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MA300.2: Cooperative Game Theory

Lecture Notes: MA300.2Cooperative Game Theory

Olivier Gossner

Contents

1 Coalitional games with transferable utility 3

2 Solution Concepts, Imputations and the Core 10

3 The Bondareva Shapley Theorem 16

4 Market Games 21

5 The Shapley Value 26

5.1 The Shapley Properties . . . . . . . . . . . . . . . . . . . . . . 26

5.2 Other properties . . . . . . . . . . . . . . . . . . . . . . . . . 27

5.3 3-player games . . . . . . . . . . . . . . . . . . . . . . . . . . 29

5.4 General games . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

6 Computing the Shapley Value 37

7 Voting 41

7.1 Election between two candidates . . . . . . . . . . . . . . . . . 41

7.2 Election between any number of candidates . . . . . . . . . . . 43

8 Social Choice 47

9 Stable Matching 53

10 Matching: Extentions 59

1

CONTENTS

Textbook:

In each lecture, we refer to chapters from the following book:

Game Theory, by Maschler, Solan and Zamir, Cambridge University Press

(2013).

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1. COALITIONAL GAMES WITH TRANSFERABLE UTILITY

1 Coalitional games with transferable utility

Reference:

Chapter:“Coalitional Games with Transferable Utility”

Introduction

Cooperative game theory models situations in which players may cooperate

to achieve their goals. It assumes that every set of players can form a coali-

tion and engage in a binding agreement that yields them a certain amount

of profit. The maximal amount that a coalition can generate through coop-

eration is called the worth of the coalition.

Assumptions of the TU-model

• The players can engage in a binding agreement.

• The players preferences are linear; therefore, they can be represented

as a real valued utility.

• Utilities of all the players can be measured in a common unit (such as

money).

• Utility can be transferred between the players.

• There are no externalities.

Example: Three entrepreneurs

Odette, Ron and Wilbur are three entrepreneurs. Odette estimates her

yearly profit to be $170, 000, Ron estimates his to be $150, 000 andWilbur

assumes that he can yield a profit of $180, 000 per year.

If they work together, they can even profit more:

Odette and Ron: $350, 000

Odette and Wilbur: $380, 000

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Ron and Wilbur: $360, 000

Odette, Ron and Wilbur: $560, 000

A natural question that arises in the above example is how to divide the

profits? Obviously, it is to their advantage to work together, but it is not

immediately clear how they ought to divide among them the profits of a

joint company, should they form one. Therefore, we first need a model and

a solution concept.

Definition 1. A coalitional game with transferable utility (TU game) is a

pair (N ; v) such that

• N = {1, 2, . . . , n} is a finite set of players. A subset of N is called a

coalition. The collection of all the coalitions is denoted by 2N .

• v : 2N → R is a function associating every coalition S with a real

number v(S) (the worth of S), with v(∅) = 0. This function is called

the coalitional function of the game.

when there is no ambiguity on the set of players we denote the game simply

by v.

The example above thus translates into:

• Players: N = {Odette,Ron,Wilbur}

• Coalitions: {∅} , {Odette} , {Ron} , {Wilbur} , {Odette,Ron} , {Ron,Wilbur} ,{Odette,Wilbur} , {Odette,Ron,Wilbur}

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• Coalitional function:

v(∅) = 0

v(Odette) = 170, 000

v(Ron) = 150, 000

v(Wilbur) = 180, 000

v(Odette,Wilbur) = 380, 000

v(Ron,Wilbur) = 360, 000

v(Odette,Ron) = 350, 000

v(Odette,Wilbur,Ron) = 560, 000.

Simple Games

Definition 2. A coalitional game (N ; v) is called simple if for each coalition

S, either v(S) = 0 or v(S) = 1.

For instance, simple games can model committee votes, including cases in

which the voting rule is not necessarily the majority rule. It can be inter-

preted as follows: if v(S) = 1, coalition S can pass a motion; if v(S) = 0,

coalition S cannot pass a motion on its own.

Example: The UN Security Council

There are 15 members, 5 of which are permanent, 10 are non-permanent

members. Adopting a resolution requires a majority of 9 members and

any permanent member can cast a veto. Thus, the coalitional function is

as follows:

v(S) =

!1 if |S| ≥ 9 and S contains all the permanent members

0 for any other coalition S.

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Example: House of Commons

The British Parliament’s House of Commons is comprised of 650 mem-

bers. A coalition requires 326 members to form a government. Suppose

there are three parties represented with 282, 260 and 108 seats. A player

in this game will correspond to a party. Denote by 1 the “worth” of being

the governing coalition and by 0 the “worth” of being in the opposition.

Observe that no single party has 326 seats or more and that every pair

of parties together has more than 326 seats. Therefore, the coalitional

function is

v(1) = v(2) = v(3) = 0;

v(12) = v(13) = v(23) = v(123) = 1.

Weighted majority games

Definition 3. A coalitional game (N ; v) is a weighted majority game if there

exists a quota q ≥ 0 and nonnegative real weights (wi)i∈N such that the worth

of each nonempty coalition S is

v(S) =

!1 if

"i∈S wi ≥ q,

0 if"

i∈S wi < q.

The game is denoted [q;w1, . . . , wn].

For instance, the example above with the three parties in the British House

of Commons is the weighted majority game [326; 282, 260, 108].

Strategic Equivalence (SE)

The descriptions of the games above are not unique. For instance, the worth

of a coalition could have been presented in an other currency. Or, one could

have incorporated external income of the players into the coalitional func-

tion. As long as the players’ income from joining a coalition is not affected

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(by currency issues, external income, etc.) one may obtain more than one

coalitional function representing the same game.

Definition 4. A coalitional game (N ;w) is strategically equivalent to the

game (N ; v) if there exists a positive number a and a vector b ∈ RN , such

that for every coalition S:

w(S) = av(S) +#

i∈S

bi = av(S) + b(S) = (av + b)(S).

Example: An equivalent game

Consider the first example with the three entrepreneurs. We will now

represent the game in British pounds (letting £1 = $2) and furthermore

incorporate external income in the coalitional function: Assume Odette

has an additional income of £15, 000 from letting a house she owns,

Ron has additional savings of £100, 000 and Wilbur own £10, 000 to his

banker. The coalitional function now looks as follows:

v(Odette) = 0.5× 170, 000 + 15, 000 = 100, 000

v(Ron) = 0.5× 150, 000 + 100, 000 = 175, 000

v(Wilbur) = 0.5× 180, 000− 10, 000 = 80, 000

v(Odette,Wilbur) = 0.5× 380, 000 + 15, 000− 10, 000 = 195, 000

v(Ron,Wilbur) = 0.5× 360, 000 + 100, 000− 10, 000 = 270, 000

v(Odette,Ron) = 0.5× 350, 000 + 15, 000 + 100, 000 = 290, 000

v(Odette,Wilbur,Ron) = 0.5× 560, 000 + 15, 000− 10, 000 + 100, 000 = 385, 000.

Theorem 1. The strategic equivalence relation is an equivalence relation.

I.e., it is reflexive, symmetric and transitive.

0-normalized games

Definition 5. The game (N ; v) is

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• (0, 1) normalized if v(i) = 0 for every player i ∈ N and v(N) = 1.

• (0, 0) normalized if v(i) = 0 for every player i ∈ N and v(N) = 0.

• (0,−1) normalized if v(i) = 0 for every player i ∈ N and v(N) = −1.

Question: Is the 3-entrepreneurs game strategically equivalent to a 0-normalized

game?

Theorem 2. A game (N ; v) is

• equivalent to a (0,−1) normalized game if v(N) >"

i∈N v(i),

• equivalent to a (0, 0) normalized game if v(N) ="

i∈N v(i),

• equivalent to a (0,−1) normalized game if v(N) <"

i∈N v(i),

In the first and third cases, the corresponding normalized game is unique.

Special families of games

Definition 6. A coalitional game (N ; v) is called super-additive, if for any

two disjoint coalitions S and T ,

v(S) + v(T ) ≤ v(S ∪ T ).

Question: Is the 3-entrepreneurs game super-additive? What about the UN

Security Council game?

Consider the following game (the dual of the UN Security Council game):

v(S) =

!1 if |S| ≥ 7 or contains a permanent member

0 for any other coalition S.

Is this game super-additive?

Definition 7. A coalitional game (N ; v) is called monotonic if for every two

coalitions S ⊆ T ,

v(S) ≤ v(T ).

Question: Are the previous examples monotonic?

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Superadditive vs Monotonic

• Superadditivity is a SE invariant: If two games are equivalent, one if

superadditive if and only if the other is superadditive. Prove

• Monotonicity is not a SE invariant. There are equivalent games such

that one is monotonic while the other isn’t. Find a counter-example to

convince yourself.

• Every game is strategically equivalent to a monotonic game (but not

all games are monotonic). Try to prove this property.

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2. SOLUTION CONCEPTS, IMPUTATIONS AND THE CORE

2 Solution Concepts, Imputations and the Core

Reference:

Chapters:“Coalitional Games with Transferable Utility”, “The Core”

We will now address the following questions:

1. What coalitions will form? Once a coalition forms, how will its mem-

bers divide its worth? Especially, when the grand coalition forms?

2. What would an arbitrator recommend that the players do?

3. What are possible stable agreements between players? When do such

agreements exist?

Solution Concepts

Definition 8. Let U be a family of games. A solution concept is a mapping

ϕ associating any game (N ; v) ∈ U with a subset of ϕ(v) ⊆ RN (possibly the

empty set). A solution concept is called a point solution if ϕ(v) is a singleton,

for every game (N ; v) ∈ U .

Imputations (a simple solution concept)

Definition 9. For a coalitional game v, an imputation is a vector x ∈ RN

satisfying

(i) (efficiency) x(N) ="

i∈N xi = v(N),

(ii) (individual rationality) xi ≥ v(i), ∀i ∈ N .

The set of all imputations is denoted X(v).

Example:

N = {1, 2, 3};

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v(1) = v(2) = v(3) = 0, v(12) = v(23) = 1, v(13) = 2, v(123) = 3

(0,3,0) (0,0,3)

(3,0,0)

(x,y,z)

Figure 1: The set of all imputations

(0,3,0) (0,0,3)

(3,0,0)

(0,2,1)

(1,2,0)

Figure 2: Imputations at which x2 ≥ 2

Theorem 3. The set of all imputations is covariant under strategic equiva-

lence. Namely,

X(av + b) = aX(v) + b,

for every game v, a > 0, and b ∈ RN .

