probabilistic bisection search 0.1cm for stochastic root …probabilistic bisection search for...

Probabilistic Bisection Searchfor Stochastic Root-Finding

Rolf Waeber Peter I. Frazier Shane G. Henderson

Operations Research & Information EngineeringCornell University, Ithaca, NY

Sunday October 14, 2012

INFORMS Annual MeetingPhoenix, AZ

This research is supported by AFOSR YIP FA9550-11-1-0083.

Rolf Waeber Probabilistic Bisection Search for Stochastic Root-Finding

Stochastic Root-Finding Problem

0 1

0

g(x)

X*

• Consider a function g : [0, 1]→ R.• Assumption: There exists a unique X∗ ∈ [0, 1] such that

• g(x) > 0 for x < X∗,• g(x) < 0 for x > X∗.

Goal: Find X∗ ∈ [0, 1].

• Can only observe Yn(Xn) = g(Xn) + εn(Xn), where εn(Xn) is anindependent noise with zero mean (median).Decisions:

• Where to place samples Xn for n = 0, 1, 2, . . .

• How to estimate X∗ after n iterations.

Stochastic Root-Finding Probabilistic Bisection Algorithm Analysis Conclusions References 2/32



0 1

0

g(x)

X*


• g(x) > 0 for x < X∗,• g(x) < 0 for x > X∗.

Goal: Find X∗ ∈ [0, 1].• Can only observe Yn(Xn) = g(Xn) + εn(Xn), where εn(Xn) is anindependent noise with zero mean (median).

Decisions:• Where to place samples Xn for n = 0, 1, 2, . . .





0 1

0

Yn(X

n)

X*


• g(x) > 0 for x < X∗,• g(x) < 0 for x > X∗.

Goal: Find X∗ ∈ [0, 1].• Can only observe Yn(Xn) = g(Xn) + εn(Xn), where εn(Xn) is anindependent noise with zero mean (median).

Decisions:• Where to place samples Xn for n = 0, 1, 2, . . .





0 1

0

Yn(X

n)

X*


• g(x) > 0 for x < X∗,• g(x) < 0 for x > X∗.

Goal: Find X∗ ∈ [0, 1].• Can only observe Yn(Xn) = g(Xn) + εn(Xn), where εn(Xn) is anindependent noise with zero mean (median).Decisions:

• Where to place samples Xn for n = 0, 1, 2, . . .

• How to estimate X∗ after n iterations.Stochastic Root-Finding Probabilistic Bisection Algorithm Analysis Conclusions References 2/32


Applications

• Simulation optimization:• g(x) as a gradient

• Sequential statistics:

• Finance:• Pricing American options• Estimating risk measures



Stochastic Approximation (Robbins and Monro, 1951)

0 1

0

g(x)

X*

1. Choose an initial estimate X0 ∈ [0, 1];

2. Determine a tuning sequence (an)n ≥ 0,∑∞

n=0 a 2n <∞, and∑∞

n=0 an =∞.(Example: an = d/n for d > 0.)

3. Xn+1 = Π[0,1](Xn + anYn(Xn)), where Π[0,1] is the projection to [0, 1].



A “Good” Tuning Sequence

0 200 400 600 800 10000

1

n

Xn

X*

an = 1/n, ε

n ~ N(0,0.5)



A “Bad” Tuning Sequence – too Large

0 200 400 600 800 10000

1

n

Xn

X*

an = 10/n, ε

n ~ N(0,0.5)



A “Bad” Tuning Sequence – too Small

0 200 400 600 800 10000

1

n

Xn

X*

an = 0.5/n, ε

n ~ N(0,0.5)



Lack of Robustness – εn with Heavy-Tails

0 200 400 600 800 10000

1

n

Xn

X*

an = 1/n, ε

n ~ t

3



A Different Approach

What about a bisection algorithm?

0 1

0

g(x)

X*

• Deterministic bisection algorithm will fail almost surely.• Need to account for the noise.



