d

U () d

D gt d ()U (p gt

19

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

the regions of the unit square such that u1 lt u2 and u1 gt u2 To this end we define the

iso-utility31 curves in terms of our transformed parameters which give the action frequencies p qsuch that u1(q) = u2(p)

)u +(q) rq + (p rq )u (q) rq + (p + rq

To be precise we abuse notation by writing u+ p q|p = u+(q) and u p q|p = u (q) and it is easy to show that p q 2 u+ [ u if and only if u1(q) = u2(p) The iso-utility curves have slopes plusmnr and since u1(q) = u2(p) = 0 they have a unique intersection at the Nash

equilibrium Necessarily since p q is interior and r gt 0 u+ and u partition the square into

four regions two in which u1 lt u2 (top and bottom) and two in which u1 gt u2 (left and

right) We illustrate this in the top left panel of Figure 5 which is drawn for a game defined by gt 1 lt 1p q and some r gt 02 2

Next we characterize regions of the unit square defined by the relative accuracies of both

players The role of these regions is less obvious but will soon become clear Recall that we defined

accuracy (7) as the probability of taking the better action In any LQRE or EQRE p q the

accuracies of players 1 and 2 are given by maxp 1 p and maxq 1 q respectively both of 1which are greater than 2 The top right panel of Figure 5 plots the iso-accuracy curves which

+we label a and a These are defined by p q such that both players are equally accurate

(maxp 1 p = maxq 1 q) and are simply the diagonals of the square Note that unlike the

iso-utility curves the iso-accuracy curves do not depend on the game Off the diagonals in the top

and bottom quadrants player 1 is more accurate maxp 1 p gt maxq 1 q In the left and

right quadrants player 2 is more accurate maxq 1 q gt maxp 1 p The bottom left panel of Figure 5 reproduces both the iso-utility and iso-accuracy curves for our

example game superimposed by the regular set (gray rectangles) and the LQRE set (solid black

curve) From McKelvey and Palfrey [1995] it is well-known that the LQRE set connects the centroid

12

1 to the Nash equilibrium p q as varies from 0 to 1 and since LQRE is regular the 2

entire LQRE set is contained within the regular set The qualitative shape of the LQRE set is known

for these games but we add to this a novel observation the LQRE set crosses all intersections of the iso-utility and iso-accuracy curves within the regular set and cannot cross the curves at any

other point This must be the case since playersrsquo accuracies are one-to-one with the products u1

and u2 Hence in any LQRE u1 = u2 if and only if maxp 1 p = maxq 1 q The

shape of the LQRE set thus depends predictably not only on the Nash equilibrium but also on + +the parameter r which controls the slopes of u and u and thus where they cross a and a In

1 2this example there are two such crossings in the regular set which we label p q1 and p q2 31ldquoIso-absolute expected utility differencerdquo would be more accurate but also a mouthful

20

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

1 2These are defined as p q1 u+ a R1 and p q2 u a+ R2 32 and in general one or

both of these points may not exist depending on the gamersquos parameters

Iso-utility Iso-accuracy 1

0 05 1

∆u1 lt ∆u2

∆u1 gt ∆u2macr macr

∆u1 lt ∆u2macr macr

u +

u minus

1

p p 05 05

0 0 0 05 1

maxp 1 minus p gt maxq 1 minus q

maxp 1 minus p gt maxq 1 minus q

maxq 1 minus q gt maxq 1 minus q gt

a

maxp 1 minus p maxp 1 minus p

+

a minus

q q LQRE and regular QRE and EQRE

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p q 1 1

R1

p lowastlowast 1 2

p lowastlowast q lowastlowast

p q 2 2 R2

1 2

1 2

1

p p 05 05

0 0 0 05 1 0 05 1

q q

Figure 5 Example iso-utility iso-accuracy and equilibrium sets The game used in this example is gt 1 lt 1such that p 2 q and r gt 0 is sufficiently low that u (whose slope is r) passes through 2

the second component of the regular set R2

Continuing with the same example the bottom right panel of Figure 5 plots the EQRE set (in

this case using c( ) = 2) on top of the LQRE set Note that since EQRE is regular the EQRE set is also contained within the regular set Note also that the the EQRE and LQRE sets agree at five

32As solutions to systems of linear equations these can be solved for explicitly

21

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

11 and p q follows from Theorem 5 (and the specific points That the models agree at analogue for LQRE) which states that these are the EQRE limits as goes from 1 to 0 (and the

2 2

LQRE limits as goes from 0 to 1) Furthermore the EQRE set ldquocrossesrdquo the LQRE set at three

points pldquoaboverdquo ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set

1 q 1 p 1 2 and p2 q2 Thus as is clear from the figure the EQRE set is ldquobelowrdquo

More generally the qualitative shape of the EQRE set relative to the LQRE set depends on how

many times the iso-utility and iso-accuracy curves cross within the regular set or in other words on

the existence of p6

1 q1 andor pCases 1 and 2 (3 and 4) involve Nash equilibria in which player 1 is less (more) accurate than

2 q2 Thus there are four cases which we illustrate in Figure

player 2 as shown by p q to the left (right) of a Cases 1 and 3 (2 and 4) involve for fixed

Nash equilibrium sufficiently low (high) ratio r such that u passes through (below) the second

component of the regular set The main result of this section establishes that the EQRE set always falls into one of the four

cases from Figure 6 for all games satisfying a technical condition In other words the models agree

at finite points (that we give explicitly) and the EQRE set deviates systematically at all other points Importantly which of the four cases applies depends on the game only not on the EQRE

cost function The result relies on several lemmas The first establishes that the relationship between the

2

2

1

1

i s the notation p q to refer to an LQRE p q that obtains for precision parameter Similarly

is an EQRE that obtains for cost parameter When a particular EQRE is clear p q p q 2

1

EQRE and LQRE sets is related to the endogenous heterogeneity of In what follows we use

from the context and are shorthand for ( u1(q) ) and ( u2(p) ) 0

1 2

Lemma 4 (i) Take any q 2 (q q (which ) with associated

0exist and are unique) p Q p Qif and only if 1 2

1

) with associated EQRE p q and LQRE p

(ii) Take any p 2 ( p20 0

2 33

1 (which exist and are unique) q Q q REQRE p q and LQRE p q if and only if

Proof See Appendix 92

1

2The main result makes use of Lemma 4 by establishing regions of the EQRE set such that gt

and 1 lt

2 Intuitively this may be difficult because

i is non-monotonic in ui by Theorem 3

for fixed and each point of the EQRE set corresponds to a different However Theorem 3 also

establishes that (1) i u

i

which is one-to-one with accuracy is strictly increasing in ui

and (2)

efor any peak precision macr() is associated with accuracy 092 Therefore if both playersrsquo 1+e

eaccuracies are on the same side of the magic number their relative accuracies imply 1+e

i sorderings of the ui

s and

0 033In part (i) satisfies 1 lt lt

2 ()

0 q lt q

p lt p 1 gt gt

2 ()q gt q

0 and p gt p and

1 2 ()

0 In = = p = p

part (ii) satisfies 1 gt gt

2 ()

1 lt lt

2 () =

1 = 2 () q = q

22

0

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

Case 1 Case 2

p q 1 1

u +

p lowastlowast 1 2

p lowastlowast q lowastlowast

a +

1 2

1 2

a minus u minus

1

p q 1 1

p lowastlowast 1 2

u +

p lowastlowast q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q Case 3 Case 4

p q lowastlowast lowastlowast p lowastlowast 1 2

u +

a +

1 1 2 2

a minus

u minus

1

p lowastlowast 1 2

p lowastlowast

u +

q lowastlowast

p q 2 2

a +

1 2

1 2

a minus

u minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

1 2Figure 6 The four cases of equilibrium sets Case 1 both p q1 and p q2 exist Case 2 only 1 2 1 2p q1 exists Case 3 only p q2 exists Case 4 neither p q1 nor p q2 exist

23

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

eLemma 5 Fix EQRE p q (i) maxp 1 p 7 maxq 1 q lt 092 implies that 1+e

eu1(q) 7 u2(p) and 7 (ii) maxp 1 p 7 maxq 1 q gt implies that u1(q) 71 2 1+e

u2(p) and 21

Proof See Appendix 92

Combining Lemma 4 with Lemma 5 gives a sufficient ldquono crossingrdquo condition

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof See Appendix 92

And finally our result which establishes the qualitative nature of the EQRE set relative to the LQRE set It is proven by establishing the ordering of s within neighborhoods of special points i

1 1p q p and 12 which determines whether the EQRE set is locally ldquobelowrdquo ldquoaboverdquo 2 2

ldquoto the right ofrdquo or ldquoto the left ofrdquo the LQRE set The result then follows by invoking the no crossing

lemma which implies that the EQRE set can only cross the LQRE set at up to three specific points 1 gt 1 1 2(and must so if conditions are met) p (if p ) and p q1 and p q2 (if they exist) 2 2

1Note that the result allows for p = which implies both that the second component of the regular 2 2set is degenerate (ie R2 = ) and that p q2 does not exist34

1 1Theorem 6 Let p 2 q lt and r be such that all LQRE p q satisfy maxp 12

p maxq 1 q lt e

09235 (I) The EQRE set crosses the LQRE set at p 1 if p gt 1 361+e

2 2 1(II) the EQRE set cannot cross the LQRE set u+ [u or a+ [a at any other points except p q1

2and p q2 (and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 1 and 2) (a) for all q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q and

1 1(b) for all q 2 (q ) p gt p 0 for EQRE p q and LQRE p

0 q2

1(ii) If p q1 does not exist (Cases 3 and 4) 1(a) for all q 2 (q ) p gt p

0 for EQRE p q and LQRE p 0 q2

2(iii) If p q2 exists (Cases 1 and 3) 2(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 2

34We have not depicted the p

exists but p 2 q

= 1 2 case in Figure 6 but it is essentially a sub-case of Case 2 in which p

2 does not Some of the games we analyze in the empirical Section 7 have this feature 35This is a restriction on equilibria but it is easy to derive sufficient conditions based on the gamersquos parameters

1 q 1

since the LQRE set is restricted in its crossings of the iso-utility and iso-accuracy curves For instance referring to

Case 1 in Figure 6 if q 2 (1 e 1+e

008 1 2 ) and p 1

2r + (p ) ltrq

1 2and r are such that u +( e) = 1+e

then the LQRE set for all q 2 (q 1 ) is bounded above by a and u + lt

e 1+e

e

lt 2 1+e

36If p 1 then R2 = and p

respectively but no ldquocrossingrdquo actually takes place 1 11 is the common limit of EQRE and LQRE as 1 and= 2 = 2 2 0

24

2

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

2(iv) If p q2 does not exist (Cases 2 and 4) (a) for all p 2 (1 p) q lt q

0 for EQRE p q and LQRE p q 0 2

Proof See Appendix 92

The result establishes that EQRE and LQRE are generically distinctndashwith predictions that overlap only on set of measure zero Also by establishing that the EQRE set cannot cross the

LQRE set iso-utility curves or iso-accuracy curves except at specific points we have bounded the

EQRE predictions Since these bounds do not depend on the cost function we have shown that the

flexibility of choosing the cost function does not allow EQRE to ldquoexplain everythingrdquo Importantly since the LQRE set serves as a bound on ldquoone siderdquo of the EQRE set depending on where the

data falls it may be that EQRE outperforms (or underperforms) LQRE for any cost function This observation will allow us to determine which model is qualitatively more consistent with data

without relying heavily on the usual measures of fit

6 Power costs in normal form games

So far none of the results have depended on a particular functional form for the EQRE cost function

c but for applications this must be specified A natural choice would be the power function c( ) = k for k gt 1 In this section we give two results for the EQRE set (8) that hold under power costs both of which will be useful for applications We show that (1) the EQRE set is independent of the units or ldquoscalerdquo of the game and (2) that the EQRE set approaches the LQRE set as k 1

(in a sense that will be made precise) These results are general to all normal form games 1It is well-known that the logit quantal response function (1) satisfies Q

i

(vi

) = Qi

( vi

)

for any scale factor gt 0 which results from the fact that only enters the expression for Qi as

the products ( vi1 v

iJ(i)) Hence scaling a gamersquos utility payoffs by some arbitrary gt 0 will leave the LQRE set unchanged Given some data this -scaling will change the best-fit parameter from ˆ to ˆ

0 = 1 ˆ but will leave the resulting prediction unchanged That the predictions do not

depend on such arbitrariness has normative appeal and is obviously important for applications Under power costs endogenous quantal response (4) satisfies a similar property Q

i (k+1) and hence the EQRE set is also unaffected by -scalings This follows from the

vi

) =

Q i ( vi

next theorem which is a simple consequence of the first-order condition (3)

kTheorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (

i ( vi

Proof See Appendix 92

k+1 Thus under power costs -scalings will change the best-fit parameter from to 0 =

but will leave the resulting prediction unchanged For general cost functions the EQRE set may

vi

)

(ie a solution to (2)) then 1 ˜ 2 k+1)

25

be affected by -scalings37 so we favor power costs for applications It is important to know how the choice of cost function effects the EQRE set and its relationship

to the LQRE set The next result shows that as k 1 the EQRE set with power costs approaches the LQRE set in the sense that (1) every LQRE is approached by a k-indexed sequence of EQRE and (2) for large k every EQRE is nearby some LQRE Hence EQRE nests LQRE as a limiting

case

kTheorem 8 Let c( ) = for k gt 1 As k 1 the EQRE set approaches the LQRE set

Proof See Appendix 92

The intuition is simple Any ˜ is associated with a particular LQRE p˜ As k 1 we construct a k-sequence of cost parameters

k

(˜) such that the marginal cost function k

(˜)k k

a step function that is vertical at ˜ and thus crosses all playersrsquo marginal benefit functions b

1 approaches 0 ( v

i

)

at approximately ˜ (independent of vi

) We then construct a k-sequence of EQRE p where

k

(˜)kp is an EQRE associated with cost function and parameter

k

(˜) By construction for

k

(˜) sufficiently large k i ˜ for all players i and p

k

(˜) p˜ as k 1 In other words the endogenous heterogeneity of s which differentiates EQRE from LQRE effectively disappears as i

k 1 Similarly as k 1 every EQRE is such that i

j for all players i and j and hence

the EQRE is nearby some LQRE

7 Data

We use experimental data from sets of 2 2 games of the form in Figure 3 to test the predictions of LQRE and EQRE We use a set of 21 games from five studies MPW 2000 (McKelvey et al [2000]) SC 2008 (Selten and Chmura [2008]) Ochs 1995 (Ochs [1995]) GH 2001 (Goeree and Holt

[2001]) and NS 2002 (Nyarko and Schotter [2002]) The parameters p q r and s that define

the relevant aspects of the games (following Section 5) and the data (aggregate action frequencies and sample size) are given in Table 2 The gamesrsquo payoff matrices and details of the experimental procedures are given in Appendix 97 We transformed some of the games relative to how they were

1presented to subjects following the discussion in Section 5 so that p (see footnote 30) We 2

denote such games with an ldquordquo next to their names 37This is a subtle issue related to the fact of multiple players To see this consider arbitrary c( ) v

i

vj and

i ( vi i) = 1i ( There may be unique

i and j (ie player-specific cost parameters) such that v

i

) =)

v

j ) but most likely it will be that Q

6

i

i ( vi i) = Q

i ( j ( vj j ) = 1j ( j ( vj j ) = Q

j (

=

j Hence for an EQRE obtained for it may be impossible to recover the same equilibrium for any -scaling

26

v

i

) and v

j ) =) Q

0 after a

71 Qualitative analysis

We begin by plotting the data in Figures 7 and 8 Each panel corresponds to a different game and

is familiar from Section 5 We plot the empirical action frequencies (black square) LQRE set (solid

black curve) regular set (gray rectangles) iso-utility curves (thin black lines) and iso-accuracy

curves (dashed lines) The plots also include the EQRE set (dotted curve) associated with power costs with k = 11

Why k = 11 From Theorem 6 all EQRE sets (with power costs or otherwise) will qualitatively

behave like those plotted in Figures 7 and 8 for k = 11 and it is clear that any point ldquobetweenrdquo this EQRE set and the LQRE set can be supported as an EQRE for some cost function In other words the region enclosed by the two sets is a subset of all possible EQRE and any other EQRE

that is not included in this region must deviate ldquoeven fartherrdquo from the LQRE set Since the

EQRE set approaches the LQRE set as k 1 (Theorem 8) we conjecture that lower values of k will give EQRE that are as far as possible from the LQRE set which is consistent with

our numerical investigations (unreported) By choosing the low value k = 1138 we exaggerate

differences between the EQRE and LQRE sets and thus we regard the region bounded by the two

sets as an approximation ldquofrom belowrdquo to the set of all possible EQRE Inspecting Figures 7 and 8 we first note that for all games the data is not contained in the region

enclosed by the LQRE and EQRE (k = 11) sets To the extent this region is a good approximation

to the set of all possible EQRE this implies that the data cannot be supported as an EQRE (ie exactly) for some cost function39 Still in many cases EQRE outperforms LQRE for any cost function As an example consider the very first panel of Figure 7 for game A Here the best-fit LQRE (by maximum likelihood) is to the left of the datapoint and we connect the two points with

a line segment Note that this segment crosses the EQRE set and in fact must cross the EQRE

set for any cost function by Theorem 6 That is for any cost function there is an EQRE that is closer to the data than is the nearest LQRE which implies that EQRE outperforms LQRE for any cost function Inspecting the plots EQRE outperforms LQRE in this sense for 15 of 21 games

(A B C D 2 3 4 6 7 9 10 12 2 3 AMP ) though in some cases it is difficult to see by

eye In game 8 the relative performance of LQRE and EQRE depends on the cost function For the remaining 5 games LQRE outperforms EQRE for all cost functions40

38Recall that k gt 1 is required to maintain properties of the cost function from Section 3 39In some cases the data is very near to but just outside the region 40Of course by Theorem 8 the EQRE set can approximate the LQRE set arbitrarily well by varying the cost

function

27

Study Game Data

p q N Parameters

p q r s

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

0500 0100 500 10 0500 0100 125 10 0500 0100 500 40 0500 0200 250 5

1 0921 0310 9600 0909 0091 100 11 2 0783 0473 9600 0818 0273 100 11 3 0837 0207 9600 0727 0091 100 11 4 0714 0264 9600 0636 0182 100 11 5 0673 0336 9600 0636 0273 100 11

SC 2008 6

7 0555 0404 9600 0859 0436 4800

0545 0364 100 11 0909 0091 100 11

8 0750 0414 4800 0818 0273 100 11 9 0746 0173 4800 0727 0091 100 11 10 0634 0301 4800 0636 0182 100 11 11 0669 0348 4800 0636 0273 100 11 12 0561 0396 4800 0545 0364 100 11

Ochs 1995 2 3

0595 0258 448 0542 0336 516

0500 0100 056 1238 0500 0200 063 1393

GH 2001 AMP RA

0960 0160 25 0080 0800 25

0500 0125 400 320 0500 0909 055 44

NS 2002 NS 0543 0610 6720 0600 0600 100 5

Table 2 Games and data

28

A B C

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q

1lowast 2lowastD

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 3lowast 4lowast 5lowast

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 6lowast 7lowast 8lowast

0 05 1

p

0

05

1

0 05 1

p

0

05

1

0 05 1

p

0

05

1

q q q

Figure 7 Plotting the data (part 1)

29

9lowast 10lowast 11lowast

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 12lowast 2 3

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q

AMP RA NSlowast

0 05 1

p

0

05

1

0 05 1

p

0

05

1

0 05 1

p

0

05

1

q q q

Figure 8 Plotting the data (part 2)

30

72 In-Sample fit

Next we explore the fit of LQRE and EQRE by choosing their parameters to maximize the log-likelihood of aggregate choice frequencies As is standard we fit LQRE by choosing For EQRE

we must choose a cost function In Section 6 we established that power costs for any k gt 1 have

nice properties so we naturally use this family of cost functions Since there is no principled way

of fixing k a priori we fit a 2-parameter EQRE where we fit both k and jointly From Theorem

8 LQRE is nested in EQRE in the sense that the equilibrium sets converge as k 1 so we

use standard likelihood ratio tests for model comparison Heuristically LQRE is simply EQRE

under the restriction that ldquok = 1rdquo We search over very fine - and -grids but for computational simplicity41 we search over a coarse k-grid k 2 11 2 5 10 201 By searching over a coarse

k-grid we are only underestimating EQRErsquos performance Prior to estimation we multiply each gamersquos payoffs by a study-specific conversion factor so that

each unit of payoff in the matrix corresponds roughly to the same real monetary payoff across all studies The conversion factors we use are from Friedman [2018] and amount to simple adjustments for inflation and currency-to-currency exchange rates42 From the results of Section 6 this is

immaterial for fitting games within a study it will effect parameter estimates but not equilibrium

predictions or fit The effect of the adjustment is to put parameter estimates ldquoon the same scalerdquo across studies It also allows in principle to fit the data pooled together from several studies though we do not pursue that here since the fit would be sensitive to the choice of factors

In addition to LQRE and EQRE we also fit a 2-parameter heterogenous LQRE which we label ldquoHQRErdquo This is simply LQRE in which each player i has a player-specific exogenous

i

We do this

for two reasons First HQRE nests both LQRE and EQRE and hence must fit the data better than

both It is standard practice to present as a benchmark the ldquoempirical likelihoodrdquo which is simply

the likelihood of any model that predicts the data exactly Since the data from any one game is 2-dimensional HQRE overfits the data from a single game to the point that its likelihood coincides with the empirical likelihood as long as the gamersquos data is consistent with some regular QRE (see

Section 2) In cases where the data cannot be rationalized by any regular QRE such as in games 2 7 and 8 (the datapoint falls outside of the gray rectangles in Figure 7) HQRE still gives the

best-possible regular QRE prediction Hence there is a sense in which the HQRE likelihood is a

better benchmark than the empirical likelihood since we are only comparing different QRE models Second we would like to compare the exogenous ˆ

i

s from fitting HQRE with our endogenous ˆ i s

from fitting EQRE Since HQRE fits the data nearly perfectly (when consistent with some regular QRE) we regard the exogenous ˆ

i

s as the ldquotruerdquo parameters and wonder if EQRErsquos performance

41To approximate a playerrsquos EQRE reaction to any mixed action of his opponents requires a numerical optimization Hence to approximate the EQRE reaction function of which EQRE is a fixed point requires many numerical optimizations On the other hand LQRE does not involve any optimization

42The factors are 1 (MPW 2000) 01560 (SC 2008) 01784 (Ochs 1995) 00973 (GH 2001) and 05 (NS 2002) See Friedman [2018] for details

31

32

Stud

y G

ame

HQ

RE

ˆ 1

ˆ 2

ln(L

)

LQR

E

ˆ ln(L

)

EQ

RE

k

ˆ 1

ˆ 2

ln(L

) 2ln(

)

MP

W 2

000

A

B C

D

042

4

00

216

75

037

1

09

145

78

006

1

42

149

38

031

7

19

792

7

530

2

288

1 0

75

147

79

200

1

603

7 7

46

817

3

11

109

10

2

3

02

540

2

264

2 1

1 2

7710

1

0

61

097

1

464

7 1

1 1

2010

1

0

78

174

1

585

9 1

1 4

6010

3

2

38

650

80

99

477

266

355

147

SC 2

008a

1

2

3

4

5

6

7

8

9 10

11

12

651

39

15

859

51

329

0

01

116

757

821

7

14

916

39

649

7

70

112

870

662

10

83

121

958

316

23

75

130

720

257

0

01

528

03

282

0

01

602

68

760

48

70

493

02

224

14

592

6 6

089

2 5

44

112

5 6

149

1 4

37

158

2 6

513

8

790

8

769

3 3

67

117

125

768

9

167

0 7

08

112

889

814

12

201

3

857

13

089

0

289

5

335

0 4

62

609

14

150

9 4

981

2 6

89

619

69

704

6

155

4 9

13

651

76

1

ndash 7

90

790

8

769

3

11

205

10

2

3

35

000

11

676

0

11

410

10

3

8

05

732

9

164

2 1

1 2

9210

3

6

79

726

11

287

7

1

ndash 8

14

814

12

201

3

11

329

10

4

12

30

102

3 13

081

1

11

508

10

2

2

52

000

5

280

4 1

1 2

3310

2

2

82

000

6

026

3 1

1 8

3210

4

15

49

180

1 4

971

7 1

1 1

6010

3

8

59

104

0 6

182

0 1

ndash

704

7

04

615

54

11

235

10

4

6

34

125

7 6

515

6

072

9

56

24 0

158

109

1

130

1

190

299

04

0

Och

s 19

95

2 3 11

07

275

6 55

85

507

40

42

684

9 18

15

561

2 15

06

691

2 1

1 2

0610

4

11

80

261

1 55

85

11

152

10

4

11

62

232

9 68

68

55

87

GH

200

1 A

MP

RA

280

0

46

152

515

0

42

195

044

17

7

101

24

6

11

189

10

0

047

0

32

168

1

ndash 1

01

101

24

6 1

8 0 N

S 20

02

NS

6

79

311

9

127

7 3

36

912

90

1

ndash 3

36

336

9

129

0 0

All

MP

W 2

000

SC 2

008

MP

W 2

000

SC 2

008

Sum

Sum

Sum

Poo

led

Poo

led

ndash ndash

11

729

62

ndash ndash

591

17

ndash ndash

10

097

88

ndash ndash

ndash

ndash ndash

ndash

ndash 11

811

62

ndash 6

187

0

ndash 10

150

54

439

6

287

2 6

37

102

121

7

ndash ndash

ndash ndash

11

785

15

ndash ndash

ndash ndash

6

124

8

ndash ndash

ndash ndash

10

131

10

11

177

10

2

ndash

ndash 622

04

1

ndash ndash

ndash

10

212

17

ndash ndash ndash13

36

0

Tabl

e 3

Par

amet

er e

stim

ates

a

Bru

nner

et

al [

2011

] poi

nt o

ut t

hat

SC 2

008

mis

repo

rts

the

empi

rica

l fre

quen

cy fo

r G

ame 3

O

ur an

alys

is is

base

d on

the

raw

data

whi

ch we

than

kT

hors

ten

Chm

ura

for

prov

idin

g

Study Game Data

p q N HQRE p q

LQRE p q

EQRE p q

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

064 024 063 024 059 026 055 033

069 012 071 022 063 011 059 021

068 012 067 021 064 012 059 023

1 0921 0310 9600 092 031 096 033 096 033 2 0783 0473 9600 078 050 081 051 079 050 3 0837 0207 9600 084 021 083 021 083 021 4 0714 0264 9600 071 026 072 026 072 026 5 0673 0336 9600 067 034 069 033 069 033

SC 2008 6

7 0555 0404 9600 0859 0436 4800

056 040 086 050

058 039 089 052

057 039 085 050

8 0750 0414 4800 075 050 083 047 075 050 9 0746 0173 4800 075 017 080 014 079 014 10 0634 0301 4800 064 030 072 026 070 024 11 0669 0348 4800 067 035 069 034 069 034 12 0561 0396 4800 056 040 058 038 057 038

Ochs 1995 2 3

0595 0258 448 0542 0336 516

060 026 054 034

065 025 062 033

060 026 058 032

GH 2001 AMP RA

0960 0160 25 0080 0800 25

096 016 008 080

083 024 035 077

089 027 035 077

NS 2002 NS 0543 0610 6720 054 061 054 062 054 062

Table 4 Best-fit Predictions

relative to LQRE is related to its ability to correctly order the playersrsquo i

s Table 3 gives the parameter estimates and log-likelihoods and Table 4 give the resulting pre-

dictions Our qualitative analysis of Section 71 is confirmed EQRE outperforms LQRE in the 15

games identified earlier (A B C D 2 3 4 6 7 9 10 12 2 3 AMP ) which are those for which EQRE would have outperformed LQRE for any cost function In particular EQRE would

have outperformed LQRE if k were chosen arbitrarily and only were optimized EQRE outper-forms LQRE in game 8 also but this requires that k and are jointly optimized In the remaining

5 games LQRE outperforms EQRE for all finite k and therefore the best-fit EQRE involves k = 1

and coincides with the best-fit LQRE Testing for statistical significance a standard likelihood ratio

test rejects the null hypothesis that LQRE performs as well as EQRE for most (but not all) games as shown in the last column of Table 343

Inspecting Table 3 more closely in 15 of the 16 games for which EQRE outperforms LQRE

(whether statistically significantly or not) the EQRE estimates ˆ and ˆ have the same ordering 1 2

243The likelihood ratio statistic is distributed as a with 1 degree of freedom (since EQRE has 1 more parameter than LQRE) under the null hypothesis that LQRE performs as well as EQRE and indicate rejection at the 010 005 and 001 levels respectively

33

as HQRE ˆ1 and ˆ2 In some cases the estimates are very close Overall though EQRE tends to

order the i

s correctly it usually underpredicts the absolute differences between the i

s In addition to individual game performance we fit the pooled data for MPW 2000 and SC 2008

the studies with more than a few games each We do not pool across studies because this would

be sensitive to the conversion factors as discussed in Section 72 We find that EQRE significantly

outperforms LQRE for MPW 2000 but not for SC 2008 in which the best-fit EQRE involves k = 1

73 Out-of-sample performance

Finally we present reduced form measures of out-of-sample performance whereby we use the indi-vidual game parameter estimates from Table 3 to make out-of-sample predictions for other games We do this for each of the games within MPW 2000 and SC 2008 and evaluate out-of-sample

performance by the squared distance between predicted and empirical action frequencies44 Table

5 presents the results as matrices for MPW 2000 and SC 2008 in which the ijth entry gives the

difference in squared distance between LQRE and EQRE in going from game i to game j A posi-tive (negative) entry indicates that EQRE outperforms (underperforms) LQRE A zero entry in the

ij-position indicates that for game i the best-fit EQRE involves k = 1 and thus coincides with

the best-fit LQRE LQRE outperforms EQRE in the majority of forecasts as indicated by the larger number of

negative entries LQRE also outperforms EQRE by the average of the entries which are -0020 and

-0011 for MPW 2000 and SC 2008 respectively We have no particular intuition for this result but believe it is more than simply an overfitting phenomenon In unreported results the matrices look similar if EQRE is estimated under the restriction that k is fixed at some k for all games (we

tried all k 2 11 2 5 10 20) Overall the results are mixed While EQRE tends to perform better in-sample it is not uni-

formly true and LQRE performs better out-of-sample

8 Conclusion

We endogenize the precision parameter of standard logit QRE (LQRE) The resulting model which

we call endogenous quantal response equilibrium (EQRE) takes as given the logit structure but endogenizes the precision parameter in a first stage In the model each player i chooses i

optimally subject to costs given correct beliefs over the opponentsrsquo second-stage actions which form a heterogenous LQRE given the vector = ( 1 )

n

Because incentives to acquire precision depend on the equilibrium behavior of opponents which

in turn depends on the payoffs of the underlying game each player generically chooses a different 44For model M with parameter the squared distance is D(M ) = (p p M ( ))2 + (q q M ( ))2 where

p M ( ) q M ( ) is the model prediction and p q is the empirical action frequency Squared distance is a com-monly used measure of fit (see for example Nyarko and Schotter [2002] and Selten and Chmura [2008])

34

MPW 2000 Game A B C D

A 39

10 3

12

10 3

30

10 3

B -10

10 1

24

10 2

-80

10 2

C -32

10 2

92

10 3

-71

10 2

D 16

10 3

23

10 3

30

10 4

SC 2008 Game 1 2 3 4 5 6 7 8 9 10 11 12

1 0 0 0 0 0 0 0 0 0 0 0

2 -74

10 3

-29

10 2

-28

10 2

-33

10 2

-10

10 2

-55

10 3

39

10 3

-48

10 2

-34

10 2

-31

10 2

-13

10 2

3 -28

10 2

-77

10 3

-39

10 4

-33

10 3

-62

10 3

32

10 3

-13

10 2

-87

10 4

-59

10 4

-27

10 3

-66

10 3

4 -43

10 3

-11

10 2

-35

10 4

-14

10 3

-36

10 3

-22

10 3

-11

10 2

36

10 3

69

10 4

-12

10 3

-40

10 3

5 0 0 0 0 0 0 0 0 0 0 0

6 -25

10 2

-19

10 2

-12

10 2

-48

10 3

-11

10 3

-64

10 2

-97

10 3

34

10 3

-23

10 4

-15

10 3

25

10 4

7 41

10 3

-52

10 2

-52

10 2

-41

10 2

-46

10 2

-94

10 3

-37

10 2

-66

10 2

-32

10 2

-43

10 2

-11

10 2

8 -19

10 2

56

10 4

-54

10 2

-40

10 2

-30

10 2

-11

10 2

-35

10 3

-77

10 2

-44

10 2

-28

10 2

-13

10 2

9 -39

10 3

-38

10 3

-14

10 3

10

10 3

81

10 4

15

10 4

-85

10 3

-42

10 3

23

10 3

11

10 3

-95

10 5

10 -17

10 3

-15

10 2

-30

10 3

-60

10 4

-57

10 5

-11

10 3

-28

10 2

-93

10 3

67

10 3

-31

10 4

-14

10 3

11 0 0 0 0 0 0 0 0 0 0 0

12 -28

10 2

-20

10 2

-13

10 2

-54

10 3

-16

10 3

32

10 4

-66

10 2

-10

10 2

24

10 3

-44

10 4

-20

10 3

Table 5 Out-of-sample performance

35

i Because of this EQRE and LQRE generically give different predictions We show that EQRE

satisfies a modified form of the regularity axioms (Goeree et al [2005]) and provide analogues to

most of the classic results for LQRE such as those for equilibrium selection Our results reveal both

similarities and differences to LQRE In the binary action case we are able to characterize behavior rather well For generalized matching pennies games we nearly have a complete characterization of the EQRE set which deviates systematically from the LQRE set and whose qualitative shape does not depend on the EQRE cost function Comparing the models in data we find that EQRE tends to outperform LQRE in-sample but performs worse out-of-sample

References

Larbi Alaoui and Antonio Penta Endogenous depth of reasoning Review of Economic Studies 2015 1

Christoph Brunner Colin Camerer and Jacob Goeree Stationary concepts for experimental 2x2

games Comment American Economic Review 2011 a

Colin Camerer Teck-Hua Ho and Juin-Kuan Chong A cognitive hierarchy model of games The

Quarterly Journal of Economics 2004 1

Andrew Caplin and Mark Dean Revealed preference rational inattention and costly information

acquisition American Economic Review 2015 10

Tommaso Denti Games with unrestricted information acquisition Working Paper 2015 1

Evan Friedman Stochastic equilibria Noise in actions or beliefs Working Paper 2018 1 11 31 72 42

Jacob Goeree and Charles Holt Ten little treasures of game theory and ten intuitive contradictions American Economic Review 2001 7 974

Jacob Goeree Charles Holt and Thomas Palfrey Regular quantal response equilibrium Experi-mental Economics 2005 1 3 4 7 22 11 31 26 8

Jacob Goeree Charles Holt and Thomas Palfrey Quantal response equilibrium Princeton Uni-versity Press 2016 1

Philip Haile Ali Hortacsu and Grigory Kosenok On the empirical content of quantal response

equilibrium American Economic Review 2008 1 22

Phillipe Jehiel Analogy-based expectations equilibrium Journal of Economic Theory 2005 1

36

Filip Matejka and Alisdair McKay Rational inattention to discrete choices A new foundation for the multinomial logit American Economic Review 2015 6

Lars-Goran Mattsson and Jorgen Weibull Probabilistic choice and procedurally bounded rationality Games and Economic Behavior 2002 1

Richard McKelvey and Thomas Palfrey Quantal response equilibria for normal form games Games

and Economic Behavior 1995 1 18 5 94 94 94 973

Richard McKelvey and Thomas Palfrey A statistical theory of equilibrium in games The Japanese

Economic Review 1996 1 23

Richard McKelvey and Thomas Palfrey Quantal response equilibria for extensive form games Experimental Economics 1998 31 94

Richard McKelvey Thomas Palfrey and Roberto Weber Endogenous rationality equilibrium Work-ing Paper 1997 1

Richard McKelvey Thomas Palfrey and Roberto Weber The effects of payoff magnitude and het-erogeneity on behavior in 2x2 games with unique mixed strategy equilibria Journal of Economic

Behavior and Organization 2000 1 7 971

Rosemarie Nagel Unraveling in guessing games an experimental study American Economic

Review 1995 1

Yaw Nyarko and Andrew Schotter An experimental study of belief learning using elicited beliefs Econometrica 2002 7 44 975

Jack Ochs An experimental study of games with unique mixed strategy equilibria Games and

Economic Behavior 1995 7 973

Reinhard Selten and Thorsten Chmura Stationary concepts for experimental 2x2 games American

Economic Review 2008 1 7 44 972

Christopher Sims Stickiness Carnegie-Rochester Conference Series on Public Policy 1998 10

Christopher Sims Implications of rational inattention Journal of Monetary Economics 2003 10

Dale Stahl Entropy control costs and entropic equilibrium International Journal of Game Theory 1990 1

Jorgen Weibull Lars-Goran Mattsson and Mark Voorneveld Better may be worse some mono-tonicity results and paradoxes in discrete choice under uncertainty Theory and Decision 2007

1 3 95

37

Michael Woodford Information-constrained state-dependent pricing Journal of Monetary Eco-nomics 2009 10

Ming Yang Coordination with flexible information acquisition Journal of Economic Theory 2015

1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

71 Qualitative analysis

We begin by plotting the data in Figures 7 and 8 Each panel corresponds to a different game and

is familiar from Section 5 We plot the empirical action frequencies (black square) LQRE set (solid

black curve) regular set (gray rectangles) iso-utility curves (thin black lines) and iso-accuracy

curves (dashed lines) The plots also include the EQRE set (dotted curve) associated with power costs with k = 11

Why k = 11 From Theorem 6 all EQRE sets (with power costs or otherwise) will qualitatively

behave like those plotted in Figures 7 and 8 for k = 11 and it is clear that any point ldquobetweenrdquo this EQRE set and the LQRE set can be supported as an EQRE for some cost function In other words the region enclosed by the two sets is a subset of all possible EQRE and any other EQRE

that is not included in this region must deviate ldquoeven fartherrdquo from the LQRE set Since the

EQRE set approaches the LQRE set as k 1 (Theorem 8) we conjecture that lower values of k will give EQRE that are as far as possible from the LQRE set which is consistent with

our numerical investigations (unreported) By choosing the low value k = 1138 we exaggerate

differences between the EQRE and LQRE sets and thus we regard the region bounded by the two

sets as an approximation ldquofrom belowrdquo to the set of all possible EQRE Inspecting Figures 7 and 8 we first note that for all games the data is not contained in the region

enclosed by the LQRE and EQRE (k = 11) sets To the extent this region is a good approximation

to the set of all possible EQRE this implies that the data cannot be supported as an EQRE (ie exactly) for some cost function39 Still in many cases EQRE outperforms LQRE for any cost function As an example consider the very first panel of Figure 7 for game A Here the best-fit LQRE (by maximum likelihood) is to the left of the datapoint and we connect the two points with

a line segment Note that this segment crosses the EQRE set and in fact must cross the EQRE

set for any cost function by Theorem 6 That is for any cost function there is an EQRE that is closer to the data than is the nearest LQRE which implies that EQRE outperforms LQRE for any cost function Inspecting the plots EQRE outperforms LQRE in this sense for 15 of 21 games

(A B C D 2 3 4 6 7 9 10 12 2 3 AMP ) though in some cases it is difficult to see by

eye In game 8 the relative performance of LQRE and EQRE depends on the cost function For the remaining 5 games LQRE outperforms EQRE for all cost functions40

38Recall that k gt 1 is required to maintain properties of the cost function from Section 3 39In some cases the data is very near to but just outside the region 40Of course by Theorem 8 the EQRE set can approximate the LQRE set arbitrarily well by varying the cost

function

27

Study Game Data

p q N Parameters

p q r s

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

0500 0100 500 10 0500 0100 125 10 0500 0100 500 40 0500 0200 250 5

1 0921 0310 9600 0909 0091 100 11 2 0783 0473 9600 0818 0273 100 11 3 0837 0207 9600 0727 0091 100 11 4 0714 0264 9600 0636 0182 100 11 5 0673 0336 9600 0636 0273 100 11

SC 2008 6

7 0555 0404 9600 0859 0436 4800

0545 0364 100 11 0909 0091 100 11

8 0750 0414 4800 0818 0273 100 11 9 0746 0173 4800 0727 0091 100 11 10 0634 0301 4800 0636 0182 100 11 11 0669 0348 4800 0636 0273 100 11 12 0561 0396 4800 0545 0364 100 11

Ochs 1995 2 3

0595 0258 448 0542 0336 516

0500 0100 056 1238 0500 0200 063 1393

GH 2001 AMP RA

0960 0160 25 0080 0800 25

0500 0125 400 320 0500 0909 055 44

NS 2002 NS 0543 0610 6720 0600 0600 100 5

Table 2 Games and data

28

A B C

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q

1lowast 2lowastD

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 3lowast 4lowast 5lowast

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 6lowast 7lowast 8lowast

0 05 1

p

0

05

1

0 05 1

p

0

05

1

0 05 1

p

0

05

1

q q q

Figure 7 Plotting the data (part 1)

29

9lowast 10lowast 11lowast

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 12lowast 2 3

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q

AMP RA NSlowast

0 05 1

p

0

05

1

0 05 1

p

0

05

1

0 05 1

p

0

05

1

q q q

Figure 8 Plotting the data (part 2)

30

72 In-Sample fit

Next we explore the fit of LQRE and EQRE by choosing their parameters to maximize the log-likelihood of aggregate choice frequencies As is standard we fit LQRE by choosing For EQRE

we must choose a cost function In Section 6 we established that power costs for any k gt 1 have

nice properties so we naturally use this family of cost functions Since there is no principled way

of fixing k a priori we fit a 2-parameter EQRE where we fit both k and jointly From Theorem

