Forward estimation for ergodic time series

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Forward estimation for ergodic time series

Gusztáv Morvai

^a

, Benjamin Weiss

^b,^∗

aResearch Group for Informatics and Electronics of the Hungarian Academy of Sciences, 1521 Goldmann György tér 3, Budapest, Hungary bHebrew University of Jerusalem, Jerusalem 91904, Israel

Received 29 August 2003; accepted 15 June 2004 Available online 5 February 2005

Abstract

The forward estimation problem for stationary and ergodic time series{Xn}^∞_n₌₀taking values from a finite alphabetXis to estimate the probability thatX_n₊₁=xbased on the observationsX_i, 0inwithout prior knowledge of the distribution of the process{Xn}. We present a simple proceduregnwhich is evaluated on the data segment(X₀, . . . , Xn)and for which, error(n)= |g_n(x)−P (X_n₊₁=x|X₀, . . . , X_n)| →0 almost surely for a subclass of all stationary and ergodic time series, while for the full class the Cesaro average of the error tends to zero almost surely and moreover, the error tends to zero in probability.

Résumé

Le problème de l’estimation future d’une série temporelle ergodique et stationnaire{Xn}^∞_n₌₀, prenant ses valeurs dans un alphabet finiX, est d’estimer la probabilité queX_n₊₁=x, connaissant lesX_ipour 0inmais sans connaissance préalable de la distribution du processus{Xi}. Nous présentons un procédé simplegn, évalué sur les données(X₀, . . . , Xn), pour lequel erreur(n)= |g_n(x)−P (X_n₊₁=x|X₀, . . . , X_n)| →0 presque sûrement pour une sous-classe de toutes les séries temporelles ergodiques et stationnaires, tandis que pour la classe entière la moyenne de Cesaro de l’erreur tend vers zéro presque sûrement.

De plus, l’erreur tend vers zéro en probabilité.

MSC: 62G05; 60G25; 60G10

Keywords: Nonparametric estimation; Stationary processes

* Corresponding author.

E-mail addresses: morvai@szit.bme.hu (G. Morvai), weiss@math.huji.ac.il (B. Weiss).

doi:10.1016/j.anihpb.2004.07.002

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1. Introduction

T. Cover [6] posed two fundamental problems concerning estimation for stationary and ergodic binary time series{X_n}^∞_n_=−∞. (Note that a stationary time series{X_n}^∞_n₌₀can be extended to be a two sided stationary time series{X_n}^∞_n_=−∞.) Cover’s first problem was on backward estimation.

Problem 1. Is there an estimation schemefnfor the valueP (X₁=1|X₋n, . . . , X₀)such thatfndepends solely on the observed data segment(X₋n, . . . , X₀)and

nlim→∞f_n(X₋_n, . . . , X₀)−P (X₁=1|X₋_n, . . . , X₀)=0

almost surely for all stationary and ergodic binary time series{Xn}^∞n=−∞?

This problem was solved by Ornstein [20] by constructing such a scheme. (See also Bailey [5].) Ornstein’s scheme is not a simple one and the proof of consistency is rather sophisticated. For an even more general case, a much simpler scheme and proof of consistency were provided by Morvai, Yakowitz and Györfi [19]. (See also Algoet [1] and Weiss [24].) Note that none of these schemes are reasonable from the data consumption point of view.

Cover’s second problem was on forward estimation.

Problem 2. Is there an estimation schemefnfor the valueP (Xn+1=1|X₀, . . . , Xn)such thatfndepends solely on the data segment(X₀, . . . , Xn)and

nlim→∞f_n(X₀, . . . , X_n)−P (X_n₊₁=1|X₀, . . . , X_n)=0

almost surely for all stationary and ergodic binary time series{Xn}^∞n=−∞?

This problem was answered by Bailey [5] in a negative way, that is, he showed that there is no such scheme.

(Also see Ryabko [22], Györfi, Morvai and Yakowitz [11] and Weiss [24].) Bailey used the technique of cutting and stacking developed by Ornstein [21] and Shields [23]. Ryabko’s construction was based on a function of an infinite state Markov-chain.