The Probabilistic Bisection Algorithm



The Probabilistic Bisection Algorithm (Horstein, 1963)

• Input: Zn(Xn) := sign(Yn(Xn)).• Assume a prior density f0 on [0, 1].

0 10

1

2

fn(x)

n = 0, Xn = 0.5, Z

n(X

n) = −1

X*

Xn





0 10

1

2

fn(x)

n = 0, Xn = 0.5, Z

n(X

n) = −1

X*

Xn

0 10

1

2

fn(x)

n = 1, Xn = 0.38462, Z

n(X

n) = −1

X*

Xn





0 10

1

2

fn(x)

n = 0, Xn = 0.5, Z

n(X

n) = −1

X*

Xn

0 10

1

2

fn(x)

n = 1, Xn = 0.38462, Z

n(X

n) = −1

X*

Xn

0 10

1

2

fn(x)

n = 2, Xn = 0.29586, Z

n(X

n) = 1

X*

Xn





0 10

1

2

fn(x)

n = 0, Xn = 0.5, Z

n(X

n) = −1

X*

Xn

0 10

1

2

fn(x)

n = 1, Xn = 0.38462, Z

n(X

n) = −1

X*

Xn

0 10

1

2

fn(x)

n = 2, Xn = 0.29586, Z

n(X

n) = 1

X*

Xn

0 10

1

2

fn(x)

n = 3, Xn = 0.36413, Z

n(X

n) = 1

X*

Xn



Stochastic Root-Finding Revisited

0 1

0

g(x)

X*

Zn(Xn) =

{sign (g(Xn)) with probability p(Xn),

−sign (g(Xn)) with probability 1− p(Xn).

• The probability of a correct sign p(·) depends on g(·) and the noise(εn)n.

• Stylized Setting:• p(·) is constant.• p(·) is known.




0 1

0

g(x)

X*

0 10

0.5

1

p(x)X*

Zn(Xn) =








0 1

0

g(x)

X*

0 10

0.5

1

p

p(x)X*

Zn(Xn) =




• Stylized Setting:• p(·) is constant.

• p(·) is known.




0 1

0

g(x)

X*

0 10

0.5

1

p

p(x)X*

Zn(Xn) =







Stylized Setting• g(x) is a step function with a jump at X∗, for example, in edgedetection applications (Castro and Nowak, 2008).

• Sample sequentially at point Xn and use Sm(Xn) =∑m

i=1 Yn,i(Xn) toconstruct an α-level test of power 1 (Siegmund, 1985):

Nn = inf{m : |Sm| ≥ [(m + 1)(log(m + 1) + 2 log(1/α))]1/2

}.

Then PXn=X∗ {Nn <∞} ≤ α, PXn 6=X∗ {Nn <∞} = 1, andPXn<X∗ {SNn(Xn) > 0} ≥ 1− α/2 = pc ,PXn>X∗ {SNn(Xn) < 0} ≥ 1− α/2 = pc .

0 5 10 15 20m

Sm

(Xn)

0 10

0.5

1

p

p(x)X*




Notation: p(·) = pc ∈ (1/2, 1] and qc = 1− pc .

1. Place a prior density f0 on the root X∗, f0 has domain [0, 1].Example: U(0, 1).

2. For n=0,1,2, . . .(a) Measure at the median Xn := F−1n (1/2).(b) Update the posterior density:

if Zn(Xn) = +1, fn+1(x) ={

2pc · fn(x), if x > Xn,

2qc · fn(x), if x ≤ Xn,

if Zn(Xn) = −1, fn+1(x) ={

2qc · fn(x), if x > Xn,

2pc · fn(x), if x ≤ Xn.