8 LQRE is nested in EQRE in the sense that the equilibrium sets converge as k 1 so we

use standard likelihood ratio tests for model comparison Heuristically LQRE is simply EQRE

under the restriction that ldquok = 1rdquo We search over very fine - and -grids but for computational simplicity41 we search over a coarse k-grid k 2 11 2 5 10 201 By searching over a coarse

k-grid we are only underestimating EQRErsquos performance Prior to estimation we multiply each gamersquos payoffs by a study-specific conversion factor so that

each unit of payoff in the matrix corresponds roughly to the same real monetary payoff across all studies The conversion factors we use are from Friedman [2018] and amount to simple adjustments for inflation and currency-to-currency exchange rates42 From the results of Section 6 this is

immaterial for fitting games within a study it will effect parameter estimates but not equilibrium

predictions or fit The effect of the adjustment is to put parameter estimates ldquoon the same scalerdquo across studies It also allows in principle to fit the data pooled together from several studies though we do not pursue that here since the fit would be sensitive to the choice of factors

In addition to LQRE and EQRE we also fit a 2-parameter heterogenous LQRE which we label ldquoHQRErdquo This is simply LQRE in which each player i has a player-specific exogenous

i

We do this

for two reasons First HQRE nests both LQRE and EQRE and hence must fit the data better than

both It is standard practice to present as a benchmark the ldquoempirical likelihoodrdquo which is simply

the likelihood of any model that predicts the data exactly Since the data from any one game is 2-dimensional HQRE overfits the data from a single game to the point that its likelihood coincides with the empirical likelihood as long as the gamersquos data is consistent with some regular QRE (see

Section 2) In cases where the data cannot be rationalized by any regular QRE such as in games 2 7 and 8 (the datapoint falls outside of the gray rectangles in Figure 7) HQRE still gives the

best-possible regular QRE prediction Hence there is a sense in which the HQRE likelihood is a

better benchmark than the empirical likelihood since we are only comparing different QRE models Second we would like to compare the exogenous ˆ

i

s from fitting HQRE with our endogenous ˆ i s

from fitting EQRE Since HQRE fits the data nearly perfectly (when consistent with some regular QRE) we regard the exogenous ˆ

i

s as the ldquotruerdquo parameters and wonder if EQRErsquos performance

41To approximate a playerrsquos EQRE reaction to any mixed action of his opponents requires a numerical optimization Hence to approximate the EQRE reaction function of which EQRE is a fixed point requires many numerical optimizations On the other hand LQRE does not involve any optimization

42The factors are 1 (MPW 2000) 01560 (SC 2008) 01784 (Ochs 1995) 00973 (GH 2001) and 05 (NS 2002) See Friedman [2018] for details

31

32

Stud

y G

ame

HQ

RE

ˆ 1

ˆ 2

ln(L

)

LQR

E

ˆ ln(L

)

EQ

RE

k

ˆ 1

ˆ 2

ln(L

) 2ln(

)

MP

W 2

000

A

B C

D

042

4

00

216

75

037

1

09

145

78

006

1

42

149

38

031

7

19

792

7

530

2

288

1 0

75

147

79

200

1

603

7 7

46

817

3

11

109

10

2

3

02

540

2

264

2 1

1 2

7710

1

0

61

097

1

464

7 1

1 1

2010

1

0

78

174

1

585

9 1

1 4

6010

3

2

38

650

80

99

477

266

355

147

SC 2

008a

1

2

3

4

5

6

7

8

9 10

11

12

651

39

15

859

51

329

0

01

116

757

821

7

14

916

39

649

7

70

112

870

662

10

83

121

958

316

23

75

130

720

257

0

01

528

03

282

0

01

602

68

760

48

70

493

02

224

14

592

6 6

089

2 5

44

112

5 6

149

1 4

37

158

2 6

513

8

790

8

769

3 3

67

117

125

768

9

167

0 7

08

112

889

814

12

201

3

857

13

089

0

289

5

335

0 4

62

609

14

150

9 4

981

2 6

89

619

69

704

6

155

4 9

13

651

76

1

ndash 7

90

790

8

769

3

11

205

10

2

3

35

000

11

676

0

11

410

10

3

8

05

732

9

164

2 1

1 2

9210

3

6

79

726

11

287

7

1

ndash 8

14

814

12

201

3

11

329

10

4

12

30

102

3 13

081

1

11

508

10

2

2

52

000

5

280

4 1

1 2

3310

2

2

82

000

6

026

3 1

1 8

3210

4

15

49

180

1 4

971

7 1

1 1

6010

3

8

59

104

0 6

182

0 1

ndash

704

7

04

615

54

11

235

10

4

6

34

125

7 6

515

6

072

9

56

24 0

158

109

1

130

1

190

299

04

0

Och

s 19

95

2 3 11

07

275

6 55

85

507

40

42

684

9 18

15

561

2 15

06

691

2 1

1 2

0610

4

11

80

261

1 55

85

11

152

10

4

11

62

232

9 68

68

55

87

GH

200

1 A

MP

RA

280

0

46

152

515

0

42

195

044

17

7

101

24

6

11

189

10

0

047

0

32

168

1

ndash 1

01

101

24

6 1

8 0 N

S 20

02

NS

6

79

311

9

127

7 3

36

912

90

1

ndash 3

36

336

9

129

0 0

All

MP

W 2

000

SC 2

008

MP

W 2

000

SC 2

008

Sum

Sum

Sum

Poo

led

Poo

led

ndash ndash

11

729

62

ndash ndash

591

17

ndash ndash

10

097

88

ndash ndash

ndash

ndash ndash

ndash

ndash 11

811

62

ndash 6

187

0

ndash 10

150

54

439

6

287

2 6

37

102

121

7

ndash ndash

ndash ndash

11

785

15

ndash ndash

ndash ndash

6

124

8

ndash ndash

ndash ndash

10

131

10

11

177

10

2

ndash

ndash 622

04

1

ndash ndash

ndash

10

212

17

ndash ndash ndash13

36

0

Tabl

e 3

Par

amet

er e

stim

ates

a

Bru

nner

et

al [

2011

] poi

nt o

ut t

hat

SC 2

008

mis

repo

rts

the

empi

rica

l fre

quen

cy fo

r G

ame 3

O

ur an

alys

is is

base

d on

the

raw

data

whi

ch we

than

kT

hors

ten

Chm

ura

for

prov

idin

g

Study Game Data

p q N HQRE p q

LQRE p q

EQRE p q

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

064 024 063 024 059 026 055 033

069 012 071 022 063 011 059 021

068 012 067 021 064 012 059 023

1 0921 0310 9600 092 031 096 033 096 033 2 0783 0473 9600 078 050 081 051 079 050 3 0837 0207 9600 084 021 083 021 083 021 4 0714 0264 9600 071 026 072 026 072 026 5 0673 0336 9600 067 034 069 033 069 033

SC 2008 6

7 0555 0404 9600 0859 0436 4800

056 040 086 050

058 039 089 052

057 039 085 050

8 0750 0414 4800 075 050 083 047 075 050 9 0746 0173 4800 075 017 080 014 079 014 10 0634 0301 4800 064 030 072 026 070 024 11 0669 0348 4800 067 035 069 034 069 034 12 0561 0396 4800 056 040 058 038 057 038

Ochs 1995 2 3

0595 0258 448 0542 0336 516

060 026 054 034

065 025 062 033

060 026 058 032

GH 2001 AMP RA

0960 0160 25 0080 0800 25

096 016 008 080

083 024 035 077

089 027 035 077

NS 2002 NS 0543 0610 6720 054 061 054 062 054 062

Table 4 Best-fit Predictions

relative to LQRE is related to its ability to correctly order the playersrsquo i

s Table 3 gives the parameter estimates and log-likelihoods and Table 4 give the resulting pre-

dictions Our qualitative analysis of Section 71 is confirmed EQRE outperforms LQRE in the 15

games identified earlier (A B C D 2 3 4 6 7 9 10 12 2 3 AMP ) which are those for which EQRE would have outperformed LQRE for any cost function In particular EQRE would

have outperformed LQRE if k were chosen arbitrarily and only were optimized EQRE outper-forms LQRE in game 8 also but this requires that k and are jointly optimized In the remaining

5 games LQRE outperforms EQRE for all finite k and therefore the best-fit EQRE involves k = 1

and coincides with the best-fit LQRE Testing for statistical significance a standard likelihood ratio

test rejects the null hypothesis that LQRE performs as well as EQRE for most (but not all) games as shown in the last column of Table 343

Inspecting Table 3 more closely in 15 of the 16 games for which EQRE outperforms LQRE

(whether statistically significantly or not) the EQRE estimates ˆ and ˆ have the same ordering 1 2

243The likelihood ratio statistic is distributed as a with 1 degree of freedom (since EQRE has 1 more parameter than LQRE) under the null hypothesis that LQRE performs as well as EQRE and indicate rejection at the 010 005 and 001 levels respectively

33

as HQRE ˆ1 and ˆ2 In some cases the estimates are very close Overall though EQRE tends to

order the i

s correctly it usually underpredicts the absolute differences between the i

s In addition to individual game performance we fit the pooled data for MPW 2000 and SC 2008

the studies with more than a few games each We do not pool across studies because this would

be sensitive to the conversion factors as discussed in Section 72 We find that EQRE significantly

outperforms LQRE for MPW 2000 but not for SC 2008 in which the best-fit EQRE involves k = 1

73 Out-of-sample performance

Finally we present reduced form measures of out-of-sample performance whereby we use the indi-vidual game parameter estimates from Table 3 to make out-of-sample predictions for other games We do this for each of the games within MPW 2000 and SC 2008 and evaluate out-of-sample

performance by the squared distance between predicted and empirical action frequencies44 Table

5 presents the results as matrices for MPW 2000 and SC 2008 in which the ijth entry gives the

difference in squared distance between LQRE and EQRE in going from game i to game j A posi-tive (negative) entry indicates that EQRE outperforms (underperforms) LQRE A zero entry in the

ij-position indicates that for game i the best-fit EQRE involves k = 1 and thus coincides with

the best-fit LQRE LQRE outperforms EQRE in the majority of forecasts as indicated by the larger number of

negative entries LQRE also outperforms EQRE by the average of the entries which are -0020 and

-0011 for MPW 2000 and SC 2008 respectively We have no particular intuition for this result but believe it is more than simply an overfitting phenomenon In unreported results the matrices look similar if EQRE is estimated under the restriction that k is fixed at some k for all games (we

tried all k 2 11 2 5 10 20) Overall the results are mixed While EQRE tends to perform better in-sample it is not uni-

formly true and LQRE performs better out-of-sample

8 Conclusion

We endogenize the precision parameter of standard logit QRE (LQRE) The resulting model which

we call endogenous quantal response equilibrium (EQRE) takes as given the logit structure but endogenizes the precision parameter in a first stage In the model each player i chooses i

optimally subject to costs given correct beliefs over the opponentsrsquo second-stage actions which form a heterogenous LQRE given the vector = ( 1 )

n

Because incentives to acquire precision depend on the equilibrium behavior of opponents which

in turn depends on the payoffs of the underlying game each player generically chooses a different 44For model M with parameter the squared distance is D(M ) = (p p M ( ))2 + (q q M ( ))2 where

p M ( ) q M ( ) is the model prediction and p q is the empirical action frequency Squared distance is a com-monly used measure of fit (see for example Nyarko and Schotter [2002] and Selten and Chmura [2008])

34

MPW 2000 Game A B C D

A 39

10 3

12

10 3

30

10 3

B -10

10 1

24

10 2

-80

10 2

C -32

10 2

92

10 3

-71

10 2

D 16

10 3

23

10 3

30

10 4

SC 2008 Game 1 2 3 4 5 6 7 8 9 10 11 12

1 0 0 0 0 0 0 0 0 0 0 0

2 -74

10 3

-29

10 2

-28

10 2

-33

10 2

-10

10 2

-55

10 3

39

10 3

-48

10 2

-34

10 2

-31

10 2

-13

10 2

3 -28

10 2

-77

10 3

-39

10 4

-33

10 3

-62

10 3

32

10 3

-13

10 2

-87

10 4

-59

10 4

-27

10 3

-66

10 3

4 -43

10 3

-11

10 2

-35

10 4

-14

10 3

-36

10 3

-22

10 3

-11

10 2

36

10 3

69

10 4

-12

10 3

-40

10 3

5 0 0 0 0 0 0 0 0 0 0 0

6 -25

10 2

-19

10 2

-12

10 2

-48

10 3

-11

10 3

-64

10 2

-97

10 3

34

10 3

-23

10 4

-15

10 3

25

10 4

7 41

10 3

-52

10 2

-52

10 2

-41

10 2

-46

10 2

-94

10 3

-37

10 2

-66

10 2

-32

10 2

-43

10 2

-11

10 2

8 -19

10 2

56

10 4

-54

10 2

-40

10 2

-30

10 2

-11

10 2

-35

10 3

-77

10 2

-44

10 2

-28

10 2

-13

10 2

9 -39

10 3

-38

10 3

-14

10 3

10

10 3

81

10 4

15

10 4

-85

10 3

-42

10 3

23

10 3

11

10 3

-95

10 5

10 -17

10 3

-15

10 2

-30

10 3

-60

10 4

-57

10 5

-11

10 3

-28

10 2

-93

10 3

67

10 3

-31

10 4

-14

10 3

11 0 0 0 0 0 0 0 0 0 0 0

12 -28

10 2

-20

10 2

-13

10 2

-54

10 3

-16

10 3

32

10 4

-66

10 2

-10

10 2

24

10 3

-44

10 4

-20

10 3

Table 5 Out-of-sample performance

35

i Because of this EQRE and LQRE generically give different predictions We show that EQRE

satisfies a modified form of the regularity axioms (Goeree et al [2005]) and provide analogues to

most of the classic results for LQRE such as those for equilibrium selection Our results reveal both

similarities and differences to LQRE In the binary action case we are able to characterize behavior rather well For generalized matching pennies games we nearly have a complete characterization of the EQRE set which deviates systematically from the LQRE set and whose qualitative shape does not depend on the EQRE cost function Comparing the models in data we find that EQRE tends to outperform LQRE in-sample but performs worse out-of-sample

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Christoph Brunner Colin Camerer and Jacob Goeree Stationary concepts for experimental 2x2

games Comment American Economic Review 2011 a

Colin Camerer Teck-Hua Ho and Juin-Kuan Chong A cognitive hierarchy model of games The

Quarterly Journal of Economics 2004 1

Andrew Caplin and Mark Dean Revealed preference rational inattention and costly information

acquisition American Economic Review 2015 10

Tommaso Denti Games with unrestricted information acquisition Working Paper 2015 1

Evan Friedman Stochastic equilibria Noise in actions or beliefs Working Paper 2018 1 11 31 72 42

Jacob Goeree and Charles Holt Ten little treasures of game theory and ten intuitive contradictions American Economic Review 2001 7 974

Jacob Goeree Charles Holt and Thomas Palfrey Regular quantal response equilibrium Experi-mental Economics 2005 1 3 4 7 22 11 31 26 8

Jacob Goeree Charles Holt and Thomas Palfrey Quantal response equilibrium Princeton Uni-versity Press 2016 1

Philip Haile Ali Hortacsu and Grigory Kosenok On the empirical content of quantal response

equilibrium American Economic Review 2008 1 22

Phillipe Jehiel Analogy-based expectations equilibrium Journal of Economic Theory 2005 1

36

Filip Matejka and Alisdair McKay Rational inattention to discrete choices A new foundation for the multinomial logit American Economic Review 2015 6

Lars-Goran Mattsson and Jorgen Weibull Probabilistic choice and procedurally bounded rationality Games and Economic Behavior 2002 1

Richard McKelvey and Thomas Palfrey Quantal response equilibria for normal form games Games

and Economic Behavior 1995 1 18 5 94 94 94 973

Richard McKelvey and Thomas Palfrey A statistical theory of equilibrium in games The Japanese

Economic Review 1996 1 23

Richard McKelvey and Thomas Palfrey Quantal response equilibria for extensive form games Experimental Economics 1998 31 94

Richard McKelvey Thomas Palfrey and Roberto Weber Endogenous rationality equilibrium Work-ing Paper 1997 1

Richard McKelvey Thomas Palfrey and Roberto Weber The effects of payoff magnitude and het-erogeneity on behavior in 2x2 games with unique mixed strategy equilibria Journal of Economic

Behavior and Organization 2000 1 7 971

Rosemarie Nagel Unraveling in guessing games an experimental study American Economic

Review 1995 1

Yaw Nyarko and Andrew Schotter An experimental study of belief learning using elicited beliefs Econometrica 2002 7 44 975

Jack Ochs An experimental study of games with unique mixed strategy equilibria Games and

Economic Behavior 1995 7 973

Reinhard Selten and Thorsten Chmura Stationary concepts for experimental 2x2 games American

Economic Review 2008 1 7 44 972

Christopher Sims Stickiness Carnegie-Rochester Conference Series on Public Policy 1998 10

Christopher Sims Implications of rational inattention Journal of Monetary Economics 2003 10

Dale Stahl Entropy control costs and entropic equilibrium International Journal of Game Theory 1990 1

Jorgen Weibull Lars-Goran Mattsson and Mark Voorneveld Better may be worse some mono-tonicity results and paradoxes in discrete choice under uncertainty Theory and Decision 2007

1 3 95

37

Michael Woodford Information-constrained state-dependent pricing Journal of Monetary Eco-nomics 2009 10

Ming Yang Coordination with flexible information acquisition Journal of Economic Theory 2015

1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

Study Game Data

p q N Parameters

p q r s

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

0500 0100 500 10 0500 0100 125 10 0500 0100 500 40 0500 0200 250 5

1 0921 0310 9600 0909 0091 100 11 2 0783 0473 9600 0818 0273 100 11 3 0837 0207 9600 0727 0091 100 11 4 0714 0264 9600 0636 0182 100 11 5 0673 0336 9600 0636 0273 100 11

SC 2008 6

7 0555 0404 9600 0859 0436 4800

0545 0364 100 11 0909 0091 100 11

8 0750 0414 4800 0818 0273 100 11 9 0746 0173 4800 0727 0091 100 11 10 0634 0301 4800 0636 0182 100 11 11 0669 0348 4800 0636 0273 100 11 12 0561 0396 4800 0545 0364 100 11

Ochs 1995 2 3

0595 0258 448 0542 0336 516

0500 0100 056 1238 0500 0200 063 1393

GH 2001 AMP RA

0960 0160 25 0080 0800 25

0500 0125 400 320 0500 0909 055 44

NS 2002 NS 0543 0610 6720 0600 0600 100 5

Table 2 Games and data

28

A B C

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q

1lowast 2lowastD

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 3lowast 4lowast 5lowast

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 6lowast 7lowast 8lowast

0 05 1

p

0

05

1

0 05 1

p

0

05

1

0 05 1

p

0

05

1

q q q

Figure 7 Plotting the data (part 1)

29

9lowast 10lowast 11lowast

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 12lowast 2 3

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q

AMP RA NSlowast

0 05 1

p

0

05

1

0 05 1

p

0

05

1

0 05 1

p

0

05

1

q q q

Figure 8 Plotting the data (part 2)

30

72 In-Sample fit

Next we explore the fit of LQRE and EQRE by choosing their parameters to maximize the log-likelihood of aggregate choice frequencies As is standard we fit LQRE by choosing For EQRE

we must choose a cost function In Section 6 we established that power costs for any k gt 1 have

nice properties so we naturally use this family of cost functions Since there is no principled way

of fixing k a priori we fit a 2-parameter EQRE where we fit both k and jointly From Theorem

8 LQRE is nested in EQRE in the sense that the equilibrium sets converge as k 1 so we

use standard likelihood ratio tests for model comparison Heuristically LQRE is simply EQRE

under the restriction that ldquok = 1rdquo We search over very fine - and -grids but for computational simplicity41 we search over a coarse k-grid k 2 11 2 5 10 201 By searching over a coarse

k-grid we are only underestimating EQRErsquos performance Prior to estimation we multiply each gamersquos payoffs by a study-specific conversion factor so that

each unit of payoff in the matrix corresponds roughly to the same real monetary payoff across all studies The conversion factors we use are from Friedman [2018] and amount to simple adjustments for inflation and currency-to-currency exchange rates42 From the results of Section 6 this is

immaterial for fitting games within a study it will effect parameter estimates but not equilibrium

predictions or fit The effect of the adjustment is to put parameter estimates ldquoon the same scalerdquo across studies It also allows in principle to fit the data pooled together from several studies though we do not pursue that here since the fit would be sensitive to the choice of factors

In addition to LQRE and EQRE we also fit a 2-parameter heterogenous LQRE which we label ldquoHQRErdquo This is simply LQRE in which each player i has a player-specific exogenous

i

We do this

for two reasons First HQRE nests both LQRE and EQRE and hence must fit the data better than

both It is standard practice to present as a benchmark the ldquoempirical likelihoodrdquo which is simply

the likelihood of any model that predicts the data exactly Since the data from any one game is 2-dimensional HQRE overfits the data from a single game to the point that its likelihood coincides with the empirical likelihood as long as the gamersquos data is consistent with some regular QRE (see

Section 2) In cases where the data cannot be rationalized by any regular QRE such as in games 2 7 and 8 (the datapoint falls outside of the gray rectangles in Figure 7) HQRE still gives the

best-possible regular QRE prediction Hence there is a sense in which the HQRE likelihood is a

better benchmark than the empirical likelihood since we are only comparing different QRE models Second we would like to compare the exogenous ˆ

i

s from fitting HQRE with our endogenous ˆ i s

from fitting EQRE Since HQRE fits the data nearly perfectly (when consistent with some regular QRE) we regard the exogenous ˆ

i

s as the ldquotruerdquo parameters and wonder if EQRErsquos performance

41To approximate a playerrsquos EQRE reaction to any mixed action of his opponents requires a numerical optimization Hence to approximate the EQRE reaction function of which EQRE is a fixed point requires many numerical optimizations On the other hand LQRE does not involve any optimization

42The factors are 1 (MPW 2000) 01560 (SC 2008) 01784 (Ochs 1995) 00973 (GH 2001) and 05 (NS 2002) See Friedman [2018] for details

31

32

Stud

y G

ame

HQ

RE

ˆ 1

ˆ 2

ln(L

)

LQR

E

ˆ ln(L

)

EQ

RE

k

ˆ 1

ˆ 2

ln(L

) 2ln(

)

MP

W 2

000

A

B C

D

042

4

00

216

75

037

1

09

145

78

006

1

42

149

38

031

7

19

792

7

530

2

288

1 0

75

147

79

200

1

603

7 7

46

817

3

11

109

10

2

3

02

540

2

264

2 1

1 2

7710

1

0

61

097

1

464

7 1

1 1

2010

1

0

78

174

1

585

9 1

1 4

6010

3

2

38

650

80

99

477

266

355

147

SC 2

008a

1

2

3

4

5

6

7

8

9 10

11

12

651

39

15

859

51

329

0

01

116

757

821

7

14

916

39

649

7

70

112

870

662

10

83

121

958

316

23

75

130

720

257

0

01

528

03

282

0

01

602

68

760

48

70

493

02

224

14

592

6 6

089

2 5

44

112

5 6

149

1 4

37

158

2 6

513

8

790

8

769

3 3

67

117

125

768

9

167

0 7

08

112

889

814

12

201

3

857

13

089

0

289

5

335

0 4

62

609

14

150

9 4

981

2 6

89

619

69

704

6

155

4 9

13

651

76

1

ndash 7

90

790

8

769

3

11

205

10

2

3

35

000

11

676

0

11

410

10

3

8

05

732

9

164

2 1

1 2

9210

3

6

79

726

11

287

7

1

ndash 8

14

814

12

201

3

11

329

10

4

12

30

102

3 13

081

1

11

508

10

2

2

52

000

5

280

4 1

1 2

3310

2

2

82

000

6

026

3 1

1 8

3210

4

15

49

180

1 4

971

7 1

1 1

6010

3

8

59

104

0 6

182

0 1

ndash

704

7

04

615

54

11

235

10

4

6

34

125

7 6

515

6

072

9

56

24 0

158

109

1

130

1

190

299

04

0

Och

s 19

95

2 3 11

07

275

6 55

85

507

40

42

684

9 18

15

561

2 15

06

691

2 1

1 2

0610

4

11

80

261

1 55

85

11

152

10

4

11

62

232

9 68

68

55

87

GH

200

1 A

MP

RA

280

0

46

152

515

0

42

195

044

17

7

101

24

6

11

189

10

0

047

0

32

168

1

ndash 1

01

101

24

6 1

8 0 N

S 20

02

NS

6

79

311

9

127

7 3

36

912

90

1

ndash 3

36

336

9

129

0 0

All

MP

W 2

000

SC 2

008

MP

W 2

000

SC 2

008

Sum

Sum

Sum

Poo

led

Poo

led

ndash ndash

11

729

62

ndash ndash

591

17

ndash ndash

10

097

88

ndash ndash

ndash

ndash ndash

ndash

ndash 11

811

62

ndash 6

187

0

ndash 10

150

54

439

6

287

2 6

37

102

121

7

ndash ndash

ndash ndash

11

785

15

ndash ndash

ndash ndash

6

124

8

ndash ndash

ndash ndash

10

131

10

11

177

10

2

ndash

ndash 622

04

1

ndash ndash

ndash

10

212

17

ndash ndash ndash13

36

0

Tabl

e 3

Par

amet

er e

stim

ates

a

Bru

nner

et

al [

2011

] poi

nt o

ut t

hat

SC 2

008

mis

repo

rts

the

empi

rica

l fre

quen

cy fo

r G

ame 3

O

ur an

alys

is is

base

d on

the

raw

data

whi

ch we

than

kT

hors

ten

Chm

ura

for

prov

idin

g

Study Game Data

p q N HQRE p q

LQRE p q

EQRE p q

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

064 024 063 024 059 026 055 033

069 012 071 022 063 011 059 021

068 012 067 021 064 012 059 023

1 0921 0310 9600 092 031 096 033 096 033 2 0783 0473 9600 078 050 081 051 079 050 3 0837 0207 9600 084 021 083 021 083 021 4 0714 0264 9600 071 026 072 026 072 026 5 0673 0336 9600 067 034 069 033 069 033

SC 2008 6

7 0555 0404 9600 0859 0436 4800

056 040 086 050

058 039 089 052

057 039 085 050

8 0750 0414 4800 075 050 083 047 075 050 9 0746 0173 4800 075 017 080 014 079 014 10 0634 0301 4800 064 030 072 026 070 024 11 0669 0348 4800 067 035 069 034 069 034 12 0561 0396 4800 056 040 058 038 057 038

Ochs 1995 2 3

0595 0258 448 0542 0336 516

060 026 054 034

065 025 062 033

060 026 058 032

GH 2001 AMP RA

0960 0160 25 0080 0800 25

096 016 008 080

083 024 035 077

089 027 035 077

NS 2002 NS 0543 0610 6720 054 061 054 062 054 062

Table 4 Best-fit Predictions

relative to LQRE is related to its ability to correctly order the playersrsquo i

s Table 3 gives the parameter estimates and log-likelihoods and Table 4 give the resulting pre-

dictions Our qualitative analysis of Section 71 is confirmed EQRE outperforms LQRE in the 15

games identified earlier (A B C D 2 3 4 6 7 9 10 12 2 3 AMP ) which are those for which EQRE would have outperformed LQRE for any cost function In particular EQRE would

have outperformed LQRE if k were chosen arbitrarily and only were optimized EQRE outper-forms LQRE in game 8 also but this requires that k and are jointly optimized In the remaining

5 games LQRE outperforms EQRE for all finite k and therefore the best-fit EQRE involves k = 1

and coincides with the best-fit LQRE Testing for statistical significance a standard likelihood ratio

test rejects the null hypothesis that LQRE performs as well as EQRE for most (but not all) games as shown in the last column of Table 343

Inspecting Table 3 more closely in 15 of the 16 games for which EQRE outperforms LQRE

(whether statistically significantly or not) the EQRE estimates ˆ and ˆ have the same ordering 1 2

243The likelihood ratio statistic is distributed as a with 1 degree of freedom (since EQRE has 1 more parameter than LQRE) under the null hypothesis that LQRE performs as well as EQRE and indicate rejection at the 010 005 and 001 levels respectively

33

as HQRE ˆ1 and ˆ2 In some cases the estimates are very close Overall though EQRE tends to

order the i

s correctly it usually underpredicts the absolute differences between the i

s In addition to individual game performance we fit the pooled data for MPW 2000 and SC 2008

the studies with more than a few games each We do not pool across studies because this would

be sensitive to the conversion factors as discussed in Section 72 We find that EQRE significantly

outperforms LQRE for MPW 2000 but not for SC 2008 in which the best-fit EQRE involves k = 1

73 Out-of-sample performance

Finally we present reduced form measures of out-of-sample performance whereby we use the indi-vidual game parameter estimates from Table 3 to make out-of-sample predictions for other games We do this for each of the games within MPW 2000 and SC 2008 and evaluate out-of-sample

performance by the squared distance between predicted and empirical action frequencies44 Table

5 presents the results as matrices for MPW 2000 and SC 2008 in which the ijth entry gives the

difference in squared distance between LQRE and EQRE in going from game i to game j A posi-tive (negative) entry indicates that EQRE outperforms (underperforms) LQRE A zero entry in the

ij-position indicates that for game i the best-fit EQRE involves k = 1 and thus coincides with

the best-fit LQRE LQRE outperforms EQRE in the majority of forecasts as indicated by the larger number of

negative entries LQRE also outperforms EQRE by the average of the entries which are -0020 and

-0011 for MPW 2000 and SC 2008 respectively We have no particular intuition for this result but believe it is more than simply an overfitting phenomenon In unreported results the matrices look similar if EQRE is estimated under the restriction that k is fixed at some k for all games (we

tried all k 2 11 2 5 10 20) Overall the results are mixed While EQRE tends to perform better in-sample it is not uni-

formly true and LQRE performs better out-of-sample

8 Conclusion

We endogenize the precision parameter of standard logit QRE (LQRE) The resulting model which

we call endogenous quantal response equilibrium (EQRE) takes as given the logit structure but endogenizes the precision parameter in a first stage In the model each player i chooses i

optimally subject to costs given correct beliefs over the opponentsrsquo second-stage actions which form a heterogenous LQRE given the vector = ( 1 )

n

Because incentives to acquire precision depend on the equilibrium behavior of opponents which

in turn depends on the payoffs of the underlying game each player generically chooses a different 44For model M with parameter the squared distance is D(M ) = (p p M ( ))2 + (q q M ( ))2 where

p M ( ) q M ( ) is the model prediction and p q is the empirical action frequency Squared distance is a com-monly used measure of fit (see for example Nyarko and Schotter [2002] and Selten and Chmura [2008])

34

MPW 2000 Game A B C D

A 39

10 3

12

10 3

30

10 3

B -10

10 1

24

10 2

-80

10 2

C -32

10 2

92

10 3

-71

10 2

D 16

10 3

23

10 3

30

10 4

SC 2008 Game 1 2 3 4 5 6 7 8 9 10 11 12

1 0 0 0 0 0 0 0 0 0 0 0

2 -74

10 3

-29

10 2

-28

10 2

-33

10 2

-10

10 2

-55

10 3

39

10 3

-48

10 2

-34

10 2

-31

10 2

-13

10 2

3 -28

10 2

-77

10 3

-39

10 4

-33

10 3

-62

10 3

32

10 3

-13

10 2

-87

10 4

-59

10 4

-27

10 3

-66

10 3

4 -43

10 3

-11

10 2

-35

10 4

-14

10 3

-36

10 3

-22

10 3

-11

10 2

36

10 3

69

10 4

-12

10 3

-40

10 3

5 0 0 0 0 0 0 0 0 0 0 0

6 -25

10 2

-19

10 2

-12

10 2

-48

10 3

-11

10 3

-64

10 2

-97

10 3

34

10 3

-23

10 4

-15

10 3

25

10 4

7 41

10 3

-52

10 2

-52

10 2

-41

10 2

-46

10 2

-94

10 3

-37

10 2

-66

10 2

-32

10 2

-43

10 2

-11

10 2

8 -19

10 2

56

10 4

-54

10 2

-40

10 2

-30

10 2

-11

10 2

-35

10 3

-77

10 2

-44

10 2

-28

10 2

-13

10 2

9 -39

10 3

-38

10 3

-14

10 3

10

10 3

81

10 4

15

10 4

-85

10 3

-42

10 3

23

10 3

11

10 3

-95

10 5

10 -17

10 3

-15

10 2

-30

10 3

-60

10 4

-57

10 5

-11

10 3

-28

10 2

-93

10 3

67

10 3

-31

10 4

-14

10 3

11 0 0 0 0 0 0 0 0 0 0 0

12 -28

10 2

-20

10 2

-13

10 2

-54

10 3

-16

10 3

32

10 4

-66

10 2

-10

10 2

24

10 3

-44

10 4

-20

10 3

Table 5 Out-of-sample performance

35

i Because of this EQRE and LQRE generically give different predictions We show that EQRE

satisfies a modified form of the regularity axioms (Goeree et al [2005]) and provide analogues to

most of the classic results for LQRE such as those for equilibrium selection Our results reveal both

similarities and differences to LQRE In the binary action case we are able to characterize behavior rather well For generalized matching pennies games we nearly have a complete characterization of the EQRE set which deviates systematically from the LQRE set and whose qualitative shape does not depend on the EQRE cost function Comparing the models in data we find that EQRE tends to outperform LQRE in-sample but performs worse out-of-sample

References

Larbi Alaoui and Antonio Penta Endogenous depth of reasoning Review of Economic Studies 2015 1

Christoph Brunner Colin Camerer and Jacob Goeree Stationary concepts for experimental 2x2

games Comment American Economic Review 2011 a

Colin Camerer Teck-Hua Ho and Juin-Kuan Chong A cognitive hierarchy model of games The

Quarterly Journal of Economics 2004 1

Andrew Caplin and Mark Dean Revealed preference rational inattention and costly information

acquisition American Economic Review 2015 10

Tommaso Denti Games with unrestricted information acquisition Working Paper 2015 1

Evan Friedman Stochastic equilibria Noise in actions or beliefs Working Paper 2018 1 11 31 72 42

Jacob Goeree and Charles Holt Ten little treasures of game theory and ten intuitive contradictions American Economic Review 2001 7 974

Jacob Goeree Charles Holt and Thomas Palfrey Regular quantal response equilibrium Experi-mental Economics 2005 1 3 4 7 22 11 31 26 8

Jacob Goeree Charles Holt and Thomas Palfrey Quantal response equilibrium Princeton Uni-versity Press 2016 1

Philip Haile Ali Hortacsu and Grigory Kosenok On the empirical content of quantal response

equilibrium American Economic Review 2008 1 22

Phillipe Jehiel Analogy-based expectations equilibrium Journal of Economic Theory 2005 1

36

Filip Matejka and Alisdair McKay Rational inattention to discrete choices A new foundation for the multinomial logit American Economic Review 2015 6

Lars-Goran Mattsson and Jorgen Weibull Probabilistic choice and procedurally bounded rationality Games and Economic Behavior 2002 1

Richard McKelvey and Thomas Palfrey Quantal response equilibria for normal form games Games

and Economic Behavior 1995 1 18 5 94 94 94 973

Richard McKelvey and Thomas Palfrey A statistical theory of equilibrium in games The Japanese

Economic Review 1996 1 23

Richard McKelvey and Thomas Palfrey Quantal response equilibria for extensive form games Experimental Economics 1998 31 94

Richard McKelvey Thomas Palfrey and Roberto Weber Endogenous rationality equilibrium Work-ing Paper 1997 1

Richard McKelvey Thomas Palfrey and Roberto Weber The effects of payoff magnitude and het-erogeneity on behavior in 2x2 games with unique mixed strategy equilibria Journal of Economic

Behavior and Organization 2000 1 7 971

Rosemarie Nagel Unraveling in guessing games an experimental study American Economic

Review 1995 1

Yaw Nyarko and Andrew Schotter An experimental study of belief learning using elicited beliefs Econometrica 2002 7 44 975

Jack Ochs An experimental study of games with unique mixed strategy equilibria Games and

Economic Behavior 1995 7 973

Reinhard Selten and Thorsten Chmura Stationary concepts for experimental 2x2 games American

Economic Review 2008 1 7 44 972

Christopher Sims Stickiness Carnegie-Rochester Conference Series on Public Policy 1998 10

Christopher Sims Implications of rational inattention Journal of Monetary Economics 2003 10

Dale Stahl Entropy control costs and entropic equilibrium International Journal of Game Theory 1990 1

Jorgen Weibull Lars-Goran Mattsson and Mark Voorneveld Better may be worse some mono-tonicity results and paradoxes in discrete choice under uncertainty Theory and Decision 2007

1 3 95

37

Michael Woodford Information-constrained state-dependent pricing Journal of Monetary Eco-nomics 2009 10

Ming Yang Coordination with flexible information acquisition Journal of Economic Theory 2015

1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

A B C

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q

1lowast 2lowastD

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 3lowast 4lowast 5lowast

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 6lowast 7lowast 8lowast

0 05 1

p

0

05

1

0 05 1

p

0

05

1

0 05 1

p

0

05

1

q q q

Figure 7 Plotting the data (part 1)

29

9lowast 10lowast 11lowast

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 12lowast 2 3

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q

AMP RA NSlowast

0 05 1

p

0

05

1

0 05 1

p

0

05

1

0 05 1

p

0

05

1

q q q

Figure 8 Plotting the data (part 2)

30

72 In-Sample fit

Next we explore the fit of LQRE and EQRE by choosing their parameters to maximize the log-likelihood of aggregate choice frequencies As is standard we fit LQRE by choosing For EQRE

we must choose a cost function In Section 6 we established that power costs for any k gt 1 have

nice properties so we naturally use this family of cost functions Since there is no principled way

of fixing k a priori we fit a 2-parameter EQRE where we fit both k and jointly From Theorem

8 LQRE is nested in EQRE in the sense that the equilibrium sets converge as k 1 so we

use standard likelihood ratio tests for model comparison Heuristically LQRE is simply EQRE

under the restriction that ldquok = 1rdquo We search over very fine - and -grids but for computational simplicity41 we search over a coarse k-grid k 2 11 2 5 10 201 By searching over a coarse

k-grid we are only underestimating EQRErsquos performance Prior to estimation we multiply each gamersquos payoffs by a study-specific conversion factor so that

each unit of payoff in the matrix corresponds roughly to the same real monetary payoff across all studies The conversion factors we use are from Friedman [2018] and amount to simple adjustments for inflation and currency-to-currency exchange rates42 From the results of Section 6 this is

immaterial for fitting games within a study it will effect parameter estimates but not equilibrium

predictions or fit The effect of the adjustment is to put parameter estimates ldquoon the same scalerdquo across studies It also allows in principle to fit the data pooled together from several studies though we do not pursue that here since the fit would be sensitive to the choice of factors

In addition to LQRE and EQRE we also fit a 2-parameter heterogenous LQRE which we label ldquoHQRErdquo This is simply LQRE in which each player i has a player-specific exogenous

i

We do this

for two reasons First HQRE nests both LQRE and EQRE and hence must fit the data better than

both It is standard practice to present as a benchmark the ldquoempirical likelihoodrdquo which is simply

the likelihood of any model that predicts the data exactly Since the data from any one game is 2-dimensional HQRE overfits the data from a single game to the point that its likelihood coincides with the empirical likelihood as long as the gamersquos data is consistent with some regular QRE (see

Section 2) In cases where the data cannot be rationalized by any regular QRE such as in games 2 7 and 8 (the datapoint falls outside of the gray rectangles in Figure 7) HQRE still gives the

best-possible regular QRE prediction Hence there is a sense in which the HQRE likelihood is a

better benchmark than the empirical likelihood since we are only comparing different QRE models Second we would like to compare the exogenous ˆ

i

s from fitting HQRE with our endogenous ˆ i s

from fitting EQRE Since HQRE fits the data nearly perfectly (when consistent with some regular QRE) we regard the exogenous ˆ

i

s as the ldquotruerdquo parameters and wonder if EQRErsquos performance

41To approximate a playerrsquos EQRE reaction to any mixed action of his opponents requires a numerical optimization Hence to approximate the EQRE reaction function of which EQRE is a fixed point requires many numerical optimizations On the other hand LQRE does not involve any optimization

42The factors are 1 (MPW 2000) 01560 (SC 2008) 01784 (Ochs 1995) 00973 (GH 2001) and 05 (NS 2002) See Friedman [2018] for details

31

32

Stud

y G

ame

HQ

RE

ˆ 1

ˆ 2

ln(L

)

LQR

E

ˆ ln(L

)

EQ

RE

k

ˆ 1

ˆ 2

ln(L

) 2ln(

)

