A precision of the sequential change point detection - arXiv

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Papers from the AMS-IMS-SIAM Summer Research Conference held at Mt. Holyoke College, South Hadley, MA, USA, July 11–16, 1992, ser. IMS Lecture.
JOURNAL OF LATEX CLASS FILES, VOL. 11, NO. 4, DECEMBER 2012

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A precision of the sequential change point detection⋆ Aleksandra Ochman-Gozdek1, Wojciech Sarnowski2 and Krzysztof J. Szajowski3

arXiv:1401.5613v1 [math.ST] 22 Jan 2014

Abstract A random sequence having two segments being the homogeneous Markov processes is registered. Each segment has his own transition probability law and the length of the segment is unknown and random. The transition probabilities of each process are known and a priori distribution of the disorder moment is given. The decision maker aim is to detect the moment of the transition probabilities change. The detection of the disorder rarely is precise. The decision maker accepts some deviation in estimation of the disorder moment. In the considered model the aim is to indicate the change point with fixed, bounded error with maximal probability. The case with various precision for over and under estimation of this point is analyzed. The case when the disorder does not appears with positive probability is also included. The results insignificantly extends range of application, explain the structure of optimal detector in various circumstances and shows new details of the solution construction. The motivation for this investigation is the modelling of the attacks in the node of networks. The objectives is to detect one of the attack immediately or in very short time before or after it appearance with highest probability. The problem is reformulated to optimal stopping of the observed sequences. The detailed analysis of the problem is presented to show the form of optimal decision function. Index Terms Bayesian approach, disorder problem, sequential detection, optimal stopping, Markov process, change point AMS Subject Classification: Primary: 60G40; Secondary: 60K99; 90D60

I. I NTRODUCTION UPPOSE that the process X = {Xn , n ∈ N}, N = {0, 1, 2, . . .}, is observed sequentially. It is obtained from Markov processes by switching between them at a random moment θ in such a way that the process after θ starts from the state Xθ−1 . It means that the state at moment n ∈ N has conditional distribution given the state at moment n − 1, where the formulae describing these distributions have the different form: one for n < θ and another for n ≥ θ. Our objective is to detect the moment θ based on observation of X. There are some papers devoted to the discrete case of such disorder detection which generalize in various directions the basic problem stated by Shiryaev in [11] (see e.g. Brodsky and Darkhovsky [5], Bojdecki [3], Bojdecki and Hosza [4], Yoshida [18], Szajowski [14], [15]). Such model of data appears in many practical problems of the quality control (see Brodsky and Darkhovsky [5], Shewhart [10] and in the collection of the papers [2]), traffic anomalies in networks (in papers Dube and Mazumdar [6], Tartakovsky et al. [16]), epidemiology models (see Baron [1], Siegmund [13]). The aim is to recognize the moment of the change over the one probabilistic characteristics to another of the phenomenon. Typically, the disorder problem is limited to the case of switching between sequences of independent random variables (see Bojdecki [3]). Some developments of the basic model can be found in the paper by Yakir [17] where the optimal detection rule of the switching moment has been obtained when the finite state-space Markov chains is disordered. Moustakides [7] formulates conditions which help to reduce the problem of the quickest detection for dependent sequences before and after the change to the case for independent random variables. Our result is a generalization of the results obtained by Bojdecki [3] and Sarnowski et al. [9]. It admits Markovian dependence structure for switched sequences (with possibly uncountable state-space). We obtain an optimal rule under probability maximizing criterion. Formulation of the problem can be found in Section II. The main result is presented in Section III.