Morvai [16] addressed a modified version of Problem 2. There one is not required to predict for all time instances rather he may refuse to predict for certain values ofn. However, he is expected to predict infinitely often. Morvai [16] proposed a sequence of stopping timesλ_nand he managed to estimate the conditional probabilityP (X_λ_n₊₁= 1|X₀, . . . , X_λ_n)in the pointwise sense, that is, for his estimator along the proposed stopping time sequence, the error tends to zero asnincreases, almost surely. Another estimator was proposed for this modified Problem 2 by Morvai and Weiss [17] for which theλ_ngrow more slowly, but the consistency only holds for a certain subclass of all stationary binary time series.

In this paper we consider the original Problem 2 but we shall impose an additional restriction on the possible time series. The conditional probabilityP (X₁=1|. . . , X₋₁, X₀)is said to be continuous if a version of it is continuous with respect to metric_∞

i=02⁻ⁱ⁻¹|x₋_i−y₋_i|, wherex₋_i, y₋_i∈ {0,1}.

Problem 3. Is there an estimation schemefnfor the valueP (Xn+1=1|X₀, . . . , Xn)such thatfndepends solely on the data segment(X₀, . . . , Xn)and

nlim→∞f_n(X₀, . . . , X_n)−P (X_n₊₁=1|X₀, . . . , X_n)=0

almost surely for all stationary and ergodic binary time series{Xn}^∞n=−∞ with continuous conditional probability P (X₁=1|. . . , X₋₁, X₀)?

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We will answer this question in the affirmative. This class includes allk-step Markov chains. It is not known if the schemes proposed by Bailey [5], Ornstein [20], Morvai, Yakowitz and Györfi [19] solve Problem 3 or not.

Problem 4. Is there an estimation schemefnfor the valueP (Xn+1=1|X₀, . . . , Xn)such thatfndepends solely on the data segment(X0, . . . , Xn)and

nlim→∞

1 n

n−1

i=0

f_i(X₀, . . . , X_i)−P (X_i₊₁=1|X₀, . . . , X_i)=0

almost surely for all stationary and ergodic binary time series{X_n}^∞_n_=−∞?

Bailey [5] (cf. Algoet [2] also) showed that any scheme that solves Problem 1 can be easily modified to solve Problem 4 (indeed, just exchange the data segment(X₋_n, . . . , X₀)for(X₀, . . . , X_n), but apparently not all solutions of Problem 4 arise in this fashion. For further reading cf. Algoet [1,3], Morvai, Yakowitz and Györfi [19], Györfi et al. [8], Györfi, Lugosi and Morvai [10], Györfi and Lugosi [9] and Weiss [24].

Problem 5. Is there an estimation schemefnfor the valueP (Xn+1=1|X0, . . . , Xn)such thatfndepends solely on the data segment(X₀, . . . , X_n)and for arbitrary >0,

nlim→∞Pf_n(X₀, . . . , Xn)−P (Xn+1=1|X₀, . . . , Xn)>

=0 for all stationary and ergodic binary time series{X_n}^∞_n_=−∞?

By stationarity, for any scheme that solves Problem 1, the shifted version of it solves Problem 5. (Just replace the data segment(X₋_n, . . . , X₀)by(X₀, . . . , X_n).)

There are existing schemes that solve Problem 4 (e.g. Bailey [5], Ornstein [20], and even for a more general case Morvai, Yakowitz and Györfi [19], Algoet [1], Györfi and Lugosi [9]) and there are schemes that solve Problem 5 (e.g. Bailey [5], Ornstein [20] and for even more general case Morvai, Yakowitz and Györfi [19], Algoet [1], Morvai, Yakowitz and Algoet [18]). In this paper we propose a reasonable, very simple algorithm that simultanously solves Problem 3, 4 and 5. Note that the schemes given by Bailey [5], Ornstein [20], Morvai, Yakowitz and Györfi [19], Algoet [1] and Weiss [24] are not reasonable at all, they consume data extremely rapidly, cf. Morvai [15] and it is not known if their schemes solve Problem 3 or not.