Sample Path of Posterior Distributions

0 1

fn(x)

n = 0, Xn = 0.5, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 1, Xn = 0.61538, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 2, Xn = 0.70414, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 3, Xn = 0.63587, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 4, Xn = 0.55589, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 5, Xn = 0.46446, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 6, Xn = 0.35727, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 7, Xn = 0.43972, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 8, Xn = 0.51955, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 9, Xn = 0.45436, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 10, Xn = 0.39721, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 11, Xn = 0.33187, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 12, Xn = 0.25529, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 13, Xn = 0.3142, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 14, Xn = 0.37124, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 15, Xn = 0.32466, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 16, Xn = 0.3632, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 17, Xn = 0.33085, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 18, Xn = 0.30024, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 19, Xn = 0.25814, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 20, Xn = 0.20046, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 21, Xn = 0.24672, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 22, Xn = 0.27985, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 23, Xn = 0.31321, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 24, Xn = 0.28617, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 25, Xn = 0.2615, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 26, Xn = 0.28243, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 27, Xn = 0.29702, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 28, Xn = 0.32015, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 29, Xn = 0.33658, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 30, Xn = 0.36118, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 31, Xn = 0.38925, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 32, Xn = 0.43944, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 33, Xn = 0.39633, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 34, Xn = 0.42932, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 35, Xn = 0.4563, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 36, Xn = 0.43531, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 37, Xn = 0.41192, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 38, Xn = 0.43177, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 39, Xn = 0.41699, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 40, Xn = 0.39722, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 41, Xn = 0.41399, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 42, Xn = 0.4015, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 43, Xn = 0.41222, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 44, Xn = 0.40403, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 45, Xn = 0.41052, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 46, Xn = 0.40553, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 47, Xn = 0.39812, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 48, Xn = 0.37339, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 49, Xn = 0.35957, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 50, Xn = 0.36904, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 60, Xn = 0.36286, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 70, Xn = 0.37119, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 80, Xn = 0.37225, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 90, Xn = 0.37336, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 100, Xn = 0.3752, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 110, Xn = 0.37371, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 120, Xn = 0.3728, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 130, Xn = 0.37269, Z

n(X

n) = −1

X*

Xn




0 1

fn(x)

n = 140, Xn = 0.37108, Z

n(X

n) = 1

X*

Xn




0 1

fn(x)

n = 150, Xn = 0.37261, Z

n(X

n) = 1

X*

Xn



Comparison to Stochastic Approximation

0 50 100 150 20010

−10

10−8

10−6

10−4

10−2

100

n

|X* − Xn|

Stochastic ApproximationProbabilistic Bisection

0 1

0

g(x)X*



Literature Review: Probabilistic Bisection Algorithm

• First introduced in Horstein (1963).

• Discretized version: Burnashev and Zigangirov (1974).

• Feige et al. (1997), Karp and Kleinberg (2007), Ben-Or andHassidim (2008), Nowak (2008), Nowak (2009), ...

• Survey paper: Castro and Nowak (2008)

“The probabilistic bisection algorithm seems to workextremely well in practice, but it is hard to analyze andthere are few theoretical guarantees for it, especiallypertaining error rates of convergence.”



Algorithm Analysis



Consistency

Setting for probabilistic bisection with power 1 tests:• X∗ ∈ [0, 1] fixed and unknown.• Xn 6= X∗ for any finite n ∈ N.• p(Xn) ≥ pc for all n ∈ N.• pc ∈ (1/2, 1) is an input parameter.

TheoremXn → X∗ almost surely as n→∞.



Analysis of Posterior Density

0 1

• If Zn = +1 :

fn+1(x) = 2qc · fn(x), x < Xn,

fn+1(x) = 2pc · fn(x), x ≥ Xn,

• If Zn = −1 :

fn+1(x) = 2pc · fn(x), x < Xn,

fn+1(x) = 2qc · fn(x), x ≥ Xn.




Xn

0 1

• If Zn = +1 :

fn+1(x) = 2qc · fn(x), x < Xn,

fn+1(x) = 2pc · fn(x), x ≥ Xn,

• If Zn = −1 :

fn+1(x) = 2pc · fn(x), x < Xn,

fn+1(x) = 2qc · fn(x), x ≥ Xn.