S

⋆ The research has been supported by grant S30103/I-18. This paper was presented in part at 59th ISI World Statistics Congress 25–30 August 2013, Hong Kong Special Administrative Region, China in the session CPS018 (see [8]), and at XXXIX COnference on Mathematical Statistics 2-6 December 2013, Wisła,Poland. 1 Aleksandra Ochman–Gozdek is with Toyota Motor Industries Poland Ltd. as HR Analyst. She prepares her PhD theses at the Institute of Mathematics and Computer Science, Wrocław University of Technology; e-mail: [email protected] 2 Wojciech Sarnowski is with Europejski Fundusz Leasingowy SA as Chief Specialist in Risk Management and Policy Rules Department, pl. Orlat ˛ Lwowskich 1, 53-605 Wrocław, Poland; e-mail: [email protected] 3 Krzysztof J. Szajowski is with Faculty of Fundamental Problems of Technology, Institute of Mathematics and Computer Science, Wrocław University of Technology, PL-50-370 Wrocław, Poland; e-mail: [email protected]

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II. F ORMULATION

OF THE PROBLEM

Let (Ω, F , P) be a probability space which supports sequence of observable random variables {Xn }n∈N generating filtration Fn = σ(X0 , X1 , ..., Xn ). Random variables Xn take values in (E, B), where E is a subset of R. Space (Ω, F , P) supports also unobservable (hence not measurable with respect to Fn ) random variable θ which has the geometric distribution: P(θ = j) = πI{j=0} + (1 − π)pj−1 qI{j≥1} ,

(1)

where q = 1 − p, π ∈ (0, 1), j = 1, 2, . . .. We introduce in (Ω, F , P) also two time homogeneous and independent Markov processes {Xn0 }n∈N and {Xn1 }n∈N taking values in (E, B) and assumed to be independent of θ. Moreover, it is assumed that {Xn0 }n∈N and {Xn1 }n∈N have transition densities with respect to a σ-finite measure µ, i.e., for i = 0, 1 and B ∈ B Z Z Pix (X1i ∈ B) = P(X1i ∈ B|X0i = x) = fxi (y)µ(dy) = µx (dy). (2) B

B

Random processes {Xn }, {Xn0 }, {Xn1 } and random variable θ are connected via the rule: conditionally on θ = k  0 Xn , if k > n, Xn = 1 Xn+1−k , if k ≤ n, 0 where {Xn1 } is started from Xk−1 (but is otherwise independent of X 0 ). Let us introduce the following notation:

xk,n Lm (xk,n )

= (xk , xk+1 , . . . , xn−1 , xn ), k ≤ n, =

n−m Y

fx1r−1 (xr ),

r=n−m+1

r=k+1

Ak,n

n Y

fx0r−1 (xr )

= ×ni=k Ai = Ak × Ak+1 × . . . × An , Ai ∈ B

Qj2 where the convention i=j xi = 1 for j1 > j2 is used. 1 Let us now define functions S· (·) and G· (·, ·) ¯ Sn (x0,n )=πLn (x0,n ) + π

n X

p

i−1

!

n

qLn−i+1 (x0,n ) + p L0 (x0,n ) ,

i=1

Gl+1 (xn−l−1,n , α)=αLl+1 (xn−l−1,n ) + (1 − α) ×

l X

(4) !

pl−i qLi+1 (xn−l−1,n ) + pl+1 L0 (xn−l−1,n ) .

i=0

n+1

(3)

where x0 , x1 , . . . , xn ∈ E , α ∈ [0, 1], 0 ≤ n − l − 1 < n. The function S(x0,n ) stands for the joint density of the vector X 0,n . For any D0,n = {ω : X 0,n ∈ B 0,n , Bi ∈ B} and any x ∈ E we have: Z Px (D 0,n ) = P(D0,n |X0 = x)= S(x0,n )µ(dx0,n ) B 0,n

The meaning of the function Gn−k+1 (xk,n , α) will be clear in the sequel. Roughly speaking our model assumes that the process {Xn } is obtained by switching at the random and unknown instant θ between two Markov processes {Xn0 } and {Xn1 }. It means that the first observation Xθ after the change depends on the 1 previous sample Xθ−1 through the transition pdf fX (Xθ ). For any fixed d1 , d2 ∈ {0, 1, 2, . . .} (the problem Dd1 d2 ) we are θ−1 ∗ looking for the stopping time τ ∈ T such that Px (−d1 ≤ θ − τ ∗ ≤ d2 ) = sup Px (−d1 ≤ θ − τ ≤ d2 )

(5)

τ ∈SX

where SX denotes the set of all stopping times with respect to the filtration {Fn }n∈N . Using parameters di , i = 1, 2, we control the precision level of detection. The problem Ddd , i.e. the case d1 = d2 = d, when π = 0 has been studied in [9].