MP

W 2

000

A

B C

D

042

4

00

216

75

037

1

09

145

78

006

1

42

149

38

031

7

19

792

7

530

2

288

1 0

75

147

79

200

1

603

7 7

46

817

3

11

109

10

2

3

02

540

2

264

2 1

1 2

7710

1

0

61

097

1

464

7 1

1 1

2010

1

0

78

174

1

585

9 1

1 4

6010

3

2

38

650

80

99

477

266

355

147

SC 2

008a

1

2

3

4

5

6

7

8

9 10

11

12

651

39

15

859

51

329

0

01

116

757

821

7

14

916

39

649

7

70

112

870

662

10

83

121

958

316

23

75

130

720

257

0

01

528

03

282

0

01

602

68

760

48

70

493

02

224

14

592

6 6

089

2 5

44

112

5 6

149

1 4

37

158

2 6

513

8

790

8

769

3 3

67

117

125

768

9

167

0 7

08

112

889

814

12

201

3

857

13

089

0

289

5

335

0 4

62

609

14

150

9 4

981

2 6

89

619

69

704

6

155

4 9

13

651

76

1

ndash 7

90

790

8

769

3

11

205

10

2

3

35

000

11

676

0

11

410

10

3

8

05

732

9

164

2 1

1 2

9210

3

6

79

726

11

287

7

1

ndash 8

14

814

12

201

3

11

329

10

4

12

30

102

3 13

081

1

11

508

10

2

2

52

000

5

280

4 1

1 2

3310

2

2

82

000

6

026

3 1

1 8

3210

4

15

49

180

1 4

971

7 1

1 1

6010

3

8

59

104

0 6

182

0 1

ndash

704

7

04

615

54

11

235

10

4

6

34

125

7 6

515

6

072

9

56

24 0

158

109

1

130

1

190

299

04

0

Och

s 19

95

2 3 11

07

275

6 55

85

507

40

42

684

9 18

15

561

2 15

06

691

2 1

1 2

0610

4

11

80

261

1 55

85

11

152

10

4

11

62

232

9 68

68

55

87

GH

200

1 A

MP

RA

280

0

46

152

515

0

42

195

044

17

7

101

24

6

11

189

10

0

047

0

32

168

1

ndash 1

01

101

24

6 1

8 0 N

S 20

02

NS

6

79

311

9

127

7 3

36

912

90

1

ndash 3

36

336

9

129

0 0

All

MP

W 2

000

SC 2

008

MP

W 2

000

SC 2

008

Sum

Sum

Sum

Poo

led

Poo

led

ndash ndash

11

729

62

ndash ndash

591

17

ndash ndash

10

097

88

ndash ndash

ndash

ndash ndash

ndash

ndash 11

811

62

ndash 6

187

0

ndash 10

150

54

439

6

287

2 6

37

102

121

7

ndash ndash

ndash ndash

11

785

15

ndash ndash

ndash ndash

6

124

8

ndash ndash

ndash ndash

10

131

10

11

177

10

2

ndash

ndash 622

04

1

ndash ndash

ndash

10

212

17

ndash ndash ndash13

36

0

Tabl

e 3

Par

amet

er e

stim

ates

a

Bru

nner

et

al [

2011

] poi

nt o

ut t

hat

SC 2

008

mis

repo

rts

the

empi

rica

l fre

quen

cy fo

r G

ame 3

O

ur an

alys

is is

base

d on

the

raw

data

whi

ch we

than

kT

hors

ten

Chm

ura

for

prov

idin

g

Study Game Data

p q N HQRE p q

LQRE p q

EQRE p q

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

064 024 063 024 059 026 055 033

069 012 071 022 063 011 059 021

068 012 067 021 064 012 059 023

1 0921 0310 9600 092 031 096 033 096 033 2 0783 0473 9600 078 050 081 051 079 050 3 0837 0207 9600 084 021 083 021 083 021 4 0714 0264 9600 071 026 072 026 072 026 5 0673 0336 9600 067 034 069 033 069 033

SC 2008 6

7 0555 0404 9600 0859 0436 4800

056 040 086 050

058 039 089 052

057 039 085 050

8 0750 0414 4800 075 050 083 047 075 050 9 0746 0173 4800 075 017 080 014 079 014 10 0634 0301 4800 064 030 072 026 070 024 11 0669 0348 4800 067 035 069 034 069 034 12 0561 0396 4800 056 040 058 038 057 038

Ochs 1995 2 3

0595 0258 448 0542 0336 516

060 026 054 034

065 025 062 033

060 026 058 032

GH 2001 AMP RA

0960 0160 25 0080 0800 25

096 016 008 080

083 024 035 077

089 027 035 077

NS 2002 NS 0543 0610 6720 054 061 054 062 054 062

Table 4 Best-fit Predictions

relative to LQRE is related to its ability to correctly order the playersrsquo i

s Table 3 gives the parameter estimates and log-likelihoods and Table 4 give the resulting pre-

dictions Our qualitative analysis of Section 71 is confirmed EQRE outperforms LQRE in the 15

games identified earlier (A B C D 2 3 4 6 7 9 10 12 2 3 AMP ) which are those for which EQRE would have outperformed LQRE for any cost function In particular EQRE would

have outperformed LQRE if k were chosen arbitrarily and only were optimized EQRE outper-forms LQRE in game 8 also but this requires that k and are jointly optimized In the remaining

5 games LQRE outperforms EQRE for all finite k and therefore the best-fit EQRE involves k = 1

and coincides with the best-fit LQRE Testing for statistical significance a standard likelihood ratio

test rejects the null hypothesis that LQRE performs as well as EQRE for most (but not all) games as shown in the last column of Table 343

Inspecting Table 3 more closely in 15 of the 16 games for which EQRE outperforms LQRE

(whether statistically significantly or not) the EQRE estimates ˆ and ˆ have the same ordering 1 2

243The likelihood ratio statistic is distributed as a with 1 degree of freedom (since EQRE has 1 more parameter than LQRE) under the null hypothesis that LQRE performs as well as EQRE and indicate rejection at the 010 005 and 001 levels respectively

33

as HQRE ˆ1 and ˆ2 In some cases the estimates are very close Overall though EQRE tends to

order the i

s correctly it usually underpredicts the absolute differences between the i

s In addition to individual game performance we fit the pooled data for MPW 2000 and SC 2008

the studies with more than a few games each We do not pool across studies because this would

be sensitive to the conversion factors as discussed in Section 72 We find that EQRE significantly

outperforms LQRE for MPW 2000 but not for SC 2008 in which the best-fit EQRE involves k = 1

73 Out-of-sample performance

Finally we present reduced form measures of out-of-sample performance whereby we use the indi-vidual game parameter estimates from Table 3 to make out-of-sample predictions for other games We do this for each of the games within MPW 2000 and SC 2008 and evaluate out-of-sample

performance by the squared distance between predicted and empirical action frequencies44 Table

5 presents the results as matrices for MPW 2000 and SC 2008 in which the ijth entry gives the

difference in squared distance between LQRE and EQRE in going from game i to game j A posi-tive (negative) entry indicates that EQRE outperforms (underperforms) LQRE A zero entry in the

ij-position indicates that for game i the best-fit EQRE involves k = 1 and thus coincides with

the best-fit LQRE LQRE outperforms EQRE in the majority of forecasts as indicated by the larger number of

negative entries LQRE also outperforms EQRE by the average of the entries which are -0020 and

-0011 for MPW 2000 and SC 2008 respectively We have no particular intuition for this result but believe it is more than simply an overfitting phenomenon In unreported results the matrices look similar if EQRE is estimated under the restriction that k is fixed at some k for all games (we

tried all k 2 11 2 5 10 20) Overall the results are mixed While EQRE tends to perform better in-sample it is not uni-

formly true and LQRE performs better out-of-sample

8 Conclusion

We endogenize the precision parameter of standard logit QRE (LQRE) The resulting model which

we call endogenous quantal response equilibrium (EQRE) takes as given the logit structure but endogenizes the precision parameter in a first stage In the model each player i chooses i

optimally subject to costs given correct beliefs over the opponentsrsquo second-stage actions which form a heterogenous LQRE given the vector = ( 1 )

n

Because incentives to acquire precision depend on the equilibrium behavior of opponents which

in turn depends on the payoffs of the underlying game each player generically chooses a different 44For model M with parameter the squared distance is D(M ) = (p p M ( ))2 + (q q M ( ))2 where

p M ( ) q M ( ) is the model prediction and p q is the empirical action frequency Squared distance is a com-monly used measure of fit (see for example Nyarko and Schotter [2002] and Selten and Chmura [2008])

34

MPW 2000 Game A B C D

A 39

10 3

12

10 3

30

10 3

B -10

10 1

24

10 2

-80

10 2

C -32

10 2

92

10 3

-71

10 2

D 16

10 3

23

10 3

30

10 4

SC 2008 Game 1 2 3 4 5 6 7 8 9 10 11 12

1 0 0 0 0 0 0 0 0 0 0 0

2 -74

10 3

-29

10 2

-28

10 2

-33

10 2

-10

10 2

-55

10 3

39

10 3

-48

10 2

-34

10 2

-31

10 2

-13

10 2

3 -28

10 2

-77

10 3

-39

10 4

-33

10 3

-62

10 3

32

10 3

-13

10 2

-87

10 4

-59

10 4

-27

10 3

-66

10 3

4 -43

10 3

-11

10 2

-35

10 4

-14

10 3

-36

10 3

-22

10 3

-11

10 2

36

10 3

69

10 4

-12

10 3

-40

10 3

5 0 0 0 0 0 0 0 0 0 0 0

6 -25

10 2

-19

10 2

-12

10 2

-48

10 3

-11

10 3

-64

10 2

-97

10 3

34

10 3

-23

10 4

-15

10 3

25

10 4

7 41

10 3

-52

10 2

-52

10 2

-41

10 2

-46

10 2

-94

10 3

-37

10 2

-66

10 2

-32

10 2

-43

10 2

-11

10 2

8 -19

10 2

56

10 4

-54

10 2

-40

10 2

-30

10 2

-11

10 2

-35

10 3

-77

10 2

-44

10 2

-28

10 2

-13

10 2

9 -39

10 3

-38

10 3

-14

10 3

10

10 3

81

10 4

15

10 4

-85

10 3

-42

10 3

23

10 3

11

10 3

-95

10 5

10 -17

10 3

-15

10 2

-30

10 3

-60

10 4

-57

10 5

-11

10 3

-28

10 2

-93

10 3

67

10 3

-31

10 4

-14

10 3

11 0 0 0 0 0 0 0 0 0 0 0

12 -28

10 2

-20

10 2

-13

10 2

-54

10 3

-16

10 3

32

10 4

-66

10 2

-10

10 2

24

10 3

-44

10 4

-20

10 3

Table 5 Out-of-sample performance

35

i Because of this EQRE and LQRE generically give different predictions We show that EQRE

satisfies a modified form of the regularity axioms (Goeree et al [2005]) and provide analogues to

most of the classic results for LQRE such as those for equilibrium selection Our results reveal both

similarities and differences to LQRE In the binary action case we are able to characterize behavior rather well For generalized matching pennies games we nearly have a complete characterization of the EQRE set which deviates systematically from the LQRE set and whose qualitative shape does not depend on the EQRE cost function Comparing the models in data we find that EQRE tends to outperform LQRE in-sample but performs worse out-of-sample

References

Larbi Alaoui and Antonio Penta Endogenous depth of reasoning Review of Economic Studies 2015 1

Christoph Brunner Colin Camerer and Jacob Goeree Stationary concepts for experimental 2x2

games Comment American Economic Review 2011 a

Colin Camerer Teck-Hua Ho and Juin-Kuan Chong A cognitive hierarchy model of games The

Quarterly Journal of Economics 2004 1

Andrew Caplin and Mark Dean Revealed preference rational inattention and costly information

acquisition American Economic Review 2015 10

Tommaso Denti Games with unrestricted information acquisition Working Paper 2015 1

Evan Friedman Stochastic equilibria Noise in actions or beliefs Working Paper 2018 1 11 31 72 42

Jacob Goeree and Charles Holt Ten little treasures of game theory and ten intuitive contradictions American Economic Review 2001 7 974

Jacob Goeree Charles Holt and Thomas Palfrey Regular quantal response equilibrium Experi-mental Economics 2005 1 3 4 7 22 11 31 26 8

Jacob Goeree Charles Holt and Thomas Palfrey Quantal response equilibrium Princeton Uni-versity Press 2016 1

Philip Haile Ali Hortacsu and Grigory Kosenok On the empirical content of quantal response

equilibrium American Economic Review 2008 1 22

Phillipe Jehiel Analogy-based expectations equilibrium Journal of Economic Theory 2005 1

36

Filip Matejka and Alisdair McKay Rational inattention to discrete choices A new foundation for the multinomial logit American Economic Review 2015 6

Lars-Goran Mattsson and Jorgen Weibull Probabilistic choice and procedurally bounded rationality Games and Economic Behavior 2002 1

Richard McKelvey and Thomas Palfrey Quantal response equilibria for normal form games Games

and Economic Behavior 1995 1 18 5 94 94 94 973

Richard McKelvey and Thomas Palfrey A statistical theory of equilibrium in games The Japanese

Economic Review 1996 1 23

Richard McKelvey and Thomas Palfrey Quantal response equilibria for extensive form games Experimental Economics 1998 31 94

Richard McKelvey Thomas Palfrey and Roberto Weber Endogenous rationality equilibrium Work-ing Paper 1997 1

Richard McKelvey Thomas Palfrey and Roberto Weber The effects of payoff magnitude and het-erogeneity on behavior in 2x2 games with unique mixed strategy equilibria Journal of Economic

Behavior and Organization 2000 1 7 971

Rosemarie Nagel Unraveling in guessing games an experimental study American Economic

Review 1995 1

Yaw Nyarko and Andrew Schotter An experimental study of belief learning using elicited beliefs Econometrica 2002 7 44 975

Jack Ochs An experimental study of games with unique mixed strategy equilibria Games and

Economic Behavior 1995 7 973

Reinhard Selten and Thorsten Chmura Stationary concepts for experimental 2x2 games American

Economic Review 2008 1 7 44 972

Christopher Sims Stickiness Carnegie-Rochester Conference Series on Public Policy 1998 10

Christopher Sims Implications of rational inattention Journal of Monetary Economics 2003 10

Dale Stahl Entropy control costs and entropic equilibrium International Journal of Game Theory 1990 1

Jorgen Weibull Lars-Goran Mattsson and Mark Voorneveld Better may be worse some mono-tonicity results and paradoxes in discrete choice under uncertainty Theory and Decision 2007

1 3 95

37

Michael Woodford Information-constrained state-dependent pricing Journal of Monetary Eco-nomics 2009 10

Ming Yang Coordination with flexible information acquisition Journal of Economic Theory 2015

1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

9lowast 10lowast 11lowast

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q 12lowast 2 3

p

0

05

1

p

0

05

1

p

0

05

1

0 05 1 0 05 1 0 05 1 q q q

AMP RA NSlowast

0 05 1

p

0

05

1

0 05 1

p

0

05

1

0 05 1

p

0

05

1

q q q

Figure 8 Plotting the data (part 2)

30

72 In-Sample fit

Next we explore the fit of LQRE and EQRE by choosing their parameters to maximize the log-likelihood of aggregate choice frequencies As is standard we fit LQRE by choosing For EQRE

we must choose a cost function In Section 6 we established that power costs for any k gt 1 have

nice properties so we naturally use this family of cost functions Since there is no principled way

of fixing k a priori we fit a 2-parameter EQRE where we fit both k and jointly From Theorem

8 LQRE is nested in EQRE in the sense that the equilibrium sets converge as k 1 so we

use standard likelihood ratio tests for model comparison Heuristically LQRE is simply EQRE

under the restriction that ldquok = 1rdquo We search over very fine - and -grids but for computational simplicity41 we search over a coarse k-grid k 2 11 2 5 10 201 By searching over a coarse

k-grid we are only underestimating EQRErsquos performance Prior to estimation we multiply each gamersquos payoffs by a study-specific conversion factor so that

each unit of payoff in the matrix corresponds roughly to the same real monetary payoff across all studies The conversion factors we use are from Friedman [2018] and amount to simple adjustments for inflation and currency-to-currency exchange rates42 From the results of Section 6 this is

immaterial for fitting games within a study it will effect parameter estimates but not equilibrium

predictions or fit The effect of the adjustment is to put parameter estimates ldquoon the same scalerdquo across studies It also allows in principle to fit the data pooled together from several studies though we do not pursue that here since the fit would be sensitive to the choice of factors

In addition to LQRE and EQRE we also fit a 2-parameter heterogenous LQRE which we label ldquoHQRErdquo This is simply LQRE in which each player i has a player-specific exogenous

i

We do this

for two reasons First HQRE nests both LQRE and EQRE and hence must fit the data better than

both It is standard practice to present as a benchmark the ldquoempirical likelihoodrdquo which is simply

the likelihood of any model that predicts the data exactly Since the data from any one game is 2-dimensional HQRE overfits the data from a single game to the point that its likelihood coincides with the empirical likelihood as long as the gamersquos data is consistent with some regular QRE (see

Section 2) In cases where the data cannot be rationalized by any regular QRE such as in games 2 7 and 8 (the datapoint falls outside of the gray rectangles in Figure 7) HQRE still gives the

best-possible regular QRE prediction Hence there is a sense in which the HQRE likelihood is a

better benchmark than the empirical likelihood since we are only comparing different QRE models Second we would like to compare the exogenous ˆ

i

s from fitting HQRE with our endogenous ˆ i s

from fitting EQRE Since HQRE fits the data nearly perfectly (when consistent with some regular QRE) we regard the exogenous ˆ

i

s as the ldquotruerdquo parameters and wonder if EQRErsquos performance

41To approximate a playerrsquos EQRE reaction to any mixed action of his opponents requires a numerical optimization Hence to approximate the EQRE reaction function of which EQRE is a fixed point requires many numerical optimizations On the other hand LQRE does not involve any optimization

42The factors are 1 (MPW 2000) 01560 (SC 2008) 01784 (Ochs 1995) 00973 (GH 2001) and 05 (NS 2002) See Friedman [2018] for details

31

32

Stud

y G

ame

HQ

RE

ˆ 1

ˆ 2

ln(L

)

LQR

E

ˆ ln(L

)

EQ

RE

k

ˆ 1

ˆ 2

ln(L

) 2ln(

)

MP

W 2

000

A

B C

D

042

4

00

216

75

037

1

09

145

78

006

1

42

149

38

031

7

19

792

7

530

2

288

1 0

75

147

79

200

1

603

7 7

46

817

3

11

109

10

2

3

02

540

2

264

2 1

1 2

7710

1

0

61

097

1

464

7 1

1 1

2010

1

0

78

174

1

585

9 1

1 4

6010

3

2

38

650

80

99

477

266

355

147

SC 2

008a

1

2

3

4

5

6

7

8

9 10

11

12

651

39

15

859

51

329

0

01

116

757

821

7

14

916

39

649

7

70

112

870

662

10

83

121

958

316

23

75

130

720

257

0

01

528

03

282

0

01

602

68

760

48

70

493

02

224

14

592

6 6

089

2 5

44

112

5 6

149

1 4

37

158

2 6

513

8

790

8

769

3 3

67

117

125

768

9

167

0 7

08

112

889

814

12

201

3

857

13

089

0

289

5

335

0 4

62

609

14

150

9 4

981

2 6

89

619

69

704

6

155

4 9

13

651

76

1

ndash 7

90

790

8

769

3

11

205

10

2

3

35

000

11

676

0

11

410

10

3

8

05

732

9

164

2 1

1 2

9210

3

6

79

726

11

287

7

1

ndash 8

14

814

12

201

3

11

329

10

4

12

30

102

3 13

081

1

11

508

10

2

2

52

000

5

280

4 1

1 2

3310

2

2

82

000

6

026

3 1

1 8

3210

4

15

49

180

1 4

971

7 1

1 1

6010

3

8

59

104

0 6

182

0 1

ndash

704

7

04

615

54

11

235

10

4

6

34

125

7 6

515

6

072

9

56

24 0

158

109

1

130

1

190

299

04

0

Och

s 19

95

2 3 11

07

275

6 55

85

507

40

42

684

9 18

15

561

2 15

06

691

2 1

1 2

0610

4

11

80

261

1 55

85

11

152

10

4

11

62

232

9 68

68

55

87

GH

200

1 A

MP

RA

280

0

46

152

515

0

42

195

044

17

7

101

24

6

11

189

10

0

047

0

32

168

1

ndash 1

01

101

24

6 1

8 0 N

S 20

02

NS

6

79

311

9

127

7 3

36

912

90

1

ndash 3

36

336

9

129

0 0

All

MP

W 2

000

SC 2

008

MP

W 2

000

SC 2

008

Sum

Sum

Sum

Poo

led

Poo

led

ndash ndash

11

729

62

ndash ndash

591

17

ndash ndash

10

097

88

ndash ndash

ndash

ndash ndash

ndash

ndash 11

811

62

ndash 6

187

0

ndash 10

150

54

439

6

287

2 6

37

102

121

7

ndash ndash

ndash ndash

11

785

15

ndash ndash

ndash ndash

6

124

8

ndash ndash

ndash ndash

10

131

10

11

177

10

2

ndash

ndash 622

04

1

ndash ndash

ndash

10

212

17

ndash ndash ndash13

36

0

Tabl

e 3

Par

amet

er e

stim

ates

a

Bru

nner

et

al [

2011

] poi

nt o

ut t

hat

SC 2

008

mis

repo

rts

the

empi

rica

l fre

quen

cy fo

r G

ame 3

O

ur an

alys

is is

base

d on

the

raw

data

whi

ch we

than

kT

hors

ten

Chm

ura

for

prov

idin

g

Study Game Data

p q N HQRE p q

LQRE p q

EQRE p q

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

064 024 063 024 059 026 055 033

069 012 071 022 063 011 059 021

068 012 067 021 064 012 059 023

1 0921 0310 9600 092 031 096 033 096 033 2 0783 0473 9600 078 050 081 051 079 050 3 0837 0207 9600 084 021 083 021 083 021 4 0714 0264 9600 071 026 072 026 072 026 5 0673 0336 9600 067 034 069 033 069 033

SC 2008 6

7 0555 0404 9600 0859 0436 4800

056 040 086 050

058 039 089 052

057 039 085 050

8 0750 0414 4800 075 050 083 047 075 050 9 0746 0173 4800 075 017 080 014 079 014 10 0634 0301 4800 064 030 072 026 070 024 11 0669 0348 4800 067 035 069 034 069 034 12 0561 0396 4800 056 040 058 038 057 038

Ochs 1995 2 3

0595 0258 448 0542 0336 516

060 026 054 034

065 025 062 033

060 026 058 032

GH 2001 AMP RA

0960 0160 25 0080 0800 25

096 016 008 080

083 024 035 077

089 027 035 077

NS 2002 NS 0543 0610 6720 054 061 054 062 054 062

Table 4 Best-fit Predictions

relative to LQRE is related to its ability to correctly order the playersrsquo i

s Table 3 gives the parameter estimates and log-likelihoods and Table 4 give the resulting pre-

dictions Our qualitative analysis of Section 71 is confirmed EQRE outperforms LQRE in the 15

games identified earlier (A B C D 2 3 4 6 7 9 10 12 2 3 AMP ) which are those for which EQRE would have outperformed LQRE for any cost function In particular EQRE would

have outperformed LQRE if k were chosen arbitrarily and only were optimized EQRE outper-forms LQRE in game 8 also but this requires that k and are jointly optimized In the remaining

5 games LQRE outperforms EQRE for all finite k and therefore the best-fit EQRE involves k = 1

and coincides with the best-fit LQRE Testing for statistical significance a standard likelihood ratio

test rejects the null hypothesis that LQRE performs as well as EQRE for most (but not all) games as shown in the last column of Table 343

Inspecting Table 3 more closely in 15 of the 16 games for which EQRE outperforms LQRE

(whether statistically significantly or not) the EQRE estimates ˆ and ˆ have the same ordering 1 2

243The likelihood ratio statistic is distributed as a with 1 degree of freedom (since EQRE has 1 more parameter than LQRE) under the null hypothesis that LQRE performs as well as EQRE and indicate rejection at the 010 005 and 001 levels respectively

33

as HQRE ˆ1 and ˆ2 In some cases the estimates are very close Overall though EQRE tends to

order the i

s correctly it usually underpredicts the absolute differences between the i

s In addition to individual game performance we fit the pooled data for MPW 2000 and SC 2008

the studies with more than a few games each We do not pool across studies because this would

be sensitive to the conversion factors as discussed in Section 72 We find that EQRE significantly

outperforms LQRE for MPW 2000 but not for SC 2008 in which the best-fit EQRE involves k = 1

73 Out-of-sample performance

Finally we present reduced form measures of out-of-sample performance whereby we use the indi-vidual game parameter estimates from Table 3 to make out-of-sample predictions for other games We do this for each of the games within MPW 2000 and SC 2008 and evaluate out-of-sample

performance by the squared distance between predicted and empirical action frequencies44 Table

5 presents the results as matrices for MPW 2000 and SC 2008 in which the ijth entry gives the

difference in squared distance between LQRE and EQRE in going from game i to game j A posi-tive (negative) entry indicates that EQRE outperforms (underperforms) LQRE A zero entry in the

ij-position indicates that for game i the best-fit EQRE involves k = 1 and thus coincides with

the best-fit LQRE LQRE outperforms EQRE in the majority of forecasts as indicated by the larger number of

negative entries LQRE also outperforms EQRE by the average of the entries which are -0020 and

-0011 for MPW 2000 and SC 2008 respectively We have no particular intuition for this result but believe it is more than simply an overfitting phenomenon In unreported results the matrices look similar if EQRE is estimated under the restriction that k is fixed at some k for all games (we

tried all k 2 11 2 5 10 20) Overall the results are mixed While EQRE tends to perform better in-sample it is not uni-

formly true and LQRE performs better out-of-sample

8 Conclusion

We endogenize the precision parameter of standard logit QRE (LQRE) The resulting model which

we call endogenous quantal response equilibrium (EQRE) takes as given the logit structure but endogenizes the precision parameter in a first stage In the model each player i chooses i

optimally subject to costs given correct beliefs over the opponentsrsquo second-stage actions which form a heterogenous LQRE given the vector = ( 1 )

n

Because incentives to acquire precision depend on the equilibrium behavior of opponents which

in turn depends on the payoffs of the underlying game each player generically chooses a different 44For model M with parameter the squared distance is D(M ) = (p p M ( ))2 + (q q M ( ))2 where

p M ( ) q M ( ) is the model prediction and p q is the empirical action frequency Squared distance is a com-monly used measure of fit (see for example Nyarko and Schotter [2002] and Selten and Chmura [2008])

34

MPW 2000 Game A B C D

A 39

10 3

12

10 3

30

10 3

B -10

10 1

24

10 2

-80

10 2

C -32

10 2

92

10 3

-71

10 2

D 16

10 3

23

10 3

30

10 4

SC 2008 Game 1 2 3 4 5 6 7 8 9 10 11 12

1 0 0 0 0 0 0 0 0 0 0 0

2 -74

10 3

-29

10 2

-28

10 2

-33

10 2

-10

10 2

-55

10 3

39

10 3

-48

10 2

-34

10 2

-31

10 2

-13

10 2

3 -28

10 2

-77

10 3

-39

10 4

-33

10 3

-62

10 3

32

10 3

-13

10 2

-87

10 4

-59

10 4

-27

10 3

-66

10 3

4 -43

10 3

-11

10 2

-35

10 4

-14

10 3

-36

10 3

-22

10 3

-11

10 2

36

10 3

69

10 4

-12

10 3

-40

10 3

5 0 0 0 0 0 0 0 0 0 0 0

6 -25

10 2

-19

10 2

-12

10 2

-48

10 3

-11

10 3

-64

10 2

-97

10 3

34

10 3

-23

10 4

-15

10 3

25

10 4

7 41

10 3

-52

10 2

-52

10 2

-41

10 2

-46

10 2

-94

10 3

-37

10 2

-66

10 2

-32

10 2

-43

10 2

-11

10 2

8 -19

10 2

56

10 4

-54

10 2

-40

10 2

-30

10 2

-11

10 2

-35

10 3

-77

10 2

-44

10 2

-28

10 2

-13

10 2

9 -39

10 3

-38

10 3

-14

10 3

10

10 3

81

10 4

15

10 4

-85

10 3

-42

10 3

23

10 3

11

10 3

-95

10 5

10 -17

10 3

-15

10 2

-30

10 3

-60

10 4

-57

10 5

-11

10 3

-28

10 2

-93

10 3

67

10 3

-31

10 4

-14

10 3

11 0 0 0 0 0 0 0 0 0 0 0

12 -28

10 2

-20

10 2

-13

10 2

-54

10 3

-16

10 3

32

10 4

-66

10 2

-10

10 2

24

10 3

-44

10 4

-20

10 3

Table 5 Out-of-sample performance

35

i Because of this EQRE and LQRE generically give different predictions We show that EQRE

satisfies a modified form of the regularity axioms (Goeree et al [2005]) and provide analogues to

most of the classic results for LQRE such as those for equilibrium selection Our results reveal both

similarities and differences to LQRE In the binary action case we are able to characterize behavior rather well For generalized matching pennies games we nearly have a complete characterization of the EQRE set which deviates systematically from the LQRE set and whose qualitative shape does not depend on the EQRE cost function Comparing the models in data we find that EQRE tends to outperform LQRE in-sample but performs worse out-of-sample

References

Larbi Alaoui and Antonio Penta Endogenous depth of reasoning Review of Economic Studies 2015 1

Christoph Brunner Colin Camerer and Jacob Goeree Stationary concepts for experimental 2x2

games Comment American Economic Review 2011 a

Colin Camerer Teck-Hua Ho and Juin-Kuan Chong A cognitive hierarchy model of games The

Quarterly Journal of Economics 2004 1

Andrew Caplin and Mark Dean Revealed preference rational inattention and costly information

acquisition American Economic Review 2015 10

Tommaso Denti Games with unrestricted information acquisition Working Paper 2015 1

Evan Friedman Stochastic equilibria Noise in actions or beliefs Working Paper 2018 1 11 31 72 42

Jacob Goeree and Charles Holt Ten little treasures of game theory and ten intuitive contradictions American Economic Review 2001 7 974

Jacob Goeree Charles Holt and Thomas Palfrey Regular quantal response equilibrium Experi-mental Economics 2005 1 3 4 7 22 11 31 26 8

Jacob Goeree Charles Holt and Thomas Palfrey Quantal response equilibrium Princeton Uni-versity Press 2016 1

Philip Haile Ali Hortacsu and Grigory Kosenok On the empirical content of quantal response

equilibrium American Economic Review 2008 1 22

Phillipe Jehiel Analogy-based expectations equilibrium Journal of Economic Theory 2005 1

36

Filip Matejka and Alisdair McKay Rational inattention to discrete choices A new foundation for the multinomial logit American Economic Review 2015 6

Lars-Goran Mattsson and Jorgen Weibull Probabilistic choice and procedurally bounded rationality Games and Economic Behavior 2002 1

Richard McKelvey and Thomas Palfrey Quantal response equilibria for normal form games Games

and Economic Behavior 1995 1 18 5 94 94 94 973

Richard McKelvey and Thomas Palfrey A statistical theory of equilibrium in games The Japanese

Economic Review 1996 1 23

Richard McKelvey and Thomas Palfrey Quantal response equilibria for extensive form games Experimental Economics 1998 31 94

Richard McKelvey Thomas Palfrey and Roberto Weber Endogenous rationality equilibrium Work-ing Paper 1997 1

Richard McKelvey Thomas Palfrey and Roberto Weber The effects of payoff magnitude and het-erogeneity on behavior in 2x2 games with unique mixed strategy equilibria Journal of Economic

Behavior and Organization 2000 1 7 971

Rosemarie Nagel Unraveling in guessing games an experimental study American Economic

Review 1995 1

Yaw Nyarko and Andrew Schotter An experimental study of belief learning using elicited beliefs Econometrica 2002 7 44 975

Jack Ochs An experimental study of games with unique mixed strategy equilibria Games and

Economic Behavior 1995 7 973

Reinhard Selten and Thorsten Chmura Stationary concepts for experimental 2x2 games American

Economic Review 2008 1 7 44 972

Christopher Sims Stickiness Carnegie-Rochester Conference Series on Public Policy 1998 10

Christopher Sims Implications of rational inattention Journal of Monetary Economics 2003 10

Dale Stahl Entropy control costs and entropic equilibrium International Journal of Game Theory 1990 1

Jorgen Weibull Lars-Goran Mattsson and Mark Voorneveld Better may be worse some mono-tonicity results and paradoxes in discrete choice under uncertainty Theory and Decision 2007

1 3 95

37

Michael Woodford Information-constrained state-dependent pricing Journal of Monetary Eco-nomics 2009 10

Ming Yang Coordination with flexible information acquisition Journal of Economic Theory 2015

1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

72 In-Sample fit

Next we explore the fit of LQRE and EQRE by choosing their parameters to maximize the log-likelihood of aggregate choice frequencies As is standard we fit LQRE by choosing For EQRE

we must choose a cost function In Section 6 we established that power costs for any k gt 1 have

nice properties so we naturally use this family of cost functions Since there is no principled way

of fixing k a priori we fit a 2-parameter EQRE where we fit both k and jointly From Theorem

8 LQRE is nested in EQRE in the sense that the equilibrium sets converge as k 1 so we

use standard likelihood ratio tests for model comparison Heuristically LQRE is simply EQRE

under the restriction that ldquok = 1rdquo We search over very fine - and -grids but for computational simplicity41 we search over a coarse k-grid k 2 11 2 5 10 201 By searching over a coarse

k-grid we are only underestimating EQRErsquos performance Prior to estimation we multiply each gamersquos payoffs by a study-specific conversion factor so that

each unit of payoff in the matrix corresponds roughly to the same real monetary payoff across all studies The conversion factors we use are from Friedman [2018] and amount to simple adjustments for inflation and currency-to-currency exchange rates42 From the results of Section 6 this is

immaterial for fitting games within a study it will effect parameter estimates but not equilibrium

predictions or fit The effect of the adjustment is to put parameter estimates ldquoon the same scalerdquo across studies It also allows in principle to fit the data pooled together from several studies though we do not pursue that here since the fit would be sensitive to the choice of factors

In addition to LQRE and EQRE we also fit a 2-parameter heterogenous LQRE which we label ldquoHQRErdquo This is simply LQRE in which each player i has a player-specific exogenous

i

We do this

for two reasons First HQRE nests both LQRE and EQRE and hence must fit the data better than

both It is standard practice to present as a benchmark the ldquoempirical likelihoodrdquo which is simply

the likelihood of any model that predicts the data exactly Since the data from any one game is 2-dimensional HQRE overfits the data from a single game to the point that its likelihood coincides with the empirical likelihood as long as the gamersquos data is consistent with some regular QRE (see

Section 2) In cases where the data cannot be rationalized by any regular QRE such as in games 2 7 and 8 (the datapoint falls outside of the gray rectangles in Figure 7) HQRE still gives the

best-possible regular QRE prediction Hence there is a sense in which the HQRE likelihood is a

better benchmark than the empirical likelihood since we are only comparing different QRE models Second we would like to compare the exogenous ˆ

i

s from fitting HQRE with our endogenous ˆ i s

from fitting EQRE Since HQRE fits the data nearly perfectly (when consistent with some regular QRE) we regard the exogenous ˆ

i

s as the ldquotruerdquo parameters and wonder if EQRErsquos performance

41To approximate a playerrsquos EQRE reaction to any mixed action of his opponents requires a numerical optimization Hence to approximate the EQRE reaction function of which EQRE is a fixed point requires many numerical optimizations On the other hand LQRE does not involve any optimization

42The factors are 1 (MPW 2000) 01560 (SC 2008) 01784 (Ochs 1995) 00973 (GH 2001) and 05 (NS 2002) See Friedman [2018] for details

31

32

Stud

y G

ame

HQ

RE

ˆ 1

ˆ 2

ln(L

)

LQR

E

ˆ ln(L

)

EQ

RE

k

ˆ 1

ˆ 2

ln(L

) 2ln(

)

MP

W 2

000

A

B C

D

042

4

00

216

75

037

1

09

145

78

006

1

42

149

38

031

7

19

792

7

530

2

288

1 0

75

147

79

200

1

603

7 7

46

817

3

11

109

10

2

3

02

540

2

264

2 1

1 2

7710

1

0

61

097

1

464

7 1

1 1

2010

1

0

78

174

1

585

9 1

1 4

6010

3

2

38

650

80

99

477

266

355

147

SC 2

008a

1

2

3

4

5

6

7

8

9 10

11

12

651

39

15

859

51

329

0

01

116

757

821

7

14

916

39

649

7

70

112

870

662

10

83

121

958

316

23

75

130

720

257

0

01

528

03

282

0

01

602

68

760

48

70

493

02

224

14

592

6 6

089

2 5

44

112

5 6

149

1 4

37

158

2 6

513

8

790

8

769

3 3

67

117

125

768

9

167

0 7

08

112

889

814

12

201

3

857

13

089

0

289

5

335

0 4

62

609

14

150

9 4

981

2 6

89

619

69

704

6

155

4 9

13

651

76

1

ndash 7

90

790

8

769

3

11

205

10

2

3

35

000

11

676

0

11

410

10

3

8

05

732

9

164

2 1

1 2

9210

3

6

79

726

11

287

7

1

ndash 8

14

814

12

201

3

11

329

10

4

12

30

102

3 13

081

1

11

508

10

2

2

52

000

5

280

4 1

1 2

3310

2

2

82

000

6

026

3 1

1 8

3210

4

15

49

180

1 4

971

7 1

1 1

6010

3

8

59

104

0 6

182

0 1

ndash

704

7

04

615

54

11

235

10

4

6

34

125

7 6

515

6

072

9

56

24 0

158

109

1

130

1

190

299

04

0

Och

s 19

95

2 3 11

07

275

6 55

85

507

40

42

684

9 18

15

561

2 15

06

691

2 1

1 2

0610

4

11

80

261

1 55

85

11

152

10

4

11

62

232

9 68

68

55

87

GH

200

1 A

MP

RA

280

0

46

152

515

0

42

195

044

17

7

101

24

6

11

189

10

0

047

0

32

168

1

ndash 1

01

101

24

6 1

8 0 N

S 20

02

NS

6

79

311

9

127

7 3

36

912

90

1

ndash 3

36

336

9

129

0 0

All

MP

W 2

000

SC 2

008

MP

W 2

000

SC 2

008

Sum

Sum

Sum

Poo

led

Poo

led

ndash ndash

11

729

62

ndash ndash

591

17

ndash ndash

10

097

88

ndash ndash

ndash

ndash ndash

ndash

ndash 11

811

62

ndash 6

187

0

ndash 10

150

54

439

6

287

2 6

37

102

121

7

ndash ndash

ndash ndash

11

785

15

ndash ndash

ndash ndash

6

124

8

ndash ndash

ndash ndash

10

131

10

11

177

10

2

ndash

ndash 622

04

1

ndash ndash

ndash

10

212

17

ndash ndash ndash13

36

0

Tabl

e 3

Par

amet

er e

stim

ates

a

Bru

nner

et

al [

2011

] poi

nt o

ut t

hat

SC 2

008

mis

repo

rts

the

empi

rica

l fre

quen

cy fo

r G

ame 3

O

ur an

alys

is is

base

d on

the

raw

data

whi

ch we

than

kT

hors

ten

Chm

ura

for

prov

idin

g

Study Game Data

p q N HQRE p q

LQRE p q

EQRE p q

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

064 024 063 024 059 026 055 033

069 012 071 022 063 011 059 021

068 012 067 021 064 012 059 023

1 0921 0310 9600 092 031 096 033 096 033 2 0783 0473 9600 078 050 081 051 079 050 3 0837 0207 9600 084 021 083 021 083 021 4 0714 0264 9600 071 026 072 026 072 026 5 0673 0336 9600 067 034 069 033 069 033

SC 2008 6

7 0555 0404 9600 0859 0436 4800

056 040 086 050

058 039 089 052

057 039 085 050

8 0750 0414 4800 075 050 083 047 075 050 9 0746 0173 4800 075 017 080 014 079 014 10 0634 0301 4800 064 030 072 026 070 024 11 0669 0348 4800 067 035 069 034 069 034 12 0561 0396 4800 056 040 058 038 057 038

Ochs 1995 2 3

0595 0258 448 0542 0336 516

060 026 054 034

065 025 062 033

060 026 058 032

GH 2001 AMP RA

0960 0160 25 0080 0800 25

096 016 008 080

083 024 035 077

089 027 035 077

NS 2002 NS 0543 0610 6720 054 061 054 062 054 062

Table 4 Best-fit Predictions

relative to LQRE is related to its ability to correctly order the playersrsquo i

s Table 3 gives the parameter estimates and log-likelihoods and Table 4 give the resulting pre-

dictions Our qualitative analysis of Section 71 is confirmed EQRE outperforms LQRE in the 15

games identified earlier (A B C D 2 3 4 6 7 9 10 12 2 3 AMP ) which are those for which EQRE would have outperformed LQRE for any cost function In particular EQRE would

have outperformed LQRE if k were chosen arbitrarily and only were optimized EQRE outper-forms LQRE in game 8 also but this requires that k and are jointly optimized In the remaining

5 games LQRE outperforms EQRE for all finite k and therefore the best-fit EQRE involves k = 1

and coincides with the best-fit LQRE Testing for statistical significance a standard likelihood ratio

test rejects the null hypothesis that LQRE performs as well as EQRE for most (but not all) games as shown in the last column of Table 343

Inspecting Table 3 more closely in 15 of the 16 games for which EQRE outperforms LQRE

(whether statistically significantly or not) the EQRE estimates ˆ and ˆ have the same ordering 1 2

243The likelihood ratio statistic is distributed as a with 1 degree of freedom (since EQRE has 1 more parameter than LQRE) under the null hypothesis that LQRE performs as well as EQRE and indicate rejection at the 010 005 and 001 levels respectively

33

as HQRE ˆ1 and ˆ2 In some cases the estimates are very close Overall though EQRE tends to

order the i

s correctly it usually underpredicts the absolute differences between the i

s In addition to individual game performance we fit the pooled data for MPW 2000 and SC 2008

the studies with more than a few games each We do not pool across studies because this would

be sensitive to the conversion factors as discussed in Section 72 We find that EQRE significantly

outperforms LQRE for MPW 2000 but not for SC 2008 in which the best-fit EQRE involves k = 1

73 Out-of-sample performance

Finally we present reduced form measures of out-of-sample performance whereby we use the indi-vidual game parameter estimates from Table 3 to make out-of-sample predictions for other games We do this for each of the games within MPW 2000 and SC 2008 and evaluate out-of-sample

performance by the squared distance between predicted and empirical action frequencies44 Table

5 presents the results as matrices for MPW 2000 and SC 2008 in which the ijth entry gives the

difference in squared distance between LQRE and EQRE in going from game i to game j A posi-tive (negative) entry indicates that EQRE outperforms (underperforms) LQRE A zero entry in the

ij-position indicates that for game i the best-fit EQRE involves k = 1 and thus coincides with

the best-fit LQRE LQRE outperforms EQRE in the majority of forecasts as indicated by the larger number of

negative entries LQRE also outperforms EQRE by the average of the entries which are -0020 and

