asymptotics for self-normalized random products of sums for mixing sequences

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Asymptotics for Self-Normalized Random Productsof Sums for Mixing SequencesWeidong Liu a & Zheng-yan Lin aa Department of Mathematics , Zhejiang University ,Hangzhou, ChinaPublished online: 19 Jun 2007.

To cite this article: Weidong Liu & Zheng-yan Lin (2007) Asymptotics for Self-Normalized Random Products of Sums for Mixing Sequences, Stochastic Analysis andApplications, 25:4, 739-762, DOI: 10.1080/07362990701419938

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Stochastic Analysis and Applications, 25: 739–762, 2007Copyright © Taylor & Francis Group, LLCISSN 0736-2994 print/1532-9356 onlineDOI: 10.1080/07362990701419938

Asymptotics for Self-Normalized RandomProducts of Sums for Mixing Sequences

Weidong Liu and Zheng-Yan LinDepartment of Mathematics, Zhejiang University, Hangzhou, China

Abstract: Let �X�Xn� n ≥ 1� be a sequence of a strictly stationary �-mixingpositive random variables, which is in the domain of attraction of the normallaw, and tn be a positive, integer random variable and denote Sn =

∑ni=1 Xi, V

2n =∑n

i=1 X2i and EX = � > 0. Under a general condition about tn and

∑�i=1 �

1/2�i� <

�, we show that the self-normalized random products of the partial sums,(∏tnj=1

Skk�

) �Vtn , is still asymptotically lognormal.

Keywords: Domain of attraction of the normal law; Lognormal distribution;Products; Self-normalized.

AMS Subject Classification (2000): 60F05.

1. INTRODUCTION AND MAIN RESULTS

Let �X�Xn� n ≥ 1� be a sequence of independent and identicallydistributed positive random variables and define the partial sumSn =

∑nj=1 Xj and V 2

n =∑ni=1 X

2i for n ≥ 1. Arnold and Villaseñor [1]

considered the limiting properties of sums of records and obtained

Received August 4, 2005; Accepted September 28, 2006The first author’s research is supported by National Natural Science

Foundation of China (No. 10671176). The second author’s research is supportedby National Natural Science Foundation of China (No. 10571159).

The authors would like to thank the referee for many valuable comments.Address correspondence to Weidong Liu, Department of Mathematics,

Zhejiang University, Hangzhou 310027, China; E-mail: [email protected]

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740 Liu and Lin

the following version of the central limit theorem for i.i.d. exponentialrandom variables with the mean one,∑n

k=1 log�Sk�− n log�n�+ n√2n

d→ � (1.1)

as n → �, here and in the sequel, � is a standard normal randomvariable. By the Stirling formulation, (1.1) can be equivalently stated as

( n∏k=1

Skk

) 1√n d→ e

√2� �

Rempala et al. [13] removed the condition that the distribution of Xi

is exponential and obtained the following theorem.

Theorem A. Let �X�Xn� n ≥ 1� be a sequence of i.i.d. positive squareintegrable random variables. Denote � = EX > 0, the coefficient of variation = /�, where 2 = VarX. Then

(∏nk=1 Skn!�n

) 1√n d→ e

√2� � (1.2)

Recently, Pang et al. [11] showed the asymptotics for self-normalizedrandom products of sums of i.i.d. random variables, and they obtained

Theorem B. Assume that the positive random variable X has mean � �>0�and is in the domain of attraction of the normal law and tn be a positiveinteger-valued random variable, and there is a positive constant sequence�bn� tending to infinity as n → � such that tn/bn

p→ �, where � is a positiverandom variable. Then

(∏tnk=1 Skn!�n

) �Vtn d→ e

√2� � (1.3)

It is well-known that the so-called self-normalized limit theorem puta totally new countenance upon classical limit theorems. We refer toBentkus et al. [3] for Berry-Esseen inequalities, Giné et al. [7] for thenecessary and sufficient condition for the asymptotic normality, Griffinet al. [8] for law of the iterated logarithm, Csörgo et al. [6] for studentizedincrements, Lin [9] for Chung-type law of the iterated logarithm, Csörgoet al. [5] for Donsker’s theorem and Shao [15–17] for large deviations.Recently, Balan and Kulik [2] extended the result of Csörgo et al. [5] tomixing sequence. In this article, we will prove an analog to Theorem Bholds for a strictly stationary �-mixing sequence of random variables.

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Products of Sums for Mixing Sequences 741

We begin to introduce some notations that will be used throughoutthis paper. A sequence �Xj�j≥1 of random variables is called �-mixing if��n� −→ 0, where

��n� �= supk≥1

�(�k

1��k+n

)�

�(�k

1��k+n

)�= sup

{�P�B �A�− P�B�� A ∈ �k1� B ∈ ��

k+n

}and �a

b denotes the -field generated by Xa�Xa+1� � � � � Xb. In the presentpaper, we consider only strictly stationary �-mixing sequences whichsatisfy

�∑n=1

�1/2�n� < �� (1.4)

We also let � �= �EX1�2/EX2

1 (=0 if EX21 = ��, l�x� = EX2

1I��X1� ≤ x� and

A2k�n� �= Var

( k∑j=1

XjI��Xj� ≤ �n�

)� B2

k�n� �= kEX21I��X1� ≤ �n�

where

�j = inf{s � s ≥ b + 1�

l�s�

s2≤ 1

j

}� j = 1� 2� 3� � � �

Sometimes, we use the notation A2n and B2

n instead of A2n�n� and

B2n�n�, respectively. It is easy to see that B2

n�n� = nl��n� ∼ �2n as n → �.From now on we suppose

A2n�n� ∼ �B2

n�n� and l�x� = EX2I��X� ≤ x� is slowly varying at �(1.5)

for some 0 < � < �. Throughout this article we use the notation an ∼ bnif an/bn → 1, an ≈ bn if an = O�bn� and bn = O�an�. We denote with C ageneric constant that may be different in each of its appearances.