2. Preliminaries and main results

Let{X_n}^∞_n_=−∞be a stationary time series taking values from a finite alphabetX. (Note that all stationary time series{X_n}^∞_n₌₀can be thought to be a two sided time series, that is,{X_n}^∞n=−∞.) For notational convenience, let X_mⁿ =(X_m, . . . , X_n), wheremn. Note that ifm > nthenX_mⁿ is the empty string.

Letg:X→(−∞,∞)be arbitrary.

Our goal is to estimate the conditional expectationE(g(X_n₊₁)|X₀ⁿ)from samplesX₀ⁿ.

Fork1 define the stopping timesτ_i^k(n)which indicate where thek-blockX_nⁿ₋_k₊₁occurs previously in the time series{X_n}. Formally we setτ₀^k(n)=0 and fori1 let

τ_i^k(n)=min

t > τ_i^k₋₁(n): Xⁿ_n⁻₋^t_k₊₁₋_t=X_nⁿ₋_k₊₁

. (1)

LetK_n1 andJ_n1 be sequences of nondecreasing positive integers tending to∞which will be fixed later.

Defineκ_nas the largest 1kK_n such that there are at leastJ_n occurrences of the blockX_nⁿ₋_k₊₁in the data segmentXⁿ₀, that is,

κn=max

1kKn: τ_J^k

n(n)n−k+1

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if there is suchkand 0 otherwise.

Defineλ_nas the number of occurrences of the blockXⁿ_n₋_κ

n+1in the data segmentX₀ⁿ, that is, λ_n=max

1j: τ_j^κⁿn−κ_n+1

(3) ifκ_n>0 and zero otherwise. Observe that ifκ_n>0 thenλ_nJ_n.

Our estimateg_nforE(g(X_n₊₁)|X₀ⁿ)is defined asg₀=0 and forn1, gn= 1

λ_n

λn

i=1

g(X_n₋_τ^κn

i (n)+1) (4)

ifκ_n>0 and zero otherwise.

LetX^∗−be the set of all one-sided sequences, that is, X^∗−=

(. . . , x₋₁, x₀): x_i∈X for all − ∞< i0 . Define the functionG:X^∗−→(−∞,∞)as

G(x_−∞⁰ )=E

g(X₁)|X⁰_−∞=x_−∞⁰ .

Note that as a conditional expectation this is only defined almost surely. E.g. ifg(x)=1_{x=z} for a fixedz∈X thenG(y_−∞⁰ )=P (X1=z|X_−∞⁰ =y_−∞⁰ ).

Define a distance onX^∗−as d^∗(x_−∞⁰ , y⁰_−∞)=^∞

i=0

2⁻ⁱ⁻¹1_{_x_−i₌_y_−i_}.

Definition. The conditional expectationG(X⁰_−∞)is said to be continuous if a version of it is continuous on the setX^∗− with respect to metricd^∗(·,·). Since this space is compact, in fact, continuity is equivalent to uniform continuity.

The processes with continuous conditional expectation are essentially the Random Markov Processes of Ka- likow [12], or the continuous g-measures studied by Mike Keane [13].

Theorem. Let{X_n}be a stationary and ergodic time series taking values from a finite alphabetX. AssumeK_n= max(1,0.1 log_|X|n)andJ_n=max(1,n^0.5 ). Then

(A) if the conditional expectationG(X_−∞⁰ )is continuous with respect to metricd^∗(·,·)then

nlim→∞g_n−E

g(X_n₊₁)|Xⁿ₀=0 almost surely, (B) without any continuity assumption,

nlim→∞

1 n

n−1

i=0

gi−E

g(Xi+1)|Xⁱ₀=0 almost surely, (C) without any continuity assumption, for arbitrary >0,

nlim→∞Pg_n−E

g(X_n₊₁)|Xⁿ₀>

=0.

Remarks. Note that these results are valid without the ergodic assumption. One may use the ergodic decomposition throughout the proofs, cf. Gray [7], p. 268.