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III. S OLUTION

OF THE PROBLEM

Let us denote: Zn(d1 ,d2 ) =Px (−d1 ≤ θ − n ≤ d2 | Fn ), n = 0, 1, 2, . . ., Vn(d1 ,d2 ) = ess sup Px (−d1 ≤ θ − τ ≤ d2 | Fn ), n = 0, 1, 2, . . ., {τ ∈SX , τ ≥n}

τ0 =inf{n : Zn(d1 ,d2 ) = Vn(d1 ,d2 ) } (d ,d2 )

Notice that, if Z∞1

(d1 ,d2 )

= 0, then Zτ

(6)

= Px (−d1 ≤ θ − τ ≤ d2 | Fτ ) for τ ∈ SX . Since Fn ⊆ Fτ (when n ≤ τ ) we have

Vn(d1 ,d2 ) =ess sup Px (−d1 ≤ θ − τ ≤ d2 | Fn ) = ess sup Ex (I{−d1 ≤θ−τ ≤d2} | Fn ) τ ≥n

τ ≥n

=ess sup Ex (Zτ(d1 ,d2 ) | Fn ). τ ≥n

The following lemma (see [3], [9]) ensures existence of the solution Lemma III.1. The stopping time τ0 defined by formula (6) is the solution of problem (5). Proof: From the theorems presented in [3] it is enough to show that lim Zn(d1 ,d2 ) = 0. For all natural numbers n, k, n→∞ ¯ we have: where n ≥ k and s, t ∈ N Zn(d1 ,d2 ) =Ex (I{−d1 ≤θ−n≤d2 } | Fn ) ≤ Ex (sup I{−d1 ≤θ−j≤d2 } | Fn ) j≥k

(d ,d ) lim supn→∞ Zn 1 2

S∞ ≤ Ex (supj≥k I{−d1 ≤θ−j≤d2 } | F∞ ) where F∞ = σ ( n=1 Fn ). It is true that: From Levy’s theorem lim supj≥k, k→∞ I{−d1 ≤θ−j≤d2 } = 0 a.s. and by the dominated convergence theorem we get lim Ex (sup I{−d1 ≤θ−j≤d2 } | F∞ ) = 0 a.s..

k→∞

j≥k

The proof of the lemma is complete. By the following lemma (see also Lemma 3.2 in [9]) we can limit the class of possible stopping rules to SX d2 +1 i.e. stopping times equal at least d2 + 1. Then rule τ˜ = max(τ, d1 + 1) is at least as good as τ . Lemma III.2. Let τ be a stopping rule in the problem (5). Then rule τ˜ = max(τ, d1 + 1) is at least as good as τ . Proof: For τ ≥ d1 + 1 the rules τ, τ˜ are the same. Let us consider the case when τ < d1 + 1. We have τ˜ = d1 + 1 and based on the fact that Px (θ ≥ 1) = 1 we get: Px (−d1 ≤ θ − τ ≤ d2 )=Px (τ − d1 ≤ θ ≤ τ + d2 ) = Px (1 ≤ θ ≤ τ + d2 ) ≤Px (1 ≤ θ ≤ d2 + d1 + 1) = Px (˜ τ − d1 ≤ θ ≤ τ˜ + d2 ) =Px (−d1 ≤ θ − τ˜ ≤ d2 ). This is the desired conclusion. For further considerations let us define the posterior process: Π0 =π, Πn =Px (θ ≤ n | Fn ) , n = 1, 2, . . . which is designed as information about the distribution of the disorder instant θ. Next lemma transforms the payoff function to the more convenient form. Lemma III.3. Let h(x0,d1 +1 , α) = where x0 , . . . , xd1 +1 ∈ E, α ∈ (0, 1), then

1−p

d2

d1 X Lm+1 (x0,d1 +1 ) +q pm L0 (x0,d1 +1 ) m=0

!

