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Comptes Rendus

Mathématique

Songqi Wu, Xiaohui Ma, Hailin Sang and Xiequan Fan

A Berry–Esseen bound of orderp1

n for martingales Volume 358, issue 6 (2020), p. 701-712.

<https://doi.org/10.5802/crmath.81>

© Académie des sciences, Paris and the authors, 2020.

Some rights reserved.

This article is licensed under the

Creative Commons Attribution4.0International License.

http://creativecommons.org/licenses/by/4.0/

LesComptes Rendus. Mathématiquesont membres du Centre Mersenne pour l’édition scientifique ouverte

www.centre-mersenne.org

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2020, 358, n6, p. 701-712

https://doi.org/10.5802/crmath.81

Probabilities /Probabilités

A Berry–Esseen bound of order p 1 n for martingales

Une borne de Berry–Esseen d’ordre p 1 n pour les martingales

Songqi Wua, Xiaohui Maa, Hailin Sangband Xiequan Fan∗,a

aCenter for Applied Mathematics, Tianjin University, Tianjin, China

bDepartment of Mathematics, The University of Mississippi, University, MS 38677, USA

E-mails:[email protected], [email protected], [email protected], [email protected]

Abstract.Renz [13] has established a rate of convergence 1/p

nin the central limit theorem for martin- gales with some restrictive conditions. In the present paper a modification of the methods, developed by Bolthausen [2] and Grama and Haeusler [6], is applied for obtaining the same convergence rate for a class of more general martingales. An application to linear processes is discussed.

Résumé.Renz [13] a établi un taux de convergence 1/p

ndans le théorème de la limite centrale pour les martingales avec certaines conditions restrictives. Dans le présent article, une modification des méthodes, développées par Bolthausen [2] et Grama et Haeusler [6], est appliquée pour obtenir le même taux de convergence pour une classe de martingales plus générales. Une application aux processus linéaires est discutée.

Funding.The work has been partially supported by the National Natural Science Foundation of China (Grant no. 11601375). The research of Hailin Sang is partial supported by the Simons Foundation (Grant no. 586789).

Manuscript received 23rd January 2020, accepted 2nd June 2020.

Corresponding author.

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1. Introduction and main result

FornN, let (ξi,Fi)i=0,...,n be a finite sequence of martingale differences defined on some probability space (Ω,F,P), whereξ0=0 and {;,Ω}=F0⊆ · · · ⊆Fn⊆F are increasingσ-fields.

Denote

X0=0, Xk=

k

X

i=1

ξi,k=1, . . . ,n.

ThenX=(Xk,Fk)k=0,...,nis a martingale. Denote by〈X〉the conditional variance ofX:

X0=0, 〈Xk=

k

X

i=1

E£ ξ2i¯

¯Fi−1

¤,k=1, . . . ,n.

Define

D(Xn)=sup

x∈R

¯¯P(Xnx)−Φ(x)¯

¯,

whereΦ(x) is the distribution function of the standard normal random variable. Denote by→P the convergence in probability asn→ ∞. According to the martingale central limit theorem, the

“conditional Lindeberg condition”

n

X

i=1

E£

ξ2i1{i|≥²}¯

¯Fi−1

¤ P

→0, for each²>0, and the “conditional normalizing condition”〈Xn

P 1 together implies asymptotic normality of Xn, that is,D(Xn)→0 asn→ ∞.

The convergence rate ofD(Xn) has attracted a lot of attentions. For instance, Bolthausen [2]

proved that if|ξi| ≤²nfor a number²nand〈Xn=1 a.s., thenD(Xn)≤3nnlogn, where, here and after,cis an absolute constant not depending on²nandn. El Machkouri and Ouchti [3] improved the factor²3nnlognin Bolthausen’s bound to²nlognunder the following more general condition

E£

i|3¯

¯Fi−1

¤≤²nE£ ξ2i¯

¯Fi−1

¤ a.s. for alli=1, 2, . . . ,n.