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We note that from the proof of Ryabko [22] and Györfi, Morvai and Yakowitz [11] it is clear that the continuity condition in the first part of the theorem cannot be relaxed. Even for the class of all stationary and ergodic binary time-series with merely almost surely continuous conditional probability P (X₁=1|. . . , X₋₁, X₀) one cannot solve Problem 2 in the Introduction. (An almost surely continuous conditional probability is such that as a function restricted to a setCwith full measure, it is continuous onC.)

We do not know if the shifted version of our proposed schemeg_nsolves Problem 1 or not. (That is, in the case wheng_nis evaluated on(X₋_n, . . . , X₀)rather than on(X₀, . . . , X_n).)

IfX is a countably infinite alphabet then there is no scheme that could achieve similar result to part (A) in the theorem for all boundedg(·), even if you assume that the resultingG(·)is continuous, and the time series is in fact a first order Markov chain. Indeed, whenever a new state appears which has not occurred before, you are unable to predict, cf. Györfi, Morvai and Yakowitz [11].

3. Auxiliary results

Fork1,n0 andj0 it will be useful to define auxiliary processes{ ˜X_i^{(k,n,j )}}^∞_i_=−∞as follows. Let X^{(k,n,j )}_i =X_n₋_τk

j(n)+i for − ∞< i <∞. (5)

For an arbitrary stationary time series{Y_n}fork1 letτ˜₀^k(Y_−∞^∞ )=0 and fori1 define

˜

τ_i^k(Y_−∞^∞ )=min

t >τ˜_i^k₋₁(Y_−∞^∞ ): Y₋^t_k₊₁₊_t =Y₋⁰_k₊₁

. (6)

If it is obvious on which time seriesτ˜_i^k(Y_−∞^∞ )is evaluated, we will writeτ˜_i^k. LetT denote the left shift, that is, (T x_−∞^∞ )_i =x_i₊₁.

We will need the next lemmas for later use.

Lemma 1. Let{X_n}be a stationary time series taking values from a finite alphabetX. Fork1,n0 andj0, the time series{X^{(k,n,j )}_i }^∞_i_=−∞has the same distribution as{X_i}^∞_i_=−∞.

Proof. Note that by (1), and (6), Tⁿ⁻^s{Xⁿ⁻^τ

k j(n)+m

n−τ_j^k(n)+l =x_l^m, τ_j^k(n)=s} = {X_l^m=x_l^m,τ˜_j^k=s}

whereτ˜_j^k is evaluated on time series{X_i}^∞_i_=−∞. Now by (5) and stationarity, P (X_l^{(k,n,j )}=xl, . . . ,Xm^{(k,n,j )}=xm)=

∞ s=0

P (Xⁿ⁻^τ

k j(n)+m

n−τ_j^k(n)+l =x_l^m, τ_j^k(n)=s)= ∞ s=0

P (X^m_l =x_l^m,τ˜_j^k=s)

=P (X_l^m=x_l^m) and the proof of Lemma 1 is complete. 2

Lemma 2. Let {X_n}be a stationary and ergodic time series taking values from a finite alphabet X. Assume K_n→ ∞,J_n→ ∞and lim_n_→∞(J_n/n)=0. Then

nlim→∞κ_n= ∞ almost surely.

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Proof. We argue by contradiction. Suppose, thatκ_n_i =K,X_nⁿⁱ

i−K =x₋⁰_K for a subsequence n_i. Then a simple frequency count (in the data segmentXⁿ₀ⁱ there are less thanJn_i occurrences of blockx₋⁰_K) yields that

P (X₋⁰_K=x₋⁰_K) lim

n→∞

Jn

n =0.