(1 − α),

  Px (−d1 ≤ θ − n ≤ d2 ) = Ex h(X n−1−d1 ,n , Πn ) .

Proof: We rewrite the initial criterion as the expectation

Px (−d1 ≤ θ − n ≤ d2 )=Ex [Px (−d1 ≤ θ − n ≤ d2 | Fn )] =Ex [Px (θ ≤ n + d2 | Fn ) − Px (θ ≤ n − d1 − 1 | Fn )]

(7)

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The probabilities under the expectation can be transformed to the convenient form using the lemmata A.1 and A.4 ( A1 and A4 of [9]). Next, with the help of Lemma A5 (ibid) (putting l = d1 ) we can express Px (θ ≤ n + d2 | Fn ) in terms of Πn . Straightforward calculations imply that: ! d1 X L (X ) m n−d −1,n 1 Px (−d1 ≤ θ − n ≤ d2 | Fn ) = 1 − pd2 + q (1 − Πn ). m L (X p ) 0 n−d 1 −1,n m=0

This proves the lemma.

Lemma III.4. The process {ηn }n≥d1 +1 , where ηn = (X n−d1 −1,n , Πn ), forms a random Markov function. Proof: According to Lemma 17 pp. 102–103 in [12] it is enough to show that ηn+1 is a function of the previous stage ηn , the variable Xn+1 and that conditional distribution of Xn+1 given Fn is a function of ηn . Let us consider, for x0 , . . . , xd1 +2 ∈ E, α ∈ (0, 1), a function ! fx1d1 +1 (xd1 +2 )(q + pα) ϕ(x0,d1 +1 , α, xd1 +2 ) = x1,d1 +2 , G(xd1 +1,d1 +2 , α) We will show that ηn+1 = ϕ(ηn , Xn+1 ). Notice that we get (see Lemma A.5(or Lemma A5 in [9]) (l = 0)) Πn+1 =

1 fX (Xn+1 )(q + pΠn ) n . G(X n,n+1 , Πn )

(8)

Hence ϕ(ηn , Xn+1 )=ϕ(X n−d1 −1,n , Πn , Xn+1 ) =  = X n−d,n+1 , Πn+1 = ηn+1 .

  f 1 (Xn+1 )(q + pΠn ) X n−d1 ,n , Xn+1 , Xn G(X n,n+1 , Πn )

Define Fˆn = σ(θ, X 0,n ). To see that the conditional distribution of Xn+1 given Fn is a function of ηn , let us consider the conditional expectation of u(Xn+1 ) for any Borel function u : E −→ R given Fn . Having Lemma A1 we get: Ex (u(Xn+1 ) | Fn )=Ex (u(Xn+1 )(1 − Πn+1 ) | Fn ) + Ex (u(Xn+1 )Πn+1 | Fn )   =Ex u(Xn+1 )I{θ>n+1} | Fn + Ex u(Xn+1 )I{θ≤n+1} | Fn     =Ex Ex (u(Xn+1 )I{θ>n+1} | Fˆn ) | Fn + Ex Ex (u(Xn+1 )I{θ≤n+1} | Fˆn ) | Fn     =Ex I{θ>n+1} Ex (u(Xn+1 ) | Fˆn ) | Fn + Ex I{θ≤n+1} Ex (u(Xn+1 ) | Fˆn ) | Fn Z Z 1 0 (y)µ(dy)Px (θ ≤ n + 1 | Fn ) = u(y)fXn (y)µ(dy)Px (θ > n + 1 | Fn ) + u(y)fX n E E Z Z 1 0 (y))µ(dy) = u(y)G(Xn , y, Πn )µ(dy) (y) + (q + pΠ )f = u(y)(p(1 − Πn )fX n Xn n E