For more related results, we refer to Ouchti [12] and Mourrat [11]. Recently, Fan [4] proved that if there exist a positive constantρand a number²n, such that

E£

i|2+ρ¯

¯Fi−1

¤≤²ρnE£ ξ2i¯

¯Fi−1

¤ a.s. for alli=1, 2, . . . ,n, and〈Xn=1 a.s., thenD(Xn)≤cρb²n, where

b²n=

(²ρn, ifρ∈(0, 1),

²n|log²n|, ifρ≥1,

andcρis a constant depending only onρ. Fan [4] also showed that this Berry–Esseen bound is optimal. In particular, if²n³1/p

n, then we have²n|log²n| ³(logn)/p

n. Thus, we cannot obtain the classical convergence rate 1/p

nfor general martingales.

However, the convergence rate 1/p

n for martingales is possible to be attained with some additional restrictive conditions. For instance, Renz [13] proved that if there exists a constant ρ>0 such that

E[ξ2i|Fi1]=1/n, E[ξ3i|Fi1]=0 and E£

i|3+ρ¯

¯Fi1¤

cn−(3+ρ)/2, a.s., (1) then it holds

D(Xn)=O µ 1

pn

. (2)

He also showed that this result is not true forρ=0. More martingale Berry–Esseen bounds of convergence rate 1/p

ncan be found in Bolthausen [2] and Kir’yanova and Rotar [10].

In this paper we are interested in extending (2) to a class of more general martingales. The following theorem is our main result.

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Theorem 1. Assume that there exist some numbersρ∈(0,+∞),²n∈(0,12]andδn∈[0,12]such that for all1≤in,

¯

¯〈Xn−1¯

¯≤δ2n, (3)

E£ ξ3i¯

¯Fi−1

¤=0 (4)

and

E£

i|3+ρ¯

¯Fi−1

¤≤²1+ρn E£ ξ2i¯

¯Fi−1

¤ a.s. (5)

Then

D(Xn)≤cρ(²nn),

where cρdepends only onρ.In addition, it holds cρ=O(ρ1),ρ→0.

Notice that under the conditions of Renz [13], the conditions of Theorem 1 are satisfied withδn =0 and²n ³1/p

n. Thus Theorem 1 extends Renz’s result to a class of more general martingales.

Thanks to the additional condition (4), the Berry–Esseen bound (6) improves the bound of Fan [4] by replacing²n|log²n|with²n.

Relaxing the condition (3), we have the following analogue estimation of Fan (cf. [4, (26)]).

Theorem 2. Assume that there exist some numbersρ∈(0,+∞)and²n ∈(0,12]such that for all 1≤in,

E£ ξ3i¯

¯Fi−1

¤=0 and

E£

i|3+ρ¯

¯Fi1¤

²1nE£ ξ2i¯

¯Fi1¤ a.s.

Then, for all p≥1,

D(Xn)≤cρ²n+cp µ

E£¯

¯〈Xn−1¯

¯

p¤ +Eh

max

1≤i≤ni|2p

1/(2p+1)

, (6)

where cρand cpdepend only onρand p, respectively.

It is easy to see that whenp→ ∞, µ

E£¯

¯〈Xn−1¯

¯

p¤

1/(2p+1)

→ k〈Xn−1k1/2 , which coincides withδnof Theorem 1.

2. Application

We first extend Theorem 1 to triangular arrays with infinity many terms in each line. FornN, let (ξn,i,Fn,i)ni=−∞be a sequence of martingale differences defined on some probability space (Ω,F,P), where the adapted filtration is {;,Ω}=F−∞⊂ · · · ⊂Fn,n−1⊂Fn,n⊂F. DenoteXn,k= Pk

i=−∞ξn,i,k≤n. Then (Xn,k,Fn,k)nk=−∞is a martingale. Let〈Xn,k=Pk

i=−∞E[ξ2n,i|Fn,i1],k≤n.

In particular, denoteXn:=Xn,nand〈Xn:= 〈Xn,n.