Corollary 1. The set of all imputations is either

(i) a (N − 1)-simplex, if v(N) >"

i∈N v(i), or

(ii) a point, if v(N) ="

i∈N v(i), or

(iii) the empty set, if v(N) <"

i∈N v(i).

The Core: a solution that ensures stability

Definition 10. For a coalitional game v, the core C(v) is the set of vectors

x ∈ RN satisfying

(i) (efficiency) x(N) = v(N),

(ii) (coalitional rationality) x(S) ="

i∈S xi ≥ v(S) ∀S ⊂ N .

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Q: What does the core look like?

• A subset of all imputations

• Can be empty

• A compact polytope: a finite intersection of closed half spaces. In

addition, it is bounded.

Examples:Consider the game (N = {1, 2, 3} ; v)with v(1) = v(2) = v(3) = 0,v(12) = v(23) = 1, v(123) = 3.Examples i) − iv) represent the coreof this game with different values forv(13). (0,3,0) (0,0,3)

(3,0,0)

(x1 ≤ 2)

(x3 ≤ 2)

i)

(x2 ≤ 1)

Figure 3: v(13) = 2

(0,3,0) (0,0,3)

(3,0,0)

(x1 ≤ 2)

(x3 ≤ 2)

ii)

(x2 ≤ 2)

Figure 4: v(13) = 1

(0,3,0) (0,0,3)

(3,0,0)

(x1 ≤ 2)

(x3 ≤ 2)

iii)

(x2 ≤ 0)

Figure 5: v(13) = 3

iv) The core with v(13) = 4 is empty.

Q: When is the core nonempty?

We consider two cases: simple games and general games.

a) Simple Games

Examples:

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(i) One seller two buyers market (aka the Gloves Games):

v(1) = v(2) = v(3) = v(23) = 0, v(12) = v(13) = v(123) = 1. C(v) =

{(1, 0, 0)}

(ii) Simple Majority:

v(1) = v(2) = v(3) = 0, v(12) = v(23) = v(13) = v(123) = 1. C(v) = ∅

Definition 11. Let v be a simple coalitional game. Player i ∈ N is called a

veto player if v(S) = 0 whenever i /∈ S.

Theorem 4. Let v be a simple coalitional game with v(N) = 1. The core

C(v) is nonempty if and only if there is at least one veto player in N .

Proof. “if”: let i be a veto player. The imputation (xi)i∈N with xi = 1,

xj = 0 (∀j ∈ N , j ∕= i) is in the core.

“only if”: Assume there are no veto players. Let x = (xi)i∈N be an impu-

tation. We show that x is not in the core. Since"

i∈N xi = 1, there must

be a player i ∈ N for whom xi > 0. Since i is not a veto player, there is a

winning coalition S ⊂ N \ {i}. Now,

x(S) ≤ x(N \ {i}) = x(N)− xi = 1− xi < v(S) = 1

b) General Games

Theorem 5. The core is covariant under strategic equivalence. Namely, for

every coalitional game v, a > 0, and b ∈ RN ,

C(av + b) = aC(v) + b.

Corollary 2. The existence of a nonempty core is invariant under strategic

equivalence.

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Balanced collections of weights

Consider a game v with |N | = 3. We establish necessary conditions for the

existence of a nonempty core: let x = (x1, x2, x3) ∈ C(v) then

x1 + x2 + x3 = v(123) (1)

x1 + x2 ≥ v(12) (2)

x2 + x3 ≥ v(23) (3)

x1 + x3 ≥ v(13) (4)

x1 ≥ v(1) (5)

x2 ≥ v(2) (6)

x3 ≥ v(3) (7)

Now, observe the follwing:

(1), (5− 7) ⇒ v(123) ≥ v(1) + v(2) + v(3) (8)

(1), (2), (7) ⇒ v(123) ≥ v(12) + v(3) (9)

(1), (3), (5) ⇒ v(123) ≥ v(1) + v(23) (10)

(1), (4), (6) ⇒ v(123) ≥ v(13) + v(2) (11)

(1), (2− 4) ⇒ v(123) ≥ 1

2v(12) +

1

2v(23) +

1

2v(13) (12)

Inequalities (8− 12) are necessary for the existence of a nonempty core of a

game v with |N | = 3. Some effort shows that they are sufficient, as well (EX

17.19). Observations (regarding inequalities (8− 12)):

• v(123) has to be large enough;

• the right hand side is a linear combination of the worths of coalitions;

• (8− 11) correspond to partitions of N , (12) does not.

In the following, we will generalize the notion of “partition”:

Definition 12. A collection of weights (δS)S∈2N−{∅,N} is called balanced if:

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1. δS ≥ 0 for every S

2. For every i ∈ S,"

S,S∋i δS = 1

Each of inequalities (8 − 12) involves a collection of coalitions D and a list

of coefficients δ = (δS)S∈D. For example:

Inequality D δ(8) {{1} , {2} , {3}} (1, 1, 1)(12) {{1, 2} , {2, 3} , {1, 3}} (1

2, 12, 12)

In each case we have a balanced collection of weights that put positive weights

on coalitions belonging to D only.

Definition 13. A coalitional game v is called balanced, if for every balanced

vector weights (δS)S,

v(N) ≥#

S

δSv(S)

Using the same logic as for games with 3 players above, it is possible to

(and we will) show that every game that has a non-empty core is balanced.

In fact, the converse is also true, as we will see in Lecture 3.

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3. THE BONDAREVA SHAPLEY THEOREM

3 The Bondareva Shapley Theorem

Reference:

Chapters: “The Core”, Appendix 23.3 (Linear Programming Duality) and

Chapter 4.12 (Zero-Sum Games).

Theorem 6. (Bondareva 1963, Shapley 1967)

A coalitional game has a nonempty core if and only if it is balanced.

Before we turn to prove the Bondareva Shapley (B-S henceforth) Theorem,

we first prove a useful lemma.

Lemma 1. A collection of weights (δS)S is balanced if and only if, for every

x ∈ RN

#

S

δSx(S) = x(N). (∗)

Proof. “if”: We assume (∗) and let i = N . We have to prove that"

S,i∈S δS =

1. Indeed, let x = (0, . . . , 1$%&'ith coordinate

, 0, . . . , 0), then

#

S,i∈S

δS =#

S

δSx(S)(∗)= x(N) = 1.

“only if”: Assume"

S:i∈S δS = 1, ∀i ∈ N , and let x ∈ RN .

#

S

δSx(S) =#

S

δS

(#

i∈S

xi

)=

#

S

#

i∈S

δSxi =#

i∈N,S:i∈S

δSxi

=#

i∈N

#

S:i∈S

δSxi =#

i∈N

xi

#

S:i∈S

δS =#

i∈N

xi · 1

= x(N)

The next Theorem will allow us to consider only 0-normalized games in the

proof of B-S Thm.

Theorem 7. Let v and u be strategically equivalent games. Then v is bal-

anced if and only if u is balanced.

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Proof. Say u = av + b, a > 0 and b ∈ RN . Suppose v is balanced and let

(δS)S be a balanced collection of weights. Now,

#

S

δSu(S) =#

S

δS(av(S) + b(S)) = a#

S

δSv(S) +#

S

δSb(S)

= a#

S

δSv(S) + b(N) ≤ av(N) + b(N)

= u(N)

The third equality follows from Lemma 1 and the inequality is due to the

fact that v is balanced. Thus, u is balanced.

The proof of the B-S Thm has two parts. The easier is to show that the B-S

condition is a necessary condition for the nonemptiness of the core.

Proof. (“ Every game whose core is nonempty is balanced”)

Let x ∈ C(v). Namely, x ∈ RN , x(N) = v(N) and x(S) ≥ v(S) ∀S ⊆ N .

Let (δS)S be a balanced vector of weights. We have

#

S

δSv(S) ≤#

S

δSx(S) = x(N) = v(N).

The inequality is true since x ∈ C(v) and the equality follows from Lemma 1.

The other direction of the proof requires more work. Since both the nonempti-

ness of the core and the B-S condition are invariant under strategic equiva-

lence, it is sufficient to consider the three types of (0,−1), (0, 0) and (0,−1)

normalized games. The (0, 0) and (0,−1) cases are relatively easy, so we

begin with the interesting case, (0, 1) normalized games.

Proof. (“Every (0, 1) normalized game whose core is empty is not balanced”)

The proof makes use of the Minmax Theorem. Our first step is to define an

appropriate zero-sum game.

Step 1: Let v be a (0, 1) normalized coalitional game whose core is empty.

Define the following zero-sum game: Player I chooses i ∈ N . Player

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II chooses a coalition whose worth is strictly positive. Pure strategy

sets are,

SI = {i : i ∈ N} , SII = {S ⊂ N : v(S) > 0} .

Note that SII is nonempty since otherwise any imputation would be in

the core. The payoff that Player II pays to Player I is

g(i.S) =

!1

v(S)if i ∈ S

0 if i /∈ S

Denote the value of this game by λ.

Step 2: “λ > 0”.

Player I can ensure a positive payoff by playing the mixed strategy

( 1n, 1n, . . . , 1

n). Indeed, for every (pure) counter strategy S of player II,

the expected payoff is1

n

|S|v(S)

> 0.

Step 3: “λ < 1”.

Let x = (xi)i∈N be an optimal strategy for player I. Since v is (0, 1)

normalized, x is an imputation of v. Now, since the core of v is empty,

there must be a coalition S ⊂ N , such that v(S) > x(S). In particular

v(S) > 0, therefore, S ∈ SII . Now, since x is optimal

λ ≤ g(x, S) =x(S)

v(S)< 1.

Step 4: “v is not balanced”

The minmax theorem ensures that player II has a mixed strategy (yS)

that guarantees a payoff of at most λ < 1, against any (pure) strategy

i ∈ N of player I. We use yS to define a balanced vector of weights

that violates the B-S condition. Define

δS =

*+,

+-

ySλv(S)

if v(S) > 0,

1−"

T :v(T )>0,i∈T δT if S = {i}0 if v(S) = 0 and S is not a singleton.

