Tracking Threshold Crossing Times of a Gaussian Random Walk Through - - PowerPoint PPT Presentation

tracking threshold crossing times of a gaussian random
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Tracking Threshold Crossing Times of a Gaussian Random Walk Through - - PowerPoint PPT Presentation

Problem formulation and background Results Conclusion Tracking Threshold Crossing Times of a Gaussian Random Walk Through Correlated Observations Aslan Tchamkerten Telecom ParisTech Joint work with Marat Burnashev, Russian Academy of


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Problem formulation and background Results Conclusion

Tracking Threshold Crossing Times

  • f a Gaussian Random Walk

Through Correlated Observations

Aslan Tchamkerten

Telecom ParisTech Joint work with Marat Burnashev, Russian Academy of Sciences (Moscow)

IWAP 2010

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Problem formulation and background Results Conclusion

Outline

1

Problem formulation and background

2

Results

3

Conclusion

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Problem formulation and background Results Conclusion

The Tracking Stopping Time problem

X = {Xt}t≥0 and stopping time τ on X Statistician has access to X through Y = {Yt}t≥0. Statistician wishes to find a stopping time η so that E|η − τ| is minimized.

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Problem formulation and background Results Conclusion

Example: monitoring

Xt: distance of an object from a barrier at time t τ: first time when Xt = 0 Yt: noisy measurements of Xt η: alarm time based on Y should be close to τ

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Problem formulation and background Results Conclusion

Example: forecasting

Xt: fatigue up to day t of a big manufacturing machine τ: first day t when Xt crosses critical fatigue threshold Machine replacement period: 10 days η: first day when new machine is operational Wanted η close to τ because {η > τ}: interrupted manufacturing process {η < τ}: storage costs ⇒ TST problem with Yt = Xt−10 if t > 10 and Xt = 0 else.

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Problem formulation and background Results Conclusion

Example: change-point problem

θ positive integers valued random variable {Yt} with Yt ∼ P0 if t < θ and Yt ∼ P1 if t ≥ θ Goal: find η such that E(η − θ)+ is minimized while P(η < θ) ≤ α Equivalent to TST problem: {Xt} with Xt = 0 if t < θ and Xt = 1 if t ≥ θ. τ = inf{t ≥ 1 : Xt = 1} (i.e., τ = θ) Goal: find η such that E(η − τ)+ minimized while P(η < τ) ≤ α

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Problem formulation and background Results Conclusion

Change-point or tracking a stopping time?

C-P and TST are not equivalent: for k > t C-P: P(θ = k|Y t = yt, θ > t) = P(θ = k|θ > t) TST: P(τ = k|Y t = yt, τ > t)=P(τ = k|τ > t) in general

i.e., for the C-P problem,

if {θ > t}, the first t samples Y t are useless for predicting θ.

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Problem formulation and background Results Conclusion

Bottom line

The TST problem formulation

A useful generalization of the change-point problem. The change-point problem is known to be hard. Any hope?

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Problem formulation and background Results Conclusion

A first step: algorithmic solution (Niesen and T, 2009)

Given finite alphabet process (X, Y) and τ ≤ b, the algorithm

  • utputs

d(α) = min

τ:P(η<τ)≤α E(η − τ)+ α ∈ [0, 1]

  • α

d(α)

the corresponding optimal stopping times. Under conditions on (X, Y) and τ, the algorithm is poly(b).

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Problem formulation and background Results Conclusion

Problem formulation

Given

X : X0 = 0 Xt =

t

  • i=1

Vi + st t ≥ 1 Y : Y0 = 0 Yt = Xt + ε

t

  • i=1

Wi t ≥ 1 with s ≥ 0, ε > 0, and Vi’s and Wi’s i.i.d. N(0, 1) τ = inf{t ≥ 0 : Xt ≥ T} T > 0 Find inf

η E|η − τ|

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Problem formulation and background Results Conclusion

Problem formulation (cont.)

100 200 300 400 500 600 10 20 30 40 50

Yt Xt

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Problem formulation and background Results Conclusion

Theorem: asymptotics

For fixed s > 0 and ε > 0 inf

η(Y ∞

0 ) E|η − τ| = inf

η E|η − τ| =

  • 2Tε2

πs3(1 + ε2) [1 + o(1)] as T → ∞. ⇒ Causality doesn’t come at the expense of delay (asym.).