The set of sequences that contain a block with zero probability has zero probability and thus Lemma 2 is proved. 2

4. Pointwise consistency

Proof of Theorem (A). By Lemma 2, for largen, g_n(x)−E

g(X_n₊₁)|Xⁿ₀= 1 λn

λ_n

j=1

g(X_n₋_τκn

j (n)+1)−E

g(X_n₊₁)|Xⁿ₀

max

J=Jn,...,n max

k=1,...,Kn

1 J

J j=1

g(X_n₋_τk

j(n)+1)−G(Xⁿ⁻^τ

k j(n)

−∞ ) +

1 λ_n

λn

j=1

G(Xⁿ⁻^τ

κn j (n)

−∞ )−E

g(Xn+1)|Xⁿ₀ . Concerning the first term, by (1), (6) and (5),

1 J

J j=1

g(X_n₋_τk

j(n)+1)−G(Xⁿ⁻^τ

k j(n)

−∞ )

= 1 J

J−1 j=0

g(X˜^{(k,n,J )}

˜

τ_j^k+1 )−G(. . . ,X˜^{(k,n,J )}

˜

τ_j^k−1 ,X^{(k,n,J )}

˜ τ_j^k )

(7) whereτ˜_j^k is evaluated on{X_i^{(k,n,J )}}^∞_i_=−∞. Since by Lemma 1

G(. . . ,X^{(k,n,J )}

˜

τ_j^k−1 ,X^{(k,n,J )}

˜

τ_j^k )=Eg(X^{(k,n,J )}

˜

τ_j^k+1 )|. . . ,X^{(k,n,J )}

˜

τ_j^k−1 ,X^{(k,n,J )}

˜ τ_j^k

,

the pair(Γj=g(X^{(k,n,J )}

˜

τ_j^k+1 )−G(. . . ,X^{(k,n,J )}

˜

τ_j^k−1 ,X^{(k,n,J )}

˜

τ_j^k ),Fj =σ (X^τ^˜

k

−∞j ))forms a martingale difference sequence (E(Γj|Fj)=0 andΓj is measurable with respect toFj+1) for which Azuma’s exponential bound (cf. Azuma [4]) yields

P

1 J

J−1

j=0

g(X^{(k,n,J )}

˜

τ_j^k+1 )−G(. . . ,X^{(k,n,J )}

˜ τ_j^k )

>

2 e⁻²^{J /B}

for anyBsuch that max_x_∈X|g(x)|< B. Now by (7) P

max

J=Jn,...,n max

k=1,...,Kn

1 J

J j=1

g(X_n₋_τk

j(n)+1)−G(Xⁿ⁻^τ

k j(n)

−∞ ) >

nK_n2 e⁻²^Jⁿ^/B

and by assumptionnK_n2 e⁻²^Jⁿ^/Bsums up and the Borel–Cantelli Lemma yields almost sure convergence to zero.

Concerning the second term,

1 λn

λ_n

j=1

G(Xⁿ⁻^τ

κn j (n)

−∞ )−E

G(Xⁿ_−∞)|X₀ⁿ

→0 almost surely

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sinceκntends to infinity by Lemma 2,Xⁿ⁻^τ

κn j (n)

n−τ_j^κn(n)−κn+1=Xⁿ_n₋_κ

n+1for 0jλn, and the conditional expectation G(·)is in fact uniformly continuous onX^∗−with respect tod^∗(·,·). The proof of Theorem (A) is complete. 2

5. Time average performance

If the process does not have continuous conditional expectations then the last step in the proof of Theorem (A) is not valid. It can be carried out for most time instancesnby using the typical behaviour of almost every realization x_−∞^∞ . More specifically, for everyδ >0, the probability of the set of thosex_−∞⁰ for which

E

g(X₁)|X⁰₋_k₊₁=x₋⁰_k₊₁

−G(x⁰_−∞)< δ for allkK

tends to one asK tends to infinity. The typical behaviour we are after is the statement that most of the times t=n−τ_j^κⁿ(n)the sequenceT^tx_−∞^t belongs to the above mentioned set. While this need not be the case for alln, it is true for mostn’s and the next lemma makes this precise. For the analysis we will fix a value ofκnatk.

Define the set of good indexesMn(δ, K)⊆ {K−1, . . . , n−1}as Mn(δ, K)=

K−1in−1:E

g(Xi+1)|Xⁱ_i₋_k₊₁

−G(X_−∞ⁱ )< δ for allkK .