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To prove that (δS)S is a balanced collection of weights, we first show

that δS ≥ 0 for every coalition S.

– For S such that v(S) > 0, it is obvious.

– For S = {i}, note that

#

T :v(T )>0,i∈T

δT =#

T :v(T )>0,i∈T

1

λ

yTv(T )

=1

λ

#

T :v(T )>0,i∈T

yT g(i, T )

=1

λ

#

T :v(T )>0

yT g(i, T )

=1

λ

#

T :T∈SII

yT g(i, T )

=g(i, y)

λ≤ λ

λ.

– For S such that v(S) = 0 and S is not a singleton v(S) = 0.

Now, (δS)S is a balanced vector of weights, since by definition

#

S:i∈S

δS = 1 ∀i ∈ N.

Finally, v is not balanced since

#

S

δSv(S) =#

S:v(S)>0

ySλ

$ %& '= 1

λ

+#

i∈N

δ{i}v(i)

$ %& '=0

=1

λ> v(N).

It remains to consider the simpler cases of (0, 0) and (0,−1) normalized

games.

Proof. (“Every (0, 0) normalized game whose core is empty is not balanced.”)

Let v be a (0, 0) normalized game whose core is empty. The only imputation

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in X(v) is the 0-vector, 0̄. Since 0̄ /∈ C(v), there must be a coalition S ⊂ N

s.t. v(S) > 0. The balanced vector of weights (δT )T where δS = 1, δi = 1 if

i ∕∈ S, and δT = 0 otherwise violates the B-S condition, since

#

T

δTv(T ) = v(S) +#

i ∕∈S

v(i) = v(S) > 0 = v(N).

Proof. (“Every (0,−1) normalized game is not balanced”)

Take the balanced vector of weights (δS)S where δS = 1 if S is a singleton

and δS = 0 otherwise:

#

T

δTv(T ) =#

i∈N

v(i) = 0 > −1 = v(N).

B-S Thm through linear programming duality

Let P(N) = 2N \ {∅}.

primal LP: ZP = min 〈(1, 1, . . . , 1), (x1, . . . , xn)〉subject to:

χP(N)

.

///0

x1

x2...xn

1

2223≥ (v(S))S∈P(N)

dual LP: ZD = max4(δS)S∈P(N), (v(S))S∈P(N)

5

subject to:

(δS)S∈P(N)χP(N) = (1, 1, . . . , 1)

and (δS ≥ 0) ∀S ∈ P(N).

Now,

C(v) ∕= ∅ ⇔ ZP ≤ v(N)

(N ; v) is balanced ⇔ ZD ≤ v(N)

LP duality ⇒ ZP = ZD

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4. MARKET GAMES

4 Market Games

Reference:

Chapter: “The Core”

In a market there are producers N = {1, 2, . . . , n} and there are commodities

L = {1, 2, . . . , l}. Each producer has a bundle of commodities ai ∈ RL+, called

“initial endowment”, and a production function ui : RL+ → R. We assume

that the production functions are continuous and concave. That is

ui(λx+ (1− λ)y) ≥ λui(x) + (1− λ)ui(y),

for every x, y ∈ RL+ and every 0 ≤ λ ≤ 1.

Example: The gin-and-tonic market.

N = {1, 2, . . . , n}, L = {1 = gin, 2 = tonic}.Each producer i has an initial endowment of gin ai,1, and and initial

endowment of tonic, ai,2.

The production function is the same for every producer

ui(x) = min {xgin, xtonic} .

Given a market 〈N,L, (ai, ui)i∈N〉 we derive a game (N ; v). That is, the

players of the game are the producers of the market. The worth of a coalition

S is given by

v(S) := max

!#

i∈S

ui(xi) : xi ∈ RL+, ∀i ∈ S,

#

i∈S

xi =#

i∈S

ai

6.

In words: the worth of S is the maximal total utility that the members of S

can produce by re-distributing their initial endowments between themselves.

c© LSE 2014–2020 /MA300.2 Page 21 of 66

4. MARKET GAMES

Example: The gin-and-tonic market (cont.).

Suppose merchant i has 2 portions of gin and 1 portion of tonic, ai =

(2, 1), and merchant j has 1 portion of gin and 2 portions of tonic, aj =

(1, 2).

v(i) = ui(ai) = min {2, 1} = 1

v(j) = uj(aj) = min {1, 2} = 1

v(i, j) = max {ui(x) = uj(y) : x+ y = (3, 3)}

Clearly, v(i, j) ≥ v(i)+ v(j), since one way to distribute the endowments

is for each player to keep his own bundle...

But they can do better! If they let one of them, say i, produce using

the entire total endowment, then he can produce 3 portions of gin and

tonic. Formally, we have an admissable allocation for the coalition {i, j}:xi = (3, 3), xj = (0, 0). Note that xi, xj ≥ 0, xi + xj = ai + aj = (3, 3).

Since ui(xi) + uj(xj) = 3 + 0, we have v(i, j) ≥ 3.

In fact, v(i, j) = 3. That is, ((3, 3), (0, 0)) is a maximizer of the function

f(xi, xj) = ui(xi) + uj(xj)

subject to

xi, xj ≥ 0,

xi + xj = (3, 3).

Further Questions

• Prove that v(i, j) = 3

• Is the above the only maximizer?

• What can be said about the set of maximizers? Is it closed? Is it

c© LSE 2014–2020 /MA300.2 Page 22 of 66

4. MARKET GAMES

convex?

Definition 14. Given a market 〈N,L, (ai, ui)i∈N〉 and a coalition ∅ ∕= S ⊂N , the set of allocations for S is denoted by

XS =7(xi)i∈S : xi ∈ RL

+ ∀i ∈ S, x(S) = a(S)8⊆ RL×S

+ .

Theorem 8. For every coalition S, XS is a compact polytop in RL×S.

Proof. XS is defined by the conjunction of weak linear inequalities. XS is

bounded, since ∀x ∈ XS, ∀i ∈ S and ∀j ∈ L, 0 ≤ xi,j ≤"

i∈S ai,j. Finally,

XS ∕= ∅, since (ai)i∈S ∈ XS.

Theorem 9. The maximum in the definition of v(S) is attained by some

xS ∈ XS.

Proof. The function f : RL×S+ → R, f : (xi)i∈S 3−→

"i∈S ui(xi) is continuous

as the sum of continuous functions. The set XS is compact and nonempty;

therefore, the maximum of f over XS is attained.

Definition 15. A game v is called a market game if it is derived from a

market.

Example: The tequila sunrise market.

N = {1, 2, 3}, L = {1 = grenadine, 2 = orange juice, 3 = tequila}.Each producer i has an initial endowment of grenadine syrup ai,1, and

orange juice, ai,2, and of tequila, ai,3.

Initial endowments are:*+,

+-

a1 = (4, 16, 0)

a2 = (0, 16, 12)

a3 = (4, 8, 12)

A tequila sunrise requires 2 units (cl.) of grenadine syrup, 8 of orange

juice, and 4 of tequila.

c© LSE 2014–2020 /MA300.2 Page 23 of 66

4. MARKET GAMES

1. Are the production functions continuous? concave?

2. What is the set of allocations for the coalition {1, 2}? For the grand

coalition?

3. Write the coalitional form of the market game.

4. For each coalition, what is an optimal allocation?

5. Show that the game satisfies conditions (8)-(12) of Chapter 2.

6. Conclude that the core is non-empty.

Theorem 10. (Shapley and Shubik, 1969).

The core of a market game is nonempty.

Proof. Let v be a game derived from a market 〈N,L, (ai, ui)i∈N〉. We use the

Bondareva-Shapley Theorem and prove that v is balanced. That is,

v(N) ≥#

S∈P(N)

δSv(S),

for every balanced vector of weights (δS)S∈P(N).

For every coalition S ∈ P(N), let xS ∈ XS be an optimal allocation for S.

Namely,

#

i∈S

ui(xSi ) = v(S), xS

i ∈ RL+, ∀i ∈ S,

xS(S) = a(S).

Let (δS)S∈P(N) be a balanced vector of weights. Define an allocation z ∈RL×N by

zi =#

S∈P(N):i∈S

δSxSi ∀i ∈ N.

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4. MARKET GAMES

Clearly, ∀i ∈ N zi ∈ RL+. Moreover, z(N) = a(N). Namely, z ∈ XN .

z(N) =#

i∈N

zi =#

i∈N

#

S∈P(N):i∈S

δSxSi

=#

S∈P(N)

δS#

i∈S

xSi =

#

S∈P(N)

δSa(S)

=#

S∈P(N)

#

i∈S

δSai =#

i∈N

ai#

S∈P(N):i∈S

δS

=#

i∈N

ai · 1 = a(N).

From the above we get

v(N) ≥#

i∈N

ui(zi). (∗)

We conclude the proof by

v(N) ≥#

i∈N

ui(zi) =#

i∈N

ui

.

0#

S∈P(N):i∈S

δSxSi

1

3

≥#

i∈N

#

S∈P(N):i∈S

δSui(xSi ) =

#

S∈P(N)

δS#

i∈S

ui(xSi )

=#

S∈P(N)

δSv(S).

The first inequality follows from (∗), the second from the fact that the uis

are concave. The last equality holds since xS are optimal allocations.

Definition 16. Let v be a coalitional game. Let S ⊂ N . The subgame of v

restricted to S is the game (S, vS), where vS(T ) := v(T ) ∀T ⊂ S.

Definition 17. A game v is totally balanced if (S; vS) is balanced, for every

∅ ! S ⊆ N .

Since a subgame of a market game is a market game (prove!), we have the

following corollary of the Shapley Shubik Theorem:

Corollary 3. Every market game is totally balanced.

The converse, whose proof is an exercise, is also true:

Theorem 11. Every totally balanced game is a market game.

c© LSE 2014–2020 /MA300.2 Page 25 of 66

5. THE SHAPLEY VALUE

5 The Shapley Value

Reference:

Chapter:“The Shapley Value”

We look for a solution concept that prescribes a well-defined allocation for

every game.

Definition 18. A point solution concept on the set of all N player games is

a function ϕ : RP(N) → RN .

The interpretation is that ϕi(v) is the value of player i in the game v, pre-

scribes by ϕ.