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Problem formulation and background Results Conclusion

Theorem: asymptotics (general)

Let 1/2 < q < 1. In the asymptotic regime where T/s ≥ 2, s T s q−1/2 − → ∞ , and T s 1−q ε2 1 + ε2 − → ∞ , we have inf

η(Y ∞

0 ) E|η − τ| = inf

η E|η − τ| =

  • 2Tε2

πs3(1 + ε2) [1 + o(1)] .

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Problem formulation and background Results Conclusion

Theorem: upper bound

Fix ε > 0, s > 0, T > 0, and let ˆ X0 = 0 ˆ Xt = st + 1 1 + ε2 (Yt − st) t ≥ 1. Then, η = inf{t ≥ 0 : ˆ Xt ≥ T} satisfies E |η − τ| ≤

  • 2Tε2

π(1 + ε2)s3 + 6 s

  • T

(2πs)3 1/4 +

  • 8(s + 2)

πs3 + 10 + 20 s

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Problem formulation and background Results Conclusion

Theorem: lower bound

Let ε > 0, s > 0, and T/s ≥ 2. Then, for any integer n such that 1 ≤ n < T/s inf

η(Y ∞

0 )

E|η − τ| ≥

  • 2nε2

πs2(1 + ε2)

  • 1 − Q
  • T − sn
  • n(1 + ε)
  • 2

πs3

  • T − sn +
  • n

2π 1/2 − 2 − 4 s .

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Problem formulation and background Results Conclusion

Theorem: case s = 0

If s = 0, ε > 0, and T > 0, then E|η − τ|p = ∞ for all η = η(Y ∞

0 ) and p ≥ 1/2.

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Problem formulation and background Results Conclusion

Extension to Brownian motion processes

Theorem X : X0 = 0 Xt = Bt + st t > 0 Y : Y0 = 0 Yt = Xt + εWt t > 0 with s ≥ 0, ε > 0, and {Bt} and {Wt} independent standard Brownian motions. Then, inf

η(Y ∞

0 ) E|η − τ| = inf

η E|η − τ| =

  • 2Tε2

πs3(1 + ε2) [1 + o(1)] .

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Problem formulation and background Results Conclusion

Bounds

Theorem for ε > 0, s > 0, and T > 0 inf

η E |η − τ| ≤

  • 2Tε2

π(1 + ε2)s3 + 6 s

  • T

(2πs)3 1/4 for ε > 0, s > 0, T/s ≥ 2 and any integer n such that 1 ≤ n < T/s inf

η(Y ∞

0 )

E|η − τ| ≥

  • 2nε2

πs2(1 + ε2)

  • 1 − Q
  • T − sn
  • n(1 + ε)
  • 2

πs3

  • T − sn +
  • n

2π 1/2 .

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Problem formulation and background Results Conclusion

Upper bound: sketch

X : X0 = 0 Xt =

t

  • i=1

Vi + st t ≥ 1 Y : Y0 = 0 Yt = Xt + ε

t

  • i=1

Wi t ≥ 1 (Xt, Yt) jointly Gaussian hence, linear estimator ˆ Xt = aYt + b minimizes E|ˆ Xt − Xt| (actually minimizes E|ˆ Xt − Xt|p for all p ≥ 1), natural tracker: η = inf{t ≥ 1 : ˆ Xt ≥ T}

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Problem formulation and background Results Conclusion

Upper bound: sketch (cont.)

to evaluate E|η − τ|, first evaluate E|ˆ Xη − Xτ| to evaluate Eˆ Xη and EXτ use bounds on overshoot connect E|ˆ Xη − Xτ| to E|η − τ| using Wald’s equation (E µ

t=1 Zt = EµEZ)

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Problem formulation and background Results Conclusion

Lower bound: sketch

Idea: reduce the problem to the estimation of Xn for n ≈ T/s. inf

η(Y ∞

0 ) E|η − τ| ≥ 1

s inf

η(Y ∞

0 ) E|η − Xn| + small

(Xn, Yn) are jointly Gaussian, thus Xn

d

= aYn + bV + c with V ∼ N(0, 1) independent of Yn, hence 1 s inf

η(Y ∞

0 ) E|η − Xn| ≈

  • 2nε2

πs2(1 + ε2)

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Problem formulation and background Results Conclusion

Summary

the TST problem formulation naturally appears in

detection prediction communication quality control information econometrics

Contribution: correlated gaussian random walks

non-asymptotic upper and lower bounds on minimum value

  • f E|η − τ|

asymptotic characterization on minimum value of E|η − τ| (extends to E|η − τ|p, p ≥ 1)

A lot remains to be done:

more general delay penalty functions Gaussianity is key for our results, what if non-gaussian?