E

This is our claim. Lemmata III.3 and III.4 are crucial for the solution of the posed problem (5). They show that the initial problem can be reduced to the problem of stopping Markov random function ηn = (X n−d1 −1,n , Πn ) with the payoff given by the equation (7). In the consequence we can use tools of the optimal stopping theory for finding the stopping time τ ∗ such that     Ex h(X τ ∗ −d1 −1,τ ∗ , Πτ ∗ ) = sup Ex h(X τ −d1 −1,τ , Πτ ) . (9) τ ∈SX d+1

To solve the reduced problem (9) for any Borel function u : Ed1 +2 × [0, 1] −→ R let us define operators:   Tu(x0,d1 +1 , α)=Ex u(X n−d1 ,n+1 , Πn+1 ) | X n−1−d1 ,n = x0,d1 +1 , Πn = α , Qu(x0,d1 +1 , α)=max{u(x0,d1 +1 , α), Tu(x0,d1 +1 , α)}. By the definition of the operator T and Q we get

Lemma III.5. For the payoff function h(x0,d1 +1 , α) characterized by (7) and for the sequence {rk }∞ k=0 : " # d1 X Lm−1 (x1,d1 +1 ) r0 (x1,d1 +1 )=p 1 − pd2 + q , pm L0 (x1,d1 +1 ) m=0 ) ( Z d1 +1 X Lm (x1,d +2 ) 0 d2 1 ; rk−1 (x2,d1 +2 ) µ(dxd1 +2 ), rk (x1,d1 +1 )=p fxd1 +1 (xd1 +2 ) max 1 − p + q pm L0 (x1,d1 +2 ) E m=1

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the following formulae hold: k

Q h1 (x1,d1 +2 , α)=(1 − α) max

(

d1 +1

1−p

d2

+q

X

m=1

) Lm (x1,d1 +2 ) ; rk−1 (x2,d1 +2 ) , k ≥ 1, pm L0 (x1,d1 +2 )

T Qk h1 (x1,d1 +2 , α)=(1 − α)rk (x2,d1 +2 ), k ≥ 0.

Proof: By the definition of the operator T and using Lemma 15 (l = 0) given that (X n−d1 −1,n , Πn ) = (x1,d1 +2 , α) we get   Th(x1,d1 +2 , α)=Ex h(X n−d1 ,n+1 , Πn+1 ) | X n−d1 −1,n = x1,d1 +2 , Πn = α # " d1 +1 X Lm (X n−d ,n+1 )  1 (1 − Π ) | X = x , Π = α =Ex 1 − p d2 + q n+1 n n−d1 −1,n 1,d1 +2 pm L0 (X n−d1 ,n+1 ) m=1 ! Z d1 +1 X Lm−1 (x2,d +2 ) fx1d +2 (xd1 +3 ) 1 1 =p(1 − α) 1 − p d2 + q m L (x 0 p ) f (xd1 +3 ) 0 x E 2,d +2 d 1 m=1 1 +2 ×

fx0d1 +2 (xd1 +3 )G(xd1 +2,d1 +3 , α)

µ(dxd1 +3 ) G(xd1 +2,d1 +3 , α) # " d1 +1 Z X Lm−1 (x2,d1 +2 ) 1 d2 fx (xd1 +3 )µ(dxd1 +3 ) =p(1 − α) 1 − p + q pm L0 (x2,d1 +2 ) d1 +2 m=1 E # " d1 +1 X Lm−1 (x2,d +2 ) d2 1 = (1 − α)r0 (x2,d1 +2 ). =(1 − α)p 1 − p + q pm L0 (x2,d1 +2 ) m=1

Directly from the definition of Q we get

 Qh(x1,d1 +2 , α)=max h(x1,d1 +2 , α); Th(x1,d1 +2 , α) ) ( d1 +1 X Lm (x1,d +2 ) d2 1 ; r0 (x2,d1 +2 ) . =(1 − α) max 1 − p + q pm L0 (x1,d1 +2 ) m=1