Xn

0 1X*

Case I: If X∗ < Xn : P(Zn = +1) = p(Xn) ≥ pc• If Zn = +1 :

fn+1(x) = 2qc · fn(x), x < Xn,

fn+1(x) = 2pc · fn(x), x ≥ Xn,

• If Zn = −1 :

fn+1(x) = 2pc · fn(x), x < Xn,

fn+1(x) = 2qc · fn(x), x ≥ Xn.




Xn

0 1X*

Case II: If X∗ > Xn : P(Zn = +1) = 1− p(Xn) ≤ pc• If Zn = +1 :

fn+1(x) = 2qc · fn(x), x < Xn,

fn+1(x) = 2pc · fn(x), x ≥ Xn,

• If Zn = −1 :

fn+1(x) = 2pc · fn(x), x < Xn,

fn+1(x) = 2qc · fn(x), x ≥ Xn.



Analysis of Posterior Density cont.

• The dynamics of fn(x) are very complicated for almost all x ∈ [0, 1].

HOWEVER, the dynamics of fn(X∗) are rather simple:

fn+1(X∗) =

{2pc · fn(X∗), with probability p(Xn) ≥ pc ,2qc · fn(X∗), with probability q(Xn) ≤ qc .

• A sample path of fn(X∗) dominates a sample path of a coupledgeometric random walk (Wn)n with dynamics

Wn+1 =

{2pc ·Wn, with probability pc ,2qc ·Wn, with probability qc .

• The process fn(X∗) behaves almost like a geometric random walkindependently of (Xn)n. The goal is then to locate this geometricrandom walk efficiently!



Analysis of Posterior Density cont.

• The dynamics of fn(x) are very complicated for almost all x ∈ [0, 1].HOWEVER, the dynamics of fn(X∗) are rather simple:

fn+1(X∗) =

{2pc · fn(X∗), with probability p(Xn) ≥ pc ,2qc · fn(X∗), with probability q(Xn) ≤ qc .

• A sample path of fn(X∗) dominates a sample path of a coupledgeometric random walk (Wn)n with dynamics

Wn+1 =

{2pc ·Wn, with probability pc ,2qc ·Wn, with probability qc .

• The process fn(X∗) behaves almost like a geometric random walkindependently of (Xn)n. The goal is then to locate this geometricrandom walk efficiently!



Confidence Intervals for X ∗

• Notation: µ = pc ln 2pc + qc ln 2qc .• For α ∈ (0, 1), define

bn = nµ− n1/2(−0.5 lnα)1/2(ln 2pc − ln 2qc).

• DefineJn = conv(x ∈ [0, 1] : fn(x) ≥ ebn).

TheoremFor α ∈ (0, 1),

P(X∗ ∈ Jn) ≥ 1− α,

for all n ∈ N.

Proof:Application of Hoeffding’s inequality.

�



Size of Confidence IntervalTheoremChoose pc ≥ 0.85, α ∈ (0, 1). For 0 < r < µ− qc ln 2pc there exists aN(pc , r , α) ∈ N, such that

P(|Jn| ≤ e−rn,X∗ ∈ Jn) ≥ 1− α,

for all n ≥ N(pc , r , α).

Proof Idea:

0 1



Size of Confidence IntervalTheoremChoose pc ≥ 0.85, α ∈ (0, 1). For 0 < r < µ− qc ln 2pc there exists aN(pc , r , α) ∈ N, such that

P(|Jn| ≤ e−rn,X∗ ∈ Jn) ≥ 1− α,

for all n ≥ N(pc , r , α).

Proof Idea:

0 1Stochastic Root-Finding Probabilistic Bisection Algorithm Analysis Conclusions References 21/32


Rate of Convergence

TheoremDefine X̂n to be any point in Jn, then there exists r > 0 such that

E[|X∗ − X̂n|] = O(e−rn).

• This is extremely fast compared to stochastic approximation:

O(e−rn) vs. O(n−1/2).

• And we have true confidence intervals for X∗.• But n is the number of measurement points, what about totalwall-clock time?



Rate of Convergence


E[|X∗ − X̂n|] = O(e−rn).


O(e−rn) vs. O(n−1/2).

• And we have true confidence intervals for X∗.