In order to find a satisfactory solution concept we take the axiomatic ap-

proach. We make a list of desirable properties that we wish our solution

concept to have, and then try to find solutions that satisfy as many of these

properties as possible.

5.1 The Shapley Properties

We want our solution concept to prescribe a system according to which the

worth of the grand coalition is divided between the players.

Definition 19. A solution concept ϕ satisfies efficiency if:#

i∈N

ϕi(v) = v(N).

for every game v.

We want symmetric players in any game to be treated the same. Given

a game v, we say that players i, j are symmetric in v, if

v(S ∪ {i}) = v(S ∪ {j}) ∀S ⊂ N \ {i, j} .

Definition 20. A solution concept ϕ satisfies symmetry if:

ϕi(v) = ϕj(v),

whenever i and j are symmetric players in v.

c© LSE 2014–2020 /MA300.2 Page 26 of 66


We want that a player who does not contribute to any coalition does

not receive anything. A player is called a null player if, for every coalition

S ⊂ N , v(S ∪ {i}) = v(S). In particular v(i) = 0 if i is a null player.

Definition 21. A solution concept ϕ satisfies the null player property if:

ϕi(v) = 0

whenever i is a null player in v.

Given two games u, v, u+v is the game in which the worth of any coalition

is the sum of its worth in u and in v. If the value of each player in u and in

v can be analysed separately, and added to obtain the value of that player

in u+ v, we way that the corresponding solution concept satisfies additivity.

More formally:s

Definition 22. A solution concept ϕ satisfies additivity if, for any two

games u and v:

ϕi(u+ v) = ϕi(u) + ϕi(v) ∀i ∈ N.

5.2 Other properties

We would like our solution to be independent of arbitrary choices in the

description of the game. This idea is captured by the following property:

• Covariance under strategic equivalence:

ϕi(av + b) = aϕi(v) + bi ∀(v), ∀a > 0, ∀b ∈ RN .

We don’t want our solution to be dependent on the names of the players.

That is, if we have a game v and a permutation π : N1:1→ N , then the game

π ◦ v, derived by permuting the players, is defined by

π ◦ v(S) = v(i ∈ N : π(i) ∈ S) = v(π−1(S)).

c© LSE 2014–2020 /MA300.2 Page 27 of 66


• Strong Symmetry:

ϕπ(i)(π ◦ v) = ϕi(v) ∀(v), ∀i ∈ N, ∀π ∈ Π(N).

Where Π(N) is the set of all permutations on N .

Another property of a similar flavor to covariance under strategic equivalence

requires that if a player has a fixed marginal contribution to any coalition,

then his value is that fixed number.

Formally, a player i ∈ N is called a dummy player in the game v, if

v(S ∪ {i}) = v(S) + v(i) ∀S ∈ N \ {i} .

• Dummy Property:

ϕi(v) = v(i),

whenever i is a dummy player in v.

Below we list a few more properties that will be “nice to have”.

• Linearity:

ϕi(au+ v) = aϕi(u) + ϕi(v) ∀i ∈ N, a ∈ R, ∀u, v.

• Monotonicity:

ϕ(v) ≥ 0,

whenever v is monotonic.

• Monotonicity of marginal contributions:

For every i ∈ N and every pair of games u, v, if

v(S ∪ {i})− v(S) ≥ u(S ∪ {i})− u(S) ∀S ⊂ N

then

ϕi(v) ≥ ϕi(u).

c© LSE 2014–2020 /MA300.2 Page 28 of 66


• Marginality:

For every i ∈ N and every pair of games u, v, if

v(S ∪ {i})− v(S) = u(S ∪ {i})− u(S) ∀S ⊂ N

then

ϕi(v) = ϕi(u).

5.3 3-player games

For the class of 3-player games, N = {1, 2, 3}, consider the following solution

concept ψ123:

ψ1231 (v) = v(1)− v(∅) = v(1)

ψ1232 (v) = v(12)− v(1)

ψ1233 (v) = v(123)− v(12)

The idea is that players enter in a room. Player 1 enters first, then player 2,

then player 3. Each player gets his “marginal contribution”, i.e. the value

of the coalition after he enters, minus the value of the coalition of players

already there when he enters.

ψ123 is a well-defined solution concept on the family of 3-player games.

Does it satisfy efficiency? additivity? the null player property? symmetry?

Instead of having player 1 entering first, followed by player 2, then by

player 3, we could have chosen another ordering of players. For instance,

132. This leads us to defining the corresponding solution concept:

ψ1321 (v) = v(1)

ψ1322 (v) = v(123)− v(13)

ψ1323 (v) = v(13)− v(1)

c© LSE 2014–2020 /MA300.2 Page 29 of 66


And similarly, we have 4 solution concepts for all 4 other orderings. This

leads us to the following 6 solution concepts, summarized in the following

table:

Ordering value for 1 value for 2 value for 3123 v(1) v(12)− v(1) v(123)− v(12)132 v(1) v(123)− v(12) v(13)− v(1)213 v(12)− v(2) v(2) v(123)− v(12)231 v(123)− v(23) v(2) v(23)− v(2)312 v(13)− v(3) v(123)− v(13) v(3)321 v(123)− v(23) v(23)− v(3) v(3)

All of these 6 solution concepts satisfy additivity, the null player property,

and efficiency. But none of them satisfies symmetry!

In order to obtain a symmetric solution concept, we are going to take

the average of all 6 solution concepts defined for different player orderings.

This way, the ordering of players should not matter anymore. This is why

we define:

ψ(v) =1

6(ψ123(v) + ψ132(v) + ψ213(v) + ψ231(v) + ψ312(v) + ψ321(v))

This gives us:

ψ1(v) =1

6(2v(1) + v(12)− v(2) + v(13)− v(3) + 2(v(123)− v(23)))

ψ2(v) =1

6(2v(2) + v(12)− v(1) + v(23)− v(3) + 2(v(123)− v(13)))

ψ3(v) =1

6(2v(3) + v(13)− v(3) + v(23)− v(2) + 2(v(123)− v(12)))

It is easy to verify that ψ satisfies additivity, efficiency and the null player

property. We now check that it satisfies symmetry. Assume for instance

that players 1 and 2 are symmetric in v. In this case v(1) = v(2), and

v(13) = v(23). We obtain:

ψ2(v) =1

6(2v(2) + v(12)− v(1) + v(23)− v(3) + 2(v(123)− v(13)))

=1

6(2v(1) + v(12)− v(2) + v(13)− v(3) + 2(v(123)− v(23)))

= ψ1(v).

c© LSE 2014–2020 /MA300.2 Page 30 of 66


Similarly, ψi(v) = ψj(v) whenever i, j are symmetric players in v.

Let us consider a couple of examples:

Example 1: Three-player simple majority game.

v(1) = v(2) = v(3) = 0; v(12) = v(13) = v(23) = v(123) = 1.

Permutation 1 2 3123 v(1) = 0 1 0132 v(1) = 0 0 1213 v(12)− v(2) = 1 0 0231 v(123)− v(23) = 0 0 1312 v(13)− v(3) = 1 0 0321 v(123)− v(23) = 0 1 0

Average 13

13

13

Example 2: Two buyers-one seller.

v(1) = v(2) = v(3) = v(23) = 0; v(12) = v(13) = v(123) = 1.

Permutation 1 2 3123 0 1 0132 0 0 1213 1 0 0231 1 0 0312 1 0 0321 1 0 0

Average 23

16

16

So, we have found a solution concept that satisfies all 4 Shapley properties

for 3 players games. But how much freedom do the Shapley properties leave

us? How many concepts are there that satisfy all 4 Shapley properties?

For instance, consider any solution concept ϕ that satisfies all 4 Shapley

properties. What can be said about ϕ(v), where v is the 3 player simple

majority game? Here, symmetry and efficiency are enough to show that

ϕ(v) = ψ(v).

c© LSE 2014–2020 /MA300.2 Page 31 of 66


What can be said about ϕ(v), where v is the 2 buyers-1 seller game? Here

things are a bit more complicated for us. We will make use of all 4 axioms,

including linearity. We will decompose the game v into a linear combination

of 3 games. Given a coalition T , we let vT (S) = 1 is T ⊆ S, and vT (S) = 0

otherwise. The table below shows that v can be written as:

v = v12 + v13 − v123.

Coalition S v(S) v12(S) v13(S) −v123(S) v12(S) + v23(S)− v123(S)123 1 1 1 1 112 1 1 0 0 113 1 0 1 0 123 0 0 0 0 01 0 0 0 0 02 1 0 0 0 03 1 0 0 0 0

Linearity imposes that:

ϕ(v) = ϕ(v12 + v13 − v123)

= ϕ(v12 + v13) + ϕ(−v123)

= ϕ(v12) + ϕ(v13) + ϕ(−v123).

So, once we figure ϕ(v12), ϕ(v13) and ϕ(−v123), we also know what ϕ(v) is!

• In v12, player 3 is dummy so ϕ3(v12) = 0. Players 1, 2 are symmetric,

and using symmetry plus efficiency we obtain ϕ1(v12) = ϕ2(v12) =12.

• In v13, player 2 is dummy so ϕ2(v23) = 0. Players 1, 3 are symmetric,

and using symmetry plus efficiency we obtain ϕ1(v13) = ϕ3(v13) =12.

• In v123, all players are symmetric. Using efficiency we obtain ϕ1(−v123) =

ϕ2(−v123) = ϕ3(−v123) = −13.

Let’s summarize this on the following table:

c© LSE 2014–2020 /MA300.2 Page 32 of 66


game ϕ1 ϕ2 ϕ3

v1212

12

0v13

12

0 12

−v123 −13

−13

−13

v 23

16

16

Here again, we obtain that ϕ(v) = ψ(v)! So any concept that satisfies all

4 Shapley properties has to coincide with the Shapley value on the 3 players

simple majority game as well as on the 2 buyers-1 seller game. This is not a

coincidence, and in the next section we will see that the Shapley value is in

fact the only concept that satisfies the 4 Shapley properties.

5.4 General games

We now generalize the ideas introduced in the previous subsection from 3

players to an arbitrary number of players. Let then the set of players be N =

{1, . . . , n}. Our first task is first to define a solution concept that satisfies

additivity, the null player property and efficiency based on an ordering of the

players. Then, by selecting the ordering randomly, we define a solution ψ

concept that satisfies all 4 Shapley properties. Finally, reasoning by necessary

conditions, we show that this concept ψ is actually the unique one that

satisfies all 4 Shapley properties.