Suppose now that Lemma III.5 holds for TQk−1 h and Qk h for some k > 1. Then, using similar transformation as in the case of k = 0, we get i h TQk h(x1,d1 +2 , α)=Ex Qk h(X n−d1 ,n+1 , Πn+1 ) | X n−d1 −1,n = x1,d1 +2 , Πn = α ( ) # Z " d1 +1 X Lm (x2,d +3 ) d2 0 1 = max 1 − p + q ; rk−1 (x3,d1 +3 ) (1 − α)pfxd +2 (xd1 +3 ) µ(dxd1 +3 ) 1 pm L0 (x2,d1 +3 ) E m=1 =(1 − α)rk (x2,d1 +2 ).

Moreover

n o Qk+1 h(x1,d1 +2 , α)=max h(x1,d1 +2 , α); TQk h(x1,d1 +2 , α) ) ( d1 +1 X Lm (x1,d +2 ) d2 1 ; rk (x2,d1 +2 ) . =(1 − α) max 1 − p + q pm L0 (x1,d1 +2 ) m=1

This completes the proof. The following theorem is the main result of the paper. Theorem III.1. (a) The solution of the problem (5) is given by: d1 +1

τ ∗ = inf{n ≥ d1 + 1 : 1 − pd2 + q

X Lm (X n−d −1,n ) 1 ≥ r ∗ (X n−d1 ,n )} pm L0 (X n−d1 −1,n ) m=1

(10)

where r ∗ (X n−d,n ) = limk−→∞ rk (X n−d,n ). (b) The value of the problem, i.e. the maximal probability for (5) given X0 = x, is equal to ) ( Z d1 +1 X Lm (x, x1,d +1 ) ∗ ∗ d1 +1 d2 1 ; r (x1,d1 +1 ) Px (−d1 ≤ θ − τ ≤ d2 )=p max 1 − p + q pm L0 (x, x1,d1 +1 ) Ed1 +1 m=1 ×L0 (x, x1,d1 +1 )µ(dx1 , . . . , dxd1 +1 ).

Proof: Part (a). According to Lemma III.2 we look for a stopping time equal at least d1 + 1. From the optimal stopping theory (c.f [12]) we know that τ0 defined by (6) can be expressed as τ0 = inf{n ≥ d1 + 1 : h(X n−1−d1 ,n , Πn ) ≥ Q∗ h(X n−1−d1 ,n , Πn )}

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where Q∗ h(X n−1−d1 ,n , Πn ) = limk−→∞ Qk h(X n−1−d1 ,n , Πn ). According to Lemma III.5: ( d1 +1 X Lm (X n−d −1,n ) 1 τ0 =inf n ≥ d1 + 1 : 1 − pd2 + q pm L0 (X n−d1 −1,n ) m=1

) Lm (X n−d1 −1,n ) ∗ ; r (X n−d1 ,n )} ≥ max{1 − p +q pm L0 (X n−d1 −1,n ) m=1 ) ( d1 +1 X Lm (X n−d −1,n ) ∗ d2 1 ≥ r (X n−d1 ,n ) =inf n ≥ d1 + 1 : 1 − p + q pm L0 (X n−d1 −1,n ) m=1 d1 +1

X

d−2

=τ ∗ .

Part (b). Basing on the known facts from the optimal stopping theory we can write:  Px ( −d1 ≤ θ − τ ∗ ≤ d2 ) = Ex h⋆1 (X 0,d1 +1 , Πd1 +1 ) )! ( d1 +1 X Lm (X 0,d +1 ) ⋆ 1 ) ; r (X =Ex (1 − Πd1 +1 ) max 1 − pd2 + q 1,d1 +1 pm L0 (X 0,d1 +1 ) m=1 )! ( d1 +1 X Lm (X 0,d +1 ) ⋆ d2 1 =Ex Ex (I{θ>d1 +1} | Fd1 +1 ) max 1 − p + q ; r (X 1,d1 +1 ) pm L0 (X 0,d1 +1 ) m=1 ( )! d1 +1 X Lm (X 0,d +1 ) d2 ⋆ 1 ; r (X 1,d1 +1 ) =Ex I{θ>d1 +1} max 1 − p + q pm L0 (X 0,d1 +1 ) m=1 ) ( Z d1 +1 X Lm (x, x1,d +1 ) ∗ d2 1 ; r (x1,d1 +1 ) =Px (θ > d1 + 1) max 1 − p + q pm L0 (x, x1,d1 +1 ) Ed1 +1 m=1 ×L0 (x, x1,d1 +1 )µ(dx1 , . . . , dxd1 +1 )