-0011 for MPW 2000 and SC 2008 respectively We have no particular intuition for this result but believe it is more than simply an overfitting phenomenon In unreported results the matrices look similar if EQRE is estimated under the restriction that k is fixed at some k for all games (we

tried all k 2 11 2 5 10 20) Overall the results are mixed While EQRE tends to perform better in-sample it is not uni-

formly true and LQRE performs better out-of-sample

8 Conclusion

We endogenize the precision parameter of standard logit QRE (LQRE) The resulting model which

we call endogenous quantal response equilibrium (EQRE) takes as given the logit structure but endogenizes the precision parameter in a first stage In the model each player i chooses i

optimally subject to costs given correct beliefs over the opponentsrsquo second-stage actions which form a heterogenous LQRE given the vector = ( 1 )

n

Because incentives to acquire precision depend on the equilibrium behavior of opponents which

in turn depends on the payoffs of the underlying game each player generically chooses a different 44For model M with parameter the squared distance is D(M ) = (p p M ( ))2 + (q q M ( ))2 where

p M ( ) q M ( ) is the model prediction and p q is the empirical action frequency Squared distance is a com-monly used measure of fit (see for example Nyarko and Schotter [2002] and Selten and Chmura [2008])

34

MPW 2000 Game A B C D

A 39

10 3

12

10 3

30

10 3

B -10

10 1

24

10 2

-80

10 2

C -32

10 2

92

10 3

-71

10 2

D 16

10 3

23

10 3

30

10 4

SC 2008 Game 1 2 3 4 5 6 7 8 9 10 11 12

1 0 0 0 0 0 0 0 0 0 0 0

2 -74

10 3

-29

10 2

-28

10 2

-33

10 2

-10

10 2

-55

10 3

39

10 3

-48

10 2

-34

10 2

-31

10 2

-13

10 2

3 -28

10 2

-77

10 3

-39

10 4

-33

10 3

-62

10 3

32

10 3

-13

10 2

-87

10 4

-59

10 4

-27

10 3

-66

10 3

4 -43

10 3

-11

10 2

-35

10 4

-14

10 3

-36

10 3

-22

10 3

-11

10 2

36

10 3

69

10 4

-12

10 3

-40

10 3

5 0 0 0 0 0 0 0 0 0 0 0

6 -25

10 2

-19

10 2

-12

10 2

-48

10 3

-11

10 3

-64

10 2

-97

10 3

34

10 3

-23

10 4

-15

10 3

25

10 4

7 41

10 3

-52

10 2

-52

10 2

-41

10 2

-46

10 2

-94

10 3

-37

10 2

-66

10 2

-32

10 2

-43

10 2

-11

10 2

8 -19

10 2

56

10 4

-54

10 2

-40

10 2

-30

10 2

-11

10 2

-35

10 3

-77

10 2

-44

10 2

-28

10 2

-13

10 2

9 -39

10 3

-38

10 3

-14

10 3

10

10 3

81

10 4

15

10 4

-85

10 3

-42

10 3

23

10 3

11

10 3

-95

10 5

10 -17

10 3

-15

10 2

-30

10 3

-60

10 4

-57

10 5

-11

10 3

-28

10 2

-93

10 3

67

10 3

-31

10 4

-14

10 3

11 0 0 0 0 0 0 0 0 0 0 0

12 -28

10 2

-20

10 2

-13

10 2

-54

10 3

-16

10 3

32

10 4

-66

10 2

-10

10 2

24

10 3

-44

10 4

-20

10 3

Table 5 Out-of-sample performance

35

i Because of this EQRE and LQRE generically give different predictions We show that EQRE

satisfies a modified form of the regularity axioms (Goeree et al [2005]) and provide analogues to

most of the classic results for LQRE such as those for equilibrium selection Our results reveal both

similarities and differences to LQRE In the binary action case we are able to characterize behavior rather well For generalized matching pennies games we nearly have a complete characterization of the EQRE set which deviates systematically from the LQRE set and whose qualitative shape does not depend on the EQRE cost function Comparing the models in data we find that EQRE tends to outperform LQRE in-sample but performs worse out-of-sample

References

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Christoph Brunner Colin Camerer and Jacob Goeree Stationary concepts for experimental 2x2

games Comment American Economic Review 2011 a

Colin Camerer Teck-Hua Ho and Juin-Kuan Chong A cognitive hierarchy model of games The

Quarterly Journal of Economics 2004 1

Andrew Caplin and Mark Dean Revealed preference rational inattention and costly information

acquisition American Economic Review 2015 10

Tommaso Denti Games with unrestricted information acquisition Working Paper 2015 1

Evan Friedman Stochastic equilibria Noise in actions or beliefs Working Paper 2018 1 11 31 72 42

Jacob Goeree and Charles Holt Ten little treasures of game theory and ten intuitive contradictions American Economic Review 2001 7 974

Jacob Goeree Charles Holt and Thomas Palfrey Regular quantal response equilibrium Experi-mental Economics 2005 1 3 4 7 22 11 31 26 8

Jacob Goeree Charles Holt and Thomas Palfrey Quantal response equilibrium Princeton Uni-versity Press 2016 1

Philip Haile Ali Hortacsu and Grigory Kosenok On the empirical content of quantal response

equilibrium American Economic Review 2008 1 22

Phillipe Jehiel Analogy-based expectations equilibrium Journal of Economic Theory 2005 1

36

Filip Matejka and Alisdair McKay Rational inattention to discrete choices A new foundation for the multinomial logit American Economic Review 2015 6

Lars-Goran Mattsson and Jorgen Weibull Probabilistic choice and procedurally bounded rationality Games and Economic Behavior 2002 1

Richard McKelvey and Thomas Palfrey Quantal response equilibria for normal form games Games

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Richard McKelvey and Thomas Palfrey A statistical theory of equilibrium in games The Japanese

Economic Review 1996 1 23

Richard McKelvey and Thomas Palfrey Quantal response equilibria for extensive form games Experimental Economics 1998 31 94

Richard McKelvey Thomas Palfrey and Roberto Weber Endogenous rationality equilibrium Work-ing Paper 1997 1

Richard McKelvey Thomas Palfrey and Roberto Weber The effects of payoff magnitude and het-erogeneity on behavior in 2x2 games with unique mixed strategy equilibria Journal of Economic

Behavior and Organization 2000 1 7 971

Rosemarie Nagel Unraveling in guessing games an experimental study American Economic

Review 1995 1

Yaw Nyarko and Andrew Schotter An experimental study of belief learning using elicited beliefs Econometrica 2002 7 44 975

Jack Ochs An experimental study of games with unique mixed strategy equilibria Games and

Economic Behavior 1995 7 973

Reinhard Selten and Thorsten Chmura Stationary concepts for experimental 2x2 games American

Economic Review 2008 1 7 44 972

Christopher Sims Stickiness Carnegie-Rochester Conference Series on Public Policy 1998 10

Christopher Sims Implications of rational inattention Journal of Monetary Economics 2003 10

Dale Stahl Entropy control costs and entropic equilibrium International Journal of Game Theory 1990 1

Jorgen Weibull Lars-Goran Mattsson and Mark Voorneveld Better may be worse some mono-tonicity results and paradoxes in discrete choice under uncertainty Theory and Decision 2007

1 3 95

37

Michael Woodford Information-constrained state-dependent pricing Journal of Monetary Eco-nomics 2009 10

Ming Yang Coordination with flexible information acquisition Journal of Economic Theory 2015

1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

32

Stud

y G

ame

HQ

RE

ˆ 1

ˆ 2

ln(L

)

LQR

E

ˆ ln(L

)

EQ

RE

k

ˆ 1

ˆ 2

ln(L

) 2ln(

)

MP

W 2

000

A

B C

D

042

4

00

216

75

037

1

09

145

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149

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031

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792

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530

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288

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603

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264

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7710

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097

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464

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1 1

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1

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585

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650

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008a

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W 2

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W 2

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Sum

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Poo

led

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Tabl

e 3

Par

amet

er e

stim

ates

a

Bru

nner

et

al [

2011

] poi

nt o

ut t

hat

SC 2

008

mis

repo

rts

the

empi

rica

l fre

quen

cy fo

r G

ame 3

O

ur an

alys

is is

base

d on

the

raw

data

whi

ch we

than

kT

hors

ten

Chm

ura

for

prov

idin

g

Study Game Data

p q N HQRE p q

LQRE p q

EQRE p q

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

064 024 063 024 059 026 055 033

069 012 071 022 063 011 059 021

068 012 067 021 064 012 059 023

1 0921 0310 9600 092 031 096 033 096 033 2 0783 0473 9600 078 050 081 051 079 050 3 0837 0207 9600 084 021 083 021 083 021 4 0714 0264 9600 071 026 072 026 072 026 5 0673 0336 9600 067 034 069 033 069 033

SC 2008 6

7 0555 0404 9600 0859 0436 4800

056 040 086 050

058 039 089 052

057 039 085 050

8 0750 0414 4800 075 050 083 047 075 050 9 0746 0173 4800 075 017 080 014 079 014 10 0634 0301 4800 064 030 072 026 070 024 11 0669 0348 4800 067 035 069 034 069 034 12 0561 0396 4800 056 040 058 038 057 038

Ochs 1995 2 3

0595 0258 448 0542 0336 516

060 026 054 034

065 025 062 033

060 026 058 032

GH 2001 AMP RA

0960 0160 25 0080 0800 25

096 016 008 080

083 024 035 077

089 027 035 077

NS 2002 NS 0543 0610 6720 054 061 054 062 054 062

Table 4 Best-fit Predictions

relative to LQRE is related to its ability to correctly order the playersrsquo i

s Table 3 gives the parameter estimates and log-likelihoods and Table 4 give the resulting pre-

dictions Our qualitative analysis of Section 71 is confirmed EQRE outperforms LQRE in the 15

games identified earlier (A B C D 2 3 4 6 7 9 10 12 2 3 AMP ) which are those for which EQRE would have outperformed LQRE for any cost function In particular EQRE would

have outperformed LQRE if k were chosen arbitrarily and only were optimized EQRE outper-forms LQRE in game 8 also but this requires that k and are jointly optimized In the remaining

5 games LQRE outperforms EQRE for all finite k and therefore the best-fit EQRE involves k = 1

and coincides with the best-fit LQRE Testing for statistical significance a standard likelihood ratio

test rejects the null hypothesis that LQRE performs as well as EQRE for most (but not all) games as shown in the last column of Table 343

Inspecting Table 3 more closely in 15 of the 16 games for which EQRE outperforms LQRE

(whether statistically significantly or not) the EQRE estimates ˆ and ˆ have the same ordering 1 2

243The likelihood ratio statistic is distributed as a with 1 degree of freedom (since EQRE has 1 more parameter than LQRE) under the null hypothesis that LQRE performs as well as EQRE and indicate rejection at the 010 005 and 001 levels respectively

33

as HQRE ˆ1 and ˆ2 In some cases the estimates are very close Overall though EQRE tends to

order the i

s correctly it usually underpredicts the absolute differences between the i

s In addition to individual game performance we fit the pooled data for MPW 2000 and SC 2008

the studies with more than a few games each We do not pool across studies because this would

be sensitive to the conversion factors as discussed in Section 72 We find that EQRE significantly

outperforms LQRE for MPW 2000 but not for SC 2008 in which the best-fit EQRE involves k = 1

73 Out-of-sample performance

Finally we present reduced form measures of out-of-sample performance whereby we use the indi-vidual game parameter estimates from Table 3 to make out-of-sample predictions for other games We do this for each of the games within MPW 2000 and SC 2008 and evaluate out-of-sample

performance by the squared distance between predicted and empirical action frequencies44 Table

5 presents the results as matrices for MPW 2000 and SC 2008 in which the ijth entry gives the

difference in squared distance between LQRE and EQRE in going from game i to game j A posi-tive (negative) entry indicates that EQRE outperforms (underperforms) LQRE A zero entry in the

ij-position indicates that for game i the best-fit EQRE involves k = 1 and thus coincides with

the best-fit LQRE LQRE outperforms EQRE in the majority of forecasts as indicated by the larger number of

negative entries LQRE also outperforms EQRE by the average of the entries which are -0020 and

-0011 for MPW 2000 and SC 2008 respectively We have no particular intuition for this result but believe it is more than simply an overfitting phenomenon In unreported results the matrices look similar if EQRE is estimated under the restriction that k is fixed at some k for all games (we

tried all k 2 11 2 5 10 20) Overall the results are mixed While EQRE tends to perform better in-sample it is not uni-

formly true and LQRE performs better out-of-sample

8 Conclusion

We endogenize the precision parameter of standard logit QRE (LQRE) The resulting model which

we call endogenous quantal response equilibrium (EQRE) takes as given the logit structure but endogenizes the precision parameter in a first stage In the model each player i chooses i

optimally subject to costs given correct beliefs over the opponentsrsquo second-stage actions which form a heterogenous LQRE given the vector = ( 1 )

n

Because incentives to acquire precision depend on the equilibrium behavior of opponents which

in turn depends on the payoffs of the underlying game each player generically chooses a different 44For model M with parameter the squared distance is D(M ) = (p p M ( ))2 + (q q M ( ))2 where

p M ( ) q M ( ) is the model prediction and p q is the empirical action frequency Squared distance is a com-monly used measure of fit (see for example Nyarko and Schotter [2002] and Selten and Chmura [2008])

34

MPW 2000 Game A B C D

A 39

10 3

12

10 3

30

10 3

B -10

10 1

24

10 2

-80

10 2

C -32

10 2

92

10 3

-71

10 2

D 16

10 3

23

10 3

30

10 4

SC 2008 Game 1 2 3 4 5 6 7 8 9 10 11 12

1 0 0 0 0 0 0 0 0 0 0 0

2 -74

10 3

-29

10 2

-28

10 2

-33

10 2

-10

10 2

-55

10 3

39

10 3

-48

10 2

-34

10 2

-31

10 2

-13

10 2

3 -28

10 2

-77

10 3

-39

10 4

-33

10 3

-62

10 3

32

10 3

-13

10 2

-87

10 4

-59

10 4

-27

10 3

-66

10 3

4 -43

10 3

-11

10 2

-35

10 4

-14

10 3

-36

10 3

-22

10 3

-11

10 2

36

10 3

69

10 4

-12

10 3

-40

10 3

5 0 0 0 0 0 0 0 0 0 0 0

6 -25

10 2

-19

10 2

-12

10 2

-48

10 3

-11

10 3

-64

10 2

-97

10 3

34

10 3

-23

10 4

-15

10 3

25

10 4

7 41

10 3

-52

10 2

-52

10 2

-41

10 2

-46

10 2

-94

10 3

-37

10 2

-66

10 2

-32

10 2

-43

10 2

-11

10 2

8 -19

10 2

56

10 4

-54

10 2

-40

10 2

-30

10 2

-11

10 2

-35

10 3

-77

10 2

-44

10 2

-28

10 2

-13

10 2

9 -39

10 3

-38

10 3

-14

10 3

10

10 3

81

10 4

15

10 4

-85

10 3

-42

10 3

23

10 3

11

10 3

-95

10 5

10 -17

10 3

-15

10 2

-30

10 3

-60

10 4

-57

10 5

-11

10 3

-28

10 2

-93

10 3

67

10 3

-31

10 4

-14

10 3

11 0 0 0 0 0 0 0 0 0 0 0

12 -28

10 2

-20

10 2

-13

10 2

-54

10 3

-16

10 3

32

10 4

-66

10 2

-10

10 2

24

10 3

-44

10 4

-20

10 3

Table 5 Out-of-sample performance

35

i Because of this EQRE and LQRE generically give different predictions We show that EQRE

satisfies a modified form of the regularity axioms (Goeree et al [2005]) and provide analogues to

most of the classic results for LQRE such as those for equilibrium selection Our results reveal both

similarities and differences to LQRE In the binary action case we are able to characterize behavior rather well For generalized matching pennies games we nearly have a complete characterization of the EQRE set which deviates systematically from the LQRE set and whose qualitative shape does not depend on the EQRE cost function Comparing the models in data we find that EQRE tends to outperform LQRE in-sample but performs worse out-of-sample

References

Larbi Alaoui and Antonio Penta Endogenous depth of reasoning Review of Economic Studies 2015 1

Christoph Brunner Colin Camerer and Jacob Goeree Stationary concepts for experimental 2x2

games Comment American Economic Review 2011 a

Colin Camerer Teck-Hua Ho and Juin-Kuan Chong A cognitive hierarchy model of games The

Quarterly Journal of Economics 2004 1

Andrew Caplin and Mark Dean Revealed preference rational inattention and costly information

acquisition American Economic Review 2015 10

Tommaso Denti Games with unrestricted information acquisition Working Paper 2015 1

Evan Friedman Stochastic equilibria Noise in actions or beliefs Working Paper 2018 1 11 31 72 42

Jacob Goeree and Charles Holt Ten little treasures of game theory and ten intuitive contradictions American Economic Review 2001 7 974

Jacob Goeree Charles Holt and Thomas Palfrey Regular quantal response equilibrium Experi-mental Economics 2005 1 3 4 7 22 11 31 26 8

Jacob Goeree Charles Holt and Thomas Palfrey Quantal response equilibrium Princeton Uni-versity Press 2016 1

Philip Haile Ali Hortacsu and Grigory Kosenok On the empirical content of quantal response

equilibrium American Economic Review 2008 1 22

Phillipe Jehiel Analogy-based expectations equilibrium Journal of Economic Theory 2005 1

36

Filip Matejka and Alisdair McKay Rational inattention to discrete choices A new foundation for the multinomial logit American Economic Review 2015 6

Lars-Goran Mattsson and Jorgen Weibull Probabilistic choice and procedurally bounded rationality Games and Economic Behavior 2002 1

Richard McKelvey and Thomas Palfrey Quantal response equilibria for normal form games Games

and Economic Behavior 1995 1 18 5 94 94 94 973

Richard McKelvey and Thomas Palfrey A statistical theory of equilibrium in games The Japanese

Economic Review 1996 1 23

Richard McKelvey and Thomas Palfrey Quantal response equilibria for extensive form games Experimental Economics 1998 31 94

Richard McKelvey Thomas Palfrey and Roberto Weber Endogenous rationality equilibrium Work-ing Paper 1997 1

Richard McKelvey Thomas Palfrey and Roberto Weber The effects of payoff magnitude and het-erogeneity on behavior in 2x2 games with unique mixed strategy equilibria Journal of Economic

Behavior and Organization 2000 1 7 971

Rosemarie Nagel Unraveling in guessing games an experimental study American Economic

Review 1995 1

Yaw Nyarko and Andrew Schotter An experimental study of belief learning using elicited beliefs Econometrica 2002 7 44 975

Jack Ochs An experimental study of games with unique mixed strategy equilibria Games and

Economic Behavior 1995 7 973

Reinhard Selten and Thorsten Chmura Stationary concepts for experimental 2x2 games American

Economic Review 2008 1 7 44 972

Christopher Sims Stickiness Carnegie-Rochester Conference Series on Public Policy 1998 10

Christopher Sims Implications of rational inattention Journal of Monetary Economics 2003 10

Dale Stahl Entropy control costs and entropic equilibrium International Journal of Game Theory 1990 1

Jorgen Weibull Lars-Goran Mattsson and Mark Voorneveld Better may be worse some mono-tonicity results and paradoxes in discrete choice under uncertainty Theory and Decision 2007

1 3 95

37

Michael Woodford Information-constrained state-dependent pricing Journal of Monetary Eco-nomics 2009 10

Ming Yang Coordination with flexible information acquisition Journal of Economic Theory 2015

1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

Study Game Data

p q N HQRE p q

LQRE p q

EQRE p q

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

064 024 063 024 059 026 055 033

069 012 071 022 063 011 059 021

068 012 067 021 064 012 059 023

1 0921 0310 9600 092 031 096 033 096 033 2 0783 0473 9600 078 050 081 051 079 050 3 0837 0207 9600 084 021 083 021 083 021 4 0714 0264 9600 071 026 072 026 072 026 5 0673 0336 9600 067 034 069 033 069 033

SC 2008 6

7 0555 0404 9600 0859 0436 4800

056 040 086 050

058 039 089 052

057 039 085 050

8 0750 0414 4800 075 050 083 047 075 050 9 0746 0173 4800 075 017 080 014 079 014 10 0634 0301 4800 064 030 072 026 070 024 11 0669 0348 4800 067 035 069 034 069 034 12 0561 0396 4800 056 040 058 038 057 038

Ochs 1995 2 3

0595 0258 448 0542 0336 516

060 026 054 034

065 025 062 033

060 026 058 032

GH 2001 AMP RA

0960 0160 25 0080 0800 25

096 016 008 080

083 024 035 077

089 027 035 077

NS 2002 NS 0543 0610 6720 054 061 054 062 054 062

Table 4 Best-fit Predictions

relative to LQRE is related to its ability to correctly order the playersrsquo i

s Table 3 gives the parameter estimates and log-likelihoods and Table 4 give the resulting pre-

dictions Our qualitative analysis of Section 71 is confirmed EQRE outperforms LQRE in the 15

games identified earlier (A B C D 2 3 4 6 7 9 10 12 2 3 AMP ) which are those for which EQRE would have outperformed LQRE for any cost function In particular EQRE would

have outperformed LQRE if k were chosen arbitrarily and only were optimized EQRE outper-forms LQRE in game 8 also but this requires that k and are jointly optimized In the remaining

5 games LQRE outperforms EQRE for all finite k and therefore the best-fit EQRE involves k = 1

and coincides with the best-fit LQRE Testing for statistical significance a standard likelihood ratio

test rejects the null hypothesis that LQRE performs as well as EQRE for most (but not all) games as shown in the last column of Table 343

Inspecting Table 3 more closely in 15 of the 16 games for which EQRE outperforms LQRE

(whether statistically significantly or not) the EQRE estimates ˆ and ˆ have the same ordering 1 2

243The likelihood ratio statistic is distributed as a with 1 degree of freedom (since EQRE has 1 more parameter than LQRE) under the null hypothesis that LQRE performs as well as EQRE and indicate rejection at the 010 005 and 001 levels respectively

33

as HQRE ˆ1 and ˆ2 In some cases the estimates are very close Overall though EQRE tends to

order the i

s correctly it usually underpredicts the absolute differences between the i

s In addition to individual game performance we fit the pooled data for MPW 2000 and SC 2008

the studies with more than a few games each We do not pool across studies because this would

be sensitive to the conversion factors as discussed in Section 72 We find that EQRE significantly

outperforms LQRE for MPW 2000 but not for SC 2008 in which the best-fit EQRE involves k = 1

73 Out-of-sample performance

Finally we present reduced form measures of out-of-sample performance whereby we use the indi-vidual game parameter estimates from Table 3 to make out-of-sample predictions for other games We do this for each of the games within MPW 2000 and SC 2008 and evaluate out-of-sample

performance by the squared distance between predicted and empirical action frequencies44 Table

5 presents the results as matrices for MPW 2000 and SC 2008 in which the ijth entry gives the

difference in squared distance between LQRE and EQRE in going from game i to game j A posi-tive (negative) entry indicates that EQRE outperforms (underperforms) LQRE A zero entry in the

ij-position indicates that for game i the best-fit EQRE involves k = 1 and thus coincides with

the best-fit LQRE LQRE outperforms EQRE in the majority of forecasts as indicated by the larger number of

negative entries LQRE also outperforms EQRE by the average of the entries which are -0020 and

-0011 for MPW 2000 and SC 2008 respectively We have no particular intuition for this result but believe it is more than simply an overfitting phenomenon In unreported results the matrices look similar if EQRE is estimated under the restriction that k is fixed at some k for all games (we

tried all k 2 11 2 5 10 20) Overall the results are mixed While EQRE tends to perform better in-sample it is not uni-

formly true and LQRE performs better out-of-sample

8 Conclusion

We endogenize the precision parameter of standard logit QRE (LQRE) The resulting model which

we call endogenous quantal response equilibrium (EQRE) takes as given the logit structure but endogenizes the precision parameter in a first stage In the model each player i chooses i

optimally subject to costs given correct beliefs over the opponentsrsquo second-stage actions which form a heterogenous LQRE given the vector = ( 1 )

n

Because incentives to acquire precision depend on the equilibrium behavior of opponents which

in turn depends on the payoffs of the underlying game each player generically chooses a different 44For model M with parameter the squared distance is D(M ) = (p p M ( ))2 + (q q M ( ))2 where

p M ( ) q M ( ) is the model prediction and p q is the empirical action frequency Squared distance is a com-monly used measure of fit (see for example Nyarko and Schotter [2002] and Selten and Chmura [2008])

34

MPW 2000 Game A B C D

A 39

10 3

12

10 3

30

10 3

B -10

10 1

24

10 2

-80

10 2

C -32

10 2

92

10 3

-71

10 2

D 16

10 3

23

10 3

30

10 4

SC 2008 Game 1 2 3 4 5 6 7 8 9 10 11 12

1 0 0 0 0 0 0 0 0 0 0 0

2 -74

10 3

-29

10 2

-28

10 2

-33

10 2

-10

10 2

-55

10 3

39

10 3

-48

10 2

-34

10 2

-31

10 2

-13

10 2

3 -28

10 2

-77

10 3

-39

10 4

-33

10 3

-62

10 3

32

10 3

-13

10 2

-87

10 4

-59

10 4

-27

10 3

-66

10 3

4 -43

10 3

-11

10 2

-35

10 4

-14

10 3

-36

10 3

-22

10 3

-11

10 2

36

10 3

69

10 4

-12

10 3

-40

10 3

5 0 0 0 0 0 0 0 0 0 0 0

6 -25

10 2

-19

10 2

-12

10 2

-48

10 3

-11

10 3

-64

10 2

-97

10 3

34

10 3

-23

10 4

-15

10 3

25

10 4

7 41

10 3

-52

10 2

-52

10 2

-41

10 2

-46

10 2

-94

10 3

-37

10 2

-66

10 2

-32

10 2

-43

10 2

-11

10 2

8 -19

10 2

56

10 4

-54

10 2

-40

10 2

-30

10 2

-11

10 2

-35

10 3

-77

10 2

-44

10 2

-28

10 2

-13

10 2

9 -39

10 3

-38

10 3

-14

10 3

10

10 3

81

10 4

15

10 4

-85

10 3

-42

10 3

23

10 3

11

10 3

-95

10 5

10 -17

10 3

-15

10 2

-30

10 3

-60

10 4

-57

10 5

-11

10 3

-28

10 2

-93

10 3

67

10 3

-31

10 4

-14

10 3

11 0 0 0 0 0 0 0 0 0 0 0

12 -28

10 2

-20

10 2

-13

10 2

-54

10 3

-16

10 3

32

10 4

-66

10 2

-10

10 2

24

10 3

-44

10 4

-20

10 3

Table 5 Out-of-sample performance

35

i Because of this EQRE and LQRE generically give different predictions We show that EQRE

satisfies a modified form of the regularity axioms (Goeree et al [2005]) and provide analogues to

most of the classic results for LQRE such as those for equilibrium selection Our results reveal both

similarities and differences to LQRE In the binary action case we are able to characterize behavior rather well For generalized matching pennies games we nearly have a complete characterization of the EQRE set which deviates systematically from the LQRE set and whose qualitative shape does not depend on the EQRE cost function Comparing the models in data we find that EQRE tends to outperform LQRE in-sample but performs worse out-of-sample

References

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1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

as HQRE ˆ1 and ˆ2 In some cases the estimates are very close Overall though EQRE tends to

order the i

s correctly it usually underpredicts the absolute differences between the i

s In addition to individual game performance we fit the pooled data for MPW 2000 and SC 2008

the studies with more than a few games each We do not pool across studies because this would

be sensitive to the conversion factors as discussed in Section 72 We find that EQRE significantly

outperforms LQRE for MPW 2000 but not for SC 2008 in which the best-fit EQRE involves k = 1

73 Out-of-sample performance

Finally we present reduced form measures of out-of-sample performance whereby we use the indi-vidual game parameter estimates from Table 3 to make out-of-sample predictions for other games We do this for each of the games within MPW 2000 and SC 2008 and evaluate out-of-sample

performance by the squared distance between predicted and empirical action frequencies44 Table

5 presents the results as matrices for MPW 2000 and SC 2008 in which the ijth entry gives the

difference in squared distance between LQRE and EQRE in going from game i to game j A posi-tive (negative) entry indicates that EQRE outperforms (underperforms) LQRE A zero entry in the

ij-position indicates that for game i the best-fit EQRE involves k = 1 and thus coincides with

the best-fit LQRE LQRE outperforms EQRE in the majority of forecasts as indicated by the larger number of

negative entries LQRE also outperforms EQRE by the average of the entries which are -0020 and

-0011 for MPW 2000 and SC 2008 respectively We have no particular intuition for this result but believe it is more than simply an overfitting phenomenon In unreported results the matrices look similar if EQRE is estimated under the restriction that k is fixed at some k for all games (we

tried all k 2 11 2 5 10 20) Overall the results are mixed While EQRE tends to perform better in-sample it is not uni-

formly true and LQRE performs better out-of-sample

8 Conclusion

We endogenize the precision parameter of standard logit QRE (LQRE) The resulting model which

we call endogenous quantal response equilibrium (EQRE) takes as given the logit structure but endogenizes the precision parameter in a first stage In the model each player i chooses i

optimally subject to costs given correct beliefs over the opponentsrsquo second-stage actions which form a heterogenous LQRE given the vector = ( 1 )

n

Because incentives to acquire precision depend on the equilibrium behavior of opponents which

in turn depends on the payoffs of the underlying game each player generically chooses a different 44For model M with parameter the squared distance is D(M ) = (p p M ( ))2 + (q q M ( ))2 where

p M ( ) q M ( ) is the model prediction and p q is the empirical action frequency Squared distance is a com-monly used measure of fit (see for example Nyarko and Schotter [2002] and Selten and Chmura [2008])

34

MPW 2000 Game A B C D

A 39

10 3

12

10 3

30

10 3

B -10

10 1

24

10 2

-80

10 2

C -32

10 2

92

10 3

-71

10 2

D 16

10 3

23

10 3

30

10 4

SC 2008 Game 1 2 3 4 5 6 7 8 9 10 11 12

1 0 0 0 0 0 0 0 0 0 0 0

2 -74

10 3

-29

10 2

-28

10 2

-33

10 2

-10

10 2

-55

10 3

39

10 3

-48

10 2

-34

10 2

-31

10 2

-13

10 2

3 -28

10 2

-77

10 3

-39

10 4

-33

10 3

-62

10 3

32

10 3

-13

10 2

-87

10 4

-59

10 4

-27

10 3

-66

10 3

4 -43

10 3

-11

10 2

-35

10 4

-14

10 3

-36

10 3

-22

10 3

-11

10 2

36

10 3

69

10 4

-12

10 3

-40

10 3

5 0 0 0 0 0 0 0 0 0 0 0

6 -25

10 2

-19

10 2

-12

10 2

-48

10 3

-11

10 3

-64

10 2

-97

10 3

34

10 3

-23

10 4

-15

10 3

25

10 4

7 41

10 3

-52

10 2

-52

10 2

-41

10 2

-46

10 2

-94

10 3

-37

10 2

-66

10 2

-32

10 2

-43

10 2

-11

10 2

8 -19

10 2

56

10 4

-54

10 2

-40

10 2

-30

10 2

-11

10 2

-35

10 3

-77

10 2

-44

10 2

-28

10 2

-13

10 2

9 -39

10 3

-38

10 3

-14

10 3

10

10 3

81

10 4

15

10 4

-85

10 3

-42

10 3

23

10 3

11

10 3

-95

10 5

10 -17

10 3

-15

10 2

-30

10 3

-60

10 4

-57

10 5

-11

10 3

-28

10 2

-93

10 3

67

10 3

-31

10 4

-14

10 3

11 0 0 0 0 0 0 0 0 0 0 0

12 -28

10 2

-20

10 2

-13

10 2

-54

10 3

-16

10 3

32

10 4

-66

10 2

-10

10 2

24

10 3

-44

10 4

-20

10 3

Table 5 Out-of-sample performance

35

i Because of this EQRE and LQRE generically give different predictions We show that EQRE

satisfies a modified form of the regularity axioms (Goeree et al [2005]) and provide analogues to

most of the classic results for LQRE such as those for equilibrium selection Our results reveal both

similarities and differences to LQRE In the binary action case we are able to characterize behavior rather well For generalized matching pennies games we nearly have a complete characterization of the EQRE set which deviates systematically from the LQRE set and whose qualitative shape does not depend on the EQRE cost function Comparing the models in data we find that EQRE tends to outperform LQRE in-sample but performs worse out-of-sample

References

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Colin Camerer Teck-Hua Ho and Juin-Kuan Chong A cognitive hierarchy model of games The

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Andrew Caplin and Mark Dean Revealed preference rational inattention and costly information

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Reinhard Selten and Thorsten Chmura Stationary concepts for experimental 2x2 games American

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1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

MPW 2000 Game A B C D

A 39

10 3

12

10 3

30

10 3

B -10

10 1

24

10 2

-80

10 2

C -32

10 2

92

10 3

-71

10 2

D 16

10 3

23

10 3

30

10 4

SC 2008 Game 1 2 3 4 5 6 7 8 9 10 11 12

1 0 0 0 0 0 0 0 0 0 0 0

2 -74

10 3

-29

10 2

-28

10 2

-33

10 2

-10

10 2

-55

10 3

39

10 3

-48

10 2

-34

10 2

-31

10 2

-13

10 2

3 -28

10 2

-77

10 3

-39

10 4

-33

10 3

-62

10 3

32

10 3

-13

10 2

-87

10 4

-59

10 4

-27

10 3

-66

10 3

4 -43

10 3

-11

10 2

-35

10 4

-14

10 3

-36

10 3

-22

10 3

-11

10 2

36

10 3

69

10 4

-12

10 3

-40

10 3

5 0 0 0 0 0 0 0 0 0 0 0

6 -25

10 2

-19

10 2

-12

10 2

-48

10 3

-11

10 3

-64

10 2

-97

10 3

34

10 3

-23

10 4

-15

10 3

25

10 4

7 41

10 3

-52

10 2

-52

10 2

-41

10 2

-46

10 2

-94

10 3

-37

10 2

-66

10 2

-32

10 2

-43

10 2

-11

10 2

8 -19

10 2

56

10 4

-54

10 2

-40

10 2

-30

10 2

-11

10 2

-35

10 3

-77

10 2

-44

10 2

-28

10 2

-13

10 2

9 -39

10 3

-38

10 3

-14

10 3

10

10 3

81

10 4

15

10 4

-85

10 3

-42

10 3

23

10 3

11

10 3

-95

10 5

10 -17

10 3

-15

10 2

-30

10 3

-60

10 4

-57

10 5

-11

10 3

-28

10 2

-93

10 3

67

10 3

-31

10 4

-14

10 3

11 0 0 0 0 0 0 0 0 0 0 0

12 -28

10 2

-20

10 2

-13

10 2

-54

10 3

-16

10 3

32

10 4

-66

10 2

-10

10 2

24

10 3

-44

10 4

-20

10 3

Table 5 Out-of-sample performance

35

i Because of this EQRE and LQRE generically give different predictions We show that EQRE

satisfies a modified form of the regularity axioms (Goeree et al [2005]) and provide analogues to

most of the classic results for LQRE such as those for equilibrium selection Our results reveal both

similarities and differences to LQRE In the binary action case we are able to characterize behavior rather well For generalized matching pennies games we nearly have a complete characterization of the EQRE set which deviates systematically from the LQRE set and whose qualitative shape does not depend on the EQRE cost function Comparing the models in data we find that EQRE tends to outperform LQRE in-sample but performs worse out-of-sample

References

Larbi Alaoui and Antonio Penta Endogenous depth of reasoning Review of Economic Studies 2015 1

Christoph Brunner Colin Camerer and Jacob Goeree Stationary concepts for experimental 2x2

games Comment American Economic Review 2011 a

Colin Camerer Teck-Hua Ho and Juin-Kuan Chong A cognitive hierarchy model of games The

Quarterly Journal of Economics 2004 1

Andrew Caplin and Mark Dean Revealed preference rational inattention and costly information

acquisition American Economic Review 2015 10

Tommaso Denti Games with unrestricted information acquisition Working Paper 2015 1

Evan Friedman Stochastic equilibria Noise in actions or beliefs Working Paper 2018 1 11 31 72 42

Jacob Goeree and Charles Holt Ten little treasures of game theory and ten intuitive contradictions American Economic Review 2001 7 974

Jacob Goeree Charles Holt and Thomas Palfrey Regular quantal response equilibrium Experi-mental Economics 2005 1 3 4 7 22 11 31 26 8

Jacob Goeree Charles Holt and Thomas Palfrey Quantal response equilibrium Princeton Uni-versity Press 2016 1

Philip Haile Ali Hortacsu and Grigory Kosenok On the empirical content of quantal response

equilibrium American Economic Review 2008 1 22

Phillipe Jehiel Analogy-based expectations equilibrium Journal of Economic Theory 2005 1

36

Filip Matejka and Alisdair McKay Rational inattention to discrete choices A new foundation for the multinomial logit American Economic Review 2015 6

Lars-Goran Mattsson and Jorgen Weibull Probabilistic choice and procedurally bounded rationality Games and Economic Behavior 2002 1

Richard McKelvey and Thomas Palfrey Quantal response equilibria for normal form games Games

and Economic Behavior 1995 1 18 5 94 94 94 973

Richard McKelvey and Thomas Palfrey A statistical theory of equilibrium in games The Japanese

Economic Review 1996 1 23

Richard McKelvey and Thomas Palfrey Quantal response equilibria for extensive form games Experimental Economics 1998 31 94

Richard McKelvey Thomas Palfrey and Roberto Weber Endogenous rationality equilibrium Work-ing Paper 1997 1

Richard McKelvey Thomas Palfrey and Roberto Weber The effects of payoff magnitude and het-erogeneity on behavior in 2x2 games with unique mixed strategy equilibria Journal of Economic

Behavior and Organization 2000 1 7 971

Rosemarie Nagel Unraveling in guessing games an experimental study American Economic

Review 1995 1

Yaw Nyarko and Andrew Schotter An experimental study of belief learning using elicited beliefs Econometrica 2002 7 44 975

Jack Ochs An experimental study of games with unique mixed strategy equilibria Games and

Economic Behavior 1995 7 973

Reinhard Selten and Thorsten Chmura Stationary concepts for experimental 2x2 games American

Economic Review 2008 1 7 44 972

Christopher Sims Stickiness Carnegie-Rochester Conference Series on Public Policy 1998 10

Christopher Sims Implications of rational inattention Journal of Monetary Economics 2003 10

Dale Stahl Entropy control costs and entropic equilibrium International Journal of Game Theory 1990 1

Jorgen Weibull Lars-Goran Mattsson and Mark Voorneveld Better may be worse some mono-tonicity results and paradoxes in discrete choice under uncertainty Theory and Decision 2007

1 3 95

37

Michael Woodford Information-constrained state-dependent pricing Journal of Monetary Eco-nomics 2009 10

Ming Yang Coordination with flexible information acquisition Journal of Economic Theory 2015

1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

i Because of this EQRE and LQRE generically give different predictions We show that EQRE

satisfies a modified form of the regularity axioms (Goeree et al [2005]) and provide analogues to

most of the classic results for LQRE such as those for equilibrium selection Our results reveal both

similarities and differences to LQRE In the binary action case we are able to characterize behavior rather well For generalized matching pennies games we nearly have a complete characterization of the EQRE set which deviates systematically from the LQRE set and whose qualitative shape does not depend on the EQRE cost function Comparing the models in data we find that EQRE tends to outperform LQRE in-sample but performs worse out-of-sample

References

Larbi Alaoui and Antonio Penta Endogenous depth of reasoning Review of Economic Studies 2015 1

Christoph Brunner Colin Camerer and Jacob Goeree Stationary concepts for experimental 2x2

games Comment American Economic Review 2011 a

Colin Camerer Teck-Hua Ho and Juin-Kuan Chong A cognitive hierarchy model of games The

Quarterly Journal of Economics 2004 1

Andrew Caplin and Mark Dean Revealed preference rational inattention and costly information

acquisition American Economic Review 2015 10

Tommaso Denti Games with unrestricted information acquisition Working Paper 2015 1

Evan Friedman Stochastic equilibria Noise in actions or beliefs Working Paper 2018 1 11 31 72 42

Jacob Goeree and Charles Holt Ten little treasures of game theory and ten intuitive contradictions American Economic Review 2001 7 974

Jacob Goeree Charles Holt and Thomas Palfrey Regular quantal response equilibrium Experi-mental Economics 2005 1 3 4 7 22 11 31 26 8

Jacob Goeree Charles Holt and Thomas Palfrey Quantal response equilibrium Princeton Uni-versity Press 2016 1

Philip Haile Ali Hortacsu and Grigory Kosenok On the empirical content of quantal response

equilibrium American Economic Review 2008 1 22

Phillipe Jehiel Analogy-based expectations equilibrium Journal of Economic Theory 2005 1

36

Filip Matejka and Alisdair McKay Rational inattention to discrete choices A new foundation for the multinomial logit American Economic Review 2015 6

Lars-Goran Mattsson and Jorgen Weibull Probabilistic choice and procedurally bounded rationality Games and Economic Behavior 2002 1

Richard McKelvey and Thomas Palfrey Quantal response equilibria for normal form games Games

and Economic Behavior 1995 1 18 5 94 94 94 973

Richard McKelvey and Thomas Palfrey A statistical theory of equilibrium in games The Japanese

Economic Review 1996 1 23

Richard McKelvey and Thomas Palfrey Quantal response equilibria for extensive form games Experimental Economics 1998 31 94

Richard McKelvey Thomas Palfrey and Roberto Weber Endogenous rationality equilibrium Work-ing Paper 1997 1

Richard McKelvey Thomas Palfrey and Roberto Weber The effects of payoff magnitude and het-erogeneity on behavior in 2x2 games with unique mixed strategy equilibria Journal of Economic

Behavior and Organization 2000 1 7 971

Rosemarie Nagel Unraveling in guessing games an experimental study American Economic