We state our results as follows.

Theorem 1.1. Assume that (1.4) and (1.5) are satisfied. � = EX > 0. Let tnbe a positive integer-valued random variable, and there is a positive constantsequence �bn� tending to infinity as n → � such that tn/bn

p→ �, where � isa positive random variable. Then(∏tn

k=1 Sktn!�tn

) �Vtn d→ e

√2�� (1.6)

Remark. From [2] we can know that if l�x� is slowly varying at � and

Sn/Vn

d→ � , then A2n�n� ∼ �B2

n�n� for some 0 < � < �. So the conditionA2

n�n� ∼ �B2n�n� seems can’t be improved under (1.4).

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742 Liu and Lin

2. SOME LEMMAS

We first provide and list some lemmas that are also of independentinterest. Lemma 2.1 is due Peligrad et al. [12].

Lemma 2.1. Let �Xni� 1 ≤ i ≤ kn� be a triangular array of randomvariables such that the following hold.

(a) Var(∑b

j=a Xnj

) ≤ C∑b

j=a Var�Xnj� for every 0 ≤ a ≤ b ≤ kn where C isa universal constant;

(b)lim infn→�

Var(∑kn

j=1 Xnj

)∑kn

j=1 Var�Xnj�> 0

(c) ∣∣∣∣Cov(exp

(it

b∑j=a

Xnj

)� exp

(it

c∑j=b+u

Xnj

))∣∣∣∣ ≤ ht�u�c∑

j=a

Var Xnj

for every 1 ≤ a ≤ b ≤ c ≤ kn where ht�u� ≥ 0,∑

ht�2i� < � and u is

of the form u = ��c − a�1−�� for a certain 0 < � < 1;(d) −2

n

∑kni=1 EX2

niI��Xni� > �n� → 0 as n −→ � for every � > 0, where 2n

denotes Var(∑kn

i=1 Xni

). Then Sn/n

d→ N�0� 1� as n → �.

Lemma 2.2. We have V 2n /B

2n�n�

p→ 1.

Proof. Note that

V 2n

B2n�n�

− 1 =∑n

i=1

(X2

i I��Xi� ≤ �n�− EX2i I��Xi� ≤ �n�

)B2n�n�

+∑n

i=1 X2i I��Xi� > �n�

B2n�n�

=� H1 +H2�

By Lemma 1 in [5] we have B−2n �n�

∑nj=1 X

2j I��Xj� > �n� ≤(

B−1n �n�

∑nj=1 �Xj�I��Xj > �n�

)2 = op�1�. Using the fact∑�

i=1 �1/2�i� < �,

we have

P��H1� ≥ �� ≤ CnEX4

1I��X1� ≤ �n�

B4n�n�

= o�1��

This implies V 2n /B

2n�n�

p→ 1.

Lemma 2.3. Assume that (1.4) and (1.5) are satisfied. � = EX > 0. Let knbe a positive constant sequence tending to infinity as n → �. Then

�√2�V 2

kn

kn∑k=1

(Skk�

− 1)

d→ � � (2.1)

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Proof. By Lemma 2.2, it is enough to prove

�√2�B2

kn�kn�

kn∑k=1

(Skk�

− 1)

d→ � (2.2)

for showing (2.1). Let X∗i �kn� = XiI��Xi� ≤ �kn� and S∗

k�kn� =∑k

i=1 X∗i �kn�.

Note that

�√2�B2

kn�kn�

kn∑k=1

(Skk�

− 1)

= 1√2�B2

kn�kn�

kn∑k=1

1k

[(S∗k�kn�− ES∗

k�kn�)+ k∑

j=1

XjI��Xj� > �kn�

−Ek∑

j=1


]� (2.3)

By using Lemma 1 [5] again, we have

P

∣∣∣∣ 1√

2�B2kn�kn�

kn∑k=1

1k

[ k∑j=1

XjI��Xj� > �kn�− Ek∑

j=1


]∣∣∣∣ > �

≤ 2knE�X�I��X� > �kn�

�√2�B2

kn�kn�

= Ckn�Bkn

�kn�· o(l��kn�

�kn

)→ 0 (2.4)

as n → �. We will show that

1√2�B2

kn�kn�

kn∑k=1

S∗k�kn�− ES∗

k�kn�

k

d→ � � (2.5)

We prove (2.5) by checking conditions (a)–(d) in Lemma 2.1. Letbi�kn =

∑knk=i 1/k and

2kn= Var

( kn∑k=1


k�kn�

k

)� Yj�kn� �= X∗

j �kn�− EX∗j �kn��

Take Xni = bi�knYi�kn� in Lemma 2.1. Note that

kn∑k=1


k�kn�

k=

kn∑i=1

bi�knYi�kn��

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744 Liu and Lin

Condition (a) is obviously satisfied since∑�

i=1 �1/2�i� < �. Now we show

that

2kn

2B2kn�kn�

−→ � as n −→ �� (2.6)