This ends the proof of the theorem.

A PPENDIX L EMMATA Lemma A.1. Let n > 0, k ≥ 0 then: Px (θ ≤ n + k | Fn )=1 − pk (1 − Πn ). Proof: It is enough to show that for D ∈ Fn Z

I{θ>n+k} dPx = D

Z

(11)

pk (1 − Πn )dPx . D

Let us define Fen = σ(Fn , I{θ>n} ). We have: Z Z Z I{θ>n+k} dPx = I{θ>n+k} I{θ>n} dPx = I{θ>n+k} dPx D D∩{θ>n} ZD Z = Ex (I{θ>n+k} | Fen )dPx = Ex (I{θ>n+k} | θ > n)dPx D∩{θ>n} D∩{θ>n} Z Z = I{θ>n} pk dPx = (1 − Πn )pk dPx D

D

Lemma A.2. For n > 0 the following equality holds: Px (θ > n | Fn ) = 1 − Πn = Proof: Put D 0,n = {ω : X 0,n ∈ A0,n , Ai ∈ B}. Then: Px (D0,n )Px ( θ > n|D 0,n ) = =

Z

A0,n

Z

π ¯ pn L0 (X 0,n ) . S(X 0,n )

I{θ>n} dPx = D0,n

Z

(12)

Px (θ > n|Fn )dPx D0,n

π ¯ pn L0 (x0,n ) S(x0,n )µ(dx0,n ) = S(x0,n )

Z

D 0,n

π ¯ pn L0 (X 0,n ) dPx S(X 0,n )

Hence, by the definition of the conditional expectation, we get the thesis. Lemma A.3. For x0,l+1 ∈ El+2 , α ∈ [0, 1] and the functions S and G given by the equations (3) and (4) we have: S(X 0,n )=S(X 0,n−l−1 )G(X n−l−1,n , Πn−l−1 )

(13)

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Proof: By (12) we have S(X 0,n−l−1 ) · G(X n−l−1,n , Πn−l−1 ) = S(X 0,n−l−1 )Πn−l−1 Ll+1 (X n−l−1,n ) + S(X 0,n−l−1 )(1 − Πn−l−1 ) ! l X l+1 l−k p qLk+1 (X n−l−1,n ) + p L0 (X n−l−1,n ) × k=0

  = Sn−l−1 (X 0,n−l−1 ) − π ¯ pn−l−1 L0 (X 0,n−l−1 ) Ll+1 (X n−l−1,n )

(12)

+¯ πp

n−l−1

l X

L0 (X 0,n−l−1 )

p

l−k

qLk+1 (X n−l−1,n ) + p

k=0

=

πLn−l−1 (X 0,n−l−1 ) + π ¯

n−l−1 X

p

k−1

+¯ π

p

n

n−k−1

qLk+1 (X 0,n ) + p L0 (X 0,n )

k−1

n

k=0

From other side we have n X

S(X 0,n )=πLn (X 0,n ) + π ¯

p

qLk (X 0,n ) + p L0 (X 0,n )

k=1 n−l−1 X

¯ =πLn (X 0,n ) + π

p

k−1

qLn−k+1 (X 0,n ) +

L0 (X n−l−1,n )

!

!

qLn−k−l (X 0,n−l−1 ) Ll+1 (X n−l−1,n )

k=1

l X

l+1

n X

!