Review 1995 1

Yaw Nyarko and Andrew Schotter An experimental study of belief learning using elicited beliefs Econometrica 2002 7 44 975

Jack Ochs An experimental study of games with unique mixed strategy equilibria Games and

Economic Behavior 1995 7 973

Reinhard Selten and Thorsten Chmura Stationary concepts for experimental 2x2 games American

Economic Review 2008 1 7 44 972

Christopher Sims Stickiness Carnegie-Rochester Conference Series on Public Policy 1998 10

Christopher Sims Implications of rational inattention Journal of Monetary Economics 2003 10

Dale Stahl Entropy control costs and entropic equilibrium International Journal of Game Theory 1990 1

Jorgen Weibull Lars-Goran Mattsson and Mark Voorneveld Better may be worse some mono-tonicity results and paradoxes in discrete choice under uncertainty Theory and Decision 2007

1 3 95

37

Michael Woodford Information-constrained state-dependent pricing Journal of Monetary Eco-nomics 2009 10

Ming Yang Coordination with flexible information acquisition Journal of Economic Theory 2015

1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

Filip Matejka and Alisdair McKay Rational inattention to discrete choices A new foundation for the multinomial logit American Economic Review 2015 6

Lars-Goran Mattsson and Jorgen Weibull Probabilistic choice and procedurally bounded rationality Games and Economic Behavior 2002 1

Richard McKelvey and Thomas Palfrey Quantal response equilibria for normal form games Games

and Economic Behavior 1995 1 18 5 94 94 94 973

Richard McKelvey and Thomas Palfrey A statistical theory of equilibrium in games The Japanese

Economic Review 1996 1 23

Richard McKelvey and Thomas Palfrey Quantal response equilibria for extensive form games Experimental Economics 1998 31 94

Richard McKelvey Thomas Palfrey and Roberto Weber Endogenous rationality equilibrium Work-ing Paper 1997 1

Richard McKelvey Thomas Palfrey and Roberto Weber The effects of payoff magnitude and het-erogeneity on behavior in 2x2 games with unique mixed strategy equilibria Journal of Economic

Behavior and Organization 2000 1 7 971

Rosemarie Nagel Unraveling in guessing games an experimental study American Economic

Review 1995 1

Yaw Nyarko and Andrew Schotter An experimental study of belief learning using elicited beliefs Econometrica 2002 7 44 975

Jack Ochs An experimental study of games with unique mixed strategy equilibria Games and

Economic Behavior 1995 7 973

Reinhard Selten and Thorsten Chmura Stationary concepts for experimental 2x2 games American

Economic Review 2008 1 7 44 972

Christopher Sims Stickiness Carnegie-Rochester Conference Series on Public Policy 1998 10

Christopher Sims Implications of rational inattention Journal of Monetary Economics 2003 10

Dale Stahl Entropy control costs and entropic equilibrium International Journal of Game Theory 1990 1

Jorgen Weibull Lars-Goran Mattsson and Mark Voorneveld Better may be worse some mono-tonicity results and paradoxes in discrete choice under uncertainty Theory and Decision 2007

1 3 95

37

Michael Woodford Information-constrained state-dependent pricing Journal of Monetary Eco-nomics 2009 10

Ming Yang Coordination with flexible information acquisition Journal of Economic Theory 2015

1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

Michael Woodford Information-constrained state-dependent pricing Journal of Monetary Eco-nomics 2009 10

Ming Yang Coordination with flexible information acquisition Journal of Economic Theory 2015

1

9 Appendix

91 Properties of b

For this section and this section only we drop the i subscript In particular we use J instead of J(i) and Q

j instead of Qij v 2 RJ is a vector with components v1 v

J indexed by j k or l Recall the maintained assumption that v

j 6 vk for some j k= Unless stated otherwise all sums go

P

from 1 to J Finally we define maxj vj and J 1v

k = as the highest utility alternative k

and the number of such alternatives respectively P

1 b(0 v) = 1 j vj and lim b( v) =

J 1

0middotvj P P P

e 1 1Qj (v 0) = P

k e0middotv

k =

J =) b(0 v) = j Qj (v 0)vj =

j (J 1 )v

j = J

j vj As is well-1known lim Q

j (v ) = 0 if vj lt v

k for some k and lim Qj (v ) =

J

Hence lim b( v) =1 1 1

1 1lim P

j Qj (v )vj =

P

j vj =

J

= J J

= 1

b( v) v)2 For all 2 [0 1 ) 2 (0 1) and 2b( 2 ( 2 3) for some 1 2 3 2 R++

2

P

P

22 v

l v

lb( v) l vl e

l vle = P P gt 0

v

k v

k k e

k e 2

X X X

2 v

l v

k v

l() vl e e v

l

e gt 0 l k l

X X

2 (vj +v

k

) (vj +v

k

)() vj e v

j vk

e gt 0 jk jk

X

2 (vj +v

k

)() (v vj v

k

)e gt 0j

jk

X

2 2 (vj +v

k

)() (v vj v

k

) + (v vk

vj )e gt 0

j k v

j gtv

k

2The last equivalence follows from the fact that (vj v

j vk

) = 0 if vj = v

k and hence it is without loss to

only consider summing over vj 6 = y= v

k

Furthermore whenever j and k are such that x = vj gt v

k

we can group this with the sum in which y = vj lt v

k = x The last inequality is easy to verify

38

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

2 2due to supermodularity of f(x y) = xy which implies that (vj v

j

vk

) + (vk v

k

vj

) gt 0 whenever vj gt v

k

b( v) b(˜v)Hence gt 0 for all 2 [0 1) It is obvious that lt 1 for any ˜ 2 [0 1)

b( v) b( v)so that

2 (0 1) for some 1 2 R++ follows from the fact that 0 as 1

(property 3 below) Similarly it is obvious from the expression for the second derivative (9) that

2b(˜v) v)1 lt

2 lt 1 for any ˜ 2 [0 1) That 2

b( 2 2 ( 2 3) for some 2 3 2 R++ follows from

v)the fact that 2b( 0 as 1 (property 3 below)

2

b( v)

2b( v) v)3 lim

= 0 lim

2 = 0 and 2

b( 2 lt 0 for sufficiently high

1 1

P

P

22 v

l v

lb( v) l vl e

l vle = P P

v

k v

k k e

k e

As 1 the terms with the largest exponents dominate and this approaches

2J2 e Je

= 2 2 = 0 Je Je

Taking another derivative

P P 2 2 (vj +v

k

+v

l

)2b( v)

j vj kl

(vj v

k + 2vk

vl 2v

j

vl

)e = P (9)

2 ( k e vk )3

As before as 1 the terms with the largest exponents dominate In the denominator the (3) 2 2term with the largest exponent is proportional to e In the numerator note that (v

j vk +

2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the term in the numerator with the largest exponent is v)proportional to e (2+micro) for some micro lt Hence

2b( 0

2

v)The denominator of 2b( is strictly positive for all The numerator is given by

2

X X

2 2 (vj +v

k

+v

l

)vj (v

j vk + 2v

k

vl 2v

j

vl

)e j

kl

v)We show that the terms with the largest exponent are negative and hence 2b( lt 0 for sufficiently

2

2 2large As noted (vj v

k + 2vk

vl 2v

j

vl

) = 0 if vj = v

k

and hence the largest exponents are

achieved exactly when (1) vj = v

l = and vk = micro and (2) when v

l = vk = and v

j = micro (where

micro lt is the second largest component of v) In these cases the terms are

(2+micro)(2 micro 2 + 2micro 22)e 2 (2+micro)micro(micro 2 + 22 2micro)e

3Summing the two gives (micro 3 +2micro2 2micro2)e (2+micro) and it is easy to show that this is negative

39

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

for any micro lt

4 b( v + eJ ) = b( v) + for all 2 R where e

J = (1 1)

As is well-known Qj (v + e

J ) = Qj (v ) (translation invariance) Hence b( v + e

J ) =P P P

Qj (v )(v

j + ) = Qj (v )v

j + Qj (v ) = b( v) + where the last step follows from

j j j

P

the fact that j Qj (v ) = 1

92 Proofs

Theorem 1 A solution to (2) exists ( i (vi ) 6= ) If 2

i (vi ) is a solution it satisfies

2P

J(i) P

J(i)2 v

il v

il 0l=1 vile

l=1 vile c ( ) = 0

P

J(i) P

J(i)v

ik v

ik k=1 e

k=1 e

If vij =6 v

ik for some j k

i (1 inf v

i

) gt 0

i (

i (2 inf v

i

) and sup

i (

vi

) are strictly decreasing in

i (3 lim 0

vi

)) = 1 and lim 1

(inf (sup vi

)) = 0

Proof The objective is continuous and defined over [0 1) Since b is bounded but c is unbound-edly increasing a solution exists and all solutions are finite Since the objective is continuously

differentiable any solution strictly greater than zero must satisfy the first-order condition (3) For the remainder of the proof assume v

ij 6

gt 0 for all 2 [0 1)= vik for some j k which implies b

Since c 0 (0) = 0 all solutions are strictly greater than zero and thus satisfy the first-order condi-

tion45 Since

b gt 0 and c 0 gt 0 any solution 2

i (vi ) is such that for any 0 0

gt = ( )

i (

obtains a strictly higher net payoff than under 0 for sufficiently small

i (

i (

Thus infvi

)) = 1 because gt 0 v

i

)b

i (

vi

)) = 0 because as 1

and sup vi

) are strictly decreasing in Since 0

gt 0 lim 0

(inf

0 0 pointwise Similarly as 0 c lim (sup c 1 1

pointwise (except for = 0)

Lemma 1 i RJ(i) (0 1) [0 ) is upper hemi-continuous 1

i (

It is obvious that the objective is jointly continuous in (vi

) To invoke the theorem all that is required in addition is a compact constraint set but this requires some preparation as the constraint set 2 [0 1) is not compact Since b(middot v

i

) is bounded for fixed vi and c(middot) is unbounded for fixed

macr for any (vi

0

0 ) and any gt 0 there exists a () such that restricting the constraint set to

45When vij = v

ik for all j k the unique solution is i (v

i

) = 0 which satisfies (3) trivially

Proof To show vi

) is upper hemi-continuous in (vi

) use Bergersquos theorem of the maximum

40

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

2 [0 ()] does not change the solution i (vi ) for all (v

i

) within an -ball of (v 0 0 i

) Thus 0 0 0 0

(vi

) is upper hemi-continuous for all (vi

) 2 B

(vi

) by Bergersquos theorem But since (vi

)i

and were chosen arbitrarily

Lemma 2 For all vi 2 RJ(i)

i (vi ) is upper hemi-continuous for all (vi

)

i (vi ) is a singleton for sufficiently low and sufficiently high

Proof This is trivial if vij = v

ik for all j k so assume vij 6 v

ik for some j k As 0=

i (vi ) ( () 1 or in other words that ) for () such that infas 0

() 1i (vi ) 1

By properties of b and c the objective is strictly concave for sufficiently high so the

objective is strictly concave over the region where the solution obtains for sufficiently low making

(vi

) is a singleton As 1i sup

i (vi ) 0 or in other words that i (vi ) (0 ())

for macr() such that macr() 0 as 1 By properties of b and c the objective is strictly concave

for sufficiently low so the objective is strictly concave over the region where the solution obtains for sufficiently high making

i (vi ) a singleton

0Lemma 4 (i) Take any q 2 (q 1 2 q (which

) with associated 0exist and are unique) p Q p Qif and only if 1

2

) with associated EQRE p q and LQRE p 1 (ii) Take any p 2 (2 p

0 (which exist and are unique) q Q q

0 1 R

2EQRE p q and LQRE p q if and only if

Proof We illustrate the proof in Figure 9 The left panel plots for some game the EQRE and

LQRE sets in the first component of the regular set with a particular EQRE as the intersection of reaction functions The right panel ldquozooms inrdquo on this component of the regular set

1 gt

1

(i) Take any EQRE p q such as the point A in the right panel of Figure 9 Suppose

First draw LQRE reaction functions for player 1 using 2 = and similarly for player

= 2 By construction these curves pass through point A but this is not an LQRE 2 using

since each player has a different Next redraw player 1rsquos reaction function by decreasing to 2 Player 1rsquos reaction function ldquopivotsrdquo clockwise as decreases until it crosses player 2rsquos reaction

function at point B This is an LQRE with = 2 and note that since the reaction functions are

strictly monotonic point B is necessarily to the south-east of point A Finally by continuity one

can increase to some value 0 2 (

2 1) such that both LQRE reaction functions pivot counter-

clockwise and intersect at the LQRE C directly below point A The argument is similar for 1 lt

2

0 0(and trivial for 1 =

2

such as points A and C ) Conversely find EQRE p q and LQRE p

Since the LQRE reactions are strictly monotonic the only way to go from

q such that p lt p

C to A is to increase player 1rsquos (pivoting his reaction counter-clockwise) and decrease player 2rsquos (pivoting his reaction clockwise) to exactly

1 and 2 respectively which satisfy

1 gt gt 2

The argument is similar for p 0 gt p (and trivial for p

0 = p) Part (ii) is essentially the same as part

(i)

41

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

1 1

A

0 05 1

A

B C

p p 05

0 05

q q

i s The left panel plots the

LQRE set (solid black curve) the EQRE set (dotted curve) and the regular set (gray rectangles) for some game Point A is a particular EQRE associated with the reaction functions that pass through it The right panel ldquozooms inrdquo on the first component of the regular set and additionally draws LQRE reaction functions as thin curves following the proof of Lemma 4 Both playersrsquo reaction functions ldquopivotrdquo counter-clockwise (clockwise) from their respective origins with an increase (decrease) in

Lemma 5 Fix EQRE p q (i) maxp 1

1

Figure 9 Relationship between LQRE EQRE and the heterogeneity of

p 7 maxq 1

p 7 maxq 1

092 implies that eq lt eq gt 1+e

u1(q) 7 7 u1(q) 7u2(p) and (ii) maxp 1 implies that 1+e

2 1 u2(p) and 2

u2 lt macr()

1 2p 7 maxq 1

v() by part (iv) of Theorem 3 u1 7Proof (i) If maxp 1

implies u1 7

then v() which eq lt 1+e

1

27That follows from part (iii) ofu2 lt

Theorem 3 Part (ii) is similar

1Lemma 6 The EQRE set cannot cross the LQRE set at any point p q 6 p such that = 2 e eeither maxp 1 p = maxq 1 q lt 092 or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Proof Suppose p q is an EQRE in which maxp 1 p 6= maxq 1 q If p q is an LQRE it u1(q) =must be that (1) macr 6 u2(p) (since some player is more accurate despite each having the same

) and (2) that

by Lemma 4 But by Lemma 5 (1) and (2) cannot occur simultaneously

i (

=1 2

e eif either maxp 1 p = maxq 1 q lt or maxp 1 p 6 q gt

6 = maxq 11+e

1+e

Theorem 3

i (0 ) = 0 and i (

i (middot ) is strictly single-peaked with macr() i (

(i) lim vi

) = 0 fv

i

1

(ii) vi

) and v() argmax vi

)max fv

i

2(01) fv

i

2(01)

42

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

(fv

i

)

(fv

i

) (fv

i

)i i i(iii)

|fv

i

=0 = 0 |0ltfv

i

ltfv() gt 0 and

|fv

i

gtfv() lt 0fv

i fv

i fv

i (iv) The product ( v

i

) vi is strictly increasing in v

i

lim ( vi

) vi = 0 and

i ifv

i

0

lim i ( v

i

) vi = 1

fv

i

1

(v) The product macr() v() 24 for all (vi) v() is strictly increasing in lim v() = 0 and lim v() = 1

0 1

Proof (i) We have already established that i (0 ) = 0 For v

i gt 0 the objective is strictly concave so i ( v

i

) gt 0 is given as the unique solution to the first order condition

2 fv

ivi e 0

( vi

) c ( ) = 0 e fv

i 2( + 1)

2 6v

ifv

i e Inspection reveals that lim = 0 which implies that lim i ( v

i

) = 0 from proper-(e 6v

i +1)2 fv

i

1 fv

i

1 ties of c

(ii) Since i (0 ) = 0 and

i ( vi

) gt 0 for vi gt 0 (i) implies

i (middot ) obtains a strictly positive

peak denoted macr() on some set v() (0 1 ) To show that v() is a singleton fix 2 (0 1 )

and use the implicit function theorem

=

vi v

i

where

vi middot e fv

i (e fv

i ( vi 2) 2 v

i

) =

3 vi (e fv

i + 1)3 fv

i ( fv

i vi e e 1) 00

= c ( )fv

i 3 (e + 1)

3

fv

i e 6v

i (e 6v

i 1) 00Note that from properties of c

= c ( ) lt 0 for all vi gt 0 and hence

(e 6v

i +1)3

( vi

) is differentiable in vi by the implicit function theorem Thus the peak is characterized

i macr fv

i (by = () and any vi that satisfy

fv

i = 0 Defining g( v

i

) (e vi 2) 2 v

i

) = 0 inspection reveals that = 0 () g( v

i

) = 0 It is easy to show that for any given fv

i

vi such that g( v

i

) = 0 is unique Thus the peak denoted by macr() is obtained at the single

point v()

(iii) Note that and vi only enter the g(middot) function above as the product x = v

i

We have macralready shown that = 0 () g(middot) = 0 () x = () v() and it easy to show that

fv

i gt 0 () g(middot) lt 0 () x lt () v() and lt 0 () g(middot) gt 0 () x gt () v()

fv

i fv

i

43

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

0 vi

0 ) v

i 0 lt macr() v() and similarly if

i (Suppose By part (iv) below x =v()vi = v

i lt 00

vi = v gt

i i ( v

00 00 i ) v gt macr()

i

v() then x = v()

(iv) Define x = vi and substitute into the first order condition

2 xvi e 0 x

(x vi

) c ( ) = 0 (ex + 1)2 v

i

Using the implicit function theorem

x =

vi v

i x

where

x 2 vi

e 00 x x = + c ( ) 2 v

i (ex + 1)2 vi v

i 2 vi e

x(ex + 1) (1 ex) 00 x 1 = c ( )

x (ex + 1)4 vi v

i

Since vi gt 0 x gt 0 ex gt 1 and c

00 (middot) gt 0 we have that gt 0 and lt 0 which implies that

fv

i x x

i 2e

x x

fv

0 gt 0 Notice that for fixed x 1 and c ( ) 0 as v

i 1 Therefore if xfv

i (ex+1)2 fv

i

were bounded as vi 1 it could not satisfy = 0 so it must be that x 1 as v

i 1

(v) As noted in part (iii) and vi only enter the g(middot) function defined in part (ii) as the product

x = vi

and g(x) (ex(x 2) 2 x) = 0 () x = 24

(vi) In the proof of part (ii) we established that v() and macr() are characterized by g( vi

)

(e fv

i ( vi 2) 2 v

i

) = 0 Using the Implicit Function Theorem

vi g g =

vi

where

g fv

i (= vi

(e vi 2) + e fv

i 1) v

i

g fv

i ( fv

i= (e vi 2) + e 1)

Since vi gt 0 and gt 0 both of these terms are strictly positive if ( v

i 2) gt 0 Inspection

reveals that this is necessarily the case as otherwise g(middot) lt 0 Hence fv

i lt 0 For all vi 2 (0 1)

( vi

) is strictly decreasing in which implies that macr() is also strictly decreasing in Thus as increases macr() decreases and v() increases For all v

i 2 (0 1) lim i ( v

i

) = 0 and 1

44

i

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

macr macrhence lim () = 0 By inspection of g(middot) if lim v() lt 1 while lim () = 0 it could not be 1 1 1

that g(middot) = 0 is satisfied Thus lim v() = 1 1

1 2 q

lt 1 2 and Theorem 6 Let p maxp 1

gt if pr be such that all LQRE p q satisfy

1 192 (I) The EQRE set crosses the LQRE set at pat any other points except

epmaxq 1 q lt

(II) the EQRE set cannot cross the LQRE set 0 21+e

2+ [ + [u u or a a

p1 q1 and p2 q2 (and must so if these points exist) and (III)

1 1 exists (Cases 1 and 2) (a) for all q 2 (q 1) p lt p

0 (i) If p q

0 q and

q q for EQRE p q and LQRE p

0 011(b) for all q 2 (q for EQRE p q and LQRE p) p gt p 21(ii) If p q1 does not exist (Cases 3 and 4)

(a) for all q 2 (q0 01 for EQRE p q and LQRE p q2) p gt p

2(iii) If p q2 exists (Cases 1 and 3) (a) for all p 2 (p ) q lt q

0 for EQRE p q and LQRE p q 0

for EQRE p q and LQRE p q 0

2 and p01

2 p2) q gt q (b) for all p 2 (

(iv) If p2 q2 does not exist (Cases 2 and 4) (a) for all p 2 ( ) q lt q

0 for EQRE p q and LQRE p q 01 p2

Proof In what follows let p

q

refer to some -indexed EQRE sequence 1 exists This implies that maxp 1 p lt maxq 1 q lt e

1)

1Suppose p As q 1+e

q and thus for EQRE p q for sufficiently low q 2 (q q lt1 2 0 p

q

p(Lemma 5) Thus for all sufficiently low q 2 (q q1) p lt p

0 for EQRE p q and LQRE p 0 q

(Lemma 4) gt 1

2 (p1 q 11 may or may not exist) For sufficiently high q 2 (qeq lt maxp 1 p lt

Suppose p 2

1

since p

q

passes through p 1+e

21

) EQRE p q for some

(Lemma 5)

must satisfy maxq 1

lt 1 ) gt 1 221

which implies that for EQRE p q for sufficiently high q 2 (q0 0Thus for all sufficiently high q 2 (q

111

for EQRE p q and LQRE p 1 exists but we do not use this fact here) In this case

q (Lemma 4)) p gt p 21Suppose p 1= q2

1 11 and

u1(

) As 1 p

q

2 2 and thus v() 1

pand

= u1(q

)

u2(p

) u2(u1(211

(which implies p) gt u2( 2) 2 2 2 2

1 2 and thus for ) By part (vi) of Theorem 3 as 1

1 2) all EQRE p q are such that

(part (iii) of Theorem 3) Thus it is still the case that for all sufficiently high q 2 (q

sufficiently large q 2 (q u2(p) lt u1(q) lt v() and 1 )gt 1 2 2

1

0 for EQRE p q and LQRE p 0

p gt p 2 may or may not exist)

e

If p 21

q (Lemma 4) (p For sufficiently R2 = so suppose p gt= q2 2

1 2

1

1

p) EQRE p q must satisfy maxq 1 q lt maxp 1 p lt 1+e

which implies that for EQRE p q for sufficiently high

(Lemma 5) Thus for all sufficiently high p 2 (

high p 2 ( since p

q

for some lt 1 2

0) 1

passes through p ) q lt q for EQRE p qp 2 (2 p gt p1 2 2

and LQRE p q 0 (Lemma 4)

45

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

1 2Suppose p gt and that p q2 exists In this case u1(1 ) lt u2(

1 ) As 12 2 2 1p

q

12 and thus u2(p

) u2(1 ) and u1(q

) u1(1 ) By part (vi) of Theorem 2 2 2

3 as 1 v() 1 and thus for sufficiently small p 2 (1 p2) all EQRE p q are such 2 that u1(p) lt u2(q) lt v() and gt (part (iii) of Theorem 3) Thus for all sufficiently 2 1

small p 2 (1 p2) q gt q 0 for EQRE p q and LQRE p q

0 (Lemma 4)2

We have established the local behavior of the EQRE set in neighborhoods of p q p 1 2 1and 1

2 We now use the path connectedness of the EQRE and LQRE sets and invoke Lemma 2

6 to show that the EQRE set can only cross the LQRE set at specific points 1If p gt then by path connectedness the EQRE set must cross the LQRE set at exactly 2

1p This shows (I)2 1If p q1 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 1 1 1with q

0 2 (q ) By But by Lemma 6 the sets can only cross at p q1 If p q1 does not 2 1exist by the same lemma the EQRE set cannot cross the LQRE set at any q

0 2 (q ) This 2

shows (III)-(i) and (III)-(ii) 2If p q2 exists by path connectedness the EQRE set must cross the LQRE set at some p

0 q

0 with p

0 2 (1 p) But by Lemma 6 the sets can only cross at p2 q2 If p1 q1 does not exist 2

2 (1 )by the same lemma the EQRE set cannot cross the LQRE set at any p 0

p This shows 2

(III)-(iii) and (III)-(iv) All that remains is to show that the EQRE set cannot cross u+ [ u or a+ [ a at any points

1 2other than p q1 and p q2 0 0 0 0 At any EQRE p q 2 u+ [u u1(q ) = u2(p ) which would imply that = making 1 2

1 2p 0 q

0

1 an LQRE but p q1 and p q2 are the only LQRE in u+ [ u Thus the EQRE 1 2set cannot cross u+ [ u except at p q1 or p q2 The EQRE set cannot cross a+ [ a at

1 2any point other than p q1 or p q2 because the EQRE set is otherwise bounded away from

a+ [ a by the LQRE set u+ [ u or boundary of the regular set This shows (II)

k Theorem 7 Let c( ) = for k gt 1 Fix vi 2 RJ(i) 2 (01) and 2 (01) If 2

i (vi ) k+1)(ie a solution to (2)) then 1 ˜ 2 i ( vi

Proof Assuming c( ) = k for k gt 1 re-write the first-order condition (3)

g( vi

) k k 1 = 0 (10)

2PJ(i) 2 P

J(i)v e v

il v

il

v

il

e

l=1 il l=1 2where g( vi

) It is easy to show that g( vi

) = g( vi

)PJ(i) P

J(i)e v

ik e v

ik k=1 k=1

Using this identity we show that ˜ solves (10) if and only if 1 ˜ solves

2 ( k+1)k k g( vi

) 1 = 0

46

which completes the proof Plugging 1 ˜ into the above

2 g((1 ˜) v

i

) ( k+1)k(1 ˜)k 1 = 0 ()

2 k g(˜ vi

) 2k˜ 1 = 0 ()k g(˜ v

i

) k˜ 1 = 0

kTheorem 8 Let c( ) = for k gt 1 As k 1 the EQRE set approaches the LQRE set

0 00 01 2Proof c ( ) = k k and c ( ) = k(k 1) k Let M gt 0 and ˜ gt 0 be arbitrary c (˜) =

M () ˜ = (M ) 11 Define

k

(˜M) as the solution to ˜ = (M ) 11 As k 1 a few k k

k k

facts about k

(˜M)c are easily verified (1) k

(˜M)c 0 ( ) M1 ˜ + 11 gt˜ (pointwise) =

k 2 00

k 1M(2)

k

(˜M)c (˜) = k

(˜M)k(k 1) 1 and (3) k

(˜M) 1 if ˜ lt 1 and

k

(˜M)k

k

(˜M) 0 if ˜ 1 Given any finite game the set of attainable expected utility vectors V = (V1 Vn

) u(p) 2

RJ p 2 A is a compact subset of RJ For any v 2 V let I1(v) (resp I0(v)) be the set of indices i such that v

im 6 = vil for all m l) Since b

00 (middot v

i

) is bounded for = vil for some m l (resp v

im

any given vi 2 RJ(i) and V is compact there is a uniform bound on b

00 (middot middot) over [01) V

i

Thus from the facts of

k

(˜M)c we can always take k sufficiently large such that for all 2 (01) v 2 V and i there exists ˜

i gt 0 such that b 0 (˜i

vi

) = c 0 (˜

(vi

) = 0 otherwise i

i

) () i (vi ) =

˜i if i 2 I1(v) and

We show that any LQRE is approached by an EQRE as k 1 Suppose p˜ is an LQRE ˜ ˜with associated expected utility vector v 2 V If I1(v ) = the result is trivial so assume

˜ 0 ˜ 0 ˜ ˜ ˜not46 Choose i 2 I1(v ) such that b (˜ vi ) b (˜ v

j ) for all j 2 I1(v ) and define k

(˜ v ) 0 ˜ ˜

k

(˜ b (˜ vi )) By construction and from the facts about

k

c (1) for the chosen i 2 I1(v )

i ( j ) =

j (

˜ ˜i ) =

k

(˜ v ˜ ˜ 0 ˜ i k(

0 0 (˜ v (˜) middot)) = ˜ for all k (2) for j 2 I1(()

˜ j k(

b ) ) b ( jk

vc v v ˜ ˜0

k

(˜ v jk where

jk ˜ as k 1 and (3) for j 2 I0()c ( jk

) ()

j k(middot)) = 0 for all k middot)) )v = v

˜ ˜This can be extended to a neighborhood of v in the sense that for all j (v

˜ ˜ ˜) for small the same

k

(˜ v ) is such that (1) for all j 2 I1(v middot)) ˜ as j (

)) as k 1 In either

2 N

(

j k()v v v˜ middot)) ˜ or possibly 0 (if j 2 I0(

j (˜ ˜case it is easy to show that Q(middot

k

(˜ v )) Q(middot ˜) uniformly over N

(v ) as k 1 Thus the ˜EQRE sequence p

k that is a fixed point of Q(middot k

(˜ v )) converges to the LQRE p˜ that is a

fixed point of Q(middot ˜) as k 1 All that remains is to show that as k 1 any EQRE p

k is nearby some LQRE p˜ but to

show this is very similar to the above and hence omitted It relies on showing that as k 1 any 146If v

il = vim for all i m and l in an EQRE or LQRE then it must be that p

il = pim =

J(i) for all i m and l (no matter the or )

k 1 and (2) for j 2 I0(

j k()v v v

47

k

k

kEQRE p

k with associated v 2 V is such that (v k

) (v k

) for all i j 2 I1(vk ) andi

i j

j

ktherefore nearby the LQRE with ˜ = mini2I1(vk ) (v

k

)i

i

93 Property 4 of Q

For this part and this part only we drop the i subscript Hence we use Qj Q

j and instead of Q

ij Q and i v 2 RJ is a vector with components v1 v

J indexed by j k l or m Unless ij

stated otherwise all sums go from 1 to J Q

(v)4 If vj = max

k

vk

gt vl for some l j gt 0 for sufficiently low and high

v

j

Assume vj = max

k

vk

Under this condition it is easy to verify that an increase in vj

ewill lead to an increase in Qj (v ) =

v

j if product x vj increases ie even if decreases P

k e v

k

by a small amount with the increase in vj Since we only consider sufficiently low and sufficiently

high we can take Q j (v ) to be continuously differentiable Hence to determine the behavior of

Q j as v

j increases we analyze the behavior of vj via the system

P

P

22 v

l v

l l vl e

l vle 0 (x v

j ) = P P c ( ) = 0 v

k v

k k e

k e

(x vj ) = x v

j = 0

These are merely the first-order condition (3) and the definition of x Using the implicit function ( ) ( )theorem we show that x = gt 0 for sufficiently low and high where

v

j (vj ) (x )

( ) ( )v

j

x = and = (v

j ) (x )v

j x

v) 00 ( )From previous results

2b( c ( ) lt 0 at optimum for sufficiently low and high so =

2 (crarr )

h i

2b( v) 00

h i0 c ( )

2

2b( v) 00

=

2 c ( ) gt 0 for sufficiently low and high as well It 1 v

j )remains to show that ( lt 0 for sufficiently low and high

(vj )

h

2b( v) 00

i

( ) v

j

2 c ( ) 2b( v) 00 = = v

j + c ( ) (v

j ) vj 2

vj

where 0

P

n o

P

1

2 2 (vj +v

k

) (vk

+v

l

)2 2 (v 2 ) e X =j

j

k 6 v

j vk +

k vk

vj + v

kl e (v

k

+v

l

))2

P

A (= P

e v

j 2 (vl

+v

k

) 2 vj + (v

j +v

k

)2 (v vl

vk

)e e =j ekl l k 6kl

48

P P

High When = 0 = (2vj 2v

k

)J2 = 2 (J 1) =j vk J2 gt 0 Byv

j k=6 j k 6

continuity

v

j gt 0 for sufficiently low which is guaranteed for sufficiently high in which case

( ) lt 0 and the result follows (v

j ) h i h i

2b( v)

2b( v)Low Note that v

j +

2 c 00 ( ) lt v

j +

2 and hence it is v

j v

j

enough to show the latter is negative for sufficiently large Reproducing the the expression 2 2 (v

m

+v

k

+v

l

)

2b( v)

P m vm

Pkl

(vm v

k

+2vk

v

l 2vm

v

l

)e = (9) and noting that terms with the largest ex-

2 (P

v

k )3 k e

ponents dominate it is easy to show that as 1

2

2b( v) 2 2micro (micro 2micro + 2) (micro )+ middot evj 2v

j 2 + ((2 micro2 + 2micro 2micro) + micro(micro 2 + 2micro 2micro))

3 (micro )= 22 + 2micro + (micro micro 2) middot e

where micro lt is the second largest component of v This approaches 0 but is negative for all lt 1 ( )Hence for sufficiently large which is guaranteed by sufficiently low lt 0 and the result (v

j )

follows

94 Proof of Theorem 5

To facilitate comparison between this proof and the analogous one for standard LQRE (Theorem 3

in McKelvey and Palfrey [1995]) we freely borrow from McKelvey and Palfrey [1995] We highlight only the most interesting differences

While parts of the proof only go through for games in which J(i) = 2 for all i we use general notation to emphasize what can be said for normal form games and where the proof breaks down The proof uses the following lemma which holds for normal form games and may be of general interest

Lemma 7 For sufficiently large p() is a singleton

Proof For any vi 2 RJ(i)

i (vi ) is a singleton for sufficiently large by Lemma 2 Since for any

finite game the set of all possible expected payoffs u(p) 2 RJ p 2 A is compact this can be extended to games for sufficiently high (u

i

(p i

) ) is a singleton for all i and p i 2 A

i

i

Thus for sufficiently large the mapping A A with components

i (ui

(p i

))uik (p

i

)e ik

(p ) = P

J(i) i (ui

(p i

))uil

(p i

) l=1 e

is a continuous function For this proof we always take sufficiently large for this to be the case Then for any (p ) 2 (p() ) if and only if p is a fixed point of

49

We will show that for sufficiently large is a contraction mapping and hence has a unique

fixed point To prove the result we use three basic facts from McKelvey and Palfrey [1995] We

also use a fourth fact which is unique to our problem For convenience we define utility differences

(ie across actions given the opponentsrsquo behavior) by uikl

(a i

) = ui

(aik

a i

) ui

(ail

a i

) and P

uikl

(p i

) = a u

ikl

(a i

)p i

(a i

) i

2A i

Fact 1 For any p q 2 A0 (the interior of A) maxik

|pik q

ik

| maxikl

|pik

pil q

ik

qil

|

xFact 2 Since the derivative of e at x = 0 is 1 then for all D gt 1 there is a such that x1whenever |x1| |x2| lt |e ex2 | D middot |x1 x2|

Fact 3 There is an M gt 0 such that maxikl

|uikl

(p i

) uikl

(q i

)| M middotmaxik

|pik q

ik

|

Fact 4 For any p q 2 A and

maxj max

j (uj (q

i (ui(q i

) )uikl

(q i

)|

j ) ) middotmaxikl

|uikl

(p i

) uikl

(q i

)|

maxikl

|i (ui(p

i

) )uikl

(p i

)

j (uj (p

j ) )

i

Since 0 Fact 4 is obvious whenever uikl

(p 6i

) for some ikl And when uikl

(p i

) =i

) = uikl

(q i (ui(p

i

) ) = i (ui(q

i

) ) = 0 for all i in which case both sides of the i

) for all ikluikl

(q

inequality equal 0 and it is satisfied trivially Pick any D gt 1 and let be defined as in Fact 2 and M defined as in Fact 3 Pick sufficiently

large such that maxip

i (ui(p

i

) ) satisfies

1 uikl

(p i

) lt for all i k l and any p and

2 lt 1(D middotM)

Letting k middot k represent the sup norm and writing middotD middotM we have that for any p q 2 A

k (p) (q)k = maxik

| ik

(p) ik

(q)|

maxikl

| ik

(p) il

(p) ik

(q) il

(q)| (u

i

(p i

))uikl

(p i

) (ui

(q i

))uikl

(qi i i

)|= maxikl

|e e

i

) )uikl

(p i (ui(q

i

) )uikl

(q D middotmaxikl

|i (ui(p

i

)i

)|

j ) ) j ) ) maxj max

j (uj (p j (uj (q middotD middotmax

ikl

|uikl

(p i

) uikl

(q i

)|

middotD middotM middotmaxik

|pik q

ik

|

= middot kp qk

where lt 1 The steps follow respectively by the definition of k middotk Fact 1 the definition of Fact

i (vi ) is2 Fact 4 Fact 3 and the definition of and the definitions of and k middot k Finally since

50

decreasing in for any fixed vi

k (p) (q)k middot k p qk for any gt Hence is a contraction

mapping and has a unique fixed point for sufficiently large

Fix game form and let U represent the set of all possible payoff functions for For each P

i 2 N define mi = J(i) 1 and M

i = 1 mi

Write N i = N i and m =

i2N mi

Define

pi0 = p

iJ(i) and X

Di = p

i 2 Rm

i pik 0 for all k 2 M

i and pik 1

k2Mi

Write Di 0 for the interior of D

i

Using the identity pi0 = p

iJ(i) = 1 P

k2Mi pik

a mixed strategy Q

pi 2 A

i can be identified by the first mi components ie by a vector in D

i

Write D = Di

i2N

and D0 for the interior of D A vector p = (p1 pn

) 2 D is referred to as a mixed profile For any u 2 U as in the proof of Lemma 7 define utility differences u

ikl

(a i

) = ui

(aik

a i

)P

ui

(ail

a i

) uik0(a

i

) = uikJ(i)(a

i

) and uikl

(p) = uikl

(p i

) = a

i

2A i uikl

(a i

)p i

(a i

) Define

X = D0 (0 1 ) and correspondence F X U Rm with components given by for any i 2 N

and k 2 Mi

pik

Fik

(p u) = i (ui10(p) u

im

i

0(p) )uik0(p) ln (11)

pi0

Note that we have redefined as a function of payoff differences relative to the utility of action i

aiJ(i) which is without loss The use of correspondence (11) is unique to our problem

For any u 2 U write fu

(p ) = F (p u) For any given u 2 U p() = p fu

(p ) = 01and the EQRE graph G = (p() ) 0 lt lt 1 is given by G = f (0) where 0 is the

u m-dimensional vector of zeros If RJ(i) 1 (0 1 ) [0 1 ) is single-valued then f

u is a i

1continuous function Then f (0) would be a closed set (preimage of a closed set) and hence Gu

would be upper hemi-continuous Hence if the game is such that is single-valued then part (i) i

of Theorem 5 is satisfied This is true in the case where J(i) = 2 for all i by Lemma 2 but it is not

true in general (see Appendix 95) When J(i) = 2 for all i replace (11) with F X U Rn (m = n in this case) defined by

pi1

Fi

(p u) = i (|ui10(p)| )ui10(p) ln

pi0

1for all i Again for any given u 2 U the EQRE correspondence is given by G = f (0) We will u

show that for any (p ) 2 X F is surjective Note that for any 2 [0 1 )

pi1

Fi

(p u) = () i (|ui10(p)| )ui10(p) ln =

pi0

pi1()

i (|ui10(p)| )ui10(p) = + ln( ) pi0

where 2 ( 1 1 ) depends on p Hence F is surjective for fixed (p ) 2 X if i (|ui10(p)| )ui10(p)

U ( 1 1 ) is surjective This follows from part (iv) of Theorem 3 after noting that ui10(p) U

51

( 1 1) is surjective for fixed p Hence F (p u) is transversal to any submanifold of Rn By the

transversality theorem (Guillemin and Pollack 1974 68) for almost all u 2 U fu

(p ) = F (p u) 1is transversal to 0 It follows (Guillemin and Pollack 1974 28) that f (0) is a manifold of codi-

u

mension n = m or dimension 1 for almost all u 2 U These particular theorems from differential topology were invoked previously in McKelvey and Palfrey [1998] but not in McKelvey and Palfrey

[1995] The rest of the proof which differs from the analogue in McKelvey and Palfrey [1995] in only

very minor ways goes through for any normal form game in which G is a manifold We have shown

this only for the binary action case but we conjecture that it is considerably more general Note that the previous argument can be extended to the case when the domain of f

u is bounded for any c = (c c) with 0 lt c lt c define X

c X by Xc = D0[c c] Then X

c is a (m+1)-dimensional 1manifold with boundary and G

c = f (0) Xc is a 1-dimensional manifold with boundary

u Now pick M gt 0 so that for all p 2 A sup

ikl

|uikl

(p)| M Define () maxip

i (ui(p i

) ) a

= e ()M and b

= e ()M Then it follows that for any (p ) 2 G that

pik() middot M log () middot M =) a

pi0 p

ik b

pi0

pi0

But

X

pik b

pi0 =) 1 p

i0 = pik b

mi

pi0

k2Mi

=) pi0 1(b

mi + 1)

and

pik a

pi0 =) p

ik a

(b

mi + 1) = c

Since a

1 it follows that for all 0 k mi (ie including k = 0) the above inequality holds

Define

W = (p ) 2 X pik c

for all i 2 N 0 k mi

Thus we have shown that G W X Similarly Gc W X

c

In other words the EQRE graph

can only ldquoexitrdquo X (ie have p on the boundary D) at the minimum and maximum values of We wish to show that in generic games the EQRE graph can be used to make a unique selection

of a Nash equilibrium To do this we use two facts First as we have already shown in Lemma

7 there is a unique solution for sufficiently large Second we will show that this branch of the

correspondence converges to a unique Nash equilibrium as goes to 0 We have now shown enough to prove part (ii) of Theorem 5 that for almost all there are an

odd number of EQREs From the argument setting c = (c c) we have shown that Gc is a compact

1-dimensional manifold with boundary which for large enough c has a unique intersection with

52

= c Any such manifold has a finite number of connected compact components each of which

must have an even number of boundary points We have also shown that any boundary point must be at = c or = c Since there is exactly one solution at = c there must be an odd number of solutions at c