At first, we can obtain that∑kn

i=1 b2i�kn

= 2kn − b1�kn (see [13]) and

2kn

= E( kn∑

i=1

bi�knYi�kn�

)2

=kn∑i=1

b2i�knVar Yi�kn�+ 2kn−1∑i=1

kn∑j=i+1

Cov(bi�knYi�kn�� bj�knYj�kn�

)�

It is easy to see that∑kni=1 b

2i�kn

Var Yi�kn�

2knl��kn�−→ 1− �� as n −→ ��

From the stationary we have

kn−1∑i=1

kn∑j=i+1


)

=kn−1∑i=1

kn∑j=i+1

bi�knbj�knEYi�kn�Yj�kn�

=kn−1∑i=1

kn−i∑j=1

bi�knbi+j�knEY1�kn�Yj+1�kn�

=kn−1∑j=1

kn−j∑i=1

bi�knbi+j�knEY1�kn�Yj+1�kn�� (2.7)

Note that

bi+j�kn=

kn∑k=i

1k−

i+j−1∑k=i

1k= bi�kn −

i+j−1∑k=i

1k�

By (2.7) we can obtain that

kn−1∑i=1

kn∑j=i+1


)

=kn−1∑j=1

kn−j∑i=1

b2i�knEY1�kn�Yj+1�kn�−kn−1∑j=1

kn−j∑i=1

bi�kn

i+j−1∑k=i

1kEY1�kn�Yj+1�kn��

=� I1 − I2�

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First we show that

I2knl��kn�

−→ 0 as n −→ �� (2.8)

Let N0 and M0 be integers, we have

�I2� ≤N0∑j=1

kn−j∑i=1

bi�kn

i+j−1∑k=i

1k�EY1�kn�Yj+1�kn��

+kn−1∑

j=N0+1

kn−j∑i=1

bi�kn

i+j−1∑k=i


≤N0∑j=1

M0∑i=1

bi�kn

i+j−1∑k=i


+N0∑j=1

kn−j∑i=M0+1

bi�kn

i+j−1∑k=i


+kn−1∑

j=N0+1

kn−j∑i=1

bi�kn

i+j−1∑k=i


=� J1 + J2 + J3�

For J1, noting that

i+j−1∑k=i

1k≤ N0�

we have

J1 ≤ N0

N0∑j=1

M0∑i=1

bi�kn �EY1�kn�Yj+1�kn�� ≤ N0M0 log knN0∑j=1

�EY1�kn�Yj+1�kn��

Since �EY1�kn�Yj+1�kn�� ≤ l��kn� by the Hölder inequality, we have

lim supn→�

J1knl��kn�

= 0� (2.9)

For J2, note that

i+j−1∑k=i

1k≤ N0

M0

�

Hence,

J2 ≤N0

M0

N0∑j=1

kn−j∑i=M0+1

bi�kn �EY1�kn�Yj+1�kn�� ≤2N 2

0

M0

knl��kn��

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746 Liu and Lin

which implies

lim supM0→�

lim supn→�

J2knl��kn�

= 0� (2.10)

For J3, note that

i+j−1∑k=i

1k≤ bi�kn �

we have

J3 ≤kn−1∑

j=N0+1

kn−j∑i=1

b2i�kn �EY1�kn�Yj+1�kn�� ≤ 2knl��kn�kn−1∑

j=N0+1

�1/2�j��

Since∑�

i=1 �1/2�i� < �, we get

lim supN0→�

lim supn→�

J3knl��kn�

= 0� (2.11)

Together with (2.9), (2.10), and (2.11), we get (2.8). Now we need toestimate I1. Recall that A2

n�n� ∼ �nl��n�. Since

A2kn�kn� = kn�l��kn�− E2X1I��X1� ≤ �kn��+ 2

kn−1∑j=1

�kn − j�EY1�kn�Yj+1�kn��

we can see that∑kn−1j=1 �kn − j�EY1�kn�Yj+1�kn�

knl��kn�−→ � − 1+ �

2� as n −→ ��

Now, we claim that

I1knl��kn�

−→ � − 1+ � as n −→ �� (2.12)

To prove this, we only need to show that

∑kn−1j=1 2�kn − j�EY1�kn�Yj+1�kn�− I1

knl��kn�−→ 0 as n −→ ��

Note that

kn−1∑j=1

2�kn − j�EY1�kn�Yj+1�kn�− I1

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=kn−1∑j=1

(2�kn − j�−

kn−j∑i=1

b2i�kn

)EY1�kn�Yj+1�kn�

=K0∑j=1

(2�kn − j�−

kn−j∑i=1

b2i�kn


+kn−j∑

j=K0+1

(2�kn − j�−

kn−j∑i=1

b2i�kn


=� Q1 +Q2�

One can easily see that Q1/�knl��kn�� → 0 since �EY1�kn�Yj+1�kn�� ≤ l��kn�and

2�kn − j�−∑kn−ji=1 b2i�kn

kn−→ 0 for any j ≤ K0� as n −→ ��

For Q2, we have

�Q2�knl��kn�

≤ 2�∑

j=K0+1

�1/2�j� −→ 0 as K0 −→ ��

This proves (2.12), which, in combination with (2.8), implies (2.6).So (b) in Lemma 2.1 is satisfied. Now we check (c) in Lemma 2.1.∣∣∣∣Cov

(exp

(it

b∑j=a

Xnj

)� exp

(it

c∑j=b+u

Xnj

))∣∣∣∣=∣∣∣∣Cov

(exp

(it

b∑j=a

Xnj

)− 1� exp

(it

c∑j=b+u

Xnj

)− 1)∣∣∣∣

≤ Ct2�1/2�u�c∑

j=a

VarXnj�

by using the inequality � exp�iu�− 1� ≤ �u� and the fact∑�

i=1 �1/2�i� < �.