With some slight modification on the proof, Theorem 1 still holds in this new setting. Now we apply Theorem 1 with this new setting to the partial sum of linear processes. Let (εi)i∈Z be a sequence of identically distributed martingale differences adapted to the filtration (Fi)i∈Z. We consider the causal linear process in the form

Yk=

k

X

j=−∞

ak−jεj, (7)

where the martingale differences have finite variance and the sequence of real coefficients satisfiesP

i=0ai2< ∞. Without loss of generality, let the variance of the martingale difference to

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be 1. We say the linear process has long memory ifP

i=0|ai| = ∞. In this case, we assume that a0=1 and

ai=`(i)i−α,i>0, with 1/2<α<1. (8) Here`(·) is a slowly varying function. On the other hand, we say the linear process has short memory ifP

i=0|ai| < ∞andP

i=0ai6=0. The third case isP

i=0|ai| < ∞andP

i=0ai=0.

The long memory linear processes covers the well-known fractional ARIMA processes (cf.

Granger and Joyeux [7]; Hosking [9]), which play an important role in financial time series modeling and application. As a special case, let 0<d<1/2 andBbe the backward shift operator withk=εk1and consider

Yk=(1−B)dεk= X i=0

aiεki, whereai= Γ(i+d) Γ(d)Γ(i+1).

For this example we have limn→∞an/nd−1=1/Γ(d). Note that these processes have long memory becauseP

j=0|aj| = ∞. The partial sumSn=Pn

k=1Ykof causal linear process (7) can be written asSn=Pn

i=−∞bn,iεi, wherebn,i =Pn−i

j=0aj for 0<in, andbn,i =Pn−i

j=1−iaj fori ≤0. The variance ofSn isBn2= var(Sn)=Pn

i=−∞b2n,i. Now let Xn,k =Pk

i=−∞bn,iεi/Bn. ThenXn = Xn,n =Sn/Bn and〈Xn = Pn

i=−∞b2n,iE[ε2i|Fi−1]/Bn2. If we assume|〈Xn−1| ≤δ2n for someδn∈[0,12],E[ε3i|Fi−1]=0 and E[i|3|Fi1]≤dρ1+ρE[ε2i|Fi1] a.s. for alliZand some constantdρ, then, by Theorem 1,

sup

x∈R|P(Sn/Bnx)−Φ(x)| ≤cρ(²n+δn), where²n=dρsupi≤n|bn,i|/Bn.

In the case thatP

i=0|ai| < ∞, supin|bn,i| ≤P

i=0|ai| < ∞and it is well known thatBn2has ordern. Hence²n has order 1/p

n in this case. In the long memory caseP

i=0|ai| = ∞, if we assume (8),Bn2has ordern3−2α`2(n) (e.g., Wu and Min [14]) and supi≤n|bn,i|has ordern1−α`(n) (see Beknazaryan et al. [1] for upper bound and Fortune et al. [5] for lower bound in the case d=1). Hence in this case²n also has order 1/p

n. In either case the Berry–Esseen bound has order 1/p

nifδn=O(n−1/2). In particular, if we in addition assume that the innovations (εi)iZ

are independent, thenδn=0 and the Berry–Esseen bound supx∈R|P(Sn/Bnx)−Φ(x)|has order 1/p

n. Here the conditionE[ε3i|Fi−1]=0 is needed to have the Berry–Esseen bound of order 1/p

n. We cannot have this order from the result of Fan [4].

3. Proofs of theorems 3.1. Preliminary lemmas

In the proofs of theorems, we need the following technical lemmas. The first two lemmas can be found in Fan [4, Lemmas 3.1 and 3.2].

Lemma 3. If there exists an s>3such that

E[i|s|Fi−1]≤²s−2n E[ξ2i|Fi−1], then, for any t∈[3,s),

E[i|t|Fi−1]≤²tn2E[ξ2i|Fi−1].