We now show part (iii) of Theorem 5 that as 0 the branch B of the manifold that passes through c (for sufficiently high c) converges to a unique Nash equilibrium

Lemma 8 Let be chosen so that p() is a singleton Then for almost all games as 0 the macrbranch B of the manifold that passes through converges to a unique Nash equilibrium

Proof It follows from the arguments above that for almost all games there exists a decreasing macrsequence

i

with i lt for all i such that if we define c

i = (i

) Xi = X

c

i and Gi = G

c

i G

macr macr1 G is a 1-dimensional manifold with a unique point say (p ) for which = and a unique macrconnected branch B that passes through (p )

2 Gi is a compact 1-dimensional manifold with boundary which has a finite number of connected

components

macr3 Letting Bi be the connected branch of G

i which begins at (p ) it follows that Bi is a

compact connected 1-dimensional manifold with boundary for all i which has a unique intersection say (p

i

i

) with = i

Now for any i define A

i = (p ) 2 B lt i

and Ai to be the closure of the projection of

Ai onto D Then A

i

is a decreasing sequence of sets We show that for almost all games i

Ai

must be a unique point First of all since D is compact and each Ai is closed and nonempty

i

Ai

cannot be empty Suppose by way of contradiction that i

Ai contains two distinct points Since

generic games contain a finite number of Nash equilibria we may assume that the game defined

by u has a finite number of Nash equilibria By Theorem 4 any point in i

Ai must be a Nash

equilibrium But if p and q are both in i

Ai

then we can construct a sequence (pi

i

) B with p2i 1 p p2i q

i 0 (ie ldquooddsrdquo approach p and ldquoevensrdquo approach q) and a

homeomorphism R B satisfying several properties In particular since B is connected it is also path connected (Guillemin and Pollack p 38 exercise 3) so can be constructed to satisfy

macr(0) = (p ) (i) = (pi

i

) and [i 1 i] [i i+1] = (i) with [i i+1] a compact 1-dimensional manifold with boundary

We have constructed an infinite sequence of compact manifolds with boundary each of whose projection on D connects a point near p with a point near q Further for any

i

at most

a finite number of these manifolds intersect with Xi (since B X

i is a compact 1-dimensional manifold with boundary which can consist of at most a finite number of components) It follows

that any separating hyperplane Ht = p 2 D p middot (p q) = t between q and p must have

a nonempty intersection with i

Ai (by compactness of H

t

) However since there are an infinity of

53

such hyperplanes there are infinitely many points in i

Ai and hence an infinite number of Nash

equilibria which is a contradiction

95 Non-concavity

0 1 2 3 4 5 6 7 -01

-005

0

005

01 θ = 002 θ = 00235 θ = 0025

λ

Figure 10 Example of non-concave objective with two pure solutions

This example due to Weibull et al [2007] results in a first-stage objective that is not-concave

and emits two pure solutions Suppose vi = (1 1 0 0) ie there are J1(i) and J0(i) high-

and low-quality alternatives respectively satisfying J0(i) + J1(i) = J(i) Denote the fraction of J1(i)high quality alternatives by = J(i) In this case it is easy to show that the benefit of precision

is given by e

b( vi

) = 1 + e

Figure 10 plots the first-stage objective b( vi

) 2 for = 001 and 2 002 00235 0025 For

54 For some critical intermediate 025 (high costs) i 03 For = 002 (low costs) = 0

i

0235 there are two optimal selections of i and the agent is willing to mix over them with 0

arbitrary probability That the optimal precision can be discontinuous in is somewhat surprising

and complicates our analysis The non-concavity of the objective in this example must depend on there being more than two

alternatives by our analysis in Section 31 To see this note that b(middot vi

) only depends on vi through

1 Thus when = the benefit function b(middot vi

) is as if there are only two alternatives vi = (1 0)2

eor vi = 1 In this case b( v

i

) = which is the same as b( vi

)|fv

i

=1 from (5)1+e

54

gt 196 Generalized matching pennies with q 2

For games of the form of Figure 3 the following is the counterpart to Theorem 6 for the case in gt 1which q 2 The proof is essentially the same as that for the Theorem 6 and hence omitted

The result is accompanied by Figure 11 which is analogous to Figure 6 Rather than redefining

every object for this case we hope their definitions are clear from the figure

1 gt 1Theorem Let p 2 q and r be such that all LQRE p q satisfy maxp 1 pmaxq 12 e gt 1q lt 092 (I) The EQRE set crosses the LQRE set at 1 q if p (II) the EQRE 1+e

2 2 1 2set cannot cross the LQRE set u+ [ u or a+ [ a at any other points except p q1 and p q2

(and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 5 and 6) 1(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p1) q gt q 0 for EQRE p q and LQRE p q

0 2 1(ii) If p q1 does not exist (Cases 7 and 8)

(a) for all p 2 (1 p) q gt q 0 for EQRE p q and LQRE p q

0 2 2(iii) If p q2 exists (Cases 5 and 7)

2(a) for all q 2 (q q) p gt p 0 for EQRE p q and LQRE p

0 q and

(b) for all q 2 (1 q2) p lt p 0 for EQRE p q and LQRE p

0 q2

2(iv) If p q2 does not exist (Cases 6 and 8) (a) for all q 2 (1 q) p gt p

0 for EQRE p q and LQRE p 0 q2

55

Case 5 Case 6 1 1

p 05

0 0

lowastlowast lowastlowast q

a minus

1 1

p q lowastlowast lowastlowast

p q 1 1

p 05

p

p q 1 1

u minus

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

u minus

a +

u +

1 2

1 2

1 lowastlowast 2 q

a minus

0 05 1 0 05 1 q q

Case 7 Case 8

u minus

p q lowastlowast lowastlowast

a +

u + 1 2

1 2

1 lowastlowast 2 q

a minus

1

u minus

lowastlowast lowastlowast

1 1

p q

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

a minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

gt 1Figure 11 The four cases of equilibrium sets (q )2

56

97 Details of Data and Procedures from Studies of 22 Games

Study Game Data

p q N Notes

Rounds (R) Subject pairs (I) Feedback

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

50 36 yes 50 24 yes 50 24 yes 50 12 yes

1 0079 0690 9600 200 48 yes 2 0217 0527 9600 200 48 yes 3 0163 0793 9600 200 48 yes 4 0286 0736 9600 200 48 yes 5 0327 0664 9600 200 48 yes

SC 2008 6 7

0445 0596 9600 0141 0564 4800

200 48 yes 200 24 yes

8 0250 0586 4800 200 24 yes 9 0254 0827 4800 200 24 yes 10 0366 0699 4800 200 24 yes 11 0331 0652 4800 200 24 yes 12 0439 0604 4800 200 24 yes

Ochs 1995 2 3

0595 0258 448 0542 0336 516

56 8 yes 64 8 yes

GH 2001 AMP RA

0960 0160 25 0080 0800 25

1 25 no 1 25 no

NS 2002 NS 0458 0390 6720 60 112 yes

Table 6 Details of Data and Procedures from Studies of 2 2 Games

971 MPW 2000 (McKelvey et al [2000])

For any one game each subject takes an action 50 times against randomly matched opponents with

feedback Subjects play multiple games and maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $010 (year 2000 Dollars)

A B C D L R

U 9 0 0 1

D 0 1 1 0

L R

U 9 0 0 4

D 0 4 1 0

L R

U 36 0 0 4

D 0 4 4 0

L R

U 4 0 0 1

D 0 1 1 0

972 SC 2008 ( Selten and Chmura [2008])

Each subject plays just one game and takes an action 200 times against randomly matched oppo-nents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of

57

the experiment The exchange rate is sect0016 (year 2008 Euros) and the experiment took place in

Germany

1

2

3

4

L R L R L R L R

10 8 9 9

0 18 10 8

8 5 6 7

0 13 9 4

10 4 7 7

0 14 8 6

9 2 5 6

0 11 7 4

U U U U

D D D D

5

6

7

8

L R L R L R L R

8 1 4 5

0 9 7 2

8 0 3 5

1 7 7 1

14 8 9 9

4 22 10 12

11 5 6 7

3 16 9 7

U U U U

D D D D

9

10

11

12

L R L R L R L R

U 13 4 7 7

3 17 8 9

U 11 2 5 6

2 13 7 6

U 10 1 4 5

2 11 7 4

U

D D D D

10 0 3 5

3 9 7 3

973 Ochs 1995 (Ochs [1995])

Each subject plays only one of the two games Those who play game 2 (game 3) take an action

56 (64) times against randomly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment Following McKelvey and Palfrey

[1995] who note that ldquoThe subjects in the Ochs experiments were paid using a lottery procedurerdquo we convert the payoffs described in Ochs [1995] to those in the matrices below before estimation The exchange rate is $001 (expected 1982 Dollars)

2 3 L R L R

11141 0 0 11141

0 11141 01238 0

11141 0 0 11141

0 11141 02785 0

U U

D D

974 GH 2001 (Goeree and Holt [2001])

Each subject takes just one action against an anonymous opponent and the exchange rate is $001

(year 2001 dollars) ldquoAMPrdquo refers to ldquoasymmetric matching penniesrdquo and ldquoRArdquo refers to ldquoreversed

asymmetryrdquo

AMP RA L R L R

320 40 40 80

40 80 80 40

44 40 40 80

40 80 80 40

U U

D D

58

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)

k

k

kEQRE p

k with associated v 2 V is such that (v k

) (v k

) for all i j 2 I1(vk ) andi

i j

j

ktherefore nearby the LQRE with ˜ = mini2I1(vk ) (v

k

)i

i

93 Property 4 of Q

For this part and this part only we drop the i subscript Hence we use Qj Q

j and instead of Q

ij Q and i v 2 RJ is a vector with components v1 v

J indexed by j k l or m Unless ij

stated otherwise all sums go from 1 to J Q

(v)4 If vj = max

k

vk

gt vl for some l j gt 0 for sufficiently low and high

v

j

Assume vj = max

k

vk

Under this condition it is easy to verify that an increase in vj

ewill lead to an increase in Qj (v ) =

v

j if product x vj increases ie even if decreases P

k e v

k

by a small amount with the increase in vj Since we only consider sufficiently low and sufficiently

high we can take Q j (v ) to be continuously differentiable Hence to determine the behavior of

Q j as v

j increases we analyze the behavior of vj via the system

P

P

22 v

l v

l l vl e

l vle 0 (x v

j ) = P P c ( ) = 0 v

k v

k k e

k e

(x vj ) = x v

j = 0

These are merely the first-order condition (3) and the definition of x Using the implicit function ( ) ( )theorem we show that x = gt 0 for sufficiently low and high where

v

j (vj ) (x )

( ) ( )v

j

x = and = (v

j ) (x )v

j x

v) 00 ( )From previous results

2b( c ( ) lt 0 at optimum for sufficiently low and high so =

2 (crarr )

h i

2b( v) 00

h i0 c ( )

2

2b( v) 00

=

2 c ( ) gt 0 for sufficiently low and high as well It 1 v

j )remains to show that ( lt 0 for sufficiently low and high

(vj )

h

2b( v) 00

i

( ) v

j

2 c ( ) 2b( v) 00 = = v

j + c ( ) (v

j ) vj 2

vj

where 0

P

n o

P

1

2 2 (vj +v

k

) (vk

+v

l

)2 2 (v 2 ) e X =j

j

k 6 v

j vk +

k vk

vj + v

kl e (v

k

+v

l

))2

P

A (= P

e v

j 2 (vl

+v

k

) 2 vj + (v

j +v

k

)2 (v vl

vk

)e e =j ekl l k 6kl

48

P P

High When = 0 = (2vj 2v

k

)J2 = 2 (J 1) =j vk J2 gt 0 Byv

j k=6 j k 6

continuity

v

j gt 0 for sufficiently low which is guaranteed for sufficiently high in which case

( ) lt 0 and the result follows (v

j ) h i h i

2b( v)

2b( v)Low Note that v

j +

2 c 00 ( ) lt v

j +

2 and hence it is v

j v

j

enough to show the latter is negative for sufficiently large Reproducing the the expression 2 2 (v

m

+v

k

+v

l

)

2b( v)

P m vm

Pkl

(vm v

k

+2vk

v

l 2vm

v

l

)e = (9) and noting that terms with the largest ex-

2 (P

v

k )3 k e

ponents dominate it is easy to show that as 1

2

2b( v) 2 2micro (micro 2micro + 2) (micro )+ middot evj 2v

j 2 + ((2 micro2 + 2micro 2micro) + micro(micro 2 + 2micro 2micro))

3 (micro )= 22 + 2micro + (micro micro 2) middot e

where micro lt is the second largest component of v This approaches 0 but is negative for all lt 1 ( )Hence for sufficiently large which is guaranteed by sufficiently low lt 0 and the result (v

j )

follows

94 Proof of Theorem 5

To facilitate comparison between this proof and the analogous one for standard LQRE (Theorem 3

in McKelvey and Palfrey [1995]) we freely borrow from McKelvey and Palfrey [1995] We highlight only the most interesting differences

While parts of the proof only go through for games in which J(i) = 2 for all i we use general notation to emphasize what can be said for normal form games and where the proof breaks down The proof uses the following lemma which holds for normal form games and may be of general interest

Lemma 7 For sufficiently large p() is a singleton

Proof For any vi 2 RJ(i)

i (vi ) is a singleton for sufficiently large by Lemma 2 Since for any

finite game the set of all possible expected payoffs u(p) 2 RJ p 2 A is compact this can be extended to games for sufficiently high (u

i

(p i

) ) is a singleton for all i and p i 2 A

i

i

Thus for sufficiently large the mapping A A with components

i (ui

(p i

))uik (p

i

)e ik

(p ) = P

J(i) i (ui

(p i

))uil

(p i

) l=1 e

is a continuous function For this proof we always take sufficiently large for this to be the case Then for any (p ) 2 (p() ) if and only if p is a fixed point of

49

We will show that for sufficiently large is a contraction mapping and hence has a unique

fixed point To prove the result we use three basic facts from McKelvey and Palfrey [1995] We

also use a fourth fact which is unique to our problem For convenience we define utility differences

(ie across actions given the opponentsrsquo behavior) by uikl

(a i

) = ui

(aik

a i

) ui

(ail

a i

) and P

uikl

(p i

) = a u

ikl

(a i

)p i

(a i

) i

2A i

Fact 1 For any p q 2 A0 (the interior of A) maxik

|pik q

ik

| maxikl

|pik

pil q

ik

qil

|

xFact 2 Since the derivative of e at x = 0 is 1 then for all D gt 1 there is a such that x1whenever |x1| |x2| lt |e ex2 | D middot |x1 x2|

Fact 3 There is an M gt 0 such that maxikl

|uikl

(p i

) uikl

(q i

)| M middotmaxik

|pik q

ik

|

Fact 4 For any p q 2 A and

maxj max

j (uj (q

i (ui(q i

) )uikl

(q i

)|

j ) ) middotmaxikl

|uikl

(p i

) uikl

(q i

)|

maxikl

|i (ui(p

i

) )uikl

(p i

)

j (uj (p

j ) )

i

Since 0 Fact 4 is obvious whenever uikl

(p 6i

) for some ikl And when uikl

(p i

) =i

) = uikl

(q i (ui(p

i

) ) = i (ui(q

i

) ) = 0 for all i in which case both sides of the i

) for all ikluikl

(q

inequality equal 0 and it is satisfied trivially Pick any D gt 1 and let be defined as in Fact 2 and M defined as in Fact 3 Pick sufficiently

large such that maxip

i (ui(p

i

) ) satisfies

1 uikl

(p i

) lt for all i k l and any p and

2 lt 1(D middotM)

Letting k middot k represent the sup norm and writing middotD middotM we have that for any p q 2 A

k (p) (q)k = maxik

| ik

(p) ik

(q)|

maxikl

| ik

(p) il

(p) ik

(q) il

(q)| (u

i

(p i

))uikl

(p i

) (ui

(q i

))uikl

(qi i i

)|= maxikl

|e e

i

) )uikl

(p i (ui(q

i

) )uikl

(q D middotmaxikl

|i (ui(p

i

)i

)|

j ) ) j ) ) maxj max

j (uj (p j (uj (q middotD middotmax

ikl

|uikl

(p i

) uikl

(q i

)|

middotD middotM middotmaxik

|pik q

ik

|

= middot kp qk

where lt 1 The steps follow respectively by the definition of k middotk Fact 1 the definition of Fact

i (vi ) is2 Fact 4 Fact 3 and the definition of and the definitions of and k middot k Finally since

50

decreasing in for any fixed vi

k (p) (q)k middot k p qk for any gt Hence is a contraction

mapping and has a unique fixed point for sufficiently large

Fix game form and let U represent the set of all possible payoff functions for For each P

i 2 N define mi = J(i) 1 and M

i = 1 mi

Write N i = N i and m =

i2N mi

Define

pi0 = p

iJ(i) and X

Di = p

i 2 Rm

i pik 0 for all k 2 M

i and pik 1

k2Mi

Write Di 0 for the interior of D

i

Using the identity pi0 = p

iJ(i) = 1 P

k2Mi pik

a mixed strategy Q

pi 2 A

i can be identified by the first mi components ie by a vector in D

i

Write D = Di

i2N

and D0 for the interior of D A vector p = (p1 pn

) 2 D is referred to as a mixed profile For any u 2 U as in the proof of Lemma 7 define utility differences u

ikl

(a i

) = ui

(aik

a i

)P

ui

(ail

a i

) uik0(a

i

) = uikJ(i)(a

i

) and uikl

(p) = uikl

(p i

) = a

i

2A i uikl

(a i

)p i

(a i

) Define

X = D0 (0 1 ) and correspondence F X U Rm with components given by for any i 2 N

and k 2 Mi

pik

Fik

(p u) = i (ui10(p) u

im

i

0(p) )uik0(p) ln (11)

pi0

Note that we have redefined as a function of payoff differences relative to the utility of action i

aiJ(i) which is without loss The use of correspondence (11) is unique to our problem

For any u 2 U write fu

(p ) = F (p u) For any given u 2 U p() = p fu

(p ) = 01and the EQRE graph G = (p() ) 0 lt lt 1 is given by G = f (0) where 0 is the

u m-dimensional vector of zeros If RJ(i) 1 (0 1 ) [0 1 ) is single-valued then f

u is a i

1continuous function Then f (0) would be a closed set (preimage of a closed set) and hence Gu

would be upper hemi-continuous Hence if the game is such that is single-valued then part (i) i

of Theorem 5 is satisfied This is true in the case where J(i) = 2 for all i by Lemma 2 but it is not

true in general (see Appendix 95) When J(i) = 2 for all i replace (11) with F X U Rn (m = n in this case) defined by

pi1

Fi

(p u) = i (|ui10(p)| )ui10(p) ln

pi0

1for all i Again for any given u 2 U the EQRE correspondence is given by G = f (0) We will u

show that for any (p ) 2 X F is surjective Note that for any 2 [0 1 )

pi1

Fi

(p u) = () i (|ui10(p)| )ui10(p) ln =

pi0

pi1()

i (|ui10(p)| )ui10(p) = + ln( ) pi0

where 2 ( 1 1 ) depends on p Hence F is surjective for fixed (p ) 2 X if i (|ui10(p)| )ui10(p)

U ( 1 1 ) is surjective This follows from part (iv) of Theorem 3 after noting that ui10(p) U

51

( 1 1) is surjective for fixed p Hence F (p u) is transversal to any submanifold of Rn By the

transversality theorem (Guillemin and Pollack 1974 68) for almost all u 2 U fu

(p ) = F (p u) 1is transversal to 0 It follows (Guillemin and Pollack 1974 28) that f (0) is a manifold of codi-

u

mension n = m or dimension 1 for almost all u 2 U These particular theorems from differential topology were invoked previously in McKelvey and Palfrey [1998] but not in McKelvey and Palfrey

[1995] The rest of the proof which differs from the analogue in McKelvey and Palfrey [1995] in only

very minor ways goes through for any normal form game in which G is a manifold We have shown

this only for the binary action case but we conjecture that it is considerably more general Note that the previous argument can be extended to the case when the domain of f

u is bounded for any c = (c c) with 0 lt c lt c define X

c X by Xc = D0[c c] Then X

c is a (m+1)-dimensional 1manifold with boundary and G

c = f (0) Xc is a 1-dimensional manifold with boundary

u Now pick M gt 0 so that for all p 2 A sup

ikl

|uikl

(p)| M Define () maxip

i (ui(p i

) ) a

= e ()M and b

= e ()M Then it follows that for any (p ) 2 G that

pik() middot M log () middot M =) a

pi0 p

ik b

pi0

pi0

But

X

pik b

pi0 =) 1 p

i0 = pik b

mi

pi0

k2Mi

=) pi0 1(b

mi + 1)

and

pik a

pi0 =) p

ik a

(b

mi + 1) = c

Since a

1 it follows that for all 0 k mi (ie including k = 0) the above inequality holds

Define

W = (p ) 2 X pik c

for all i 2 N 0 k mi

Thus we have shown that G W X Similarly Gc W X

c

In other words the EQRE graph

can only ldquoexitrdquo X (ie have p on the boundary D) at the minimum and maximum values of We wish to show that in generic games the EQRE graph can be used to make a unique selection

of a Nash equilibrium To do this we use two facts First as we have already shown in Lemma

7 there is a unique solution for sufficiently large Second we will show that this branch of the

correspondence converges to a unique Nash equilibrium as goes to 0 We have now shown enough to prove part (ii) of Theorem 5 that for almost all there are an

odd number of EQREs From the argument setting c = (c c) we have shown that Gc is a compact

1-dimensional manifold with boundary which for large enough c has a unique intersection with

52

= c Any such manifold has a finite number of connected compact components each of which

must have an even number of boundary points We have also shown that any boundary point must be at = c or = c Since there is exactly one solution at = c there must be an odd number of solutions at c

We now show part (iii) of Theorem 5 that as 0 the branch B of the manifold that passes through c (for sufficiently high c) converges to a unique Nash equilibrium

Lemma 8 Let be chosen so that p() is a singleton Then for almost all games as 0 the macrbranch B of the manifold that passes through converges to a unique Nash equilibrium

Proof It follows from the arguments above that for almost all games there exists a decreasing macrsequence

i

with i lt for all i such that if we define c

i = (i

) Xi = X

c

i and Gi = G

c

i G

macr macr1 G is a 1-dimensional manifold with a unique point say (p ) for which = and a unique macrconnected branch B that passes through (p )

2 Gi is a compact 1-dimensional manifold with boundary which has a finite number of connected

components

macr3 Letting Bi be the connected branch of G

i which begins at (p ) it follows that Bi is a

compact connected 1-dimensional manifold with boundary for all i which has a unique intersection say (p

i

i

) with = i

Now for any i define A

i = (p ) 2 B lt i

and Ai to be the closure of the projection of

Ai onto D Then A

i

is a decreasing sequence of sets We show that for almost all games i

Ai

must be a unique point First of all since D is compact and each Ai is closed and nonempty

i

Ai

cannot be empty Suppose by way of contradiction that i

Ai contains two distinct points Since

generic games contain a finite number of Nash equilibria we may assume that the game defined

by u has a finite number of Nash equilibria By Theorem 4 any point in i

Ai must be a Nash

equilibrium But if p and q are both in i

Ai

then we can construct a sequence (pi

i

) B with p2i 1 p p2i q

i 0 (ie ldquooddsrdquo approach p and ldquoevensrdquo approach q) and a

homeomorphism R B satisfying several properties In particular since B is connected it is also path connected (Guillemin and Pollack p 38 exercise 3) so can be constructed to satisfy

macr(0) = (p ) (i) = (pi

i

) and [i 1 i] [i i+1] = (i) with [i i+1] a compact 1-dimensional manifold with boundary

We have constructed an infinite sequence of compact manifolds with boundary each of whose projection on D connects a point near p with a point near q Further for any

i

at most

a finite number of these manifolds intersect with Xi (since B X

i is a compact 1-dimensional manifold with boundary which can consist of at most a finite number of components) It follows

that any separating hyperplane Ht = p 2 D p middot (p q) = t between q and p must have

a nonempty intersection with i

Ai (by compactness of H

t

) However since there are an infinity of

53

such hyperplanes there are infinitely many points in i

Ai and hence an infinite number of Nash

equilibria which is a contradiction

95 Non-concavity

0 1 2 3 4 5 6 7 -01

-005

0

005

01 θ = 002 θ = 00235 θ = 0025

λ

Figure 10 Example of non-concave objective with two pure solutions

This example due to Weibull et al [2007] results in a first-stage objective that is not-concave

and emits two pure solutions Suppose vi = (1 1 0 0) ie there are J1(i) and J0(i) high-

and low-quality alternatives respectively satisfying J0(i) + J1(i) = J(i) Denote the fraction of J1(i)high quality alternatives by = J(i) In this case it is easy to show that the benefit of precision

is given by e

b( vi

) = 1 + e

Figure 10 plots the first-stage objective b( vi

) 2 for = 001 and 2 002 00235 0025 For

54 For some critical intermediate 025 (high costs) i 03 For = 002 (low costs) = 0

i

0235 there are two optimal selections of i and the agent is willing to mix over them with 0

arbitrary probability That the optimal precision can be discontinuous in is somewhat surprising

and complicates our analysis The non-concavity of the objective in this example must depend on there being more than two

alternatives by our analysis in Section 31 To see this note that b(middot vi

) only depends on vi through

1 Thus when = the benefit function b(middot vi

) is as if there are only two alternatives vi = (1 0)2

eor vi = 1 In this case b( v

i

) = which is the same as b( vi

)|fv

i

=1 from (5)1+e

54

gt 196 Generalized matching pennies with q 2

For games of the form of Figure 3 the following is the counterpart to Theorem 6 for the case in gt 1which q 2 The proof is essentially the same as that for the Theorem 6 and hence omitted

The result is accompanied by Figure 11 which is analogous to Figure 6 Rather than redefining

every object for this case we hope their definitions are clear from the figure

1 gt 1Theorem Let p 2 q and r be such that all LQRE p q satisfy maxp 1 pmaxq 12 e gt 1q lt 092 (I) The EQRE set crosses the LQRE set at 1 q if p (II) the EQRE 1+e

2 2 1 2set cannot cross the LQRE set u+ [ u or a+ [ a at any other points except p q1 and p q2

(and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 5 and 6) 1(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p1) q gt q 0 for EQRE p q and LQRE p q

0 2 1(ii) If p q1 does not exist (Cases 7 and 8)

(a) for all p 2 (1 p) q gt q 0 for EQRE p q and LQRE p q

0 2 2(iii) If p q2 exists (Cases 5 and 7)

2(a) for all q 2 (q q) p gt p 0 for EQRE p q and LQRE p

0 q and

(b) for all q 2 (1 q2) p lt p 0 for EQRE p q and LQRE p

0 q2

2(iv) If p q2 does not exist (Cases 6 and 8) (a) for all q 2 (1 q) p gt p

0 for EQRE p q and LQRE p 0 q2

55

Case 5 Case 6 1 1

p 05

0 0

lowastlowast lowastlowast q

a minus

1 1

p q lowastlowast lowastlowast

p q 1 1

p 05

p

p q 1 1

u minus

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

u minus

a +

u +

1 2

1 2

1 lowastlowast 2 q

a minus

0 05 1 0 05 1 q q

Case 7 Case 8

u minus

p q lowastlowast lowastlowast

a +

u + 1 2

1 2

1 lowastlowast 2 q

a minus

1

u minus

lowastlowast lowastlowast

1 1

p q

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

a minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

gt 1Figure 11 The four cases of equilibrium sets (q )2

56

97 Details of Data and Procedures from Studies of 22 Games

Study Game Data

p q N Notes

Rounds (R) Subject pairs (I) Feedback

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

50 36 yes 50 24 yes 50 24 yes 50 12 yes

1 0079 0690 9600 200 48 yes 2 0217 0527 9600 200 48 yes 3 0163 0793 9600 200 48 yes 4 0286 0736 9600 200 48 yes 5 0327 0664 9600 200 48 yes

SC 2008 6 7

0445 0596 9600 0141 0564 4800

200 48 yes 200 24 yes

8 0250 0586 4800 200 24 yes 9 0254 0827 4800 200 24 yes 10 0366 0699 4800 200 24 yes 11 0331 0652 4800 200 24 yes 12 0439 0604 4800 200 24 yes

Ochs 1995 2 3

0595 0258 448 0542 0336 516

56 8 yes 64 8 yes

GH 2001 AMP RA

0960 0160 25 0080 0800 25

1 25 no 1 25 no

NS 2002 NS 0458 0390 6720 60 112 yes

Table 6 Details of Data and Procedures from Studies of 2 2 Games

971 MPW 2000 (McKelvey et al [2000])

For any one game each subject takes an action 50 times against randomly matched opponents with

feedback Subjects play multiple games and maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $010 (year 2000 Dollars)

A B C D L R

U 9 0 0 1

D 0 1 1 0

L R

U 9 0 0 4

D 0 4 1 0

L R

U 36 0 0 4

D 0 4 4 0

L R

U 4 0 0 1

D 0 1 1 0

972 SC 2008 ( Selten and Chmura [2008])

Each subject plays just one game and takes an action 200 times against randomly matched oppo-nents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of

57

the experiment The exchange rate is sect0016 (year 2008 Euros) and the experiment took place in

Germany

1

2

3

4

L R L R L R L R

10 8 9 9

0 18 10 8

8 5 6 7

0 13 9 4

10 4 7 7

0 14 8 6

9 2 5 6

0 11 7 4

U U U U

D D D D

5

6

7

8

L R L R L R L R

8 1 4 5

0 9 7 2

8 0 3 5

1 7 7 1

14 8 9 9

4 22 10 12

11 5 6 7

3 16 9 7

U U U U

D D D D

9

10

11

12

L R L R L R L R

U 13 4 7 7

3 17 8 9

U 11 2 5 6

2 13 7 6

U 10 1 4 5

2 11 7 4

U

D D D D

10 0 3 5

3 9 7 3

973 Ochs 1995 (Ochs [1995])

Each subject plays only one of the two games Those who play game 2 (game 3) take an action

56 (64) times against randomly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment Following McKelvey and Palfrey

[1995] who note that ldquoThe subjects in the Ochs experiments were paid using a lottery procedurerdquo we convert the payoffs described in Ochs [1995] to those in the matrices below before estimation The exchange rate is $001 (expected 1982 Dollars)

2 3 L R L R

11141 0 0 11141

0 11141 01238 0

11141 0 0 11141

0 11141 02785 0

U U

D D

974 GH 2001 (Goeree and Holt [2001])

Each subject takes just one action against an anonymous opponent and the exchange rate is $001

(year 2001 dollars) ldquoAMPrdquo refers to ldquoasymmetric matching penniesrdquo and ldquoRArdquo refers to ldquoreversed

asymmetryrdquo

AMP RA L R L R

320 40 40 80

40 80 80 40

44 40 40 80

40 80 80 40

U U

D D

58

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)

P P

High When = 0 = (2vj 2v

k

)J2 = 2 (J 1) =j vk J2 gt 0 Byv

j k=6 j k 6

continuity

v

j gt 0 for sufficiently low which is guaranteed for sufficiently high in which case

( ) lt 0 and the result follows (v

j ) h i h i

2b( v)

2b( v)Low Note that v

j +

2 c 00 ( ) lt v

j +

2 and hence it is v

j v

j

enough to show the latter is negative for sufficiently large Reproducing the the expression 2 2 (v

m

+v

k

+v

l

)

2b( v)

P m vm

Pkl

(vm v

k

+2vk

v

l 2vm

v

l

)e = (9) and noting that terms with the largest ex-

2 (P

v

k )3 k e

ponents dominate it is easy to show that as 1

2

2b( v) 2 2micro (micro 2micro + 2) (micro )+ middot evj 2v

j 2 + ((2 micro2 + 2micro 2micro) + micro(micro 2 + 2micro 2micro))

3 (micro )= 22 + 2micro + (micro micro 2) middot e

where micro lt is the second largest component of v This approaches 0 but is negative for all lt 1 ( )Hence for sufficiently large which is guaranteed by sufficiently low lt 0 and the result (v

j )

follows

94 Proof of Theorem 5

To facilitate comparison between this proof and the analogous one for standard LQRE (Theorem 3

in McKelvey and Palfrey [1995]) we freely borrow from McKelvey and Palfrey [1995] We highlight only the most interesting differences

While parts of the proof only go through for games in which J(i) = 2 for all i we use general notation to emphasize what can be said for normal form games and where the proof breaks down The proof uses the following lemma which holds for normal form games and may be of general interest

Lemma 7 For sufficiently large p() is a singleton

Proof For any vi 2 RJ(i)

i (vi ) is a singleton for sufficiently large by Lemma 2 Since for any

finite game the set of all possible expected payoffs u(p) 2 RJ p 2 A is compact this can be extended to games for sufficiently high (u

i

(p i

) ) is a singleton for all i and p i 2 A

i

i

Thus for sufficiently large the mapping A A with components

i (ui

(p i

))uik (p

i

)e ik

(p ) = P

J(i) i (ui

(p i

))uil

(p i

) l=1 e

is a continuous function For this proof we always take sufficiently large for this to be the case Then for any (p ) 2 (p() ) if and only if p is a fixed point of

49

We will show that for sufficiently large is a contraction mapping and hence has a unique

fixed point To prove the result we use three basic facts from McKelvey and Palfrey [1995] We

also use a fourth fact which is unique to our problem For convenience we define utility differences

(ie across actions given the opponentsrsquo behavior) by uikl

(a i

) = ui

(aik

a i

) ui

(ail

a i

) and P

uikl

(p i

) = a u

ikl

(a i

)p i

(a i

) i

2A i

Fact 1 For any p q 2 A0 (the interior of A) maxik

|pik q

ik

| maxikl

|pik

pil q

ik

qil

|

xFact 2 Since the derivative of e at x = 0 is 1 then for all D gt 1 there is a such that x1whenever |x1| |x2| lt |e ex2 | D middot |x1 x2|

Fact 3 There is an M gt 0 such that maxikl

|uikl

(p i

) uikl

(q i

)| M middotmaxik

|pik q

ik

|

Fact 4 For any p q 2 A and

maxj max

j (uj (q

i (ui(q i

) )uikl

(q i

)|

j ) ) middotmaxikl

|uikl

(p i

) uikl

(q i

)|

maxikl

|i (ui(p

i

) )uikl

(p i

)

j (uj (p

j ) )

i

Since 0 Fact 4 is obvious whenever uikl

(p 6i

) for some ikl And when uikl

(p i

) =i

) = uikl

(q i (ui(p

i

) ) = i (ui(q

i

) ) = 0 for all i in which case both sides of the i

) for all ikluikl

(q

inequality equal 0 and it is satisfied trivially Pick any D gt 1 and let be defined as in Fact 2 and M defined as in Fact 3 Pick sufficiently

large such that maxip

i (ui(p

i

) ) satisfies

1 uikl

(p i

) lt for all i k l and any p and

2 lt 1(D middotM)

Letting k middot k represent the sup norm and writing middotD middotM we have that for any p q 2 A

k (p) (q)k = maxik

| ik

(p) ik

(q)|

maxikl

| ik

(p) il

(p) ik

(q) il

(q)| (u

i

(p i

))uikl

(p i

) (ui

(q i

))uikl

(qi i i

)|= maxikl

|e e

i

) )uikl

(p i (ui(q

i

) )uikl

(q D middotmaxikl

|i (ui(p

i

)i

)|

j ) ) j ) ) maxj max

j (uj (p j (uj (q middotD middotmax

ikl

|uikl

(p i

) uikl

(q i

)|

middotD middotM middotmaxik

|pik q

ik

|

= middot kp qk

where lt 1 The steps follow respectively by the definition of k middotk Fact 1 the definition of Fact

i (vi ) is2 Fact 4 Fact 3 and the definition of and the definitions of and k middot k Finally since

50

decreasing in for any fixed vi

k (p) (q)k middot k p qk for any gt Hence is a contraction

mapping and has a unique fixed point for sufficiently large

Fix game form and let U represent the set of all possible payoff functions for For each P

i 2 N define mi = J(i) 1 and M

i = 1 mi

Write N i = N i and m =

i2N mi

Define

pi0 = p

iJ(i) and X

Di = p

i 2 Rm

i pik 0 for all k 2 M

i and pik 1

k2Mi

Write Di 0 for the interior of D

i

Using the identity pi0 = p

iJ(i) = 1 P

k2Mi pik

a mixed strategy Q

pi 2 A

i can be identified by the first mi components ie by a vector in D

i

Write D = Di

i2N

and D0 for the interior of D A vector p = (p1 pn

) 2 D is referred to as a mixed profile For any u 2 U as in the proof of Lemma 7 define utility differences u

ikl

(a i

) = ui

(aik

a i

)P

ui

(ail

a i

) uik0(a

i

) = uikJ(i)(a

i

) and uikl

(p) = uikl

(p i

) = a

i

2A i uikl

(a i

)p i

(a i

) Define

X = D0 (0 1 ) and correspondence F X U Rm with components given by for any i 2 N

and k 2 Mi

pik

Fik

(p u) = i (ui10(p) u

im

i

0(p) )uik0(p) ln (11)

pi0

Note that we have redefined as a function of payoff differences relative to the utility of action i

aiJ(i) which is without loss The use of correspondence (11) is unique to our problem

For any u 2 U write fu

(p ) = F (p u) For any given u 2 U p() = p fu

(p ) = 01and the EQRE graph G = (p() ) 0 lt lt 1 is given by G = f (0) where 0 is the

u m-dimensional vector of zeros If RJ(i) 1 (0 1 ) [0 1 ) is single-valued then f

u is a i

1continuous function Then f (0) would be a closed set (preimage of a closed set) and hence Gu

would be upper hemi-continuous Hence if the game is such that is single-valued then part (i) i

of Theorem 5 is satisfied This is true in the case where J(i) = 2 for all i by Lemma 2 but it is not

true in general (see Appendix 95) When J(i) = 2 for all i replace (11) with F X U Rn (m = n in this case) defined by

pi1

Fi

(p u) = i (|ui10(p)| )ui10(p) ln

pi0

1for all i Again for any given u 2 U the EQRE correspondence is given by G = f (0) We will u

show that for any (p ) 2 X F is surjective Note that for any 2 [0 1 )

pi1

Fi

(p u) = () i (|ui10(p)| )ui10(p) ln =

pi0

pi1()

i (|ui10(p)| )ui10(p) = + ln( ) pi0

where 2 ( 1 1 ) depends on p Hence F is surjective for fixed (p ) 2 X if i (|ui10(p)| )ui10(p)

U ( 1 1 ) is surjective This follows from part (iv) of Theorem 3 after noting that ui10(p) U

51

( 1 1) is surjective for fixed p Hence F (p u) is transversal to any submanifold of Rn By the

transversality theorem (Guillemin and Pollack 1974 68) for almost all u 2 U fu

(p ) = F (p u) 1is transversal to 0 It follows (Guillemin and Pollack 1974 28) that f (0) is a manifold of codi-

u

mension n = m or dimension 1 for almost all u 2 U These particular theorems from differential topology were invoked previously in McKelvey and Palfrey [1998] but not in McKelvey and Palfrey

[1995] The rest of the proof which differs from the analogue in McKelvey and Palfrey [1995] in only

very minor ways goes through for any normal form game in which G is a manifold We have shown

this only for the binary action case but we conjecture that it is considerably more general Note that the previous argument can be extended to the case when the domain of f

u is bounded for any c = (c c) with 0 lt c lt c define X

c X by Xc = D0[c c] Then X

c is a (m+1)-dimensional 1manifold with boundary and G

c = f (0) Xc is a 1-dimensional manifold with boundary

u Now pick M gt 0 so that for all p 2 A sup

ikl

|uikl

(p)| M Define () maxip

i (ui(p i

) ) a

= e ()M and b

= e ()M Then it follows that for any (p ) 2 G that

pik() middot M log () middot M =) a

pi0 p

ik b

pi0

pi0

But

X

pik b

pi0 =) 1 p

i0 = pik b

mi

pi0

k2Mi

=) pi0 1(b

mi + 1)

and

pik a

pi0 =) p

ik a

(b

mi + 1) = c

Since a

1 it follows that for all 0 k mi (ie including k = 0) the above inequality holds

Define

W = (p ) 2 X pik c

for all i 2 N 0 k mi

Thus we have shown that G W X Similarly Gc W X

c

In other words the EQRE graph

can only ldquoexitrdquo X (ie have p on the boundary D) at the minimum and maximum values of We wish to show that in generic games the EQRE graph can be used to make a unique selection

of a Nash equilibrium To do this we use two facts First as we have already shown in Lemma

7 there is a unique solution for sufficiently large Second we will show that this branch of the

correspondence converges to a unique Nash equilibrium as goes to 0 We have now shown enough to prove part (ii) of Theorem 5 that for almost all there are an

odd number of EQREs From the argument setting c = (c c) we have shown that Gc is a compact

1-dimensional manifold with boundary which for large enough c has a unique intersection with

52

= c Any such manifold has a finite number of connected compact components each of which

must have an even number of boundary points We have also shown that any boundary point must be at = c or = c Since there is exactly one solution at = c there must be an odd number of solutions at c

We now show part (iii) of Theorem 5 that as 0 the branch B of the manifold that passes through c (for sufficiently high c) converges to a unique Nash equilibrium

Lemma 8 Let be chosen so that p() is a singleton Then for almost all games as 0 the macrbranch B of the manifold that passes through converges to a unique Nash equilibrium

Proof It follows from the arguments above that for almost all games there exists a decreasing macrsequence

i

with i lt for all i such that if we define c

i = (i

) Xi = X

c

i and Gi = G

c

i G

macr macr1 G is a 1-dimensional manifold with a unique point say (p ) for which = and a unique macrconnected branch B that passes through (p )