We will analyze the behaviour of our algorithm for κn=k for each in by first dividing up the indices {1,2, . . . , n}according to the value ofXⁱ_i₋_k₊₁=y₋⁰_k₊₁, and considering what happens for each of these.

Lety₋⁰_k₊₁∈X^k. Define the set of indexes I_n^k(y₋⁰_k₊₁)⊆ {k−1, . . . , n−1}, where you can find the pattern y₋⁰_k₊₁, that is,

I_n^k(y₋⁰_k₊₁)=

k−1in−1:Xⁱ_i₋_k₊₁=y₋⁰_k₊₁ . DefineD_k(i)as

Dk(i)=

{τ_j^k(i): τ_j^k(i)i−k+1 and 1ji+1} ifτ_J^k

i(i)i−k+1,

∅ otherwise.

LetE_n^k(δ, K)be defined as E_n^k(δ, K)=

0in−1:D_k(i)∩M_n(δ, K)> (1−δ^0.5)|D_k(i)| .

If the number of occurrences ofy₋⁰_k₊₁prior toiwas not enough for our algorithm thenDk(i)will be empty. This is rare, and can be expressed as follows: Let

F_n^k=

0in−1:Dk(i)= ∅ . It is immediate that

|F_n^k||X|^kJ_n. (8)

Lemma 3. Assume|M_n(δ, K)|(1−δ)n. Then

0ln−1: l /∈E_n^k(δ, K)andl /∈F_n^kδ^0.5n.

Proof. Fixδ,K,kandx∈X. Temporarily fix alsoy₋⁰_k₊₁∈X^k. Letz= |I_n^k(y₋⁰_k₊₁)|and letki1i2· · ·iz

denote the elements ofI_n^k(y₋⁰_k₊₁). Leti_j(y₋⁰_k₊₁)be the largest elementi_jofI_n^k(y₋⁰_k₊₁)such thatD_k(i_j)= ∅and 0ln−1: l∈Dk(i_j)andl /∈Mn(δ, K)δ^0.5Dk(i_j).

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DefineSto be the set of these indexes asy₋⁰_k₊₁varies over all elementX^k. It is clear that ifi, j∈S,i=j then Dk(i)∩Dk(j )= ∅since different blocksy₋⁰_k₊₁are involved. It follows from the construction that{Dk(i)}i∈S is a disjoint cover of{0ln−1: l /∈E_n^k(δ, K)andl /∈F_n^k}. It follows that

nδ0ln−1: l /∈Mn(δ, K)

i∈S

0ln−1: l∈Dk(i)andl /∈Mn(δ, K)

δ^0.5

i∈S

Dk(i)=δ^0.5

i∈S

Dk(i) . Now

0ln−1: l /∈E_n^k(δ, K)andl /∈F_n^k

i∈S

D_k(i) δ^0.5n and the proof of Lemma 3 is complete. 2

Proof of Theorem (B). Consider 1

n

n−1

i=0

gi(x)−E

g(Xi+1)|Xⁱ₀ |0−E(g(X₁)|X₀)|

n +1

n

n−1

i=1

1_{_κ_i_<K_i_}+1 n

n−1

i=1

max

J=J_i,...,i

1 J

J j=1

g(Xi−τ_j^Ki(i)+1)−G(Xⁱ⁻^τ

Ki j (i)

−∞ ) +1

n

n−1

i=1

1 λ_i

λ_i

j=1

G(Xⁱ⁻^τ

Ki j (i)

−∞ )−E

g(X_i₊₁)|X_iⁱ₋_K

i+1

1_{_κ_i₌_K_i_} +1

n

n−1

i=1

E

g(X_i₊₁)|X_iⁱ₋_K

i+1

−E

g(X_i₊₁)|Xⁱ₀.

The first term tends to zero. The second term tends to zero since by (8)|F_n^Kⁿ|/n|X|^KⁿJn/n→0.