Lemma 4. If there exists an s>3such that

E[i|s|Fi−1]≤²s−2n E[ξ2i|Fi−1], then

E[ξ2i|Fi−1]≤²2n.

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The next two technical lemmas are due to Bolthausen (cf. [2, Lemmas 1 and 2]).

Lemma 5. Let X and Y be random variables. Then sup

u

¯¯P¡ Xu¢

−Φ(u)¯

¯≤c1sup

u

¯¯P¡

X+Yu¢

−Φ(u)¯

¯+c2°

°E[Y2|X

°

1/2

, where c1and c2are two positive constants.

Lemma 6. Let G(x)be an integrable function onRof bounded variationkGkV, X be a random variable and a,b6=0are real numbers. Then

E

· G

µX+a b

¶¸

≤ kGkVsup

u

¯¯P¡ Xu¢

−Φ(u)¯

¯+ kGk1|b|, wherekGk1is the L1(R)norm of G(x).

In the proof of Theorem 2, we also need the following lemma of El Machkouri and Ouchti [3].

Lemma 7. Let X and Y be two random variables. Then, for p≥1, D(X+Y)≤2D(X)+3°

°E£

Y2p|X¤°

°

1/(2p+1)

1 . (9)

3.2. Proof of Theorem 1

By Lemma 3, we only need to consider the case ofρ∈(0, 1]. We follow the method of Grama and Haeusler [6]. LetT=1+δ2n. We introduce a modification of the conditional variance〈Xnas follows:

Vk= 〈Xk1{k<n}+T1{k=n}. (10)

It is easy to see thatV0=0,Vn=T, and that (Vk,Fk)k=0,...,nis a predictable process. Set γ=²n+δn.

Letcbe some positive and sufficient large constant. Define the following non-increasing dis- crete time predictable process

Ak=c2γ2+TVk, k=1, . . . ,n. (11) Obviously, we haveA0=c2γ2+T andAn=c2γ2. In addition, foru,xR, andy>0, denote

Φu(x,y)=Φ µux

py

. (12)

LetN =N(0, 1) be a standard normal random variable, which is independent ofXn. Using a smoothing procedure, by Lemma 5, we deduce that

sup

u

¯¯P¡ Xnu¢

−Φ(u)¯

¯≤c1sup

u

¯¯P¡

Xn+cγNu¢

−Φ(u)¯

¯+c2γ

=c1sup

u

¯

¯E£ Φu¡

Xn,An¢¤

−Φ(u)¯

¯+c2γ

c1sup

u

¯¯E£ Φu

¡Xn,An¢¤

E£ Φu

¡X0,A0¢¤¯

¯ +c1sup

u

¯

¯E£ Φu¡

X0,A0¢¤

−Φ(u)¯

¯+c2γ

=c1sup

u

¯¯E£ Φu

¡Xn,An¢¤

E£ Φu

¡X0,A0¢¤¯

¯

+c1sup

u

¯

¯

¯

¯Φ

µ u

q c2γ2+T

−Φ(u)

¯

¯

¯

¯+c2γ. (13) It is obvious that ¯

¯

¯

¯Φ

µ u

q c2γ2+T

−Φ(u)

¯

¯

¯

¯≤c3

¯

¯

¯

¯ 1 q

c2γ2+T

−1

¯

¯

¯

¯≤c4γ. (14)

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Returning to (13), we get sup

u

¯¯P¡ Xnu¢

−Φ(u)¯

¯≤c1sup

u

¯¯E£ Φu

¡Xn,An¢¤

E£ Φu

¡X0,A0¢¤¯

¯+c5γ. (15) By a simple telescoping, we know that

E£ Φu

¡Xn,An¢¤

E£ Φu

¡X0,A0¢¤

=E

· n

X

k=1

¡Φu

¡Xk,Ak¢

−Φu

¡Xk1,Ak1¢¢

¸

. (16)