• But n is the number of measurement points, what about totalwall-clock time?



Rate of Convergence


E[|X∗ − X̂n|] = O(e−rn).


O(e−rn) vs. O(n−1/2).

• And we have true confidence intervals for X∗.• But n is the number of measurement points, what about totalwall-clock time?



Wall-Clock TimeAt each iteration of the Probabilistic Bisection Algorithm:

• Sample sequentially at point Xn and observeSm(Xn) =

∑mi=1 Yn,i(Xn), until

Nn = inf{m : |Sm| ≥ [(m + 1)(log(m + 1) + 2 log(1/α))]1/2

},

then PXn=X∗ {Nn <∞} ≤ α, PXn 6=X∗ {Nn <∞} = 1, and

PXn<X∗ {SNn(Xn) > 0} ≥ 1− α/2 = pc ,PXn>X∗ {SNn(Xn) < 0} ≥ 1− α/2 = pc .

• Wall-clock time: Tn =∑n

i=1 Nn.

0 5 10 15 20m

Sm

(Xn)

0 10

0.5

1

p

p(x)X*



Sample Paths

100

101

102

103

104

105

−0.5

0

0.5

Tn

X* − Xn

Robbin−Monro, an = 1/n, ε

n ~ N(0,1)

100

101

102

103

104

105

−0.5

0

0.5

Tn

X* − Xn


n ~ N(0,1)

100

101

102

103

104

105

−0.5

0

0.5

Tn

X* − Xn


n ~ N(0,1)

100

101

102

103

104

105

−0.5

0

0.5

Tn

X* − Xn

Bisection, p = 0.75, εn ~ N(0,1)

100

101

102

103

104

105

−0.5

0

0.5

Tn

X* − Xn


100

101

102

103

104

105

−0.5

0

0.5

Tn

X* − Xn




Numerical Comparison

100

101

102

103

104

105

10−4

10−3

10−2

10−1

100

Tn

E[|X

* −

Xn|]

Robbins−Monro (an=1/n)

Bisection, Siegmund (pc=0.85)



Rate of Convergence in Wall-Clock Time?• Farrell (1964):

Eg(x)[N] ∼ (1/g(x))2 log log(1/|g(x)|) as g(x)→ 0,

and for all tests of power one, if P0(N =∞) > 0, then

limg(x)→0

g(x)2Eg(x)[N] =∞.

Theorem

1. limn→∞ E[|X∗ − Xn|(Tn)1/2] =∞.

2. (|X∗ − Xn|(Tn)1/2)n is not tight.

• Ifg(x)→ 0 as x → X∗,

and if we use Xn as the best estimate of X∗ then the ProbabilisticBisection Algorithm with power one tests is asymptotically slowerthan Stochastic Approximation.



Conjecture

• Xn might not be the best estimate for X∗ when we use power onetests.

• Intuitively, observations where we spend more time should also becloser to X∗, hence an estimator of the form

X̃n =1Tn

n∑i=1

NiXi

should perform better.

• Conjecture: For any β > 0 it holds that

E[|X̃n − X∗|] = O(T− 1

2(1+β)n ),

(if g satisfies some growth conditions).• Sufficient Condition: |Xn − X∗| = O(e−rn) for some r > 0.



Conjecture



X̃n =1Tn

n∑i=1

NiXi

should perform better.• Conjecture: For any β > 0 it holds that

E[|X̃n − X∗|] = O(T− 1

2(1+β)n ),

(if g satisfies some growth conditions).

• Sufficient Condition: |Xn − X∗| = O(e−rn) for some r > 0.



Conjecture



X̃n =1Tn

n∑i=1

NiXi

should perform better.• Conjecture: For any β > 0 it holds that

E[|X̃n − X∗|] = O(T− 1

2(1+β)n ),

(if g satisfies some growth conditions).• Sufficient Condition: |Xn − X∗| = O(e−rn) for some r > 0.



Numerical Comparison Cont.