Suppose the players enter a room in some order

i1, i2, . . . , in.

Each one of the players contributes something relative to the worth of the

coalitions of the players already in the room. The contribution of the k− th

player is

v(i1, . . . , ik)− v(i1, . . . , ik−1).

More formally, a permutation π : N1:1→ N prescribes an order

(π(1), π(2), . . . , π(n)) .

c© LSE 2014–2020 /MA300.2 Page 33 of 66


Player π(1) enters first, followed by π(2), and so on . . . The set of players

that precede player i according to that order is defined as:

Pi(π) =7π(j) : j < π−1(i)

8.

The marginal contribution of player i according to π is defined by

ψπi (v) = v(Pi(π) ∪ {i})− v(Pi(π)).

Claim 1. For every π : N1:1→ N , ψπ is a point solution concept that satisfies

all of the Shapley properties except for symmetry.

Question: How can we modify ψπ to obtain a symmetric solution concept

while preserving the other properties?

Answer: We choose the permutation randomly.

Definition 23. The Shapley value is defined by

Shi(v) =1

n!

#

π∈Π(N)

ψπi (v)

where Π(N) is the set of all permutations on N.

Theorem 12. The Shapley value is the unique solution concept that satisfies

symmetry, null player property, additivity and efficiency.

Remark 1. In fact, the Shapley value satisfies all of the properties listed in

Section 5.2 as well. It means that all of these properties are implied by the

conjunction of symmetry, null player property, additivity and efficiency.

We now turn to prove Theorem 10. In one direction we have to check that

the Shapley value satisfies the 4 axioms. This direction is relatively easy, and

it is left as an exercise to the reader. In the other direction, we have to prove

that there is only one point solution concept that satisfies the 4 axioms.

Step 1. A basis for the space of games.

Given a set of players N , the set of all games over N is the vector space

RP(N).

c© LSE 2014–2020 /MA300.2 Page 34 of 66


Lemma 2. The games (uS)S∈P(N) defined by

uS(T ) =

!1 S ⊆ T

0 S ⊈ T

form a basis for RP(N). The game uS is called the unanimity game with

carrier S.

Proof. The dimension of RP(N) is equal to the number of unanimity games,

2|N |−1; therefore, it is sufficient to verify that the (uS)S∈P(N) are independant.

Assume by negation that there are coefficients (αS ∈ R)S∈P(N), not all αS =

0, s.t."

αSuS = 0. Let T be a smallest coalition s.t. αT ∕= 0. It follows

0 =

.

0#

S∈P(N)

αSuS

1

3 (T ) =#

S∈P(N)

αSuS(T )

=#

S∈P(N):S⊈T

αS · 0 +#

S∈P(N):S"T

0 · uS(T ) + αTuT (T )

= αT .

Contradiction!

Step 2. Let ϕ be a point solution concept that satisfies efficiency, null player

property and symmetry. Let α ∈ R and S ∈ P(N).

ϕi(αuS) =

!α|S| if i ∈ S

0 if i /∈ S.

Proof. If i /∈ S, then i is a null player. All the players in S are sym-

metric; thus, they have the same value x. Finally, by efficiency,

α = αuS(N) =#

j∈S

ϕj(αuS) = |S| · x.

Step 3. There is only one solution satisfying the 4 axioms.

c© LSE 2014–2020 /MA300.2 Page 35 of 66


Proof. Let (v) be a game. There are unique coefficients (αS)S∈P(N) s.t.

v = αSuS. Let ϕ be any solution concept satisfying the 4 axioms:

ϕi(v) = ϕi(#

S∈P(N)

αSuS) =#

S∈P(N)

ϕi(αSuS) =#

S∈P(N):i∈S

αS

|S|

where the first equality follows from Step 1, the second holds due to

additivity and the third follows from step 2.

c© LSE 2014–2020 /MA300.2 Page 36 of 66

6. COMPUTING THE SHAPLEY VALUE

6 Computing the Shapley Value

Reference:

Chapter:“The Shapley Value”

Recall that the Shapley value is given by

Shi(v) =1

n!

#

π∈Π(N)

v(Pi(π) ∪ {i})− v(Pi(π)).

The number of summands is huge: n!. Each summand repeats many times,as

for a given S ⊂ N \ {i} there are many ordering π such that Pi(π) = S.

Taking a random ordering π the probability that S is realized as Pi(π) de-

pends only on the size of S. The probability that player j is in position

|S|+ 1 is the same for any player j, therefore,

Pπ(|Pi(π)| = |S|) = 1

n.

Given that |Pi(π)| = |S| (equivalently, player i is in position |S| + 1), the

probability that Pi(π) = T is the same for any T ⊂ N \ {i} of size |S|,therefore,

Pπ(Pi(π) = S) =1

n· 19

n−1|S|

: .

The computation gives the formula

Shi(v) =#

S⊂N\{i}

1

n· 19

n−1|S|

: (v(S ∪ {i})− v(S))

(don’t forget the empty set!).

This formula has the practical advantage that it has “only” 2n−1 summands,

as opposed to n!. It has another advantage: it makes it easy to verify the

strong symmetry property.

Let π ∈ Π(N).

Shπ(i)(π ◦ v) =#

S∈N\{π(i)}

1

n· 19

n−1|S|

:9v(π−1(S ∪ π(i)))− v(π−1(S))

:

=#

S∈N\{π(i)}

1

n· 19

n−1|S|

:9v(π−1(S) ∪ {i})− v(π−1(S))

:

c© LSE 2014–2020 /MA300.2 Page 37 of 66


Substituting T = π−1(S), noting |T | = |S|,

Shπ(i)(π ◦ v) =#

T⊂N\{i}

1

n· 19

n−1|T |

: (v(T ∪ {i})− v(T )) = Shi(v).

Strong symmetry is very useful in computing the Shapley value for games of

symmetric structure. Consider the following example:

Example: 4-player simple monotonic game

4 3

21

Minimal winning coalitions:{1, 2}, {2, 3}, {3, 4}, {1, 4}. Since all the play-ers have the same role, it seems obvious that theyall have the same value, but one should be carefulwhen making such an argument.

Strictly speaking, player 1 and 2 are not symmetric players in the above

example, as 1 = v(14) ∕= v(24) = 0. However, there is a certain symmetry

that relates player 1 to player 2: Consider the rotation

π : i 3→ i+ 1 mod 4.

This permutation maps 1 to 2 while preserving the structure of the game.

Such a permutation is called a “symmetry” of the game.

Definition 24. Let v be a game. A permutation π ∈ Π(N) is called a

symmetry of v if v = π ◦ v: namely, v(S) = v(π−1(S)) ∀S ⊂ N .

Lemma 3. Let v be a game and i, j ∈ N . If there is a symmetry of v that

maps i to j, then i and j have the same Shapley value in v.

Proof. Suppose π is a symmetry such that π(i) = j. By the strong symmetry

property

Shj(v) = Shπ(i)(v) = Shπ(i)(π ◦ v) = Shi(v).

c© LSE 2014–2020 /MA300.2 Page 38 of 66


Example: k sellers-k buyers game

N = S ∪B, |S| = |B|.

v(T ) = min {|S ∩ T |, |B ∩ T |} .

The buyers are all symmetric, and so are the sellers. If |S| > 1, then no

pair of buyer and seller is symmetric. Nevertheless, any 1 : 1 correspon-

dence between the sellers and the buyers is a symmetry; therefore, the

sellers and the buyers have the same value 12(why is it 1

2?)

A coalitional game derived from a graph

So far, we didn’t use additivity to compute the Shapley value. In the follow-

ing, we do:

1

23

4

8

765

Given a graph G = (N ;E) (like the one above), we define a game v by

v(S) = | {(i, j) ∈ E : i, j ∈ S} |

Namely, v(S) is the number of edges contained in S.

How can we compute the Shapley value of this game?

For every edge {i, j} ∈ E, consider the unanimity game u(i,j) with carrier

{i, j}. The crux is to notice that

v =#

e∈E

ue.

c© LSE 2014–2020 /MA300.2 Page 39 of 66


By additivity, we have

Shi(v) =#

e∈E

Shi(ue) =#

e∈E

1

2· 1{i∈e} =

1

2· deg(i).

Namely, the value of i is 12the number of neighbours of i.

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7. VOTING

7 Voting

Reference: Chapter:“Voting”

Jonathan K. Hodge and Richard E. Klima. The Mathematics of Voting and

Elections: A Hands-On Approach. American Mathematical Society, Provi-

dence, R.I., 2005.

In this chapter we study election systems. What is a good, or “demo-

cratic” system? Can we define this mathematically? We shall see that the

situation is quite different depending on the number of candidates. With

two candidates, our life is relatively easy. Whereas with three candidates or

more, it is quite difficult (perhaps even impossible!) to find a system with

all desired properties.

7.1 Election between two candidates

Suppose the class has to decide between two restaurants for dinner: Chinese

or Indian. How should we make this collective choice?

Let us first look at some funny rules first, then discuss why we

wouldn’t like to rely on them:

• Dictatorship One of the students takes the lead and decides for

everyone else. Not exactly fair.

• Imposed rule No matter what students prefer, the Indian restau-

rant gets selected. Not exactly representative of student’s prefer-

ences.

• Minority rule All the students vote, and the choice with the least

vote gets selected.

All of these are voting systems, i.e. the way in which the winner of an

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7. VOTING

election is determined from the individual ballots. However, none of these

systems would be considered a good one, for legitimate reasons.

The data of the problem is as follows. We have a set of alternatives {a, b}with two elements. This is the set of alternatives the group has to decide

on. These can be electoral parties, restaurants, binary referendum options,

. . . There is a group of N agents, and each agent i ∈ N either prefers a to b,

or prefers b to a. Since each agent has two possible preferences, there are in

total 2N preference profiles, describing the preferences of all the agents. We

represent agent i’s preference by his or her preferred candidate, a or b. A

voting system V consists of a mapping the set {a, b}N of possible preference

profiles to the set {a, b}. Hence, it prescribes a collective decision, a or b, for

each possible preference profile.