We get (c) by taking ht�u� = t2�1/2�u�. So we only need to check (d) now.Recall that 2

kn∼ 2�knl��kn� ∼ 2��2kn .∑kn

i=1 Eb2i�knY2i �kn�I��bi�knYi�kn�� > ��kn�

knl��kn�

=∑kn/M

i=1 Eb2i�knY2i �kn�I��Yi�kn�� > ��kn/bi�kn�

knl��kn�

+∑kn

i=kn/M+1 Eb2i�knY2i �kn�I��Yi�kn�� > ��kn/bi�kn�

knl��kn�

=� U1 + U2�

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748 Liu and Lin

We estimate U2 first. When i ≥ kn/M , we have bi�kn ≤ M . Together withE�X1� < �, we can get

U2 ≤ CEX2I��′�kn < X < �kn�

l��kn�

= Cl��kn�− l��′�kn�

l��kn�−→ 0 as n −→ ��

Note that U1 ≤ k−1n

∑kn/Mi=1 b2i�kn . Just like the proof of Lemma 1 in [13], we

can construct a sequence of i.i.d. random variables �Yi� 1 ≤ i ≤ kn� withEY1 = 0 and EY 2

1 = 1. And define Zi�kn= bi�knYi. Therefore∑kn/M

i=1 b2i�knkn

= Var(∑kn/M

i=1 Zi�kn

)kn

≤ 2Var(∑kn/M

k=11k

∑ki=1 Yi

)+ 2Var(∑kn

k=kn/M+11k

∑kn/Mi=1 Yi

)kn

≤ 4∑kn/M

k=1 1/k+ 2∑kn/M

k=2

∑k−1l=1 1/k+ �kn log

2 M�/M

kn

≤ 121M

+ 4log2 MM

−→ 0 as M −→ ��

This proves (d) and hence (2.5) holds. The proof is completed.

Lemma 2.4. Let An �n = 1� 2� � � � � be a sequence of events such that

P�An�Ak�− P�An� → 0 as n → � for every k ≥ 1�

Then, for any event A, we have

limn→�

(P�An�A�− P�An�

) = 0

where we set P�An�A� = P�An� if P�A� = 0.

Proof. We have P�AkAn�− P�Ak�P�An� → 0 as n → �� Let fn =In −P�An� where In is the set characteristic function of An. We clearlyhave limn→� E�fkfn� = 0. It follows from Lemma 1 in [14] that

limn→� P�AAn�− P�A�P�An� = 0�

Hence,

limn→� P�An�A�− P�An� = 0�

Lemma 2.5. Let tn and �m�n be sequences of positive integer-valuedrandom variables for any fixed m, and there is a positive constant sequence

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�bn� tending to infinity as n → � such that

tn�mbn

P→ 1 as n → � and m → ��

�m�n

bn

P→ �m as n → � for any fixed m�

where �m is a positive random variable having a discrete distribution foreach m ≥ 1. Let km�n = ��mbn�. Then

V 2tn

B2km�n

�km�n�

P→ 1V 2tn

V 2km�n

P→ 1 andV 2�m�n

V 2tn

p→ 1

as n → � and m → �.

Remark. Here, the sequences of random variables �Xn� and Ym�n satisfy

Xn/Ym�n

p→ 1 as n → � and m → �, means,

limm→� lim sup

n→�P(∣∣∣ Xn

Ym�n

− 1∣∣∣ ≥ �

)= 0

for every � > 0.

Proof. First we will show that for any �′ > 0,

B2�′n��

′n�B2n�n�

−→ �′� (2.13)

W.l.o.g, we assume that �n ≤ ��′n. By the define of B2n, it suffice to prove

that ��′n ≤ C�n when n large. This follows from(��′n�n

)2

≤ Cl��′n�

l��n�≤ C

(��′n�n

)�

�

where 0 < � < 2 by a well-known property of slowly varying function.For proving the lemma, we take �mbn as integers for every m and n

for the sake of convenience. Write

V 2tn

B2km�n

�km�n�= V 2

km�n

B2km�n

�km�n�+ V 2

tn− V 2

km�n

B2km�n

�km�n��

Let lm�k�0 < lm�1 < lm�2 < · · · � denote the values taken on by �m withpositive probability. Then we have

P(∣∣∣∣ V 2

km�n

B2km�n

�km�n�− 1

∣∣∣∣ ≥ �

)