100

101

102

103

104

105

10−4

10−3

10−2

10−1

100

Tn

E[|X

* −

Xn|]

Robbins−Monro (an=1/n)

Bisection, Siegmund (pc=0.85)

Polyak−RuppertBisection Averaging



Conclusions



ConclusionsPositive:

• Provides true confidence interval of the root X∗.

• Works extremely well if there is a jump at g(X∗) (geometric rate ofconvergence).

• Only one tuning parameter.

Drawbacks:• Seems to be asymptotically slower than Stochastic Approximation(but by not much).

• Higher computational cost.

Future Research:• Robustness of algorithm.

• Use parallel computing (very little switching of (Xn)n).

• Extension to higher dimensions.Stochastic Root-Finding Probabilistic Bisection Algorithm Analysis Conclusions References 30/32


THANK YOU!



M. Ben-Or and A. Hassidim (2008): The Bayesian learner is optimal for noisy binary search. In49th Annual IEEE Symposium on Foundations of Computer Science. IEEE, pp. 221–230.

M. Burnashev and K. Zigangirov (1974): An interval estimation problem for controlledobservations. Problemy Peredachi Informatsii 10:51–61.

R. Castro and R. Nowak (2008): Active learning and sampling. In Foundations andApplications of Sensor Management, A. O. Hero, D. A. Castañón, D. Cochran and K. Kastella,editors. Springer, pp. 177–200.

R. H. Farrell (1964): Asymptotic behavior of expected sample size in certain one sided tests.The Annals of Mathematical Statistics 35:36–72.

U. Feige, P. Raghavan, D. Peleg and E. Upfal (1997): Computing with noisy information.SIAM Journal of Computing :1001–1018.

M. Horstein (1963): Sequential transmission using noiseless feedback. IEEE Transactions onInformation Theory 9:136–143.

R. Karp and R. Kleinberg (2007): Noisy binary search and its applications. In Proceedings ofthe eighteenth annual ACM-SIAM symposium on Discrete algorithms. Society for Industrial andApplied Mathematics, pp. 881–890.

R. Nowak (2008): Generalized binary search. In 46th Annual Allerton Conference onCommunication, Control, and Computing . pp. 568–574.

R. Nowak (2009): Noisy generalized binary search. In Advances in neural information processingsystems, volume 22. pp. 1366–1374.

H. Robbins and S. Monro (1951): A stochastic approximation method. The Annals ofMathematical Statistics 22:400–407.

D. Siegmund (1985): Sequential Analysis: tests and confidence intervals. Springer.Stochastic Root-Finding Probabilistic Bisection Algorithm Analysis Conclusions References 32/32


Tuning Parameters

100

101

102

103

104

105

10−4

10−3

10−2

10−1

100

n

Input Parameters d to Robbins−Monro (stepsize d/n), X* ~ U(0,1)

E[|X* − Xn|]

0.10.250.50.7511.5251020



Tuning Parameters

100

101

102

103

104

105

10−4

10−3

10−2

10−1

100

n

Input Parameters pc to Bisection Algorithm, X* ~ U(0,1)

E[|X* − Xn|]

0.5250.5750.6250.6750.7250.7750.8250.8750.9250.975



Higher Dimensions

Boundary detection:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1n = 0, Z = −1



Higher Dimensions

Boundary detection:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1n = 1, Z = −1



Higher Dimensions

Boundary detection:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1n = 2, Z = 1



Higher Dimensions

Boundary detection:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1n = 3, Z = 1



Higher Dimensions

Boundary detection:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1n = 4, Z = −1



Higher Dimensions

Boundary detection:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1n = 5, Z = 1



Higher Dimensions

Boundary detection:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1n = 6, Z = 1



Higher Dimensions

Boundary detection:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1n = 7, Z = −1



Higher Dimensions

Boundary detection:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1n = 8, Z = 1



Higher Dimensions

Boundary detection:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1n = 9, Z = 1



Higher Dimensions

Boundary detection:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1n = 10, Z = 1


probabilistic bisection search 0.1cm for stochastic root …probabilistic bisection search for...

Documents