! p

k−1

!

n

qLn−k+1 (X 0,n ) + p L0 (X 0,n ) .

k=n−l

k=1

This establishes the formula (13). Lemma A.4. For n > l ≥ 0 the following equation is satisfied: Px (θ ≤ n − l − 1 | Fn )= Proof: Let D 0,n = {ω : X 0,n ∈ A0,n , Ai ∈ B}. Then Px ( D0,n )Px (θ > n − l − 1|D 0,n ) = =

Z

=

Z

Pn

= (13)

=

(12)

=

Z

Z

I{θ>n−l−1} dPx = D0,n

Z

Px (θ > n − l − 1|Fn )dPx D0,n

Px (θ = k)Ln−k+1 (x0,n ) + Px (θ > n)L0 (x0,n ) S(x0,n )µ(dx0,n ) S(x0,n ) P  l l−k qLk+1 (xn−l−1,n ) + pl+1 L0 (xn−l−1,n ) π ¯ pn−l−1 L0 (x0,n−l−1 ) k=0 p k=n−l

A0,n

A0,n

S(x0,n )

×S(x0,n )µ(dx0,n ) Z

Z

Πn−l−1 Ll+1 (X n−l−1,n ) . G(X n−l−1,n , Πn−l−1 )

π ¯ pn−l−1 L0 (x0,n−l−1 ) D0,n

π ¯ pn−l−1 L0 (x0,n−l−1 ) D0,n

(1 − Πn−l−1 ) D0,n

Pl

k=0

P P

l k=0

pl−k qLk+1 (X n−l−1,n ) + pl+1 L0 (X n−l−1,n ) S(X 0,n )

l k=0

pl−k qLk+1 (X n−l−1,n ) + pl+1 L0 (X n−l−1,n )

S(X 0,n−l−1 )G(X n−l−1,n , Πn−l−1 )

 

dPx

dPx

pl−k qLk+1 (X n−l−1,n ) + pl+1 L0 (X n−l−1,n ) dPx G(X n−l−1,n , Πn−l−1 )

This implies that: Px (θ > n − l − 1|Fn )=(1 − Πn−l−1 ) Pl pl−k qLk+1 (X n−l−1,n ) + pl+1 L0 (X n−l−1,n ) × k=0 G(X n−l−1,n , Πn−l−1 )

(14)

Simple transformations of (14) lead to the thesis. Lemma A.5. For n > l ≥ 0 the recursive equation holds: Πn =

P Πn−l−1 Ll+1 (X n−l−1,n ) + (1 − Πn−l−1 )q lk=0 pl−k Lk+1 (X n−l−1,n ) G(X n−l−1,n , Πn−l−1 )

(15)

JOURNAL OF LATEX CLASS FILES, VOL. 11, NO. 4, DECEMBER 2012

8

Proof: With the aid of (12) we get: pn L0 (X 0,n ) S(X 0,n−l−1 ) pl+1 L0 (X n−l−1,n ) 1 − Πn = = 1 − Πn−l−1 S(X 0,n ) pn−l−1 L0 (X 0,n−l−1 ) G(X n−l−1,n , Πn−l−1 ) Hence G(X n−l−1,n , Πn−l−1 ) − pn−l−1 L0 (X 0,n−l−1 )(1 − Πn−l−1 ) G(X n−l−1,n , Πn−l−1 ) P Πn−l−1 Ll+1 (X n−l−1,n ) + (1 − Πn−l−1 )q lk=0 pl−k Lk+1 (X n−l−1,n ) . = G(X n−l−1,n , Πn−l−1 )

Πn =

This establishes the formula (15).

F INAL

REMARKS

The presented analysis of the problem when the acceptable error for stopping before or after the disorder allows to see which protection is more difficult to control. When we admit that it is possible sequence of observation without disorder it is interesting question how to detect not only that we observe the second kind of data but that there were no at all data of the first kind. It can be verified by standard testing procedure when we stop very early (τ ≤ min{d1 , d2 }).

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