Concerning the third term, by (7) and by Azuma’s exponential bound (cf. Azuma [4]) i

J=J_i

P

1 J

J j=1

g(X^(Kⁱ^{,i,J )}

˜

τ_j^Ki+1 )−G(. . . ,X^(Kⁱ^{,i,J )}

˜

τ_j^Ki−1 ,X^(Kⁱ^{,i,J )}

˜

τ_j^Ki ) >

2ie⁻²^Jⁱ^/B

(whereBis any real such that 2 max_x_∈X|g(x)|< B) and the right-hand side is summable, hence the Borel–Cantelli Lemma yields almost sure convergence to zero. By Toeplitz lemma the average also converges to zero.

Now we deal with the fourth term. Let 0< <1 be arbitrary. Choose the integerd large enough such that

n =

1 almost surely.) Now letN₁N₀ be so large thatKn−d+2K and|X|^10(Kⁿ⁻^d⁺¹⁾N₀ for allnN₁. AssumenN₁. The sum

1 n

n−1

i=1

1 λ_i

λi

j=1

G(Xⁱ⁻^τ

Ki j (i)

−∞ )−E

g(X_i₊₁)|Xⁱ_i₋_K

i+1

1_{_κ_i₌_K_i_}

that we are trying to estimate will be divided into blocks according to the value of Ki. In fact only values in the range [Kn−d +2, Kn] need be considered since the sum up to |X|^10Kⁿ⁻^d⁺¹ can be estimated

(9)

by |X|^10(Kⁿ⁻^d⁺¹⁾2 max_y_∈X|g(y)| and so by our assumption on d, after dividing by n this will be at most 2 max_y_∈X|g(y)|. For i in the range [|X|^10(k⁻¹⁾,|X|^10k) for Kn −δ +2 kKn, and κi =Ki, if i ∈ E^k⁻¹

|X|^10k(δ, K)then we get for more than(1−√

δ)|D_k(i)|terms an upper bound ofδwhile for the rest we may use 2 max_y_∈X|g(y)|. This gives an upper bound of

δ|D_k(i)| +√

δ|D_k(i)|2 max_y_∈X|g(y)|

|Dk(i)| .

Using Lemma 3 we can estimate the sum over alliin the interval[|X|^10(k⁻¹⁾,|X|^10k)by n

δ+√

δ2 max

y∈Xg(y)+√ δn2 max

y∈Xg(y). Dividing byn, we have an upper bound:

δ+√ δ2 max

y∈Xg(y)+√ δ2 max

y∈Xg(y).

The same argument yields the same upper bound for thei’s in the range[|X|^10Kⁿ, n).

Summing overkin the range[Kn−d+2, Kn+1]yields an upper bound:

dδ+d√ δ2 max

y∈Xg(y)+d√ δ2 max

y∈Xg(y).

Recall that√ δd=√

and this yields an upper bound:

+√ 2 max

y∈Xg(y)+√ 2 max

y∈Xg(y).

Sincewas arbitrary, the fourth term tends to zero.

Now we deal with the last term. Since by the martingale convergence theorem,E(g(X₁)|X⁰₋_i)→G(X_−∞⁰ ) almost surely, thus

ilim→∞E

g(X₁)|X₋⁰_K

i+1

−E

g(X₁)|X₋⁰_i=0

and applying Breiman’s generalized ergodic theorem, cf. Maker [14] (or Algoet [2]),

nlim→∞

1 n

n−1

i=0

E

g(X_i₊₁)|X_iⁱ₋_K

i+1

−E

g(X_i₊₁)|Xⁱ₀=0 almost surely and the proof of Theorem (B) is complete. 2

6. Weak consistency

Proof of Theorem (C). In order to show that for all ergodic stationary processes our estimateg_n converges in probability we follow the steps in the proof of Theorem (A). The probability that

g_n(x)−E

g(X_n₊₁)|Xⁿ₀>3

can be estimated as the sum of the probability of several sets, P

max

J=Jn,...,n max

k=1,...,Kn

1 J

J j=1

g(X_n₋_τk

j(n)+1)−G(Xⁿ⁻^τ

k j(n)

−∞ ) >

, P (κn< Kn),

PE

g(Xn+1)|Xⁿ₀

−E

g(Xn+1)|Xⁿ_n₋_K

n+1>