=�∑k=1

P(∣∣∣∣ V 2

lm�kbn

B2lm�kbn

�lm�kbn�− 1

∣∣∣∣ ≥ � � �m = lm�k

)P��m = lm�k��

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Set

An �={∣∣∣∣ V 2

lm�kbn

B2lm�kbn

�lm�kbn�− 1

∣∣∣∣ ≥ �

}�

Since B2lm�kbn

�lm�kbn� → � as n → �, by Lemma 2.2 it is easy to show that

limn→� P�An�Aj� = lim

n→� P�An� = 0

for any fixed j. By Lemma 2.4,

limn→� P

(∣∣∣∣ V 2lm�kbn

B2lm�kbn

�lm�kbn�− 1

∣∣∣∣ ≥ � � �m = lm�k

)= 0�

which implies V 2km�n

/B2km�n

�km�n�P→ 1. Now we need to show

(V 2tn− V 2

km�n

)/

B2km�n

�km�n�P→ 0 as n→� and m→�. It suffice to show that for 10�<�,

P(∣∣∣∣V

2tn− V 2

km�n

B2km�n

�km�n�

∣∣∣∣ ≥ �� 1− � ≤ tn�mbn

≤ 1+ �

)

≤ P(∣∣∣∣∑�1+��km�n

k=�1−��km�nX2

k

B2km�n

�km�n�

∣∣∣∣ ≥ �

)→ 0 as n → � and m → �� (2.14)

Since

P(∣∣∣∣∑�1+��km�n

k=�1−��knX2

k

B2kn�km�n�

∣∣∣∣ ≥ �

)

=�∑j=1

P(∣∣∣∣∑�1+��lm�jbn

k=�1−��lm�jbnX2

k

B2lm�jbn

�lm�jbn�

∣∣∣∣ ≥ � � �m = lm�j

)P��m = lm�j�

and

P(∣∣∣∣∑�1+��lm�jbn

k=�1−��lm�jbnX2

k

B2lm�jbn

�lm�jbn�

∣∣∣∣ ≥ �

)≤ P

(∣∣∣∣∑2�lm�jbn

k=1 X2k

B2lm�jbn

�lm�jbn�

∣∣∣∣ ≥ �

)→ 0 as n → ��

by Lemma 2.2 and (2.13), we get (2.14) from Lemma 2.4. This

implies V 2tn/B2

km�n�km�n�

P→ 1. Noting that V 2km�n

/B2km�n

�km�n�P→ 1, we have

V 2tn/V 2

km�n

P→ 1. Obviously if V 2tn/bn

P→ �m for any fixed m as n → �,

then by the proof above, we have V 2tn/V 2

km�n

P→ 1 as n → �. Therefore

V 2�m�n

/V 2km�n

p→ 1 as n → � for any fixed m, which implies V 2�m�n

/V 2tn

p→ 1 asn → � and m → �. The proof is completed.

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Lemma 2.6. Assume that (1.4) and (1.5) are satisfied. � = EX > 0. Lettn be a sequence of positive integer-valued random variables, and thereis a positive constant sequence �bn� tending to infinity as n → � suchthat tn/bn

p→ �, where � is a positive random variable having a discretedistribution. Then

�√2�V 2

tn

tn∑k=1

(Skk�

− 1)

d→ � �

Proof. By Lemma 2.5, it is enough to prove

�√2�B2

��bn�

tn∑k=1

(Skk�

− 1)

d→ � � (2.15)

Note that

�√2�B2

��bn�

tn∑k=1

(Skk�

− 1)

= �√2�B2

��bn�

��bn�∑k=1

(Skk�

− 1)+ �√

2�B2��bn�

( tn∑k=1

(Skk�

− 1)−

��bn�∑k=1

(Skk�

− 1))

=� D�bn+ E�bn

�

We first prove that D�bn

d→ � . We have

P�D�bn≤ x� =

�∑j=−�

P�Dljbn≤ x � � = lj�P�� = lj�

where lj �0 < · · · < l−1 < l0 < l1 < l2 < · · · � denote the values taken onby � with positive probability. Hence, it suffices to show that for any j,

P�Dljbn≤ x � � = lj� → ��x� as n → �� (2.16)

where ��x� denotes the standard normal distribution function. ByLemma 2.2 and the fact ��n� → 0, one can easily get that for any m ≥ 1,

P�Dljbn≤ x �Dljbm

≤ x� → ��x� as n → ��

Therefore, we get (2.16) by Lemma 2.4. This proves D�bn

d→ � . Now, we

show that E�bn

P→ 0. We have, for kn = ��1− ��bn/M� (� and M to bespecialized later),

√2�E�bn

= 1√B2��bn�

( tn∑k=1

1k

(S∗k�kn�− ES∗

k�kn�)− ��bn�∑

k=1

1k

(S∗k�kn�− ES∗

k�kn�))

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+ 1√B2��bn�

( tn∑k=1

1k

k∑j=1

(XjI��Xj� > �kn�− EXjI��Xj� > �kn�

)

−��bn�∑k=1

1k

k∑j=1

(XjI��Xj� > �kn�− EXjI��Xj� > �kn�

))

=� F�bn+G�bn

�

Let pn = �1+ ��Mbn. So kn ≈ pn ≈ bn. We have

P��G�bn� ≥ ��

≤ P

1√

B2kn�kn�

pn∑k=1

1k

∣∣∣∣k∑

j=1

XjI��Xj� > �kn�− Ek∑

j=1


∣∣∣∣ ≥ �

+P(∣∣∣∣ tn

��bn�− 1

∣∣∣∣ ≥ �

)+ P�� ≥ M�+ P�� ≤ 1/M�

=� P11 + P12 + P13 + P14� (2.17)