In order for us to objectively determine what a “good” voting system

should look like, let us determine a series of mathematical properties such a

system should have.

First, a voting system should treat all of the voters equally. So if any two

voters trade ballots, the outcome of the election should remain the same.

This means that the system should be anonymous. In particular, such a

system should depend on the number of people who prefer one alternative to

the other, but not on their names. In practice, this means that the outcome

of V should only depend on the number of agents who favour a over b and

on the number of agents who favour b over a.

Second, a voting system should not favour one option over the other.

Both options should be treated symmetrically. Mathematically, we say that

a voting system is neutral if, whenever for a certain profile of preferences, op-

tion x is chosen, then for the opposite preference profile (where all preferences

are reversed), option the other option is chosen.

Third, voting in favour of an option can only make it more likely to be

chosen. More precisely, if for a certain preference profile where agent i likes

option x better, y is chosen, then in the modified preference profil in which

i likes option y better, y should also be chosen. If this property is true, we

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7. VOTING

say that V is monotone.

Consider all three “funny” systems above. Which ones are anony-

mous, neutral, monotone?

Consider now the simple majority system, for odd values of N . We count

the number of agents who prefer a over b, and the number of agents who

prefer b over a. We choose then choose the option which is preferred by the

most agents. Which ones of the three properties is satisfied by the simple

majority rule?

Theorem 13 (May’s Theorem). In a two-candidate election with an odd

number of voters, simple majority rule is the only voting system that is anony-

mous, neutral, and monotone.

Proof. Since the system is anonymous, we know that the outcome of the

voting system only depends on the number, say k of agents who prefer a over

b. We need to show that V outputs a when k > N/2, and b otherwise.

Let us ask this simple question, what is the outcome of V when k = N+12

?

Can this be b? In this case by symmetry, V outputs a when k = N−12

. But

this contradicts the fact that V is monotone.

We have then established that the outcome of V when k = N+12

is a. Since

V is monotone, this means that the outcome of V is also a when k > N+12

.

Now, by symmetry, the outcome of V is b when k < N−12

.

By necessary conditions, we have shown that a voting system that is

anonymous, neutral, and monotone, can only be simple majority rule. Since

we already know that simple majority has all these properties, we have es-

tablished May’s theorem.

7.2 Election between any number of candidates

Consider the following situation: A committee of 21 members has to select

one out of three candidates A, B, C. Their preferences are shown in Table 1.

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7. VOTING

# committee members 1st choice 2nd choice 3rd choice1 A B C7 A C B7 B C A6 C B A

Table 1: Preferences of a committee with 21 members

How should the winner be selected? Here again, there are several possi-

bilities. For instance.

• Dictatorship: one of the members decides.

• Democracy: the majority decides. How? (≥ 2 alternatives). Many

rules are possible.

• Pairwise majority vote (Condorcet method): candidate C beats any

other alternative. She is called the Condorcet winner.

Consider now a different committee that again has to select a candidate out of

three alternatives A, B, C (with preferences represented in Table 2). Observe

# committee members 1st choice 2nd choice 3rd choice23 A B C2 B A C17 B C A10 C A B8 C B A

Table 2: Preferences of committee with 60 members

that A beats B, B beats C and C beats A in pairwise comparisons. Thus,

pairwise comparison can be cyclic, which means that there is no Condorcet

winner in this case. This type of situation is called a Condorcet paradox.

Let’s go back to the preferences depicted in Table 2. Let’s examine dif-

ferent voting systems and their results for the preferences in Table 1.

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7. VOTING

• 1st choice majority: In this system, for each of the candidates, we

count the number of voters who rank this candidate as first in order of

preferences. The candidate with the highest number (if unique) is then

declared the winner. With this system, A wins.

• 2-round majority: In the first stage, each agent votes for his/her pre-

ferred candidate, and two candidates with the most votes make it to

second round. Then, in the second round, simple majority voting takes

place between the two remaining candidates. In our example, C is

eliminated in the first round, and A wins gets elected in the second

round.

• (Borda method) Everyone gives a number of points to each candidate:

2 points for their preferred candidate, 1 point for the median one, and 0

for the least preferred one. With this system, A comes first 8 times and

last 13 times, gets 16 points. B comes first 7 times, second 7 times, and

third 7 times, gets 21 points. C comes first 6 times, second 14 times,

and third 1 time, get 26 points. C wins.

Now consider the 1st choice majority vote. Can committee members ben-

efit from misrepresenting their preferences? Yes! If everyone reports their

true preferences then A wins. If the 8 “CBA” members report “BCA”, then

B will be selected.

We now formalize a couple of definitions.

Definition 25. A strict preference relation P is a binary relation over A

such that ∀a, b, c ∈ A

• a >P b, b >P c ⇒ a >P c (transitivity),

• a ≯P a (irreflexivity),

• a >P b or b >P a or a = b (completeness).

Definition 26. A (weak) preference relation over A is a binary relation P ∗


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7. VOTING

• a >P ∗ b, b >P ∗ c ⇒ a >P ∗ c (transitivity),

• a ≥P ∗ a (reflexivity),

• a ≥P ∗ b or b ≥P ∗ a (completeness).

If P is a strict preference relation over A we denote

a ≥P b

to say that “a >P b or a = b”. If P ∗ is a (weak) preference relation over A

we denote

a >P ∗ b

to say that “a ≥P ∗ b and b ≱P ∗ a”.

Definition 27. The components of a social choice problem are

• A - finite set of alternatives

• N - set of individuals

• (Pi)i∈N - a profile of strict preference relations over A.

Definition 28. Given a social choice problem (Pi)i∈N , an alternative a ∈ A

is a Condorcet winner if for every alternative b ∕= a,

|{i, a >Pib}| > N/2.

Given a social choice problem (Pi)i∈N . According to the Borda method,

each agent i attributes to alternative a a number of points equal to

ni(a) = |{b, a >Pib}|,

thus no points to the least liked alternative, one point to the second disliked,

and so on... and |A| − 1 points to the preferred alternative. The total

number of points to alternative a is: N(a) ="

i∈N ni(a), and the Borda

method selects the alternative that maximizes N(a) when it exists.

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8. SOCIAL CHOICE

8 Social Choice

Chapter:“Social Choice”

Main Question:

Is there a function that outputs the “preference of society” taking

as input the preferences of all the individuals and aggregating them

in a way that avoids such paradoxes?

Definition 29. A strict preference relation P is a binary relation over A


• a >P b, b >P c ⇒ a >P c (transitivity),

• a ≯P a (irreflexivity),

• a >P b or b >P a or a = b (completeness).

Definition 30. A (weak) preference relation over A is a binary relation P ∗


• a >P ∗ b, b >P ∗ c ⇒ a >P ∗ c (transitivity),

• a ≥P ∗ a (reflexivity),

• a ≥P ∗ b or b ≥P ∗ a (completeness).

If P is a strict preference relation over A we denote

a ≥P b

to say that “a >P b or a = b”. If P ∗ is a (weak) preference relation over A

we denote

a >P ∗ b

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8. SOCIAL CHOICE

to say that “a ≥P ∗ b and b ≱P ∗ a”.

A list of strict preference relations PN = (Pi)i∈N , one for each individual, is

called a strict preference profile. The set of all strict preference relations over

A is denoted by P (A). The set of all strict preference profiles is (P (A))N .

Similarly, the set of all weak preference relations is denoted by P ∗(A).

A function F : (P (A))N → P ∗(A) is called a social welfare function (SWF).

Example: the Borda SWF. According to the Borda SWF, each agent

i assigns to each alternative a number of points equal to

ni(a) = |{b, bPia}|.

The total number of points received by alternative a is then N(a) ="

i∈N ni(a). The social choice function F is such that a >F (P ) b if and

only if N(a) > N(b).

Let’s try to rank the best food types in the world. Consider everyone’s

preferences in the class, can you think of a way to rank food types from

each of these preferences?

For instance, each member can vote for his or her preferred food type.

Each food type gets a number of points equal to the number of votes in

its favor, and then food types are ranked by decreasing number of points.

Is this a good system? Typically, one problem encountered with this

type of systems, as with many others, is that the ranking between two

food types depends on the presence of a third food type in the contest or

not.

Assume that 6 members have preferences Chinese > Indian > French,

5 have preferences are Indian > Chinese > French, and 2 have preferences

French > Indian > Chinese. The proposed system gives the ranking

Chinese > Indian > French. But what happens if French food, which is

not a very serious contender in the competition, is taken out? We see

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8. SOCIAL CHOICE

that a majority prefers Indian to Chinese, so that Indian > Chinese.

One idea in looking for a ranking that aggregates individual prefer-

ences is that the ranking between to alternatives should not depend on

the presence of a third one. It should depend only on how agents rank

these two alternatives themselves. This idea is called Independence of

Irrelevant Alternatives.

Properties of a SWF

A list of strict preference relations PN = (Pi)i∈N , one for each individual, is

called a strict preference profile. The set of all strict preference relations over

A is denoted by P (A). The set of all strict preference profiles is (P (A))N .

Similarly, the set of all weak preference relations is denoted by P ∗(A).

• A social welfare function F is dictatorial is there is an individual i ∈ N

such that

F (PN) = Pi ∀PN ∈ (P (A))N .

Player i is called a dictator.

• A SWF F satisfies unanimity, if for every a, b ∈ A and every profile

PN ∈ (P (A))N :

(a >Pib ∀i ∈ N) ⇒ a >F (PN ) b.

• A SWF F satisfies independence of irrelevant alternatives (IIA) if:

∀a, b ∈ A ∀PN , QN ∈ (P (A))N

(a >Pib ⇔ a >Qi

b ∀i ∈ N) ⇒ (a ≥F (PN ) b ⇔ a ≥F (QN ) b).

In other words: the way PN relates to a to b depends only on the way

the individuals relate a to b (and not on how they relate a or b to any

other alternative).

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8. SOCIAL CHOICE

Theorem 14. (Arrow, 1953) Suppose |A| ≥ 3. Any SWF that satisfies

unanimity and IIA is dictatorial.

The proof of Arrow’s impossibility theorem above is not straightforward at

all. The idea of the proof is to generalize the notion of a dictator to sets that

dictate the outcome, called “decisive sets”, and then to study the structure

of the collection of all decisive sets.