Taking into account the fact that

2

∂x2Φu(x,y)=2

∂yΦu(x,y), we get

E£ Φu

¡Xn,An¢¤

E£ Φu

¡X0,A0¢¤

=J1+J2J3, (17) where

J1=E

· n

X

k=1

µ

Φu(Xk,Ak)−Φu(Xk−1,Ak)−

∂xΦu(Xk−1,Ak)ξk

−1 2

2

∂x2Φu(Xk−1,Ak)ξ2k−1 6

3

∂x3Φu(Xk−1,Ak)ξ3k

¶¸

, (18)

J2=1 2E

· n X

k=1

2

∂x2Φu(Xk−1,Ak

4 〈Xk− 4Vk¢

¸

, (19)

J3=E

· n

X

k=1

µ

Φu(Xk−1,Ak−1)−Φu(Xk−1,Ak)−

∂yΦu(Xk−1,Ak)4Vk

¶¸

, (20)

where4〈Xk= 〈Xk− 〈Xk1.

Now, we need to give some estimates ofJ1,J2andJ3. To this end, we introduce some notations.

Denote byϑi some random variables satisfying 0≤ϑi≤1, which may represent different values at different places. For the rest of the paper,ϕstands for the density function of the standard normal random variable.

Control ofJ1. For convenience’s sake, letTk−1=(u−Xk−1)/p

Ak,k=1, 2, . . . ,n. It is easy to see that

Bk=:Φu(Xk,Ak)−Φu(Xk−1,Ak)−

∂xΦu(Xk−1,Ak)ξk

−1 2

2

∂x2Φu(Xk−1,Ak)ξ2k−1 6

3

∂x3Φu(Xk−1,Ak)ξ3k

=Φ µ

Tk−1ξk

pAk

−Φ(Tk−1)+Φ0(Tk−1) ξk

pAk

−1

00(Tk−1) µ ξk

pAk

2

+1

000(Tk−1) µ ξk

pAk

3

. To estimate the right hand side of the last equality, we distinguish two cases.

Case 1:k/p

Ak| ≤2+ |Tk−1|/2. By a four-term Taylor expansion, it is obvious that if|ξk/p Ak| ≤ 1, then

¯

¯Bk¯

¯=

¯

¯

¯

¯ 1 24Φ(4)

µ

Tk−1ϑ ξk

pAk

¶¯

¯

¯

¯ ξk

pAk

¯

¯

¯

¯

4¯

¯

¯

¯

¯

¯

¯

¯Φ(4) µ

Tk1ϑ ξk

pAk

¶¯

¯

¯

¯

¯

¯

¯

¯ ξk

pAk

¯

¯

¯

¯

3+ρ

.

(8)

If|ξk/p

Ak| >1, by a three-term Taylor expansion, then

¯¯Bk¯

¯≤1 2

µ¯

¯

¯

¯Φ000 µ

Tk−1ϑ ξk

pAk

¶¯

¯

¯

¯+

¯

¯

¯

¯Φ000(Tk−1)

¯

¯

¯

¯

¶¯

¯

¯

¯ ξk

pAk

¯

¯

¯

¯

3

¯

¯

¯

¯Φ000 µ

Tk1ϑ0 ξk

pAk

¶¯

¯

¯

¯

¯

¯

¯

¯ ξk

pAk

¯

¯

¯

¯

3

¯

¯

¯

¯Φ000 µ

Tk1ϑ0 ξk

pAk

¶¯

¯

¯

¯

¯

¯

¯

¯ ξk

pAk

¯

¯

¯

¯

3+ρ

, where

ϑ0=

ϑ, if¯

¯Φ000¡

Tk−1−ϑpξk

Ak

¢¯

¯≥ |Φ000(Tk−1)|, 0, if¯

¯Φ000¡

Tk−1−ϑpξk

Ak

¢¯

¯< |Φ000(Tk−1)|. Using the inequality max{|Φ000(t)|,|Φ0000(t)|}≤ϕ(t)(2+t4), we find that

¯¯Bk1{

k/p

Ak|≤2+|Tk1|/2}

¯

¯≤ϕ µ

Tk1−ϑ1 ξk

pAk

¶µ 2+

µ

Tk1ϑ1 ξk

pAk

4¶¯

¯

¯

¯ ξk

pAk

¯

¯

¯

¯

3+ρ

g1(Tk−1)

¯

¯

¯

¯ ξk

pAk

¯

¯

¯

¯

3+ρ

, (21)

where

g1(z)= sup

|tz|≤2+|z|/2ϕ(t)(2+t4).

