Convergence Rate of Probabilistic Bisection in Stochastic Root - - PowerPoint PPT Presentation

Convergence Rate of Probabilistic Bisection in Stochastic Root Finding

Shane G. Henderson

Operations Research & Information Engineering Cornell University, Ithaca, NY

June 1, 2015 Banff, Canada Joint work with Peter I. Frazier and Rolf Waeber Research supported by AFOSR YIP FA9550-11-1-0083, NSF CMMI 1200315

Probabilistic Bisection Search for Stochastic Root Finding

Stochastic Root Finding

• Suppose g : [0, 1] → R is decreasing, has unique root.
• g(x) is observed with noise, Y (x) = g(x) + ǫ(x)
• Goal: Locate the root
• Instead of stochastic approximation, use probabilistic

bisection

• Assumes oracle indicates direction of root from any x and

is correct with probability p > 1/2 (independent of x)

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Probabilistic Bisection Search for Stochastic Root Finding

The Probabilistic Bisection Algorithm Horstein 63

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

1 1 2 fn(x) n = 0, Xn = 0.5, Zn(Xn) = −1 X* Xn

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Probabilistic Bisection Search for Stochastic Root Finding

The Probabilistic Bisection Algorithm Horstein 63

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

1 1 2 fn(x) n = 0, Xn = 0.5, Zn(Xn) = −1 X* Xn 1 1 2 fn(x) n = 1, Xn = 0.38462, Zn(Xn) = −1 X* Xn

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Probabilistic Bisection Search for Stochastic Root Finding

The Probabilistic Bisection Algorithm Horstein 63

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

1 1 2 fn(x) n = 0, Xn = 0.5, Zn(Xn) = −1 X* Xn 1 1 2 fn(x) n = 1, Xn = 0.38462, Zn(Xn) = −1 X* Xn 1 1 2 fn(x) n = 2, Xn = 0.29586, Zn(Xn) = 1 X* Xn

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Probabilistic Bisection Search for Stochastic Root Finding

The Probabilistic Bisection Algorithm Horstein 63

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

1 1 2 fn(x) n = 0, Xn = 0.5, Zn(Xn) = −1 X* Xn 1 1 2 fn(x) n = 1, Xn = 0.38462, Zn(Xn) = −1 X* Xn 1 1 2 fn(x) n = 2, Xn = 0.29586, Zn(Xn) = 1 X* Xn 1 1 2 fn(x) n = 3, Xn = 0.36413, Zn(Xn) = 1 X* Xn

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Probabilistic Bisection Search for Stochastic Root Finding

Stochastic Root-Finding Revisited

1 g(x) X*

Zn(Xn) =

• sign (g(Xn))

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

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Probabilistic Bisection Search for Stochastic Root Finding

Stochastic Root-Finding Revisited

1 g(x) X* 1 0.5 1 p(x) X*

Zn(Xn) =

• sign (g(Xn))

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

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Probabilistic Bisection Search for Stochastic Root Finding

Stochastic Root-Finding Revisited

1 g(x) X* 1 0.5 1 p p(x) X*

Zn(Xn) =

• sign (g(Xn))

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

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Probabilistic Bisection Search for Stochastic Root Finding

Sequential Tests

• Generate a new signal Z ′(x) using a sequential test of

power one

• For x = x∗, test halts in finite time and is correct with

user-specifed probability at least pc > 1/2

• Use PBA with such a pc
• Exponential convergence in n...
• At x, expected running time θ−2| ln | ln θ||, θ = p(x) − 1/2
• Slows down as x → x∗

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Probabilistic Bisection Search for Stochastic Root Finding

Performance

10 10

1

10

2

10

3

10

4

10

5

−0.5 0.5 Tn X* − Xn Bisection, p = 0.75, εn ~ N(0,1)

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Probabilistic Bisection Search for Stochastic Root Finding

Convergence

For any fixed ǫ ∈ (0, 1/2), T 1/2−ǫ

n+1 ( ˆ

Xn − x∗) ⇒ 0 as n → ∞, where ˆ Xn = 1

n

i=0 N 1/2−ǫ i n

• i=0

N 1/2−ǫ

i

Xi. What about the more natural (and empirically better) 1 Tn

n−1

• i=0

NiXi?

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