Definition 31. Let F be a SWF, and a, b ∈ A. A set (coalition) S ⊂ N is

called decisive for a over b (relative to F) if for every PN ∈ (P (A))N :

if a >Pib ∀i ∈ S

and b >Pja ∀j /∈ S,

then: a >F (PN ) b.

S is called strongly decisive if it is decisive for all a over b for all a, b ∈ A.

Lemma 4. If a SWF F satisfies unanimity then N is strongly decisive rel-

ative to F and ∅ is not.

Proof. Immediate from the unanimity property.

Lemma 5. If a SWF F satisfies unanimity and IIA, and |A| ≥ 3, then

S ⊂ N is strongly decisive if and only if it is decisive for a over b for some

a, b ∈ A.

Proof. “only if” is obvious.

Suppose S is decisive for a over b, and let c ∈ A \ {a, b}.1) We first show that S is decisive for a over c. Consider the profile of

preferences P below. By unanimity we have b >F (P ) c, and since S is decisive

S a b cN \ S b c aF (P ) a b c

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8. SOCIAL CHOICE

for a over b, a >F (P ) b.

2) We now show that S is decisive for b over c. Consider the profile of

preferences P below.

S b a cN \ S c b aF (P ) b a c

We already have shown S is decisive for a over c, thus a >F (P ) c. Since

F satisfies unanimity, b >F (P ) a, and by transitivity a >F (P ) c. By IIA,

whenever all elements of S prefer b to c and elements not in S prefer c to b,

F (P ) prefers b to c.

Now, let x, y ∈ A be any two distinct alternatives.

• If x = a, by 1) we deduce that S is decisive for a over y, therefore for

x over y.

• If x ∕= a, y ∕= a, by 1) since S is decisive for a over b, it is also is decisive

for a over x. This by 2) implies that S is decisive for x over y.

• If x ∕= a, y = a, then let z ∈ A \ {x, y}. From 1), S is decisive for y

over z. From 2) S is decisive for z over x, and still by 2) S is decisive

for x over y.

This completes the proof of Lemma 5.

Throughout the rest of the proof we assume that F satisfies unanimity

and IIA, and |A| ≥ 3. Our plan is to prove 3 properties of strongly decisive

sets:

(i) ∅ is not strongly decisive, N is strongly decisive;

(ii) If V is strongly decisive and V ⊂ U , then U is strongly decisive.

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8. SOCIAL CHOICE

(iii) If U and V are not strongly decisive then U∪V is not strongly decisive.

We have already shown (i). We next prove (ii).

Claim 2. If V is strongly decisive and V ⊂ U , then U is strongly decisive.

Proof. Let a, b, c ∈ A be three distinct alternatives. Consider the profile P

below. By unanimity we have a >F (P ) b, and since V is strongly decisive

V a b cU \ V a c bN \ V c a bF (P ) a b c

b >F (P ) c. If follows that a >F (P ) c therefore, U is decisive for a over c.

Therefore, U is strongly decisive.

Claim 3. If U and V are not strongly decisive then W = U ∪ V is not

strongly decisive.

Proof. Let U ′ = U ∩ V c, by (ii) we have that U ′ is not strongly decisive.

We also have W = U ′ ∪ V , and U ′ ∩ V = ∅. Now consider the profile P

below. Since U ′ is not strongly decisive, we have a ≥F (P ) b and since V is

U ′ b c aV c a b

N \W a b cF (P ) a b c

not strongly decisive we have b ≥F (P ) c. It follows that a ≥F (P ) c, but this

shows that W is not decisive for c over a.

Having proved (i)− (iii) we are now ready to conclude the proof: The collec-

tion of strongly decisive coalitions is not empty, by (i). It contains a smallest

coalition which must be a singleton, by (iii). Suppose {j} is a minimal

coalition, then by (ii) j is a dictator.

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9. STABLE MATCHING

9 Stable Matching

Reference:

Chapter :“Stable Matching”

Examples of Matchings

• Matching doctors and hospitals

• Matching students and high schools

• Matching kidneys and patients

• Marriage and dating markets

• Job assignments in firms

Stable Marriage

There are n single men and n single women. Each person ranks those of the

opposite sex in accordance to his or her preferences for a marriage partner.

Problem: How should they match?

Definition 32. A matching with the requirement that there is no pair of a

man and a woman such that each one of them prefers the other to his actual

mate is called a stable matching.

Question: Is it possible to find a stable matching, for any pattern of prefer-

ences?

Consider the following lists of preferences: The men are denoted by α, β

and γ, the women by A, B, C. An element (i, j) in the Table indicates

the position of the woman/man in the preference list of the corresponding

man/woman. For instance, the first element in the first row, (1, 3), means

that A is on the first position in the preference list of α, and that α is on the

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9. STABLE MATCHING

A B Cα (1,3) (2,2) (3,1)β (3,1) (1,3) (2,2)γ (2,2) (3,1) (1,3)

Table 3: Lists of preferences of three men and three women

last position in the preference list of A.

Examples of matchings:

• All men get their first choice: (α, A), (β, B), (γ, C)

• All women get their first choice: (α, C), (β, A), (γ, B)

• Everybody gets their second choice: (α, B), (β, C), (γ, A)

Consider a different example with 4 men and 4 women: A unique stable

A B C Dα (1,3) (2,3) (3,2) (4,3)β (1,4) (4,1) (3,3) (2,2)γ (2,2) (1,4) (3,4) (4,1)δ (4,1) (2,2) (3,1) (1,4)

Table 4: Lists of preferences of four men and four women

matching in this example would be the following:

(α, C), (β, D), (γ, A), (δ, B).

Theorem 15. There always exists a stable matching.

In order to prove the result, we present an algorithm that eventually yields

a stable matching:

Deferred Acceptance Algorithm

• In the beginning, no one is matched.

• As long as there are unmatched men, each man proposes to his favorite

woman among those to whom he has not proposed previously.

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9. STABLE MATCHING

• Each woman retains the best proposal among the previous man she was

tentatively matched if any and her new proposals. Other proposals are

dismissed, and becomes unmatched, again.

Let’s apply the algorithm to the matching problem in Table 4. In a first

step, α proposes to A, β proposes to A, γ proposes to B and δ proposes

to D. A dismisses β, so β is unmatched again and proposes to D. Now,

D dismisses δ, δ becomes unmatched and proposes to B, etc. This goes on

until no man is dismissed anymore and the following (stable) matching is

established: (α, C), (β, D), (γ, A), (δ, B).

Why does the algorithm work?

• The algorithm terminates:

In every step a new entry of the matrix is marked; therefore the algo-

rithm must terminate within n2 steps.

• In the end, everyone is matched:

If there is an unmatched man, it is because he has been rejected by

every woman. A woman rejects a man only if she is matched to another

man, so that means that the women must all be matched and since the

number of women is equal to the number of men, the men must all be

matched, as well. Contradiction!

• The resulting matching is stable:

Suppose there is a man and a woman, say John and Marry, who prefer

each other to their proposed mates. Since the men propose to women

in descending preference order, John must have proposed to Marry at

some stage before the procedure terminated. Since John and Marry are

not matched, she must have rejected him in favour of a more preferable

man, replacing the latter only with an even more preferable man, and

thus ending up with someone she prefers to John. Contradiction!

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9. STABLE MATCHING

Example of DA in action: clinical psychology

U.S. clinical psychologists are employed as interns after they complete

their doctoral degrees. About 500 sites offer 2,000 positions each year.

The market fully clears on one day. On selection day, market opens

at 9 am, closes at 4 pm.

While market is open, offers are made and accepted according to a

version of the DA algorithm, except that it is a human version where

people make phone calls.

Offers can be accepted early and programs often ask students to in-

dicate in advance their willingness to accept an offer. Roth and Xing

(1994) describe their visit to a site in 1993.

The program at this given site had 5 positions, 71 applicants, gave 29

interviews. Directors had ranked 20 applicants, and knew 6 would say

yes if asked. Their strategy: “don’t tie up offers with people who will

hold them”.

The selection day unfolded like this:

• At 9:00, calls are placed to candidates 1, 2, 3, 5, and 12. Candidates

3, 5, 12 accept.

• Candidate 1 is reached at 9:05, holds until 9:13, rejects.

• In the interim, candidate 8 calls, says she will accept if an offer is

made to her.

• When 1 rejects, a call is placed to applicant 8, who accepts.

• While call is in progress, candidates 2 calls to reject the offer.

• A call is placed to 10 (who’d indicated acceptance), who accepts at

9:21.

• By 9:35, remaining candidates were informed of non-offer.

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9. STABLE MATCHING

The deferred acceptance algorithm ensures stability, but it does not guar-

antee happiness and uniqueness. When there are several stable matchings,

individuals prefer some matching to another.

Definition 33. A stable matching µ is called men (resp. women) optimal if

for every stable matching µ′ and every man (resp. woman) who is matched

differently in µ and µ′, that man (resp. woman) prefers his (her) mate in µ

to his (her) mate in µ′.

Question: Are there men (women) optimal matchings?

Theorem 16. There always exists a man optimal stable matching and the

man courtship algorithm produces it.

Proof. In order to prove the above theorem, we introduce the following defi-

nition:

Definition 34. Woman A is possible for a man α, if there exists a stable

matching in which A and α are matched.

It now remains to prove that men are rejected by impossible women only.

Assume the claim was false. Let α be the first man, in order of applying the

algorithm, who is rejected by a possible woman A. When α is rejected, he is

replaced by another man β whom A prefers to α.

Since α is the first man to be rejected by a possible woman, the women

that β ranks higher than A are all impossible for β.

Now let µ be a matching in which A and α are matched, and assume that

in this matching, B is matched to β. Since β is possible for B, β prefers A

to B. But we also know that A prefers α to β. This is a contradiction.

Corollary 4. There always exists a woman optimal stable matching.

Theorem 17. Among all stable matchings, the men optimal matching is the

worst for all women. Similarly, the women optimal matching is the worst for

men among all stable matchings.

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9. STABLE MATCHING

Proof. Assume that woman A is matched to α in the man optimal matching.

Consider any other matching, µ, in which A is matched to some man β, and

α is matched to some woman B. Since α prefers A to B, and (α, A) cannot

object in µ, it must be the case that B prefers β to α.

