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 Sciences (Moscow) IWAP 2010

Problem formulation and background Results Conclusion Outline Problem formulation and background 1 Results 2 Conclusion 3

Problem formulation and background Results Conclusion The T racking S topping T ime problem X = { X t } t ≥ 0 and stopping time τ on X Statistician has access to X through Y = { Y t } t ≥ 0 . Statistician wishes to find a stopping time η so that E | η − τ | is minimized.

Problem formulation and background Results Conclusion Example: monitoring X t : distance of an object from a barrier at time t τ : first time when X t = 0 Y t : noisy measurements of X t η : alarm time based on Y should be close to τ

Problem formulation and background Results Conclusion Example: forecasting X t : fatigue up to day t of a big manufacturing machine τ : first day t when X t 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 Y t = X t − 10 if t > 10 and X t = 0 else.

Problem formulation and background Results Conclusion Example: change-point problem θ positive integers valued random variable { Y t } with Y t ∼ P 0 if t < θ and Y t ∼ P 1 if t ≥ θ Goal: find η such that E ( η − θ ) + is minimized while P ( η < θ ) ≤ α Equivalent to TST problem: { X t } with X t = 0 if t < θ and X t = 1 if t ≥ θ . τ = inf { t ≥ 1 : X t = 1 } (i.e., τ = θ ) Goal: find η such that E ( η − τ ) + minimized while P ( η < τ ) ≤ α

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 = y t , θ > t ) = P ( θ = k | θ > t ) TST: P ( τ = k | Y t = y t , 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 .

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?

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 outputs d ( α ) = τ : P ( η<τ ) ≤ α E ( η − τ ) + α ∈ [ 0 , 1 ] min d ( α ) � � � � � � � � � � � � α � � the corresponding optimal stopping times. Under conditions on ( X , Y ) and τ , the algorithm is poly( b ).

Problem formulation and background Results Conclusion Problem formulation Given t � X : X 0 = 0 X t = V i + st t ≥ 1 i = 1 t � Y : Y 0 = 0 Y t = X t + ε W i t ≥ 1 i = 1 with s ≥ 0, ε > 0, and V i ’s and W i ’s i.i.d. N ( 0 , 1 ) τ = inf { t ≥ 0 : X t ≥ T } T > 0 Find inf η E | η − τ |

Problem formulation and background Results Conclusion Problem formulation (cont.) 50 40 Y t 30 20 X t 10 100 200 300 400 500 600

Problem formulation and background Results Conclusion Theorem: asymptotics For fixed s > 0 and ε > 0 � 2 T ε 2 inf 0 ) E | η − τ | = inf η E | η − τ | = π s 3 ( 1 + ε 2 ) [ 1 + o ( 1 )] η ( Y ∞ as T → ∞ . ⇒ Causality doesn’t come at the expense of delay (asym.).

Problem formulation and background Results Conclusion Theorem: asymptotics (general) Let 1 / 2 < q < 1. In the asymptotic regime where T / s ≥ 2, � q − 1 / 2 � T s − → ∞ , s and � 1 − q ε 2 � T 1 + ε 2 − → ∞ , s we have � 2 T ε 2 inf 0 ) E | η − τ | = inf η E | η − τ | = π s 3 ( 1 + ε 2 ) [ 1 + o ( 1 )] . η ( Y ∞

Problem formulation and background Results Conclusion Theorem: upper bound Fix ε > 0, s > 0, T > 0, and let 1 ˆ ˆ X 0 = 0 X t = st + 1 + ε 2 ( Y t − st ) t ≥ 1 . Then, η = inf { t ≥ 0 : ˆ X t ≥ T } satisfies � � 1 / 4 � 2 T ε 2 π ( 1 + ε 2 ) s 3 + 6 � T 8 ( s + 2 ) + 10 + 20 E | η − τ | ≤ + ( 2 π s ) 3 π s 3 s s

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 � � � �� 2 n ε 2 T − sn E | η − τ | ≥ 1 − Q inf π s 2 ( 1 + ε 2 ) � n ( 1 + ε ) η ( Y ∞ 0 ) � 1 / 2 � � � 2 n − 2 − 4 − T − sn + s . π s 3 2 π

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.

Problem formulation and background Results Conclusion Extension to Brownian motion processes Theorem X : X 0 = 0 X t = B t + st t > 0 Y : Y 0 = 0 Y t = X t + ε W t t > 0 with s ≥ 0, ε > 0, and { B t } and { W t } independent standard Brownian motions. Then, � 2 T ε 2 0 ) E | η − τ | = inf η E | η − τ | = π s 3 ( 1 + ε 2 ) [ 1 + o ( 1 )] . inf η ( Y ∞

Problem formulation and background Results Conclusion Bounds Theorem for ε > 0, s > 0, and T > 0 � � 1 / 4 2 T ε 2 � π ( 1 + ε 2 ) s 3 + 6 T inf η E | η − τ | ≤ ( 2 π s ) 3 s for ε > 0, s > 0, T / s ≥ 2 and any integer n such that 1 ≤ n < T / s � � � �� 2 n ε 2 T − sn E | η − τ | ≥ 1 − Q inf π s 2 ( 1 + ε 2 ) � n ( 1 + ε ) η ( Y ∞ 0 ) � 1 / 2 � � � 2 n − T − sn + . π s 3 2 π

Problem formulation and background Results Conclusion Upper bound: sketch t � X : X 0 = 0 X t = V i + st t ≥ 1 i = 1 t � Y : Y 0 = 0 Y t = X t + ε W i t ≥ 1 i = 1 ( X t , Y t ) jointly Gaussian hence, linear estimator ˆ X t = aY t + b minimizes E | ˆ X t − X t | X t − X t | p for all p ≥ 1), (actually minimizes E | ˆ natural tracker: η = inf { t ≥ 1 : ˆ X t ≥ T }

Problem formulation and background Results Conclusion Upper bound: sketch (cont.) to evaluate E | η − τ | , first evaluate E | ˆ X η − X τ | to evaluate E ˆ X η and E X τ use bounds on overshoot connect E | ˆ X η − X τ | to E | η − τ | using Wald’s equation ( E � µ t = 1 Z t = E µ E Z )

Problem formulation and background Results Conclusion Lower bound: sketch Idea: reduce the problem to the estimation of X n for n ≈ T / s . 0 ) E | η − τ | ≥ 1 0 ) E | η − X n | + small inf inf s η ( Y ∞ η ( Y ∞ d ( X n , Y n ) are jointly Gaussian, thus X n = aY n + bV + c with V ∼ N ( 0 , 1 ) independent of Y n , hence � 2 n ε 2 1 inf 0 ) E | η − X n | ≈ s π s 2 ( 1 + ε 2 ) η ( Y ∞

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 of 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?