Case 2:k/p

Ak| >2+ |Tk−1|/2. It is obvious that, for|4x| >1+ |x|/2,

¯

¯

¯

¯Φ(x− 4x)−Φ(x)+Φ0(x)4x−1

00(x)(4x)2+1

000(x)(4x)3

¯

¯

¯

¯

≤ µ¯

¯

¯

¯

Φ(x− 4x)−Φ(x)

|4x|3

¯

¯

¯

¯+ |Φ0(x)| + |Φ00(x)| + |Φ000(x)|

|4x|3

≤ µ

8

¯

¯

¯

¯

Φ(x− 4x)−Φ(x) (2+ |x|)3

¯

¯

¯

¯+ |Φ0(x)| + |Φ00(x)| + |Φ000(x)|

|4x|3

≤ µ

ce

(2+ |x|)3+ |Φ0(x)| + |Φ00(x)| + |Φ000(x)|

|4x|3

cb

(2+ |x|)3|4x|3

cb

(2+ |x|)3|4x|3+ρ. Hence, we have

¯¯Bk1{

k/p

Ak|>2+|Tk−1|/2}

¯

¯≤g2(Tk−1)

¯

¯

¯

¯ ξk

pAk

¯

¯

¯

¯

3+ρ

, (22)

where

g2(z)= cb (2+ |z|)3. Denote

G(z)=g1(z)+g2(z).

Combining (21) and (22) together, we get

|Bk| ≤G(Tk1)

¯

¯

¯

¯ ξk

pAk

¯

¯

¯

¯

3+ρ

. (23)

(9)

Therefore,

¯¯J1¯

¯=

¯

¯

¯

¯ E

· n X

k=1

Bk

¸¯

¯

¯

¯≤E

· n X

k=1

G(Tk1)

¯

¯

¯

¯ ξk

pAk

¯

¯

¯

¯

3¸

. (24)

Next, we consider conditional expectation of|ξk|3+ρ. By condition (5), we get E£

k|3+ρ¯

¯Fk1

¤≤²1+ρn 4 〈Xk, (25)

where4〈Xk= 〈Xk− 〈Xk1and we know that

4 〈Xk= 4Vk=VkVk−1, 1≤k<n,4〈Xn≤ 4Vn, (26) then

E£

k|3+ρ¯

¯Fk−1

¤≤²1n4Vk. (27)

By (24) and (27), we obtain

|J1| ≤R1:=²1n

· n X

k=1

G(Tk−1) A(3k)/24Vk

¸

. (28)

To estimateR1, we introduce the time changeτtas follow: for any realt∈[0,T],

τt=min{k≤n:Vkt}, where min; =n. (29) Obviously, for anyt∈[0,T], the stopping timeτtis predictable. In addition, (σk)k=1,...,n+1(with σ1 = 0) stands for the increasing sequence of moments when the increasing and stepwise functionτt,t ∈[0,T], has jumps. It is easy to see that4Vk=R

[σk,σk+1)dt, and that k =τt for t∈[σk,σk+1). SinceτT =n, we have

n

X

k=1

G(Tk−1) A(3+ρk )/24Vk=

n

X

k=1

Z

kk+1)

G(Tτt−1) A(3+ρτt )/2

dt= ZT

0

G(Tτt−1) A(3+ρτt )/2

dt. (30)