Therefore, any woman is better off in any other matching than in the

man optimal one.

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10. MATCHING: EXTENTIONS

10 Matching: Extentions

The possibility to remain single

Men and women may remain single. All men have preferences on the set

W ∪ {∅} where ∅ represents remaining single. All women similarly have

preferences on the set M ∪ {∅}. The number of men is not necessarily equal

to the number of women.

A matching is a mapping µ : M → W ∪ {∅} so that for every w ∈ W ,

µ−1(w) is either the empty set or a singleton.

A matching is stable if no pair m,w would rather be together than with

their respective matches and that no one would rather remain remain single

than being with their match. We thus say that

• A pair (m,w) objects to µ if both m >w µ−1(w) and w >m µ(m)

• A man m objects to µ if ∅ >m µ(m)

• A woman w objects to µ if ∅ >w µ−1(w)

and the matching µ is stable if no pair (m,w), no man m, no woman w

objects to it.

We say that a man is acceptable for a woman if she prefers this man to

being single, and similarly that a woman is acceptable for a man if he prefers

this woman to being single.

We modify the men courtship algorithm to take into account the possi-

bility to remain single.

Deferred Acceptance Algorithm with single people

• In the beginning, no one is matched.

• Each man proposes to his favorite acceptable woman among those to

whom he has not proposed previously.

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• Each woman retains the best acceptable man the previous man she

was tentatively matched and her new proposals. Other proposals are

dismissed, and becomes unmatched again.

• The algorithm terminates when there are no unmatched men who have

not yet made offers to acceptable women.

Example. Consider 3 men α, β, γ and two women A,B. We represent,

for each man and woman, his or her preferences among acceptable matches.

α : A,B

β : A

γ : B,A

A : γ, β,α

B : α, γ.

The resulting match is (α, B), (β, ∅), (γ, A) and it is stable.

The following results extend the case of 1-1 matchings:

Definition 35. A stable matching µ is called men (resp. women) optimal if

for every stable matching µ′ and every man (resp. woman) who is matched

differently in µ and µ′, that man (resp. woman) prefers his (her) mate in µ

to his (her) mate in µ′.

Theorem 18. There always exists a man optimal stable matching and the

man courtship algorithm produces it.

Corollary 5. There always exists a woman optimal stable matching.

Theorem 19. Among all stable matchings, the man optimal matching is the

worst for all women. Similarly, the women optimal matching is the worst for

men among all stable matchings.

We have a new result about the set of men and women who stay un-

matched.

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Theorem 20 (Rural Hospital Theorem). The set of women and men that

are unmatched is the same for all stable matching.

The name comes from the following consideration about the allocation

of residents in rural hospitals. Hospitals in rural areas have extra difficulties

filling positions for residents, and some people have argued that the matching

mechanism used should chosen in order to have more doctors in rural hospi-

tals. The result shows that this is impossible to do if one wants to implement

a stable matching.

Another way to look at the result is that some students may or may

not be allocated to schools depending on the mechanism. The result says

that one should not worry about this as long as only stable matchings are

considered.

Proof. Let µm be the man’s optimal matching, and let µ be any other match-

ing. Let Mm (Wm) be the set of matched men (women) under the man’s

optimal matching, and let Mµ (W µ) be the set of matched men (women)

under µ.

Since µm is optimal for men, the men who are matched under µ are also

matched under µm. Indeed, if a man is matched under µ, this man prefers

his match under µ than being single, and cannot be worse-off under µ than

under µm, thus this man is also matched under µ. Therefore Mµ ⊆ Mm.

Similarly, since µm is the worst for women, the women matched men

under µm are also matched under µ, Wm ⊆ W µ.

Thus,

|Mµ| ≤ |Mm|, |Wm| ≤ |W µ|.

But note that |Mµ| = |W µ| and |Mm| = |Wm|. Hence,

|Mµ| ≤ |Mm| ≤ |Mµ|,

the number of matched men and matched women under µ and under µm are

the same: |Mµ| = |Mm|. Since Mµ ⊆ Mm, the set of matched men and

matched women are also the same under µ and µm.

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Strategic considerations

Consider the list of preferences of table 5 with 3 men and 3 women.

• What is the resulting matching using the men’s courtship algorithm?

• What is the resulting matching using the women’s courtship algorithm?

• Assume that A behaves strategically and behaves as if her preferences

were γ >A β >A α. What is then the result of the men’s courtship

algorithm?

A B Cα (1,2) (2,1) (3,2)β (1,3) (3,3) (2,1)γ (2,1) (1,2) (3,3)

Table 5: Lists of preferences of 3 men and 3 women

Theorem 21. In the man’s courtship algorithm, no man can get a better

match by mis-representing his preferences. In particular, if there are more

women than men, a man cannot get matched by mis-representing his prefer-

ences.

Proof. Assume that α can gain by misrepresenting his preferences, and >α′

are the preferences that allow α to end with the best woman possible among

all preferences that he can report. Denote by µ′ the matching when α reports

>α′ . Assume that α is matched to A when reporting his true preferences,

and to B when reporting >α′ .

We show that, under the true preferences, µ′ is a stable matching. No

man except α can be part of a objecting pair, since µ′ is stable under their

preferences. Assume that (α, C) object. This means that α could have been

matched to C by proposing to C before B, which contradicts the fact that

B is the best possible match for him under all possible strategies.

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Finally, since µ′ is a stable matching, and µ is the man-optimal matching,

it must be the case that α prefers A to B. Hence no man can be better off

by mis-representing his preferences.

Consider the situation in which women and men engage in the men’s

courtship algorithm and men represent their preferences accurately. Women

may misrepresent their preferences. This is a game played by women in which

women decide which preferences to reveal, and each woman’s objective is to

get matched to to the best man according to their preferences. We call this

game the men’s courtship game with strategic women.

Theorem 22. All Nash equilibria of the men’s courtship game with strategic

women lead to a stable matching. There exists a Nash equilibrium in which

all the women get matched to their women optimal stable matching.

Proof. We first prove that if women’s strategies form a Nash equilibrium,

then the resulting matching is stable under the true preferences. Consider

a situation in which women’s strategies lead to woman A being matched to

a man α, while there exists a man β who gets matched to a woman B such

that β >A α and A >β B. Then it must be the case that β proposes to A

before B, hence by revealing her true preferences A would get matched to β

or better for her. The original women’s strategies therefore cannot constitute

a Nash equilibrium.

Now consider the woman’s strategies in which all women announce that

they prefer their woman’s optimal match to remaining single, and remaining

single to any other man. All men prefer their woman’s optimal match to

remaining single. Therefore each man gets dismissed until he proposes to her

woman optimal match, if any, and the algorithm produces the woman optimal

match. We now show that this is a Nash equilibrium. If a woman w mis-

represents her preferences, all men she prefers to her woman optimal match

either prefers to remain single, or prefer another woman to w. Therefore,

no matter what preferences w uses, these men are either going to remain

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single or to be matched to other women. There is no profitable deviation for

woman w.

Strategic proofness

We saw that the man courtship algorithm is such that men do not want to

mis-represent their preferences, but women might want to. We are asking if

we can find a way to create stable matchings such that neither side way want

to mis-represent their preferences.

A mechanism is a rule that inputs preferences for men and women and

produces a matching. A mechanism is strategy-proof if neither men nor

women can benefit by mis-representing their preferences, no matter what

profile of preferences other agents report. A mechanism is stable if, for any

profile of preferences reported, it produces a matching that is stable for this

profile of preferences.

The man courtship algorithm is stable, but it is not strategy-proof. Unfor-

tunately, we cannot overcome the difficulty by considering other mechanisms.

Theorem 23. There is no mechanism that is stable and strategy-proof.

Proof. To prove the Theorem it suffices to produce an example showing im-

possibility. Consider two men α, β and two women A,B. Each person prefers

to be matched than to remain single, and preferences among possible matches

are as follows.

α : A,B

β : B,A

A : β,α

B : α, β

Under these preferences, there are two stable matchings: One is (α, A), (β, B)

and the other one is (α, B), (β, A).

Assume the mechanism produces (α, A), (β, B) if men and women an-

nounce their true preferences. Then, A could announce that she prefers β to

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being single, and remaining single to α. The profile announced in this case

is:

α : A,B

β : B,A

A : β

B : α, β

For these preferences, there is only one stable matching which is (α, B), (β, A),

thus it must be produced by the mechanism. Hence, A has incentives to de-

viate, the mechanism is not strategy proof.

Similarly, if, under the true preferences, the mechanism produces (α, B), (β, A),

α could announce that he prefers A to remaining single and remaining single

to B. In this case, the announced preferences would be:

α : A

β : B,A

A : β,α

B : α, β

One can check that the only stable matching in this case is (α, A), (β, B).

Since this matching is preferred by A to the one produced under the true

preferences, the mechanism is not stable!

The core of the marriage game

Now, we would like to relate the marriage game to familiar notions of coali-

tional games. To this end, we need a a formal definition of the matching

game. We use the model of non-transferable utility (NTU) games.

Let N be the set of players, X the set of outcomes. Each player i ∈ N has

a complete preference relation, ≼i, over the outcomes. Each coalition S can

bring about a set of outcomes v(S) ⊆ X. A NTU game is then represented

by the tuple

〈N,X, (≼i)i∈N , v : P(N) → X〉 .

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The matching game

• The players are the men and women (not necessarily of equal numbers).

• A set of outcomes is the set of (partial) matchings. Some men and

women are matched and the others are not. We consider the unmatched

players as being matched to themselves.

• Each person has a complete strict ranking of the members of the oppo-

site sex and himself. This ranking extends to a preference over partial

matchings.

• The matchings that a coalition can bring about by itself are those that

involve only pairs in that coalition.

The core of NTU games

The core consists of the outcomes that the grand coalition can bring about

(efficiency) and no coalition can block (coalitional rationality). Formally,

core = {x ∈ v(N) : ∀S, ∀y ∈ v(S), ∃i ∈ S s.t. y ≼i x} .

(Or, using a different terminology: y does not dominate x via S.)

Theorem 24. The core of the matching game is the set of all stable match-

ings.

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Written by: Professor Olivier Gossner

ma300.2: cooperative game theory - pku.edu.cn

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