Clearly, we have P12 → 0 as n → �. From (2.4) we have

P11 → 0 as n → �� (2.18)

Let us denote by Cn�� the event �tn/��bn�− 1� ≤ �. Then we have

P��F�bn� ≥ �� ≤

�∑j=−�

P��Fljbn� ≥ �� = lj� Cn��+ P�Cc

n��

By Markov inequality and (1.4)

P��Fljbn� ≥ �� = lj� Cn��

≤ P(

1√B2�ljbn�

�1+��ljbn∑k=�1−��ljbn

1k

∣∣S∗k�kn�− ES∗

k�kn�∣∣ ≥ �

)

≤ C

√�l��kn�√l��ljbn�

� (2.19)

when n large. Here C does not depend on �, M and n. Let � > 0 bearbitrarily small; let us choose first N so large that

∑�j�≥N

P�� = lj� ≤ �� (2.20)

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Now we choose M so large that

P13 + P14 ≤ �� (2.21)

From the fact l�x� is slowly varying and (2.17) to (2.21), we have

limn→� P��F�bn

� ≥ ��+ P��G�bn� ≥ �� ≤ 2� + CN

√�� (2.22)

Let � → 0 and � → 0, we get limn→� P��F�bn� ≥ ��+ P��G�bn

� ≥ �� = 0.

This implies E�bn

P→ 0. The proof is now completed.

The next lemma is due to Blum et al. [4].

Lemma 2.7. Let Wn, Xm�n, Y �j�m�n, and Z�j�

m�n be random variables form�n= 1� 2� � � � and j = 1� � � � � k. Suppose that

Wn = Xm�n +k∑

j=1

Y �j�m�nZ

�j�m�n

and

A) limm→� lim supn→� P��Y �j�m�n�>��= 0 for every �> 0 and j= 1� � � � � k;

B) limM→� lim supm→� lim supn→� P��Z�j�m�n� > M� = 0 for j = 1� � � � � k;

C) the distributions of �Xm�n� converge to the distribution function F foreach fixed m. Then the distribution functions of �Wn� converge to F .

By using Lemmas 2.4–2.7, we will show (2.1) still holds under theconditions of Theorem 1.1.

Lemma 2.8. Under the conditions of Theorem 1.1, we have

�√2�V 2

tn

tn∑k=1

(Skk�

− 1)

d→ � � (2.23)

Proof. Let m� k be positive integers, define �m = k/2m when�k− 1�/2m ≤ � < k/2m and

�m�n = tn + �bn��m − ��

Note that �m is discrete for each m� 0 < �m − � ≤ 1/2m and

�m�n

bn= tn

bn+ �bn��m − ��

bn

p→ �m > � (2.24)

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as n → �. Put S′j =

∑jk=1

Sk−k�

k� X∗

j ��mbn� = XjI��Xj� ≤ ��mbn� andS∗k��mbn� =

∑kj=1 X

∗j ��mbn�. then

�√2�V 2

tn

tn∑k=1

(Skk�

− 1)

= S′�m�n√

2�V 2�m�n

+ S′tn− S′

�m�n√B2�mbn

��mbn�

√B2�mbn

��mbn�

2�V 2tn

+√V 2�m�n

−√V 2tn√

V 2tn

S′�m�n√

2�V 2�m�n

=� Xm�n + Y �1�m�nZ

�1�m�n + Y �2�

m�nZ�2�m�n� (2.25)

It follows from Lemma 2.6 that for each fixed m, Xm�n = Z�2�m�n

d→ � asn → �, so it is easy to see that Z�2�

m�n satisfies condition B of Lemma 2.7.Moreover, P�� < m/2m� → 0 as m → � and for m/2m ≤ � we have

limn→�

∣∣∣∣�m�n − tntn

∣∣∣∣ p= �m

�− 1 ≤ �1+ 1/m�− 1 → 0 (2.26)

as m → �, which together with (2.24) imply that

tn�mbn

p→ 1 (2.27)

as n → � and m → �. By Lemma 2.5, we get

V 2�m�n

V 2tn

p→ 1�V 2tn

V 2�mbn

p→ 1�V 2tn

B2�mbn

��mbn�

p→ 1 (2.28)

as n → � and m → �. (2.28) implies that Y �2�m�n and Z�1�

m�n satisfy theconditions A and B of Lemma 2.7, respectively. Next, we only need toshow Y �1�

m�n

p→ 0 as n → � and m → � for showing (2.23). Note that

Y �1�m�n = S′

tn− S′

�mbn√B2�mbn

��mbn�− S′

�m�n− S′

�mbn√B2�mbn

��mbn�

=� Zm�n�1 − Zm�n�2�

Since �m�n/bnp→ �m as n → �, and �m is discrete for each m, by the

proof of Lemma 2.6, we can see that Zm�n�2

p→ 0 as n → � for any

fixed m. So we only need prove Zm�n�1

p→ 0 as n → � and m → �. LetJn��m� = 1 ∨ �log��1+ ��mbn�� where � is a constant to be specializedlater. We have