Letat=c2γ2+Tt. Because of4Vτt≤2²2n+2δ2n(cf. Lemma 4), we know that

tVτt=Vτt1+ 4Vτtt+2²2n+2δ2n, t∈[0,T]. (31) Assumec≥2, then we have

1

2atAτt=c2γ2+TVτtat, t∈[0,T]. (32) Note thatG(z) is symmetric and is non-increasing inz≥0. The last bound implies that

R1≤2(3)/2²1+ρn

Z T 0

1 at(3+ρ)/2

E

· G

µuXτt1 a1/2t

¶¸

dt. (33)

Note also thatG(z) is a symmetric integrable function of bounded variation. By Lemma 6, it is obvious that

E

· G

µuXτt−1 a1/2t

¶¸

c6sup

z

¯¯P¡

Xτt−1z¢

−Φ(z)¯

¯+c7p

at. (34)

Because ofc≥2,Vτt−1=Vτt− 4Vτt,Vτttand4Vτt≤2²2n+2δ2n, we obtain

VnVτt1=VnVτt+ 4Vτt≤2²2n+2δ2n+Ttat. (35) Therefore

E£¡

XnXτt−1¢2¯

¯Fτt−1

¤=E

· n X

k=τt

E£ ξ2k¯

¯Fk1

¤

¯

¯

¯

¯Fτt−1

¸

=E£

Xn− 〈Xτt−1

¯

¯Fτt−1

¤

E[VnVτt−1|Fτt−1]

at.

(10)

Then, by Lemma 5, we deduce that for anyt∈[0,T], sup

z

¯

¯P¡

Xτt1z¢

−Φ(z)¯

¯≤c8sup

z

¯

¯P¡ Xnz¢

−Φ(z)¯

¯+c9p

at. (36)

Combining (28), (33), (34) and (36) together, we get

|J1| ≤c10²1n

ZT 0

1 a(3t)/2

dtsup

z

¯

¯P¡ Xnz¢

−Φ(z)¯

¯+c11²1n

Z T 0

1 a1t/2

dt. (37)

Taking some elementary computations, it follows that ZT

0

1 a(3+ρt )/2

dt= ZT

0

1

(c2γ2+Tt)(3+ρ)/2dt≤ 2

c1+ρ(1+ρ)γ1 (38) and

ZT 0

1 a1t/2

dt= Z T

0

1

(c2γ2+Tt)1+ρ/2dt≤ 2

cρργρ. (39) This yields

¯¯J1¯

¯≤ c12

c1sup

z

¯¯P¡ Xnz¢

−Φ(z)¯

¯+cρ,1²n

ρ . (40)

Control ofJ2. Since 0≤ 4Vk− 4〈Xk≤2δ21{k=n}, we have

|J2| ≤E

· 1 2An

¯

¯ϕ0(Tn1)(4Vn− 4〈Xn

¯

¸ .

DenoteG(z)e =sup|z−t|≤10(t)|, and then|ϕ0(z)| ≤G(ze ) for any realz. SinceAn=c2γ2, then we get the following estimation:

|J2| ≤ 1 c2E£

G(Te n1)¤ .

Note thatGeis non-increasing inz ≥0, and thus it has bounded variation onR. By Lemma 6, we get

|J2| ≤c13 c2 sup

z

¯¯P¡

Xn−1z¢

−Φ(z)¯

¯+c∗,2(²n+δn). (41) Then, by Lemma 5, we deduce that

sup

z

¯¯P¡

Xn−1z¢

−Φ(z)¯

¯≤c14sup

z

¯¯P¡ Xnz¢

−Φ(z)¯

¯+c15²n. (42) This yields

|J2| ≤c16 c2 sup

z

¯

¯P¡ Xnz¢

−Φ(z)¯

¯+cρ,2(²n+δn). (43) Control ofJ3. By a two-term Taylor expansion, it follows that

|J3| =1 8E

· n

X

k=1

1

(Akϑk4Ak)2ϕ000

µ uXk−1 pAkϑk4Ak

¶ (4Ak)2

¸ . Note thatc≥2,4Ak≤0 and, by Lemma 4,|4Ak| = 4Vk≤2²2n+2δ2n. We obtain

AkAkϑk4Akc2γ2+TVk+2²2n+2δ2n≤2Ak. (44) DenoteG(z)b =sup|t−z|≤2000(t)|. ThenG(z) is symmetric, and is non-increasing inb z ≥0. Us- ing (44), we get

|J3| ≤(2²2n+2δ2n)E

· n

X

k=1

1 A2kGb

µTk−1 p2

¶ 4Vk

¸

. (45)