2 Gi is a compact 1-dimensional manifold with boundary which has a finite number of connected

components

macr3 Letting Bi be the connected branch of G

i which begins at (p ) it follows that Bi is a

compact connected 1-dimensional manifold with boundary for all i which has a unique intersection say (p

i

i

) with = i

Now for any i define A

i = (p ) 2 B lt i

and Ai to be the closure of the projection of

Ai onto D Then A

i

is a decreasing sequence of sets We show that for almost all games i

Ai

must be a unique point First of all since D is compact and each Ai is closed and nonempty

i

Ai

cannot be empty Suppose by way of contradiction that i

Ai contains two distinct points Since

generic games contain a finite number of Nash equilibria we may assume that the game defined

by u has a finite number of Nash equilibria By Theorem 4 any point in i

Ai must be a Nash

equilibrium But if p and q are both in i

Ai

then we can construct a sequence (pi

i

) B with p2i 1 p p2i q

i 0 (ie ldquooddsrdquo approach p and ldquoevensrdquo approach q) and a

homeomorphism R B satisfying several properties In particular since B is connected it is also path connected (Guillemin and Pollack p 38 exercise 3) so can be constructed to satisfy

macr(0) = (p ) (i) = (pi

i

) and [i 1 i] [i i+1] = (i) with [i i+1] a compact 1-dimensional manifold with boundary

We have constructed an infinite sequence of compact manifolds with boundary each of whose projection on D connects a point near p with a point near q Further for any

i

at most

a finite number of these manifolds intersect with Xi (since B X

i is a compact 1-dimensional manifold with boundary which can consist of at most a finite number of components) It follows

that any separating hyperplane Ht = p 2 D p middot (p q) = t between q and p must have

a nonempty intersection with i

Ai (by compactness of H

t

) However since there are an infinity of

53

such hyperplanes there are infinitely many points in i

Ai and hence an infinite number of Nash

equilibria which is a contradiction

95 Non-concavity

0 1 2 3 4 5 6 7 -01

-005

0

005

01 θ = 002 θ = 00235 θ = 0025

λ

Figure 10 Example of non-concave objective with two pure solutions

This example due to Weibull et al [2007] results in a first-stage objective that is not-concave

and emits two pure solutions Suppose vi = (1 1 0 0) ie there are J1(i) and J0(i) high-

and low-quality alternatives respectively satisfying J0(i) + J1(i) = J(i) Denote the fraction of J1(i)high quality alternatives by = J(i) In this case it is easy to show that the benefit of precision

is given by e

b( vi

) = 1 + e

Figure 10 plots the first-stage objective b( vi

) 2 for = 001 and 2 002 00235 0025 For

54 For some critical intermediate 025 (high costs) i 03 For = 002 (low costs) = 0

i

0235 there are two optimal selections of i and the agent is willing to mix over them with 0

arbitrary probability That the optimal precision can be discontinuous in is somewhat surprising

and complicates our analysis The non-concavity of the objective in this example must depend on there being more than two

alternatives by our analysis in Section 31 To see this note that b(middot vi

) only depends on vi through

1 Thus when = the benefit function b(middot vi

) is as if there are only two alternatives vi = (1 0)2

eor vi = 1 In this case b( v

i

) = which is the same as b( vi

)|fv

i

=1 from (5)1+e

54

gt 196 Generalized matching pennies with q 2

For games of the form of Figure 3 the following is the counterpart to Theorem 6 for the case in gt 1which q 2 The proof is essentially the same as that for the Theorem 6 and hence omitted

The result is accompanied by Figure 11 which is analogous to Figure 6 Rather than redefining

every object for this case we hope their definitions are clear from the figure

1 gt 1Theorem Let p 2 q and r be such that all LQRE p q satisfy maxp 1 pmaxq 12 e gt 1q lt 092 (I) The EQRE set crosses the LQRE set at 1 q if p (II) the EQRE 1+e

2 2 1 2set cannot cross the LQRE set u+ [ u or a+ [ a at any other points except p q1 and p q2

(and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 5 and 6) 1(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p1) q gt q 0 for EQRE p q and LQRE p q

0 2 1(ii) If p q1 does not exist (Cases 7 and 8)

(a) for all p 2 (1 p) q gt q 0 for EQRE p q and LQRE p q

0 2 2(iii) If p q2 exists (Cases 5 and 7)

2(a) for all q 2 (q q) p gt p 0 for EQRE p q and LQRE p

0 q and

(b) for all q 2 (1 q2) p lt p 0 for EQRE p q and LQRE p

0 q2

2(iv) If p q2 does not exist (Cases 6 and 8) (a) for all q 2 (1 q) p gt p

0 for EQRE p q and LQRE p 0 q2

55

Case 5 Case 6 1 1

p 05

0 0

lowastlowast lowastlowast q

a minus

1 1

p q lowastlowast lowastlowast

p q 1 1

p 05

p

p q 1 1

u minus

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

u minus

a +

u +

1 2

1 2

1 lowastlowast 2 q

a minus

0 05 1 0 05 1 q q

Case 7 Case 8

u minus

p q lowastlowast lowastlowast

a +

u + 1 2

1 2

1 lowastlowast 2 q

a minus

1

u minus

lowastlowast lowastlowast

1 1

p q

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

a minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

gt 1Figure 11 The four cases of equilibrium sets (q )2

56

97 Details of Data and Procedures from Studies of 22 Games

Study Game Data

p q N Notes

Rounds (R) Subject pairs (I) Feedback

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

50 36 yes 50 24 yes 50 24 yes 50 12 yes

1 0079 0690 9600 200 48 yes 2 0217 0527 9600 200 48 yes 3 0163 0793 9600 200 48 yes 4 0286 0736 9600 200 48 yes 5 0327 0664 9600 200 48 yes

SC 2008 6 7

0445 0596 9600 0141 0564 4800

200 48 yes 200 24 yes

8 0250 0586 4800 200 24 yes 9 0254 0827 4800 200 24 yes 10 0366 0699 4800 200 24 yes 11 0331 0652 4800 200 24 yes 12 0439 0604 4800 200 24 yes

Ochs 1995 2 3

0595 0258 448 0542 0336 516

56 8 yes 64 8 yes

GH 2001 AMP RA

0960 0160 25 0080 0800 25

1 25 no 1 25 no

NS 2002 NS 0458 0390 6720 60 112 yes

Table 6 Details of Data and Procedures from Studies of 2 2 Games

971 MPW 2000 (McKelvey et al [2000])

For any one game each subject takes an action 50 times against randomly matched opponents with

feedback Subjects play multiple games and maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $010 (year 2000 Dollars)

A B C D L R

U 9 0 0 1

D 0 1 1 0

L R

U 9 0 0 4

D 0 4 1 0

L R

U 36 0 0 4

D 0 4 4 0

L R

U 4 0 0 1

D 0 1 1 0

972 SC 2008 ( Selten and Chmura [2008])

Each subject plays just one game and takes an action 200 times against randomly matched oppo-nents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of

57

the experiment The exchange rate is sect0016 (year 2008 Euros) and the experiment took place in

Germany

1

2

3

4

L R L R L R L R

10 8 9 9

0 18 10 8

8 5 6 7

0 13 9 4

10 4 7 7

0 14 8 6

9 2 5 6

0 11 7 4

U U U U

D D D D

5

6

7

8

L R L R L R L R

8 1 4 5

0 9 7 2

8 0 3 5

1 7 7 1

14 8 9 9

4 22 10 12

11 5 6 7

3 16 9 7

U U U U

D D D D

9

10

11

12

L R L R L R L R

U 13 4 7 7

3 17 8 9

U 11 2 5 6

2 13 7 6

U 10 1 4 5

2 11 7 4

U

D D D D

10 0 3 5

3 9 7 3

973 Ochs 1995 (Ochs [1995])

Each subject plays only one of the two games Those who play game 2 (game 3) take an action

56 (64) times against randomly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment Following McKelvey and Palfrey

[1995] who note that ldquoThe subjects in the Ochs experiments were paid using a lottery procedurerdquo we convert the payoffs described in Ochs [1995] to those in the matrices below before estimation The exchange rate is $001 (expected 1982 Dollars)

2 3 L R L R

11141 0 0 11141

0 11141 01238 0

11141 0 0 11141

0 11141 02785 0

U U

D D

974 GH 2001 (Goeree and Holt [2001])

Each subject takes just one action against an anonymous opponent and the exchange rate is $001

(year 2001 dollars) ldquoAMPrdquo refers to ldquoasymmetric matching penniesrdquo and ldquoRArdquo refers to ldquoreversed

asymmetryrdquo

AMP RA L R L R

320 40 40 80

40 80 80 40

44 40 40 80

40 80 80 40

U U

D D

58

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)

We will show that for sufficiently large is a contraction mapping and hence has a unique

fixed point To prove the result we use three basic facts from McKelvey and Palfrey [1995] We

also use a fourth fact which is unique to our problem For convenience we define utility differences

(ie across actions given the opponentsrsquo behavior) by uikl

(a i

) = ui

(aik

a i

) ui

(ail

a i

) and P

uikl

(p i

) = a u

ikl

(a i

)p i

(a i

) i

2A i

Fact 1 For any p q 2 A0 (the interior of A) maxik

|pik q

ik

| maxikl

|pik

pil q

ik

qil

|

xFact 2 Since the derivative of e at x = 0 is 1 then for all D gt 1 there is a such that x1whenever |x1| |x2| lt |e ex2 | D middot |x1 x2|

Fact 3 There is an M gt 0 such that maxikl

|uikl

(p i

) uikl

(q i

)| M middotmaxik

|pik q

ik

|

Fact 4 For any p q 2 A and

maxj max

j (uj (q

i (ui(q i

) )uikl

(q i

)|

j ) ) middotmaxikl

|uikl

(p i

) uikl

(q i

)|

maxikl

|i (ui(p

i

) )uikl

(p i

)

j (uj (p

j ) )

i

Since 0 Fact 4 is obvious whenever uikl

(p 6i

) for some ikl And when uikl

(p i

) =i

) = uikl

(q i (ui(p

i

) ) = i (ui(q

i

) ) = 0 for all i in which case both sides of the i

) for all ikluikl

(q

inequality equal 0 and it is satisfied trivially Pick any D gt 1 and let be defined as in Fact 2 and M defined as in Fact 3 Pick sufficiently

large such that maxip

i (ui(p

i

) ) satisfies

1 uikl

(p i

) lt for all i k l and any p and

2 lt 1(D middotM)

Letting k middot k represent the sup norm and writing middotD middotM we have that for any p q 2 A

k (p) (q)k = maxik

| ik

(p) ik

(q)|

maxikl

| ik

(p) il

(p) ik

(q) il

(q)| (u

i

(p i

))uikl

(p i

) (ui

(q i

))uikl

(qi i i

)|= maxikl

|e e

i

) )uikl

(p i (ui(q

i

) )uikl

(q D middotmaxikl

|i (ui(p

i

)i

)|

j ) ) j ) ) maxj max

j (uj (p j (uj (q middotD middotmax

ikl

|uikl

(p i

) uikl

(q i

)|

middotD middotM middotmaxik

|pik q

ik

|

= middot kp qk

where lt 1 The steps follow respectively by the definition of k middotk Fact 1 the definition of Fact

i (vi ) is2 Fact 4 Fact 3 and the definition of and the definitions of and k middot k Finally since

50

decreasing in for any fixed vi

k (p) (q)k middot k p qk for any gt Hence is a contraction

mapping and has a unique fixed point for sufficiently large

Fix game form and let U represent the set of all possible payoff functions for For each P

i 2 N define mi = J(i) 1 and M

i = 1 mi

Write N i = N i and m =

i2N mi

Define

pi0 = p

iJ(i) and X

Di = p

i 2 Rm

i pik 0 for all k 2 M

i and pik 1

k2Mi

Write Di 0 for the interior of D

i

Using the identity pi0 = p

iJ(i) = 1 P

k2Mi pik

a mixed strategy Q

pi 2 A

i can be identified by the first mi components ie by a vector in D

i

Write D = Di

i2N

and D0 for the interior of D A vector p = (p1 pn

) 2 D is referred to as a mixed profile For any u 2 U as in the proof of Lemma 7 define utility differences u

ikl

(a i

) = ui

(aik

a i

)P

ui

(ail

a i

) uik0(a

i

) = uikJ(i)(a

i

) and uikl

(p) = uikl

(p i

) = a

i

2A i uikl

(a i

)p i

(a i

) Define

X = D0 (0 1 ) and correspondence F X U Rm with components given by for any i 2 N

and k 2 Mi

pik

Fik

(p u) = i (ui10(p) u

im

i

0(p) )uik0(p) ln (11)

pi0

Note that we have redefined as a function of payoff differences relative to the utility of action i

aiJ(i) which is without loss The use of correspondence (11) is unique to our problem

For any u 2 U write fu

(p ) = F (p u) For any given u 2 U p() = p fu

(p ) = 01and the EQRE graph G = (p() ) 0 lt lt 1 is given by G = f (0) where 0 is the

u m-dimensional vector of zeros If RJ(i) 1 (0 1 ) [0 1 ) is single-valued then f

u is a i

1continuous function Then f (0) would be a closed set (preimage of a closed set) and hence Gu

would be upper hemi-continuous Hence if the game is such that is single-valued then part (i) i

of Theorem 5 is satisfied This is true in the case where J(i) = 2 for all i by Lemma 2 but it is not

true in general (see Appendix 95) When J(i) = 2 for all i replace (11) with F X U Rn (m = n in this case) defined by

pi1

Fi

(p u) = i (|ui10(p)| )ui10(p) ln

pi0

1for all i Again for any given u 2 U the EQRE correspondence is given by G = f (0) We will u

show that for any (p ) 2 X F is surjective Note that for any 2 [0 1 )

pi1

Fi

(p u) = () i (|ui10(p)| )ui10(p) ln =

pi0

pi1()

i (|ui10(p)| )ui10(p) = + ln( ) pi0

where 2 ( 1 1 ) depends on p Hence F is surjective for fixed (p ) 2 X if i (|ui10(p)| )ui10(p)

U ( 1 1 ) is surjective This follows from part (iv) of Theorem 3 after noting that ui10(p) U

51

( 1 1) is surjective for fixed p Hence F (p u) is transversal to any submanifold of Rn By the

transversality theorem (Guillemin and Pollack 1974 68) for almost all u 2 U fu

(p ) = F (p u) 1is transversal to 0 It follows (Guillemin and Pollack 1974 28) that f (0) is a manifold of codi-

u

mension n = m or dimension 1 for almost all u 2 U These particular theorems from differential topology were invoked previously in McKelvey and Palfrey [1998] but not in McKelvey and Palfrey

[1995] The rest of the proof which differs from the analogue in McKelvey and Palfrey [1995] in only

very minor ways goes through for any normal form game in which G is a manifold We have shown

this only for the binary action case but we conjecture that it is considerably more general Note that the previous argument can be extended to the case when the domain of f

u is bounded for any c = (c c) with 0 lt c lt c define X

c X by Xc = D0[c c] Then X

c is a (m+1)-dimensional 1manifold with boundary and G

c = f (0) Xc is a 1-dimensional manifold with boundary

u Now pick M gt 0 so that for all p 2 A sup

ikl

|uikl

(p)| M Define () maxip

i (ui(p i

) ) a

= e ()M and b

= e ()M Then it follows that for any (p ) 2 G that

pik() middot M log () middot M =) a

pi0 p

ik b

pi0

pi0

But

X

pik b

pi0 =) 1 p

i0 = pik b

mi

pi0

k2Mi

=) pi0 1(b

mi + 1)

and

pik a

pi0 =) p

ik a

(b

mi + 1) = c

Since a

1 it follows that for all 0 k mi (ie including k = 0) the above inequality holds

Define

W = (p ) 2 X pik c

for all i 2 N 0 k mi

Thus we have shown that G W X Similarly Gc W X

c

In other words the EQRE graph

can only ldquoexitrdquo X (ie have p on the boundary D) at the minimum and maximum values of We wish to show that in generic games the EQRE graph can be used to make a unique selection

of a Nash equilibrium To do this we use two facts First as we have already shown in Lemma

7 there is a unique solution for sufficiently large Second we will show that this branch of the

correspondence converges to a unique Nash equilibrium as goes to 0 We have now shown enough to prove part (ii) of Theorem 5 that for almost all there are an

odd number of EQREs From the argument setting c = (c c) we have shown that Gc is a compact

1-dimensional manifold with boundary which for large enough c has a unique intersection with

52

= c Any such manifold has a finite number of connected compact components each of which

must have an even number of boundary points We have also shown that any boundary point must be at = c or = c Since there is exactly one solution at = c there must be an odd number of solutions at c

We now show part (iii) of Theorem 5 that as 0 the branch B of the manifold that passes through c (for sufficiently high c) converges to a unique Nash equilibrium

Lemma 8 Let be chosen so that p() is a singleton Then for almost all games as 0 the macrbranch B of the manifold that passes through converges to a unique Nash equilibrium

Proof It follows from the arguments above that for almost all games there exists a decreasing macrsequence

i

with i lt for all i such that if we define c

i = (i

) Xi = X

c

i and Gi = G

c

i G

macr macr1 G is a 1-dimensional manifold with a unique point say (p ) for which = and a unique macrconnected branch B that passes through (p )

2 Gi is a compact 1-dimensional manifold with boundary which has a finite number of connected

components

macr3 Letting Bi be the connected branch of G

i which begins at (p ) it follows that Bi is a

compact connected 1-dimensional manifold with boundary for all i which has a unique intersection say (p

i

i

) with = i

Now for any i define A

i = (p ) 2 B lt i

and Ai to be the closure of the projection of

Ai onto D Then A

i

is a decreasing sequence of sets We show that for almost all games i

Ai

must be a unique point First of all since D is compact and each Ai is closed and nonempty

i

Ai

cannot be empty Suppose by way of contradiction that i

Ai contains two distinct points Since

generic games contain a finite number of Nash equilibria we may assume that the game defined

by u has a finite number of Nash equilibria By Theorem 4 any point in i

Ai must be a Nash

equilibrium But if p and q are both in i

Ai

then we can construct a sequence (pi

i

) B with p2i 1 p p2i q

i 0 (ie ldquooddsrdquo approach p and ldquoevensrdquo approach q) and a

homeomorphism R B satisfying several properties In particular since B is connected it is also path connected (Guillemin and Pollack p 38 exercise 3) so can be constructed to satisfy

macr(0) = (p ) (i) = (pi

i

) and [i 1 i] [i i+1] = (i) with [i i+1] a compact 1-dimensional manifold with boundary

We have constructed an infinite sequence of compact manifolds with boundary each of whose projection on D connects a point near p with a point near q Further for any

i

at most

a finite number of these manifolds intersect with Xi (since B X

i is a compact 1-dimensional manifold with boundary which can consist of at most a finite number of components) It follows

that any separating hyperplane Ht = p 2 D p middot (p q) = t between q and p must have

a nonempty intersection with i

Ai (by compactness of H

t

) However since there are an infinity of

53

such hyperplanes there are infinitely many points in i

Ai and hence an infinite number of Nash

equilibria which is a contradiction

95 Non-concavity

0 1 2 3 4 5 6 7 -01

-005

0

005

01 θ = 002 θ = 00235 θ = 0025

λ

Figure 10 Example of non-concave objective with two pure solutions

This example due to Weibull et al [2007] results in a first-stage objective that is not-concave

and emits two pure solutions Suppose vi = (1 1 0 0) ie there are J1(i) and J0(i) high-

and low-quality alternatives respectively satisfying J0(i) + J1(i) = J(i) Denote the fraction of J1(i)high quality alternatives by = J(i) In this case it is easy to show that the benefit of precision

is given by e

b( vi

) = 1 + e

Figure 10 plots the first-stage objective b( vi

) 2 for = 001 and 2 002 00235 0025 For

54 For some critical intermediate 025 (high costs) i 03 For = 002 (low costs) = 0

i

0235 there are two optimal selections of i and the agent is willing to mix over them with 0

arbitrary probability That the optimal precision can be discontinuous in is somewhat surprising

and complicates our analysis The non-concavity of the objective in this example must depend on there being more than two

alternatives by our analysis in Section 31 To see this note that b(middot vi

) only depends on vi through

1 Thus when = the benefit function b(middot vi

) is as if there are only two alternatives vi = (1 0)2

eor vi = 1 In this case b( v

i

) = which is the same as b( vi

)|fv

i

=1 from (5)1+e

54

gt 196 Generalized matching pennies with q 2

For games of the form of Figure 3 the following is the counterpart to Theorem 6 for the case in gt 1which q 2 The proof is essentially the same as that for the Theorem 6 and hence omitted

The result is accompanied by Figure 11 which is analogous to Figure 6 Rather than redefining

every object for this case we hope their definitions are clear from the figure

1 gt 1Theorem Let p 2 q and r be such that all LQRE p q satisfy maxp 1 pmaxq 12 e gt 1q lt 092 (I) The EQRE set crosses the LQRE set at 1 q if p (II) the EQRE 1+e

2 2 1 2set cannot cross the LQRE set u+ [ u or a+ [ a at any other points except p q1 and p q2

(and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 5 and 6) 1(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p1) q gt q 0 for EQRE p q and LQRE p q

0 2 1(ii) If p q1 does not exist (Cases 7 and 8)

(a) for all p 2 (1 p) q gt q 0 for EQRE p q and LQRE p q

0 2 2(iii) If p q2 exists (Cases 5 and 7)

2(a) for all q 2 (q q) p gt p 0 for EQRE p q and LQRE p

0 q and

(b) for all q 2 (1 q2) p lt p 0 for EQRE p q and LQRE p

0 q2

2(iv) If p q2 does not exist (Cases 6 and 8) (a) for all q 2 (1 q) p gt p

0 for EQRE p q and LQRE p 0 q2

55

Case 5 Case 6 1 1

p 05

0 0

lowastlowast lowastlowast q

a minus

1 1

p q lowastlowast lowastlowast

p q 1 1

p 05

p

p q 1 1

u minus

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

u minus

a +

u +

1 2

1 2

1 lowastlowast 2 q

a minus

0 05 1 0 05 1 q q

Case 7 Case 8

u minus

p q lowastlowast lowastlowast

a +

u + 1 2

1 2

1 lowastlowast 2 q

a minus

1

u minus

lowastlowast lowastlowast

1 1

p q

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

a minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

gt 1Figure 11 The four cases of equilibrium sets (q )2

56

97 Details of Data and Procedures from Studies of 22 Games

Study Game Data

p q N Notes

Rounds (R) Subject pairs (I) Feedback

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

50 36 yes 50 24 yes 50 24 yes 50 12 yes

1 0079 0690 9600 200 48 yes 2 0217 0527 9600 200 48 yes 3 0163 0793 9600 200 48 yes 4 0286 0736 9600 200 48 yes 5 0327 0664 9600 200 48 yes

SC 2008 6 7

0445 0596 9600 0141 0564 4800

200 48 yes 200 24 yes

8 0250 0586 4800 200 24 yes 9 0254 0827 4800 200 24 yes 10 0366 0699 4800 200 24 yes 11 0331 0652 4800 200 24 yes 12 0439 0604 4800 200 24 yes

Ochs 1995 2 3

0595 0258 448 0542 0336 516

56 8 yes 64 8 yes

GH 2001 AMP RA

0960 0160 25 0080 0800 25

1 25 no 1 25 no

NS 2002 NS 0458 0390 6720 60 112 yes

Table 6 Details of Data and Procedures from Studies of 2 2 Games

971 MPW 2000 (McKelvey et al [2000])

For any one game each subject takes an action 50 times against randomly matched opponents with

feedback Subjects play multiple games and maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $010 (year 2000 Dollars)

A B C D L R

U 9 0 0 1

D 0 1 1 0

L R

U 9 0 0 4

D 0 4 1 0

L R

U 36 0 0 4

D 0 4 4 0

L R

U 4 0 0 1

D 0 1 1 0

972 SC 2008 ( Selten and Chmura [2008])

Each subject plays just one game and takes an action 200 times against randomly matched oppo-nents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of

57

the experiment The exchange rate is sect0016 (year 2008 Euros) and the experiment took place in

Germany

1

2

3

4

L R L R L R L R

10 8 9 9

0 18 10 8

8 5 6 7

0 13 9 4

10 4 7 7

0 14 8 6

9 2 5 6

0 11 7 4

U U U U

D D D D

5

6

7

8

L R L R L R L R

8 1 4 5

0 9 7 2

8 0 3 5

1 7 7 1

14 8 9 9

4 22 10 12

11 5 6 7

3 16 9 7

U U U U

D D D D

9

10

11

12

L R L R L R L R

U 13 4 7 7

3 17 8 9

U 11 2 5 6

2 13 7 6

U 10 1 4 5

2 11 7 4

U

D D D D

10 0 3 5

3 9 7 3

973 Ochs 1995 (Ochs [1995])

Each subject plays only one of the two games Those who play game 2 (game 3) take an action

56 (64) times against randomly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment Following McKelvey and Palfrey

[1995] who note that ldquoThe subjects in the Ochs experiments were paid using a lottery procedurerdquo we convert the payoffs described in Ochs [1995] to those in the matrices below before estimation The exchange rate is $001 (expected 1982 Dollars)

2 3 L R L R

11141 0 0 11141

0 11141 01238 0

11141 0 0 11141

0 11141 02785 0

U U

D D

974 GH 2001 (Goeree and Holt [2001])

Each subject takes just one action against an anonymous opponent and the exchange rate is $001

(year 2001 dollars) ldquoAMPrdquo refers to ldquoasymmetric matching penniesrdquo and ldquoRArdquo refers to ldquoreversed

asymmetryrdquo

AMP RA L R L R

320 40 40 80

40 80 80 40

44 40 40 80

40 80 80 40

U U

D D

58

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)

decreasing in for any fixed vi

k (p) (q)k middot k p qk for any gt Hence is a contraction

mapping and has a unique fixed point for sufficiently large

Fix game form and let U represent the set of all possible payoff functions for For each P

i 2 N define mi = J(i) 1 and M

i = 1 mi

Write N i = N i and m =

i2N mi

Define

pi0 = p

iJ(i) and X

Di = p

i 2 Rm

i pik 0 for all k 2 M

i and pik 1

k2Mi

Write Di 0 for the interior of D

i

Using the identity pi0 = p

iJ(i) = 1 P

k2Mi pik

a mixed strategy Q

pi 2 A

i can be identified by the first mi components ie by a vector in D

i

Write D = Di

i2N

and D0 for the interior of D A vector p = (p1 pn

) 2 D is referred to as a mixed profile For any u 2 U as in the proof of Lemma 7 define utility differences u

ikl

(a i

) = ui

(aik

a i

)P

ui

(ail

a i

) uik0(a

i

) = uikJ(i)(a

i

) and uikl

(p) = uikl

(p i

) = a

i

2A i uikl

(a i

)p i

(a i

) Define

X = D0 (0 1 ) and correspondence F X U Rm with components given by for any i 2 N

and k 2 Mi

pik

Fik

(p u) = i (ui10(p) u

im

i

0(p) )uik0(p) ln (11)

pi0

Note that we have redefined as a function of payoff differences relative to the utility of action i

aiJ(i) which is without loss The use of correspondence (11) is unique to our problem

For any u 2 U write fu

(p ) = F (p u) For any given u 2 U p() = p fu

(p ) = 01and the EQRE graph G = (p() ) 0 lt lt 1 is given by G = f (0) where 0 is the

u m-dimensional vector of zeros If RJ(i) 1 (0 1 ) [0 1 ) is single-valued then f

u is a i

1continuous function Then f (0) would be a closed set (preimage of a closed set) and hence Gu

would be upper hemi-continuous Hence if the game is such that is single-valued then part (i) i

of Theorem 5 is satisfied This is true in the case where J(i) = 2 for all i by Lemma 2 but it is not

true in general (see Appendix 95) When J(i) = 2 for all i replace (11) with F X U Rn (m = n in this case) defined by

pi1

Fi

(p u) = i (|ui10(p)| )ui10(p) ln

pi0

1for all i Again for any given u 2 U the EQRE correspondence is given by G = f (0) We will u

show that for any (p ) 2 X F is surjective Note that for any 2 [0 1 )

pi1

Fi

(p u) = () i (|ui10(p)| )ui10(p) ln =

pi0

pi1()

i (|ui10(p)| )ui10(p) = + ln( ) pi0

where 2 ( 1 1 ) depends on p Hence F is surjective for fixed (p ) 2 X if i (|ui10(p)| )ui10(p)

U ( 1 1 ) is surjective This follows from part (iv) of Theorem 3 after noting that ui10(p) U

51

( 1 1) is surjective for fixed p Hence F (p u) is transversal to any submanifold of Rn By the

transversality theorem (Guillemin and Pollack 1974 68) for almost all u 2 U fu

(p ) = F (p u) 1is transversal to 0 It follows (Guillemin and Pollack 1974 28) that f (0) is a manifold of codi-

u

mension n = m or dimension 1 for almost all u 2 U These particular theorems from differential topology were invoked previously in McKelvey and Palfrey [1998] but not in McKelvey and Palfrey

[1995] The rest of the proof which differs from the analogue in McKelvey and Palfrey [1995] in only

very minor ways goes through for any normal form game in which G is a manifold We have shown

this only for the binary action case but we conjecture that it is considerably more general Note that the previous argument can be extended to the case when the domain of f

u is bounded for any c = (c c) with 0 lt c lt c define X

c X by Xc = D0[c c] Then X

c is a (m+1)-dimensional 1manifold with boundary and G

c = f (0) Xc is a 1-dimensional manifold with boundary

u Now pick M gt 0 so that for all p 2 A sup

ikl

|uikl

(p)| M Define () maxip

i (ui(p i

) ) a

= e ()M and b

= e ()M Then it follows that for any (p ) 2 G that

pik() middot M log () middot M =) a

pi0 p

ik b

pi0

pi0

But

X

pik b

pi0 =) 1 p

i0 = pik b

mi

pi0

k2Mi

=) pi0 1(b

mi + 1)

and

pik a

pi0 =) p

ik a

(b

mi + 1) = c

Since a

1 it follows that for all 0 k mi (ie including k = 0) the above inequality holds

Define

W = (p ) 2 X pik c

for all i 2 N 0 k mi

Thus we have shown that G W X Similarly Gc W X

c

In other words the EQRE graph

can only ldquoexitrdquo X (ie have p on the boundary D) at the minimum and maximum values of We wish to show that in generic games the EQRE graph can be used to make a unique selection

of a Nash equilibrium To do this we use two facts First as we have already shown in Lemma

7 there is a unique solution for sufficiently large Second we will show that this branch of the

correspondence converges to a unique Nash equilibrium as goes to 0 We have now shown enough to prove part (ii) of Theorem 5 that for almost all there are an

odd number of EQREs From the argument setting c = (c c) we have shown that Gc is a compact

1-dimensional manifold with boundary which for large enough c has a unique intersection with

52

= c Any such manifold has a finite number of connected compact components each of which

must have an even number of boundary points We have also shown that any boundary point must be at = c or = c Since there is exactly one solution at = c there must be an odd number of solutions at c

We now show part (iii) of Theorem 5 that as 0 the branch B of the manifold that passes through c (for sufficiently high c) converges to a unique Nash equilibrium

Lemma 8 Let be chosen so that p() is a singleton Then for almost all games as 0 the macrbranch B of the manifold that passes through converges to a unique Nash equilibrium

Proof It follows from the arguments above that for almost all games there exists a decreasing macrsequence

i

with i lt for all i such that if we define c

i = (i

) Xi = X

c

i and Gi = G

c

i G

macr macr1 G is a 1-dimensional manifold with a unique point say (p ) for which = and a unique macrconnected branch B that passes through (p )

2 Gi is a compact 1-dimensional manifold with boundary which has a finite number of connected

components

macr3 Letting Bi be the connected branch of G

i which begins at (p ) it follows that Bi is a

compact connected 1-dimensional manifold with boundary for all i which has a unique intersection say (p

i

i

) with = i

Now for any i define A

i = (p ) 2 B lt i

and Ai to be the closure of the projection of

Ai onto D Then A

i

is a decreasing sequence of sets We show that for almost all games i

Ai

must be a unique point First of all since D is compact and each Ai is closed and nonempty

i

Ai

cannot be empty Suppose by way of contradiction that i

Ai contains two distinct points Since

generic games contain a finite number of Nash equilibria we may assume that the game defined

by u has a finite number of Nash equilibria By Theorem 4 any point in i

Ai must be a Nash

equilibrium But if p and q are both in i

Ai

then we can construct a sequence (pi

i

) B with p2i 1 p p2i q

i 0 (ie ldquooddsrdquo approach p and ldquoevensrdquo approach q) and a

homeomorphism R B satisfying several properties In particular since B is connected it is also path connected (Guillemin and Pollack p 38 exercise 3) so can be constructed to satisfy

macr(0) = (p ) (i) = (pi

i

) and [i 1 i] [i i+1] = (i) with [i i+1] a compact 1-dimensional manifold with boundary

We have constructed an infinite sequence of compact manifolds with boundary each of whose projection on D connects a point near p with a point near q Further for any

i

at most

a finite number of these manifolds intersect with Xi (since B X

i is a compact 1-dimensional manifold with boundary which can consist of at most a finite number of components) It follows

that any separating hyperplane Ht = p 2 D p middot (p q) = t between q and p must have

a nonempty intersection with i

Ai (by compactness of H

t

) However since there are an infinity of

53

such hyperplanes there are infinitely many points in i

Ai and hence an infinite number of Nash

equilibria which is a contradiction

95 Non-concavity

0 1 2 3 4 5 6 7 -01

-005

0

005

01 θ = 002 θ = 00235 θ = 0025

λ

Figure 10 Example of non-concave objective with two pure solutions

This example due to Weibull et al [2007] results in a first-stage objective that is not-concave

and emits two pure solutions Suppose vi = (1 1 0 0) ie there are J1(i) and J0(i) high-

and low-quality alternatives respectively satisfying J0(i) + J1(i) = J(i) Denote the fraction of J1(i)high quality alternatives by = J(i) In this case it is easy to show that the benefit of precision

is given by e

b( vi

) = 1 + e

Figure 10 plots the first-stage objective b( vi

) 2 for = 001 and 2 002 00235 0025 For

54 For some critical intermediate 025 (high costs) i 03 For = 002 (low costs) = 0

i

0235 there are two optimal selections of i and the agent is willing to mix over them with 0

arbitrary probability That the optimal precision can be discontinuous in is somewhat surprising

and complicates our analysis The non-concavity of the objective in this example must depend on there being more than two

alternatives by our analysis in Section 31 To see this note that b(middot vi

) only depends on vi through

1 Thus when = the benefit function b(middot vi

) is as if there are only two alternatives vi = (1 0)2

eor vi = 1 In this case b( v

i

) = which is the same as b( vi

)|fv

i

=1 from (5)1+e

54

gt 196 Generalized matching pennies with q 2

For games of the form of Figure 3 the following is the counterpart to Theorem 6 for the case in gt 1which q 2 The proof is essentially the same as that for the Theorem 6 and hence omitted

The result is accompanied by Figure 11 which is analogous to Figure 6 Rather than redefining

every object for this case we hope their definitions are clear from the figure

1 gt 1Theorem Let p 2 q and r be such that all LQRE p q satisfy maxp 1 pmaxq 12 e gt 1q lt 092 (I) The EQRE set crosses the LQRE set at 1 q if p (II) the EQRE 1+e

2 2 1 2set cannot cross the LQRE set u+ [ u or a+ [ a at any other points except p q1 and p q2

(and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 5 and 6) 1(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p1) q gt q 0 for EQRE p q and LQRE p q

0 2 1(ii) If p q1 does not exist (Cases 7 and 8)

(a) for all p 2 (1 p) q gt q 0 for EQRE p q and LQRE p q

0 2 2(iii) If p q2 exists (Cases 5 and 7)

2(a) for all q 2 (q q) p gt p 0 for EQRE p q and LQRE p

0 q and

(b) for all q 2 (1 q2) p lt p 0 for EQRE p q and LQRE p

0 q2

2(iv) If p q2 does not exist (Cases 6 and 8) (a) for all q 2 (1 q) p gt p

0 for EQRE p q and LQRE p 0 q2

55

Case 5 Case 6 1 1

p 05

0 0

lowastlowast lowastlowast q

a minus

1 1

p q lowastlowast lowastlowast

p q 1 1

p 05

p

p q 1 1

u minus

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

u minus

a +

u +

1 2

1 2

1 lowastlowast 2 q

a minus

0 05 1 0 05 1 q q

Case 7 Case 8

u minus

p q lowastlowast lowastlowast

a +

u + 1 2

1 2

1 lowastlowast 2 q

a minus

1

u minus

lowastlowast lowastlowast

1 1

p q

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

a minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

gt 1Figure 11 The four cases of equilibrium sets (q )2

56

97 Details of Data and Procedures from Studies of 22 Games

Study Game Data

p q N Notes

Rounds (R) Subject pairs (I) Feedback

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

50 36 yes 50 24 yes 50 24 yes 50 12 yes

1 0079 0690 9600 200 48 yes 2 0217 0527 9600 200 48 yes 3 0163 0793 9600 200 48 yes 4 0286 0736 9600 200 48 yes 5 0327 0664 9600 200 48 yes

SC 2008 6 7

0445 0596 9600 0141 0564 4800

200 48 yes 200 24 yes

8 0250 0586 4800 200 24 yes 9 0254 0827 4800 200 24 yes 10 0366 0699 4800 200 24 yes 11 0331 0652 4800 200 24 yes 12 0439 0604 4800 200 24 yes

Ochs 1995 2 3

0595 0258 448 0542 0336 516

56 8 yes 64 8 yes

GH 2001 AMP RA

0960 0160 25 0080 0800 25

1 25 no 1 25 no

NS 2002 NS 0458 0390 6720 60 112 yes

Table 6 Details of Data and Procedures from Studies of 2 2 Games

971 MPW 2000 (McKelvey et al [2000])

For any one game each subject takes an action 50 times against randomly matched opponents with

feedback Subjects play multiple games and maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $010 (year 2000 Dollars)

A B C D L R

U 9 0 0 1

D 0 1 1 0

L R

U 9 0 0 4

D 0 4 1 0

L R

U 36 0 0 4

D 0 4 4 0

L R

U 4 0 0 1

D 0 1 1 0

972 SC 2008 ( Selten and Chmura [2008])

Each subject plays just one game and takes an action 200 times against randomly matched oppo-nents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of

57

the experiment The exchange rate is sect0016 (year 2008 Euros) and the experiment took place in

Germany

1

2

3

4

L R L R L R L R

10 8 9 9

0 18 10 8

8 5 6 7

0 13 9 4

10 4 7 7

0 14 8 6

9 2 5 6

0 11 7 4

U U U U

D D D D

5

6

7

8

L R L R L R L R

8 1 4 5

0 9 7 2

8 0 3 5

1 7 7 1

14 8 9 9

4 22 10 12

11 5 6 7

3 16 9 7

U U U U

D D D D

9

10

11

12

L R L R L R L R

U 13 4 7 7

3 17 8 9

U 11 2 5 6

2 13 7 6

U 10 1 4 5

2 11 7 4

U

D D D D

10 0 3 5

3 9 7 3

973 Ochs 1995 (Ochs [1995])

Each subject plays only one of the two games Those who play game 2 (game 3) take an action

56 (64) times against randomly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment Following McKelvey and Palfrey

[1995] who note that ldquoThe subjects in the Ochs experiments were paid using a lottery procedurerdquo we convert the payoffs described in Ochs [1995] to those in the matrices below before estimation The exchange rate is $001 (expected 1982 Dollars)

2 3 L R L R

11141 0 0 11141

0 11141 01238 0

11141 0 0 11141

0 11141 02785 0

U U

D D

974 GH 2001 (Goeree and Holt [2001])

Each subject takes just one action against an anonymous opponent and the exchange rate is $001

(year 2001 dollars) ldquoAMPrdquo refers to ldquoasymmetric matching penniesrdquo and ldquoRArdquo refers to ldquoreversed

asymmetryrdquo

AMP RA L R L R

320 40 40 80

40 80 80 40

44 40 40 80

40 80 80 40

U U

D D

58

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)

( 1 1) is surjective for fixed p Hence F (p u) is transversal to any submanifold of Rn By the

transversality theorem (Guillemin and Pollack 1974 68) for almost all u 2 U fu

(p ) = F (p u) 1is transversal to 0 It follows (Guillemin and Pollack 1974 28) that f (0) is a manifold of codi-

u

mension n = m or dimension 1 for almost all u 2 U These particular theorems from differential topology were invoked previously in McKelvey and Palfrey [1998] but not in McKelvey and Palfrey

[1995] The rest of the proof which differs from the analogue in McKelvey and Palfrey [1995] in only

very minor ways goes through for any normal form game in which G is a manifold We have shown

this only for the binary action case but we conjecture that it is considerably more general Note that the previous argument can be extended to the case when the domain of f

u is bounded for any c = (c c) with 0 lt c lt c define X

c X by Xc = D0[c c] Then X

c is a (m+1)-dimensional 1manifold with boundary and G

c = f (0) Xc is a 1-dimensional manifold with boundary

u Now pick M gt 0 so that for all p 2 A sup

ikl

|uikl

(p)| M Define () maxip

i (ui(p i

) ) a

= e ()M and b

= e ()M Then it follows that for any (p ) 2 G that

pik() middot M log () middot M =) a

pi0 p

ik b

pi0

pi0

But

X

pik b

pi0 =) 1 p

i0 = pik b

mi

pi0

k2Mi

=) pi0 1(b

mi + 1)

and

pik a

pi0 =) p

ik a

(b

mi + 1) = c

Since a

1 it follows that for all 0 k mi (ie including k = 0) the above inequality holds

Define

W = (p ) 2 X pik c

for all i 2 N 0 k mi

Thus we have shown that G W X Similarly Gc W X

c

In other words the EQRE graph

can only ldquoexitrdquo X (ie have p on the boundary D) at the minimum and maximum values of We wish to show that in generic games the EQRE graph can be used to make a unique selection

of a Nash equilibrium To do this we use two facts First as we have already shown in Lemma

7 there is a unique solution for sufficiently large Second we will show that this branch of the

correspondence converges to a unique Nash equilibrium as goes to 0 We have now shown enough to prove part (ii) of Theorem 5 that for almost all there are an

odd number of EQREs From the argument setting c = (c c) we have shown that Gc is a compact