S′tn

=tn∑

i=Jn��m�

bi�tn �Xi − ��+Jn��m�−1∑

i=1

bi�tn �Xi − �� =� S′1�tn

+ S′2�tn

�

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S′�mbn

=�mbn∑

i=Jn��m�

bi��mbn�Xi − ��+Jn��m�−1∑

i=1

bi��mbn�Xi − ��

=� S′1��mbn

+ S′2��mbn

�

And

P��Zm�n�1� ≥ 2�� ≤ P

∣∣∣∣ S

′1�tn

− S′1��mbn√

B2�mbn

��mbn�

∣∣∣∣ ≥ ��

∣∣∣∣ tn�mbn

− 1

∣∣∣∣ ≤ �

+ P

∣∣∣∣ S

′2�tn

− S′2��mbn√

B2�mbn

��mbn�

∣∣∣∣ ≥ ��


− 1

∣∣∣∣ ≤ �

+ P(∣∣∣∣ tn

�mbn− 1

∣∣∣∣ ≥ �

)=� P21 + P22 + P23�

One can easily show that P22 → 0 as n → � for any fixed m. We beginto estimate P21. Define

Mk��m� =k∑

i=Jn��m�

�Xi − ��

=k∑

i=Jn��m�

�XiI��Xi� ≤ �bn�− EXiI��Xi� ≤ �bn��

+k∑

i=Jn��m�

�XiI��Xi� > �bn�− EXiI��Xi� > �bn��

=� M1�k��m�+M2�k��m�

Then we have

P21 = P

1√

B2�mbn

��mbn�

∣∣∣∣tn∑

k=Jn��m�

1kMk��m�−

�mbn∑k=Jn��n�

1kMk��m�

∣∣∣∣ ≥ ��


− 1

∣∣∣∣ ≤ �

≤ P

1√

B2�mbn

��mbn�

�1+��m�bn∑k=�1−��m�bn

1k

∣∣Mk��m�∣∣ ≥ �

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≤ P

1√

B2�mbn

��mbn�

�1+��m�bn∑k=�1−��m�bn

1k

∣∣M1�k��m�∣∣ ≥ �/2

+P

1√

B2�mbn

��mbn�

�1+��m�bn∑k=�1−��m�bn

1k

∣∣M2�k��m�∣∣ ≥ �/2

=� P31 + P32�

It follows from (2.4) that P32 → 0 as n → �. For P31, we have

P31 ≤N∑

j=−N

P

1√

B2ljbn

�ljbn�


1k

∣∣M1�k�lj�∣∣ ≥ �/2

∣∣ �m = lj

×P��m = lj�+�∑

�j�≥N

P��m = lj��

If we set the event

An =

1√B2ljbn

�ljbn�


1k

∣∣M1�k�lj�∣∣ ≥ �/2

�

then, since Jn�lj� → � as n → � for any j, together with the fact��n� → 0, we have

limn→�

(P�An�Ak�− P�An�

) = 0�

By Lemma 2.4, we get

lim supn→�

P�An��m = lj� = lim supn→�

P�An��

But

P�An� ≤ C

√�l��bn�√l��ljbn�

(see (2.19)),

where C is a constant does not depend on m, �, j, and n. We have

lim supn→�

P31 ≤ C√�+ ∑

�j�≥N

P��m = lj��

Let N → � so that we can get lim supn→� P31 ≤ C√�. Hence,

lim supn→�

P��Zm�n�1� ≥ 2�� ≤ C√�+ lim sup

n→�P(∣∣∣∣ tn

�mbn− 1

∣∣∣∣ ≥ �

)�

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We let m → �, then let � → 0, we get Zm�n�1

p→ 0. The proof iscompleted.

3. PROOF OF THEOREM 1.1

The Proof of Theorem 1�1. The proof relies on the delta-methodexpansion used in [13]. Denote Tk = Sk

k�, k = 1� 2� � � � . By the strong law

of large numbers for �-mixing sequence (cf. [10]), it follows that for any� > 0 there exists a positive integer R such that

P(supk≥R

�Tk − 1� > �)< ��

Consequently, there exist two sequences ��m� ↓ 0 ��1 = 1/2�, �Rm� ↑ �such that

P(supk≥Rm

�Tk − 1� > �m

)< �m�

We first show that

lim supn→�

∣∣∣∣∣∣∣P �√

2�V 2tn

tn∑k=1

log�Tk� ≤ x

−��x�

∣∣∣∣∣∣∣ = 0� (3.1)

where ��x� denotes the distribution function of a standard normalrandom variable. For any real x,

P

�√

2�V 2tn

tn∑k=1

log�Tk� ≤ x

= P

�√

2�V 2tn

tn∑k=1

log�Tk� ≤ x� supk≥Rm

�Tk − 1� > �m

+ P

�√

2�V 2tn

tn∑k=1

log�Tk� ≤ x� supk≥Rm

�Tk − 1� ≤ �m

=� P4�n�m�+ P5�n�m�� (3.2)

and P4�n�m� < �m. So it suffices to prove that

lim supm→�

lim supn→�

�P5�n�m�−��x�� = 0� (3.3)

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To estimate P5 we will use the expansion log�1+ x� = x + x2