(11)

By an argument similar to that of (40), we get

|J3| ≤c17(2²2n+2δ2n) c2γ2 sup

z

¯¯P¡ Xnz¢

−Φ(z)¯

¯+2c18(2²2n+2δ2n) cγ

c19

c2 sup

z

¯¯P¡ Xnz¢

−Φ(z)¯

¯+4c18(²nn)2 cγ

c19 c2 sup

z

¯

¯P¡ Xnz¢

−Φ(z)¯

¯+cρ,3(²n+δn). (46)

Combining (17), (40), (43) and (46) together, we get

¯¯E£ Φu

¡Xn,An¢¤

E£ Φu

¡X0,A0¢¤¯

¯≤ c20 c1+ρ

sup

z

¯¯P¡ Xnz¢

−Φ(z)¯

¯+bcρ

ρ(²nn), By (15), we know that

sup

z

¯¯P¡ Xnz¢

−Φ(z)¯

¯≤ c21 c1+ρ

sup

z

¯¯P¡ Xnz¢

−Φ(z)¯

¯+ceρ

ρ(²n+δn), from which, choosingc1=max {2c21, 21+ρ}, we get

sup

z

¯

¯P¡ Xnz¢

−Φ(z)¯

¯≤2ceρ(²n+δn)

ρ . (47)

3.3. Proof of Theorem 2

Following the method of Bolthausen [2], we enlarge the sequence (ξi,Fi)1≤i≤n to¡ ξbi,Fci

¢

1iN

such that­ X

N:=PN i=1E£

ξb2i|Fci1¤

=1 a.s., and then apply Theorem 1 to the enlarged sequence.

Consider the stopping time

τ=sup{k≤n:〈Xk≤1}. (48)

Assume that 0≤ε²n. Letr1−〈X〉τ

ε2

¦, wherebxcdenotes the “integer part” ofx. It is easy to see thatr≤¥1

ε2

¦. SetN =n+r+1. Let (ζi)i≥1be a sequence of independent Rademacher random variables, which is independent of the martingale differences (ξi)1in. Consider the random variables¡

ξbi,Fci¢

1≤i≤Ndefined as follows:

ξbi=









ξi a.s., ifiτ,

εζi a.s., ifτ+1≤iτ+r,

¡1− 〈Xτ2¢1/2

ζi a.s., ifi=τ+r+1,

0 a.s., ifτ+r+1≤iN,

andFci=σ¡

ξb1,ξb2, . . . ,ξbi

¢. Clearly,¡

ξbi,Fci

¢

1iNstill forms a martingale difference sequence with respect to the enlarged filtration. ThenXbk=Pk

i=1ξbi,k=0, . . . ,N, withXb0=0, is also a martingale. Moreover, it holds that

­ X

N=1,E£ bξ3i¯

¯ cFi−1

¤=0 and E£¯

¯ bξi

¯

¯3+ρ

¯

¯ cFi−1

¤≤²1+ρn E£ ξb2i¯

¯ cFi−1

¤, a.s.

By Theorem 1, we have

D¡ XbN¢

cρ²n

ρ . (49)

Using Lemma 7, we obtain that D(Xn)≤2D¡

XbN¢ +3°

°E£¯

¯XnXbN¯

¯

2p¯

¯ bXN¤°

°1/(2p+1)

1 ≤2cρ²n

ρ +3¡ E£¯

¯ bXNXn¯

¯

2p¤¢1/(2p+1) . (50)

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