1-dimensional manifold with boundary which for large enough c has a unique intersection with

52

= c Any such manifold has a finite number of connected compact components each of which

must have an even number of boundary points We have also shown that any boundary point must be at = c or = c Since there is exactly one solution at = c there must be an odd number of solutions at c

We now show part (iii) of Theorem 5 that as 0 the branch B of the manifold that passes through c (for sufficiently high c) converges to a unique Nash equilibrium

Lemma 8 Let be chosen so that p() is a singleton Then for almost all games as 0 the macrbranch B of the manifold that passes through converges to a unique Nash equilibrium

Proof It follows from the arguments above that for almost all games there exists a decreasing macrsequence

i

with i lt for all i such that if we define c

i = (i

) Xi = X

c

i and Gi = G

c

i G

macr macr1 G is a 1-dimensional manifold with a unique point say (p ) for which = and a unique macrconnected branch B that passes through (p )

2 Gi is a compact 1-dimensional manifold with boundary which has a finite number of connected

components

macr3 Letting Bi be the connected branch of G

i which begins at (p ) it follows that Bi is a

compact connected 1-dimensional manifold with boundary for all i which has a unique intersection say (p

i

i

) with = i

Now for any i define A

i = (p ) 2 B lt i

and Ai to be the closure of the projection of

Ai onto D Then A

i

is a decreasing sequence of sets We show that for almost all games i

Ai

must be a unique point First of all since D is compact and each Ai is closed and nonempty

i

Ai

cannot be empty Suppose by way of contradiction that i

Ai contains two distinct points Since

generic games contain a finite number of Nash equilibria we may assume that the game defined

by u has a finite number of Nash equilibria By Theorem 4 any point in i

Ai must be a Nash

equilibrium But if p and q are both in i

Ai

then we can construct a sequence (pi

i

) B with p2i 1 p p2i q

i 0 (ie ldquooddsrdquo approach p and ldquoevensrdquo approach q) and a

homeomorphism R B satisfying several properties In particular since B is connected it is also path connected (Guillemin and Pollack p 38 exercise 3) so can be constructed to satisfy

macr(0) = (p ) (i) = (pi

i

) and [i 1 i] [i i+1] = (i) with [i i+1] a compact 1-dimensional manifold with boundary

We have constructed an infinite sequence of compact manifolds with boundary each of whose projection on D connects a point near p with a point near q Further for any

i

at most

a finite number of these manifolds intersect with Xi (since B X

i is a compact 1-dimensional manifold with boundary which can consist of at most a finite number of components) It follows

that any separating hyperplane Ht = p 2 D p middot (p q) = t between q and p must have

a nonempty intersection with i

Ai (by compactness of H

t

) However since there are an infinity of

53

such hyperplanes there are infinitely many points in i

Ai and hence an infinite number of Nash

equilibria which is a contradiction

95 Non-concavity

0 1 2 3 4 5 6 7 -01

-005

0

005

01 θ = 002 θ = 00235 θ = 0025

λ

Figure 10 Example of non-concave objective with two pure solutions

This example due to Weibull et al [2007] results in a first-stage objective that is not-concave

and emits two pure solutions Suppose vi = (1 1 0 0) ie there are J1(i) and J0(i) high-

and low-quality alternatives respectively satisfying J0(i) + J1(i) = J(i) Denote the fraction of J1(i)high quality alternatives by = J(i) In this case it is easy to show that the benefit of precision

is given by e

b( vi

) = 1 + e

Figure 10 plots the first-stage objective b( vi

) 2 for = 001 and 2 002 00235 0025 For

54 For some critical intermediate 025 (high costs) i 03 For = 002 (low costs) = 0

i

0235 there are two optimal selections of i and the agent is willing to mix over them with 0

arbitrary probability That the optimal precision can be discontinuous in is somewhat surprising

and complicates our analysis The non-concavity of the objective in this example must depend on there being more than two

alternatives by our analysis in Section 31 To see this note that b(middot vi

) only depends on vi through

1 Thus when = the benefit function b(middot vi

) is as if there are only two alternatives vi = (1 0)2

eor vi = 1 In this case b( v

i

) = which is the same as b( vi

)|fv

i

=1 from (5)1+e

54

gt 196 Generalized matching pennies with q 2

For games of the form of Figure 3 the following is the counterpart to Theorem 6 for the case in gt 1which q 2 The proof is essentially the same as that for the Theorem 6 and hence omitted

The result is accompanied by Figure 11 which is analogous to Figure 6 Rather than redefining

every object for this case we hope their definitions are clear from the figure

1 gt 1Theorem Let p 2 q and r be such that all LQRE p q satisfy maxp 1 pmaxq 12 e gt 1q lt 092 (I) The EQRE set crosses the LQRE set at 1 q if p (II) the EQRE 1+e

2 2 1 2set cannot cross the LQRE set u+ [ u or a+ [ a at any other points except p q1 and p q2

(and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 5 and 6) 1(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p1) q gt q 0 for EQRE p q and LQRE p q

0 2 1(ii) If p q1 does not exist (Cases 7 and 8)

(a) for all p 2 (1 p) q gt q 0 for EQRE p q and LQRE p q

0 2 2(iii) If p q2 exists (Cases 5 and 7)

2(a) for all q 2 (q q) p gt p 0 for EQRE p q and LQRE p

0 q and

(b) for all q 2 (1 q2) p lt p 0 for EQRE p q and LQRE p

0 q2

2(iv) If p q2 does not exist (Cases 6 and 8) (a) for all q 2 (1 q) p gt p

0 for EQRE p q and LQRE p 0 q2

55

Case 5 Case 6 1 1

p 05

0 0

lowastlowast lowastlowast q

a minus

1 1

p q lowastlowast lowastlowast

p q 1 1

p 05

p

p q 1 1

u minus

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

u minus

a +

u +

1 2

1 2

1 lowastlowast 2 q

a minus

0 05 1 0 05 1 q q

Case 7 Case 8

u minus

p q lowastlowast lowastlowast

a +

u + 1 2

1 2

1 lowastlowast 2 q

a minus

1

u minus

lowastlowast lowastlowast

1 1

p q

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

a minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

gt 1Figure 11 The four cases of equilibrium sets (q )2

56

97 Details of Data and Procedures from Studies of 22 Games

Study Game Data

p q N Notes

Rounds (R) Subject pairs (I) Feedback

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

50 36 yes 50 24 yes 50 24 yes 50 12 yes

1 0079 0690 9600 200 48 yes 2 0217 0527 9600 200 48 yes 3 0163 0793 9600 200 48 yes 4 0286 0736 9600 200 48 yes 5 0327 0664 9600 200 48 yes

SC 2008 6 7

0445 0596 9600 0141 0564 4800

200 48 yes 200 24 yes

8 0250 0586 4800 200 24 yes 9 0254 0827 4800 200 24 yes 10 0366 0699 4800 200 24 yes 11 0331 0652 4800 200 24 yes 12 0439 0604 4800 200 24 yes

Ochs 1995 2 3

0595 0258 448 0542 0336 516

56 8 yes 64 8 yes

GH 2001 AMP RA

0960 0160 25 0080 0800 25

1 25 no 1 25 no

NS 2002 NS 0458 0390 6720 60 112 yes

Table 6 Details of Data and Procedures from Studies of 2 2 Games

971 MPW 2000 (McKelvey et al [2000])

For any one game each subject takes an action 50 times against randomly matched opponents with

feedback Subjects play multiple games and maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $010 (year 2000 Dollars)

A B C D L R

U 9 0 0 1

D 0 1 1 0

L R

U 9 0 0 4

D 0 4 1 0

L R

U 36 0 0 4

D 0 4 4 0

L R

U 4 0 0 1

D 0 1 1 0

972 SC 2008 ( Selten and Chmura [2008])

Each subject plays just one game and takes an action 200 times against randomly matched oppo-nents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of

57

the experiment The exchange rate is sect0016 (year 2008 Euros) and the experiment took place in

Germany

1

2

3

4

L R L R L R L R

10 8 9 9

0 18 10 8

8 5 6 7

0 13 9 4

10 4 7 7

0 14 8 6

9 2 5 6

0 11 7 4

U U U U

D D D D

5

6

7

8

L R L R L R L R

8 1 4 5

0 9 7 2

8 0 3 5

1 7 7 1

14 8 9 9

4 22 10 12

11 5 6 7

3 16 9 7

U U U U

D D D D

9

10

11

12

L R L R L R L R

U 13 4 7 7

3 17 8 9

U 11 2 5 6

2 13 7 6

U 10 1 4 5

2 11 7 4

U

D D D D

10 0 3 5

3 9 7 3

973 Ochs 1995 (Ochs [1995])

Each subject plays only one of the two games Those who play game 2 (game 3) take an action

56 (64) times against randomly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment Following McKelvey and Palfrey

[1995] who note that ldquoThe subjects in the Ochs experiments were paid using a lottery procedurerdquo we convert the payoffs described in Ochs [1995] to those in the matrices below before estimation The exchange rate is $001 (expected 1982 Dollars)

2 3 L R L R

11141 0 0 11141

0 11141 01238 0

11141 0 0 11141

0 11141 02785 0

U U

D D

974 GH 2001 (Goeree and Holt [2001])

Each subject takes just one action against an anonymous opponent and the exchange rate is $001

(year 2001 dollars) ldquoAMPrdquo refers to ldquoasymmetric matching penniesrdquo and ldquoRArdquo refers to ldquoreversed

asymmetryrdquo

AMP RA L R L R

320 40 40 80

40 80 80 40

44 40 40 80

40 80 80 40

U U

D D

58

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)

= c Any such manifold has a finite number of connected compact components each of which

must have an even number of boundary points We have also shown that any boundary point must be at = c or = c Since there is exactly one solution at = c there must be an odd number of solutions at c

We now show part (iii) of Theorem 5 that as 0 the branch B of the manifold that passes through c (for sufficiently high c) converges to a unique Nash equilibrium

Lemma 8 Let be chosen so that p() is a singleton Then for almost all games as 0 the macrbranch B of the manifold that passes through converges to a unique Nash equilibrium

Proof It follows from the arguments above that for almost all games there exists a decreasing macrsequence

i

with i lt for all i such that if we define c

i = (i

) Xi = X

c

i and Gi = G

c

i G

macr macr1 G is a 1-dimensional manifold with a unique point say (p ) for which = and a unique macrconnected branch B that passes through (p )

2 Gi is a compact 1-dimensional manifold with boundary which has a finite number of connected

components

macr3 Letting Bi be the connected branch of G

i which begins at (p ) it follows that Bi is a

compact connected 1-dimensional manifold with boundary for all i which has a unique intersection say (p

i

i

) with = i

Now for any i define A

i = (p ) 2 B lt i

and Ai to be the closure of the projection of

Ai onto D Then A

i

is a decreasing sequence of sets We show that for almost all games i

Ai

must be a unique point First of all since D is compact and each Ai is closed and nonempty

i

Ai

cannot be empty Suppose by way of contradiction that i

Ai contains two distinct points Since

generic games contain a finite number of Nash equilibria we may assume that the game defined

by u has a finite number of Nash equilibria By Theorem 4 any point in i

Ai must be a Nash

equilibrium But if p and q are both in i

Ai

then we can construct a sequence (pi

i

) B with p2i 1 p p2i q

i 0 (ie ldquooddsrdquo approach p and ldquoevensrdquo approach q) and a

homeomorphism R B satisfying several properties In particular since B is connected it is also path connected (Guillemin and Pollack p 38 exercise 3) so can be constructed to satisfy

macr(0) = (p ) (i) = (pi

i

) and [i 1 i] [i i+1] = (i) with [i i+1] a compact 1-dimensional manifold with boundary

We have constructed an infinite sequence of compact manifolds with boundary each of whose projection on D connects a point near p with a point near q Further for any

i

at most

a finite number of these manifolds intersect with Xi (since B X

i is a compact 1-dimensional manifold with boundary which can consist of at most a finite number of components) It follows

that any separating hyperplane Ht = p 2 D p middot (p q) = t between q and p must have

a nonempty intersection with i

Ai (by compactness of H

t

) However since there are an infinity of

53

such hyperplanes there are infinitely many points in i

Ai and hence an infinite number of Nash

equilibria which is a contradiction

95 Non-concavity

0 1 2 3 4 5 6 7 -01

-005

0

005

01 θ = 002 θ = 00235 θ = 0025

λ

Figure 10 Example of non-concave objective with two pure solutions

This example due to Weibull et al [2007] results in a first-stage objective that is not-concave

and emits two pure solutions Suppose vi = (1 1 0 0) ie there are J1(i) and J0(i) high-

and low-quality alternatives respectively satisfying J0(i) + J1(i) = J(i) Denote the fraction of J1(i)high quality alternatives by = J(i) In this case it is easy to show that the benefit of precision

is given by e

b( vi

) = 1 + e

Figure 10 plots the first-stage objective b( vi

) 2 for = 001 and 2 002 00235 0025 For

54 For some critical intermediate 025 (high costs) i 03 For = 002 (low costs) = 0

i

0235 there are two optimal selections of i and the agent is willing to mix over them with 0

arbitrary probability That the optimal precision can be discontinuous in is somewhat surprising

and complicates our analysis The non-concavity of the objective in this example must depend on there being more than two

alternatives by our analysis in Section 31 To see this note that b(middot vi

) only depends on vi through

1 Thus when = the benefit function b(middot vi

) is as if there are only two alternatives vi = (1 0)2

eor vi = 1 In this case b( v

i

) = which is the same as b( vi

)|fv

i

=1 from (5)1+e

54

gt 196 Generalized matching pennies with q 2

For games of the form of Figure 3 the following is the counterpart to Theorem 6 for the case in gt 1which q 2 The proof is essentially the same as that for the Theorem 6 and hence omitted

The result is accompanied by Figure 11 which is analogous to Figure 6 Rather than redefining

every object for this case we hope their definitions are clear from the figure

1 gt 1Theorem Let p 2 q and r be such that all LQRE p q satisfy maxp 1 pmaxq 12 e gt 1q lt 092 (I) The EQRE set crosses the LQRE set at 1 q if p (II) the EQRE 1+e

2 2 1 2set cannot cross the LQRE set u+ [ u or a+ [ a at any other points except p q1 and p q2

(and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 5 and 6) 1(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p1) q gt q 0 for EQRE p q and LQRE p q

0 2 1(ii) If p q1 does not exist (Cases 7 and 8)

(a) for all p 2 (1 p) q gt q 0 for EQRE p q and LQRE p q

0 2 2(iii) If p q2 exists (Cases 5 and 7)

2(a) for all q 2 (q q) p gt p 0 for EQRE p q and LQRE p

0 q and

(b) for all q 2 (1 q2) p lt p 0 for EQRE p q and LQRE p

0 q2

2(iv) If p q2 does not exist (Cases 6 and 8) (a) for all q 2 (1 q) p gt p

0 for EQRE p q and LQRE p 0 q2

55

Case 5 Case 6 1 1

p 05

0 0

lowastlowast lowastlowast q

a minus

1 1

p q lowastlowast lowastlowast

p q 1 1

p 05

p

p q 1 1

u minus

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

u minus

a +

u +

1 2

1 2

1 lowastlowast 2 q

a minus

0 05 1 0 05 1 q q

Case 7 Case 8

u minus

p q lowastlowast lowastlowast

a +

u + 1 2

1 2

1 lowastlowast 2 q

a minus

1

u minus

lowastlowast lowastlowast

1 1

p q

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

a minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

gt 1Figure 11 The four cases of equilibrium sets (q )2

56

97 Details of Data and Procedures from Studies of 22 Games

Study Game Data

p q N Notes

Rounds (R) Subject pairs (I) Feedback

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

50 36 yes 50 24 yes 50 24 yes 50 12 yes

1 0079 0690 9600 200 48 yes 2 0217 0527 9600 200 48 yes 3 0163 0793 9600 200 48 yes 4 0286 0736 9600 200 48 yes 5 0327 0664 9600 200 48 yes

SC 2008 6 7

0445 0596 9600 0141 0564 4800

200 48 yes 200 24 yes

8 0250 0586 4800 200 24 yes 9 0254 0827 4800 200 24 yes 10 0366 0699 4800 200 24 yes 11 0331 0652 4800 200 24 yes 12 0439 0604 4800 200 24 yes

Ochs 1995 2 3

0595 0258 448 0542 0336 516

56 8 yes 64 8 yes

GH 2001 AMP RA

0960 0160 25 0080 0800 25

1 25 no 1 25 no

NS 2002 NS 0458 0390 6720 60 112 yes

Table 6 Details of Data and Procedures from Studies of 2 2 Games

971 MPW 2000 (McKelvey et al [2000])

For any one game each subject takes an action 50 times against randomly matched opponents with

feedback Subjects play multiple games and maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $010 (year 2000 Dollars)

A B C D L R

U 9 0 0 1

D 0 1 1 0

L R

U 9 0 0 4

D 0 4 1 0

L R

U 36 0 0 4

D 0 4 4 0

L R

U 4 0 0 1

D 0 1 1 0

972 SC 2008 ( Selten and Chmura [2008])

Each subject plays just one game and takes an action 200 times against randomly matched oppo-nents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of

57

the experiment The exchange rate is sect0016 (year 2008 Euros) and the experiment took place in

Germany

1

2

3

4

L R L R L R L R

10 8 9 9

0 18 10 8

8 5 6 7

0 13 9 4

10 4 7 7

0 14 8 6

9 2 5 6

0 11 7 4

U U U U

D D D D

5

6

7

8

L R L R L R L R

8 1 4 5

0 9 7 2

8 0 3 5

1 7 7 1

14 8 9 9

4 22 10 12

11 5 6 7

3 16 9 7

U U U U

D D D D

9

10

11

12

L R L R L R L R

U 13 4 7 7

3 17 8 9

U 11 2 5 6

2 13 7 6

U 10 1 4 5

2 11 7 4

U

D D D D

10 0 3 5

3 9 7 3

973 Ochs 1995 (Ochs [1995])

Each subject plays only one of the two games Those who play game 2 (game 3) take an action

56 (64) times against randomly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment Following McKelvey and Palfrey

[1995] who note that ldquoThe subjects in the Ochs experiments were paid using a lottery procedurerdquo we convert the payoffs described in Ochs [1995] to those in the matrices below before estimation The exchange rate is $001 (expected 1982 Dollars)

2 3 L R L R

11141 0 0 11141

0 11141 01238 0

11141 0 0 11141

0 11141 02785 0

U U

D D

974 GH 2001 (Goeree and Holt [2001])

Each subject takes just one action against an anonymous opponent and the exchange rate is $001

(year 2001 dollars) ldquoAMPrdquo refers to ldquoasymmetric matching penniesrdquo and ldquoRArdquo refers to ldquoreversed

asymmetryrdquo

AMP RA L R L R

320 40 40 80

40 80 80 40

44 40 40 80

40 80 80 40

U U

D D

58

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)

such hyperplanes there are infinitely many points in i

Ai and hence an infinite number of Nash

equilibria which is a contradiction

95 Non-concavity

0 1 2 3 4 5 6 7 -01

-005

0

005

01 θ = 002 θ = 00235 θ = 0025

λ

Figure 10 Example of non-concave objective with two pure solutions

This example due to Weibull et al [2007] results in a first-stage objective that is not-concave

and emits two pure solutions Suppose vi = (1 1 0 0) ie there are J1(i) and J0(i) high-

and low-quality alternatives respectively satisfying J0(i) + J1(i) = J(i) Denote the fraction of J1(i)high quality alternatives by = J(i) In this case it is easy to show that the benefit of precision

is given by e

b( vi

) = 1 + e

Figure 10 plots the first-stage objective b( vi

) 2 for = 001 and 2 002 00235 0025 For

54 For some critical intermediate 025 (high costs) i 03 For = 002 (low costs) = 0

i

0235 there are two optimal selections of i and the agent is willing to mix over them with 0

arbitrary probability That the optimal precision can be discontinuous in is somewhat surprising

and complicates our analysis The non-concavity of the objective in this example must depend on there being more than two

alternatives by our analysis in Section 31 To see this note that b(middot vi

) only depends on vi through

1 Thus when = the benefit function b(middot vi

) is as if there are only two alternatives vi = (1 0)2

eor vi = 1 In this case b( v

i

) = which is the same as b( vi

)|fv

i

=1 from (5)1+e

54

gt 196 Generalized matching pennies with q 2

For games of the form of Figure 3 the following is the counterpart to Theorem 6 for the case in gt 1which q 2 The proof is essentially the same as that for the Theorem 6 and hence omitted

The result is accompanied by Figure 11 which is analogous to Figure 6 Rather than redefining

every object for this case we hope their definitions are clear from the figure

1 gt 1Theorem Let p 2 q and r be such that all LQRE p q satisfy maxp 1 pmaxq 12 e gt 1q lt 092 (I) The EQRE set crosses the LQRE set at 1 q if p (II) the EQRE 1+e

2 2 1 2set cannot cross the LQRE set u+ [ u or a+ [ a at any other points except p q1 and p q2

(and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 5 and 6) 1(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p1) q gt q 0 for EQRE p q and LQRE p q

0 2 1(ii) If p q1 does not exist (Cases 7 and 8)

(a) for all p 2 (1 p) q gt q 0 for EQRE p q and LQRE p q

0 2 2(iii) If p q2 exists (Cases 5 and 7)

2(a) for all q 2 (q q) p gt p 0 for EQRE p q and LQRE p

0 q and

(b) for all q 2 (1 q2) p lt p 0 for EQRE p q and LQRE p

0 q2

2(iv) If p q2 does not exist (Cases 6 and 8) (a) for all q 2 (1 q) p gt p

0 for EQRE p q and LQRE p 0 q2

55

Case 5 Case 6 1 1

p 05

0 0

lowastlowast lowastlowast q

a minus

1 1

p q lowastlowast lowastlowast

p q 1 1

p 05

p

p q 1 1

u minus

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

u minus

a +

u +

1 2

1 2

1 lowastlowast 2 q

a minus

0 05 1 0 05 1 q q

Case 7 Case 8

u minus

p q lowastlowast lowastlowast

a +

u + 1 2

1 2

1 lowastlowast 2 q

a minus

1

u minus

lowastlowast lowastlowast

1 1

p q

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

a minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

gt 1Figure 11 The four cases of equilibrium sets (q )2

56

97 Details of Data and Procedures from Studies of 22 Games

Study Game Data

p q N Notes

Rounds (R) Subject pairs (I) Feedback

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

50 36 yes 50 24 yes 50 24 yes 50 12 yes

1 0079 0690 9600 200 48 yes 2 0217 0527 9600 200 48 yes 3 0163 0793 9600 200 48 yes 4 0286 0736 9600 200 48 yes 5 0327 0664 9600 200 48 yes

SC 2008 6 7

0445 0596 9600 0141 0564 4800

200 48 yes 200 24 yes

8 0250 0586 4800 200 24 yes 9 0254 0827 4800 200 24 yes 10 0366 0699 4800 200 24 yes 11 0331 0652 4800 200 24 yes 12 0439 0604 4800 200 24 yes

Ochs 1995 2 3

0595 0258 448 0542 0336 516

56 8 yes 64 8 yes

GH 2001 AMP RA

0960 0160 25 0080 0800 25

1 25 no 1 25 no

NS 2002 NS 0458 0390 6720 60 112 yes

Table 6 Details of Data and Procedures from Studies of 2 2 Games

971 MPW 2000 (McKelvey et al [2000])

For any one game each subject takes an action 50 times against randomly matched opponents with

feedback Subjects play multiple games and maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $010 (year 2000 Dollars)

A B C D L R

U 9 0 0 1

D 0 1 1 0

L R

U 9 0 0 4

D 0 4 1 0

L R

U 36 0 0 4

D 0 4 4 0

L R

U 4 0 0 1

D 0 1 1 0

972 SC 2008 ( Selten and Chmura [2008])

Each subject plays just one game and takes an action 200 times against randomly matched oppo-nents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of

57

the experiment The exchange rate is sect0016 (year 2008 Euros) and the experiment took place in

Germany

1

2

3

4

L R L R L R L R

10 8 9 9

0 18 10 8

8 5 6 7

0 13 9 4

10 4 7 7

0 14 8 6

9 2 5 6

0 11 7 4

U U U U

D D D D

5

6

7

8

L R L R L R L R

8 1 4 5

0 9 7 2

8 0 3 5

1 7 7 1

14 8 9 9

4 22 10 12

11 5 6 7

3 16 9 7

U U U U

D D D D

9

10

11

12

L R L R L R L R

U 13 4 7 7

3 17 8 9

U 11 2 5 6

2 13 7 6

U 10 1 4 5

2 11 7 4

U

D D D D

10 0 3 5

3 9 7 3

973 Ochs 1995 (Ochs [1995])

Each subject plays only one of the two games Those who play game 2 (game 3) take an action

56 (64) times against randomly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment Following McKelvey and Palfrey

[1995] who note that ldquoThe subjects in the Ochs experiments were paid using a lottery procedurerdquo we convert the payoffs described in Ochs [1995] to those in the matrices below before estimation The exchange rate is $001 (expected 1982 Dollars)

2 3 L R L R

11141 0 0 11141

0 11141 01238 0

11141 0 0 11141

0 11141 02785 0

U U

D D

974 GH 2001 (Goeree and Holt [2001])

Each subject takes just one action against an anonymous opponent and the exchange rate is $001

(year 2001 dollars) ldquoAMPrdquo refers to ldquoasymmetric matching penniesrdquo and ldquoRArdquo refers to ldquoreversed

asymmetryrdquo

AMP RA L R L R

320 40 40 80

40 80 80 40

44 40 40 80

40 80 80 40

U U

D D

58

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)

gt 196 Generalized matching pennies with q 2

For games of the form of Figure 3 the following is the counterpart to Theorem 6 for the case in gt 1which q 2 The proof is essentially the same as that for the Theorem 6 and hence omitted

The result is accompanied by Figure 11 which is analogous to Figure 6 Rather than redefining

every object for this case we hope their definitions are clear from the figure

1 gt 1Theorem Let p 2 q and r be such that all LQRE p q satisfy maxp 1 pmaxq 12 e gt 1q lt 092 (I) The EQRE set crosses the LQRE set at 1 q if p (II) the EQRE 1+e

2 2 1 2set cannot cross the LQRE set u+ [ u or a+ [ a at any other points except p q1 and p q2

(and must so if these points exist) and (III)

1(i) If p q1 exists (Cases 5 and 6) 1(a) for all p 2 (p p) q lt q

0 for EQRE p q and LQRE p q 0 and

(b) for all p 2 (1 p1) q gt q 0 for EQRE p q and LQRE p q

0 2 1(ii) If p q1 does not exist (Cases 7 and 8)

(a) for all p 2 (1 p) q gt q 0 for EQRE p q and LQRE p q

0 2 2(iii) If p q2 exists (Cases 5 and 7)

2(a) for all q 2 (q q) p gt p 0 for EQRE p q and LQRE p

0 q and

(b) for all q 2 (1 q2) p lt p 0 for EQRE p q and LQRE p

0 q2

2(iv) If p q2 does not exist (Cases 6 and 8) (a) for all q 2 (1 q) p gt p

0 for EQRE p q and LQRE p 0 q2

55

Case 5 Case 6 1 1

p 05

0 0

lowastlowast lowastlowast q

a minus

1 1

p q lowastlowast lowastlowast

p q 1 1

p 05

p

p q 1 1

u minus

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

u minus

a +

u +

1 2

1 2

1 lowastlowast 2 q

a minus

0 05 1 0 05 1 q q

Case 7 Case 8

u minus

p q lowastlowast lowastlowast

a +

u + 1 2

1 2

1 lowastlowast 2 q

a minus

1

u minus

lowastlowast lowastlowast

1 1

p q

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

a minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

gt 1Figure 11 The four cases of equilibrium sets (q )2

56

97 Details of Data and Procedures from Studies of 22 Games

Study Game Data

p q N Notes

Rounds (R) Subject pairs (I) Feedback

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

50 36 yes 50 24 yes 50 24 yes 50 12 yes

1 0079 0690 9600 200 48 yes 2 0217 0527 9600 200 48 yes 3 0163 0793 9600 200 48 yes 4 0286 0736 9600 200 48 yes 5 0327 0664 9600 200 48 yes

SC 2008 6 7

0445 0596 9600 0141 0564 4800

200 48 yes 200 24 yes

8 0250 0586 4800 200 24 yes 9 0254 0827 4800 200 24 yes 10 0366 0699 4800 200 24 yes 11 0331 0652 4800 200 24 yes 12 0439 0604 4800 200 24 yes

Ochs 1995 2 3

0595 0258 448 0542 0336 516

56 8 yes 64 8 yes

GH 2001 AMP RA

0960 0160 25 0080 0800 25

1 25 no 1 25 no

NS 2002 NS 0458 0390 6720 60 112 yes

Table 6 Details of Data and Procedures from Studies of 2 2 Games

971 MPW 2000 (McKelvey et al [2000])

For any one game each subject takes an action 50 times against randomly matched opponents with

feedback Subjects play multiple games and maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $010 (year 2000 Dollars)

A B C D L R

U 9 0 0 1

D 0 1 1 0

L R

U 9 0 0 4

D 0 4 1 0

L R

U 36 0 0 4

D 0 4 4 0

L R

U 4 0 0 1

D 0 1 1 0

972 SC 2008 ( Selten and Chmura [2008])

Each subject plays just one game and takes an action 200 times against randomly matched oppo-nents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of

57

the experiment The exchange rate is sect0016 (year 2008 Euros) and the experiment took place in

Germany

1

2

3

4

L R L R L R L R

10 8 9 9

0 18 10 8

8 5 6 7

0 13 9 4

10 4 7 7

0 14 8 6

9 2 5 6

0 11 7 4

U U U U

D D D D

5

6

7

8

L R L R L R L R

8 1 4 5

0 9 7 2

8 0 3 5

1 7 7 1

14 8 9 9

4 22 10 12

11 5 6 7

3 16 9 7

U U U U

D D D D

9

10

11

12

L R L R L R L R

U 13 4 7 7

3 17 8 9

U 11 2 5 6

2 13 7 6

U 10 1 4 5

2 11 7 4

U

D D D D

10 0 3 5

3 9 7 3

973 Ochs 1995 (Ochs [1995])

Each subject plays only one of the two games Those who play game 2 (game 3) take an action

56 (64) times against randomly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment Following McKelvey and Palfrey

[1995] who note that ldquoThe subjects in the Ochs experiments were paid using a lottery procedurerdquo we convert the payoffs described in Ochs [1995] to those in the matrices below before estimation The exchange rate is $001 (expected 1982 Dollars)

2 3 L R L R

11141 0 0 11141

0 11141 01238 0

11141 0 0 11141

0 11141 02785 0

U U

D D

974 GH 2001 (Goeree and Holt [2001])

Each subject takes just one action against an anonymous opponent and the exchange rate is $001

(year 2001 dollars) ldquoAMPrdquo refers to ldquoasymmetric matching penniesrdquo and ldquoRArdquo refers to ldquoreversed

asymmetryrdquo

AMP RA L R L R

320 40 40 80

40 80 80 40

44 40 40 80

40 80 80 40

U U

D D

58

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)

Case 5 Case 6 1 1

p 05

0 0

lowastlowast lowastlowast q

a minus

1 1

p q lowastlowast lowastlowast

p q 1 1

p 05

p

p q 1 1

u minus

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

u minus

a +

u +

1 2

1 2

1 lowastlowast 2 q

a minus

0 05 1 0 05 1 q q

Case 7 Case 8

u minus

p q lowastlowast lowastlowast

a +

u + 1 2

1 2

1 lowastlowast 2 q

a minus

1

u minus

lowastlowast lowastlowast

1 1

p q

a +

2 2

u +

p q 2 2

1 lowastlowast 2 q

a minus

1

p p 05 05

0 0 0 05 1 0 05 1

q q

gt 1Figure 11 The four cases of equilibrium sets (q )2

56

97 Details of Data and Procedures from Studies of 22 Games

Study Game Data

p q N Notes

Rounds (R) Subject pairs (I) Feedback

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

50 36 yes 50 24 yes 50 24 yes 50 12 yes

1 0079 0690 9600 200 48 yes 2 0217 0527 9600 200 48 yes 3 0163 0793 9600 200 48 yes 4 0286 0736 9600 200 48 yes 5 0327 0664 9600 200 48 yes

SC 2008 6 7

0445 0596 9600 0141 0564 4800

200 48 yes 200 24 yes

8 0250 0586 4800 200 24 yes 9 0254 0827 4800 200 24 yes 10 0366 0699 4800 200 24 yes 11 0331 0652 4800 200 24 yes 12 0439 0604 4800 200 24 yes

Ochs 1995 2 3

0595 0258 448 0542 0336 516

56 8 yes 64 8 yes

GH 2001 AMP RA

0960 0160 25 0080 0800 25

1 25 no 1 25 no

NS 2002 NS 0458 0390 6720 60 112 yes

Table 6 Details of Data and Procedures from Studies of 2 2 Games

971 MPW 2000 (McKelvey et al [2000])

For any one game each subject takes an action 50 times against randomly matched opponents with

feedback Subjects play multiple games and maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $010 (year 2000 Dollars)

A B C D L R

U 9 0 0 1

D 0 1 1 0

L R

U 9 0 0 4

D 0 4 1 0

L R

U 36 0 0 4

D 0 4 4 0

L R

U 4 0 0 1

D 0 1 1 0

972 SC 2008 ( Selten and Chmura [2008])

Each subject plays just one game and takes an action 200 times against randomly matched oppo-nents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of

57

the experiment The exchange rate is sect0016 (year 2008 Euros) and the experiment took place in

Germany

1

2

3

4

L R L R L R L R

10 8 9 9

0 18 10 8

8 5 6 7

0 13 9 4

10 4 7 7

0 14 8 6

9 2 5 6

0 11 7 4

U U U U

D D D D

5

6

7

8

L R L R L R L R

8 1 4 5

0 9 7 2

8 0 3 5

1 7 7 1

14 8 9 9

4 22 10 12

11 5 6 7

3 16 9 7

U U U U

D D D D

9

10

11

12

L R L R L R L R

U 13 4 7 7

3 17 8 9

U 11 2 5 6

2 13 7 6

U 10 1 4 5

2 11 7 4

U

D D D D

10 0 3 5

3 9 7 3

973 Ochs 1995 (Ochs [1995])

Each subject plays only one of the two games Those who play game 2 (game 3) take an action

56 (64) times against randomly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment Following McKelvey and Palfrey

[1995] who note that ldquoThe subjects in the Ochs experiments were paid using a lottery procedurerdquo we convert the payoffs described in Ochs [1995] to those in the matrices below before estimation The exchange rate is $001 (expected 1982 Dollars)

2 3 L R L R

11141 0 0 11141

0 11141 01238 0

11141 0 0 11141

0 11141 02785 0

U U

D D

974 GH 2001 (Goeree and Holt [2001])

Each subject takes just one action against an anonymous opponent and the exchange rate is $001

(year 2001 dollars) ldquoAMPrdquo refers to ldquoasymmetric matching penniesrdquo and ldquoRArdquo refers to ldquoreversed

asymmetryrdquo

AMP RA L R L R

320 40 40 80

40 80 80 40

44 40 40 80

40 80 80 40

U U

D D

58

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)

97 Details of Data and Procedures from Studies of 22 Games

Study Game Data

p q N Notes

Rounds (R) Subject pairs (I) Feedback

MPW 2000

A B C D

0643 0241 1800 0630 0244 1200 0594 0257 1200 0550 0328 600

50 36 yes 50 24 yes 50 24 yes 50 12 yes

1 0079 0690 9600 200 48 yes 2 0217 0527 9600 200 48 yes 3 0163 0793 9600 200 48 yes 4 0286 0736 9600 200 48 yes 5 0327 0664 9600 200 48 yes

SC 2008 6 7

0445 0596 9600 0141 0564 4800

200 48 yes 200 24 yes

8 0250 0586 4800 200 24 yes 9 0254 0827 4800 200 24 yes 10 0366 0699 4800 200 24 yes 11 0331 0652 4800 200 24 yes 12 0439 0604 4800 200 24 yes

Ochs 1995 2 3

0595 0258 448 0542 0336 516

56 8 yes 64 8 yes

GH 2001 AMP RA

0960 0160 25 0080 0800 25

1 25 no 1 25 no

NS 2002 NS 0458 0390 6720 60 112 yes

Table 6 Details of Data and Procedures from Studies of 2 2 Games

971 MPW 2000 (McKelvey et al [2000])

For any one game each subject takes an action 50 times against randomly matched opponents with

feedback Subjects play multiple games and maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $010 (year 2000 Dollars)

A B C D L R

U 9 0 0 1

D 0 1 1 0

L R

U 9 0 0 4

D 0 4 1 0

L R

U 36 0 0 4

D 0 4 4 0

L R

U 4 0 0 1

D 0 1 1 0

972 SC 2008 ( Selten and Chmura [2008])

Each subject plays just one game and takes an action 200 times against randomly matched oppo-nents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of

57

the experiment The exchange rate is sect0016 (year 2008 Euros) and the experiment took place in

Germany

1

2

3

4

L R L R L R L R

10 8 9 9

0 18 10 8

8 5 6 7

0 13 9 4

10 4 7 7

0 14 8 6

9 2 5 6

0 11 7 4

U U U U

D D D D

5

6

7

8

L R L R L R L R

8 1 4 5

0 9 7 2

8 0 3 5

1 7 7 1

14 8 9 9

4 22 10 12

11 5 6 7

3 16 9 7

U U U U

D D D D

9

10

11

12

L R L R L R L R

U 13 4 7 7

3 17 8 9

U 11 2 5 6

2 13 7 6

U 10 1 4 5

2 11 7 4

U

D D D D

10 0 3 5

3 9 7 3

973 Ochs 1995 (Ochs [1995])

Each subject plays only one of the two games Those who play game 2 (game 3) take an action

56 (64) times against randomly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment Following McKelvey and Palfrey

[1995] who note that ldquoThe subjects in the Ochs experiments were paid using a lottery procedurerdquo we convert the payoffs described in Ochs [1995] to those in the matrices below before estimation The exchange rate is $001 (expected 1982 Dollars)

2 3 L R L R

11141 0 0 11141

0 11141 01238 0

11141 0 0 11141

0 11141 02785 0

U U

D D

974 GH 2001 (Goeree and Holt [2001])

Each subject takes just one action against an anonymous opponent and the exchange rate is $001

(year 2001 dollars) ldquoAMPrdquo refers to ldquoasymmetric matching penniesrdquo and ldquoRArdquo refers to ldquoreversed

asymmetryrdquo

AMP RA L R L R

320 40 40 80

40 80 80 40

44 40 40 80

40 80 80 40

U U

D D

58

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)

the experiment The exchange rate is sect0016 (year 2008 Euros) and the experiment took place in

Germany

1

2

3

4

L R L R L R L R

10 8 9 9

0 18 10 8

8 5 6 7

0 13 9 4

10 4 7 7

0 14 8 6

9 2 5 6

0 11 7 4

U U U U

D D D D

5

6

7

8

L R L R L R L R

8 1 4 5

0 9 7 2

8 0 3 5

1 7 7 1

14 8 9 9

4 22 10 12

11 5 6 7

3 16 9 7

U U U U

D D D D

9

10

11

12

L R L R L R L R

U 13 4 7 7

3 17 8 9

U 11 2 5 6

2 13 7 6

U 10 1 4 5

2 11 7 4

U

D D D D

10 0 3 5

3 9 7 3

973 Ochs 1995 (Ochs [1995])

Each subject plays only one of the two games Those who play game 2 (game 3) take an action

56 (64) times against randomly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment Following McKelvey and Palfrey

[1995] who note that ldquoThe subjects in the Ochs experiments were paid using a lottery procedurerdquo we convert the payoffs described in Ochs [1995] to those in the matrices below before estimation The exchange rate is $001 (expected 1982 Dollars)

2 3 L R L R

11141 0 0 11141

0 11141 01238 0

11141 0 0 11141

0 11141 02785 0

U U

D D

974 GH 2001 (Goeree and Holt [2001])

Each subject takes just one action against an anonymous opponent and the exchange rate is $001

(year 2001 dollars) ldquoAMPrdquo refers to ldquoasymmetric matching penniesrdquo and ldquoRArdquo refers to ldquoreversed

asymmetryrdquo

AMP RA L R L R

320 40 40 80

40 80 80 40

44 40 40 80

40 80 80 40

U U

D D

58

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)

975 NS 2002 (Nyarko and Schotter [2002])

Each subject plays one of four treatments which differ in terms of details regarding a belief-elicitation procedure Since there are no significant differences in empirical frequencies of actions across treatments we pool all data together Each subject takes an action 60 times against ran-domly matched opponents with feedback Subjects maintain their role as either player 1 or player 2 for the duration of the experiment The exchange rate is $005 (year 2000 Dollars)

NS

L R

U

D

5 3 3 5

3 5 6 2

59

• 1 Introduction
• 2 Quantal response equilibrium
• • 21 Normal form games
  • 22 Quantal response equilibrium
  • • 3 Endogenous quantal response equilibrium
    • • 31 Binary actions
      • • 4 The structure of equilibria and equilibrium selection
        • 5 Generalized matching pennies
        • 6 Power costs in normal form games
        • 7 Data
        • • 71 Qualitative analysis
          • 72 In-Sample fit
          • 73 Out-of-sample performance
          • • 8 Conclusion
            • 9 Appendix
            • • 91 Properties of b
              • 92 Proofs
              • 93 Property 4 of Q
              • 94 Proof of Theorem 5
              • 95 Non-concavity
              • 96 Generalized matching pennies with qgt12
              • 97 Details of Data and Procedures from Studies of 22 Games
              • • 971 MPW 2000 (McKelvey2000)
                • 972 SC 2008 (Selten2008)
                • 973 Ochs 1995 (Ochs1995)
                • 974 GH 2001 (Goeree2001)
                • 975 NS 2002 (NS2002)