�1+�x�2, where

� ∈ �0� 1� depends on x ∈ �−1� 1�. Then

P5�n�m� = P

�√

2�V 2tn

Rm∧�tn−1�∑k=1

log�Tk�

+ �√2�V 2

tn

tn∑k=�Rm∧�tn−1��+1

log�1+ Tk − 1� ≤ x�

supk≥Rm

�Tk − 1� ≤ �m

= P

�√

2�V 2tn

Rm∧�tn−1�∑k=1

log�Tk�+�√2�V 2

tn

tn∑k=�Rm∧�tn−1��+1

�Tk − 1�

+ �√2�V 2

tn

tn∑k=�Rm∧�tn−1��+1

�Tk − 1�2

�1+ �Tk − 1��k�2≤ x�

supk≥Rm

�Tk − 1� ≤ �m

= P

�√

2�V 2tn

Rm∧�tn−1�∑k=1


tn

tn∑k=�Rm∧�tn−1��+1

�Tk − 1�

+ �√

2�V 2tn

tn∑k=�Rm∧�tn−1��+1

�Tk − 1�2

�1+ �Tk − 1��k�2

× I{supk≥Rm

�Tk − 1� ≤ �m

}≤ x

−P

�√

2�V 2tn

Rm∧�tn−1�∑k=1


tn

tn∑k=�Rm∧�tn−1��+1

�Tk − 1� ≤ x�

supk≥Rm

�Tk − 1� > �m

)

=� P51�n�m�− P52�n�m� (3.4)

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and P52�n�m� < �m. Moreover,

P51�n�m� = P

�√

2�V 2tn

Rm∧�tn−1�∑k=1

�log�Tk�− Tk + 1�+ �√2�V 2

tn

tn∑k=1

�Tk − 1�

+ �√

2�V 2tn

tn∑k=�Rm∧�tn−1��+1

�Tk − 1�2

�1+ �Tk − 1��k�2

× I{supk≥Rm

�Tk − 1� ≤ �m

}≤ x

� (3.5)

It is easy to know for every � > 0,

lim supm→�

lim supn→�

P

∣∣∣∣∣∣

�√2�V 2

tn

Rm∧�tn−1�∑k=1

�log�Tk�− Tk + 1�

∣∣∣∣∣∣ ≥ �

= 0 (3.6)

by noting that V 2tn

p→ �. If Rm ≥ tn − 1, then as n large enough, �√

2�V 2tn

tn∑k=�Rm∧�tn−1��+1

�Tk − 1�2

�1+ �Tk − 1��k�2

I{supk≥Rm

�Tk − 1� ≤ �m

}

≤ �√2�V 2

tn

�Ttn− 1�2

�1+ �Ttn− 1��tn�

2

a�s�≤ C√2�V 2

tn

p→ 0 (3.7)

as n → �. If Rm < tn − 1, then Rm + 1 < tn. Set

W�n�m� �= �√

V 2tn

tn∑k=Rm+1

�Tk − 1�2

�1+ �Tk − 1��k�2

× I(supk≥Rm

�Tk − 1� ≤ �m

)I�Rm + 1 < tn��

We have

P�W�n�m� ≥ �� ≤ P(W�n�m� ≥ �� 1− � ≤ tn

�bn≤ 1+ �

)

+P(

tn�bn

> 1+ �

)+ P

(tn�bn

< 1− �

)

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and

P(W�n�m� ≥ �� 1− � ≤ tn

�bn≤ 1+ �

)

≤ P(W�n�m� ≥ �� 1− � ≤ tn

�bn≤ 1+ ��

1M

≤ � ≤ M

)

+P�� > M�+ P(� <

1M

)=� P61�n�m�+ P62 + P63�

Let kn = M−1�1− ��bn and pn = M�1+ ��bn, then pn ≈ kn. We have

P61�n�m� ≤ P

�√

V 2kn

pn∑k=Rm

�Tk − 1�2

�1+ �Tk − 1��k�2

)I(supk≥Rm

�Tk − 1� ≤ �m

)≥ �

≤ P

2�√

B2kn�kn�

pn∑k=Rm

�Tk − 1�2

�1+ �Tk − 1��k�2

)I(supk≥Rm

�Tk − 1� ≤ �m

)≥ �

+P(B2kn

V 2kn

≥ 4)

≤ P

C�m√

B2kn�kn�

pn∑k=Rm+1

1k

∣∣S∗k�kn�− ES∗

k�kn�∣∣ ≥ �/2

+P

C�m√

B2kn�kn�

pn∑k=Rm+1

1k

∣∣∣∣k∑

j=1


− EXjI��Xj� > �kn�

∣∣∣∣ ≥ �/2

+ P

(B2kn�kn�

V 2kn

≥ 4)

=� P611�n�m�+ P612�n�m�+ P(B2kn�kn�

V 2kn

≥ 4)�

From (2.4) we get P612�n�m� → 0 as n → �. By Markov inequality wehave

P611�n�m� ≤ C�m√knl��kn�

pn∑k=1

√l��kn�

k≤ C�m�

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which implies P611 → 0 as m → �, uniformly in n. Hence by lettingn→�, m → � and M → �, we have

lim supm→�

lim supn→�

P��W�n�m�� > �� = 0 (3.8)

for any � > 0. By Lemma 2.8, (3.4)–(3.8), we can get (3.3), which implies(3.1). We complete the proof of the theorem by noting that

P(log( tn∏

k=1

Skk�

) �Vtn ≤ √2�x

)= P

(�√2�V 2

tn

tn∑k=1

log Tk ≤ x

)� (3.9)

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asymptotics for self-normalized random products of sums for mixing sequences

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