# Precise rates in complete moment convergence for ρ-mixing sequences

## Precise rates in complete moment convergence for ρ-mixing sequences

J. Math. Anal. Appl. 339 (2008) 553–565 www.elsevier.com/locate/jmaa Precise rates in complete moment convergence for ρ-mixing sequences ✩ Yuexu Zhao...

J. Math. Anal. Appl. 339 (2008) 553–565 www.elsevier.com/locate/jmaa

Precise rates in complete moment convergence for ρ-mixing sequences ✩ Yuexu Zhao Institute of Applied Mathematics and Engineering Computation, Hangzhou Dianzi University, Hangzhou 310018, PR China Received 27 September 2006 Available online 19 July 2007 Submitted by M. Peligrad

Abstract with mean zeros and positive, finite variances, Let X1 , X2 , . . . be a strictly stationary sequence of ρ-mixing random  variables 2/q (2n ) < ∞, where q > 2δ + 2. We prove that, if set Sn = X1 + · · · + Xn . Suppose that limn→∞ ESn2 /n = σ 2 > 0, ∞ n=1 ρ EX12 (log+ |X1 |)δ < ∞ for any 0 < δ  1, then lim  2δ

↓0

∞ 2δ+2   (log n)δ−1 2 I |S |  σ n log n  = E|N | ES , n n δ n2

n=2

where N is the standard normal random variable. © 2007 Elsevier Inc. All rights reserved. Keywords: Precise rates; Complete moment convergence; ρ-Mixing; Mixing rates

1. Introduction and the main results Suppose that {Xn : n  1} is a sequence of random variables on a probability space (Ω, F , P ), set Fn− = σ (Xi : 1  i  n), Fn+ = σ (Xi : i  n), ρ(n) := sup

sup

sup

k1 X∈L2 (F − ) Y ∈L2 (F + ) k k+n



|EXY − EXEY | E(X − EX)2 E(Y − EY )2

,

(1.1)

the sequence {Xn : n  1} is said to be ρ-mixing, if ρ(n) → 0 as n → ∞. This definition was introduced by Kolmogorov and Rozanov , and the limiting behaviors of ρ-mixing sequences have received more and more attention recently. Under appropriate mixing rates, lots of limit theorems have been obtained. The central limit theorem (CLT) was proved by Ibragimov , Bradley [1,2], Dehling, Denker and Philipp , Peligrad [12–14]. The functional central limit theorem (FCLT) is due to Shao . Further results are probability in✩

This work is supported by the Natural Science Foundation of Department of Education of Zhejiang Province (Grant No. 20060237). E-mail address: [email protected]

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Y. Zhao / J. Math. Anal. Appl. 339 (2008) 553–565

equalities (see, e.g., [20,22]), invariance principle (cf. [10,15,16,18,21]), the complete convergence (see, e.g., [11,19]), and so forth. Note that in the above-mentioned limit theorems, the precise rates in the complete moment convergence for ρmixing sequences are little known. The aim of the present work is to investigate this asymptotic behavior, of course, our results have the connection with complete moment convergence. It is well known that a lot of beautiful results have been established for independent random variables. Let {Xn : n  1} be i.i.d. (independent and identically distributed) nondegenerate random variables. Chow  obtained a result as follows. Theorem A. Suppose that EX1 = 0 and E(|X1 |r + |X1 | log(1 + |X1 |)) < ∞, for r > p, 1  p < 2. Then ∞ 

  nr/p−2−1/p E |Sn | − n1/p + < ∞,

 > 0.

(1.2)

n=1

A natural question is that how larger the convergence rate of (1.2) is as  ↓ 0? As we know this is an interesting question, one of the answers to moving-average process reads as follows: lim  2(r−p)/(2−p)−1 ↓0

∞ 

  nr/p−2−1/p E |Sn | − n1/p + =

n=1

p(2 − p) E|N |2(r−p)/(2−p) , (r − p)(2r − p − 2)

(1.3)

where N is the standard normal random variable. For more details, refer to [9, Theorem 1.1]. The following theorem for i.i.d. random variables was recently proved by Liu and Lin . Theorem B. Suppose that EX1 = 0,

EX12 = σ 2

 α and EX12 log+ |X1 | < ∞,

(1.4)

for 0 < α  1. Then lim  2α ↓0

∞  (log n)α−1 n=2

n2

   σ 2α+2 E|N |2α+2 . ESn2 I |Sn |   n log n = α

(1.5)

Conversely, if (1.5) is true, then (1.4) holds. Should (1.5) be true for ρ-mixing sequences under appropriate conditions? Furthermore, if Sn is replaced by max1kn |Sk |, what can be said about the precise asymptotics? The present work gives the answers. From now, it is very convenient to adopt the following notations throughout  the paper: let X1 , X2 , . . . be strictly stationary ρmixing sequences with EX1 = 0 and EX12 < ∞, σ 2 = EX12 + 2 ∞ n=2 EX1 Xn > 0, and set Sn = X1 + · · · + Xn , Mn = max1kn |Sk |, log x = loge (x ∨ e), [z] denotes the largest integer which is not lager than z, the letter C with subscripts denotes some finite and positive universal constants not important in our investigations, it may take different values in each appearance. The organization of the paper is as follows. We first introduce our main results, after which the proofs are exposed in Sections 2 and 3. Theorem 1.1. Suppose that limn→∞ ESn2 /n = σ 2 > 0, and EX12 (log |X1 |)δ < ∞ for any 0 < δ  1. Then lim  2δ ↓0

∞  (log n)δ−1 n=2

n2

∞

n=1 ρ

2/q (2n )

< ∞, where q > 2δ + 2,

   E|N |2δ+2 . ESn2 I |Sn |  σ n log n = δ

(1.6)

Theorem 1.2. Under the conditions of Theorem 1.1, for any 0 < δ  1, lim  2δ ↓0

∞  (log n)δ−1 n=2

n2

∞    2E|N |2δ+2  (−1)n EMn2 I Mn  σ n log n = . δ (2n + 1)2δ+2 n=0

(1.7)

Y. Zhao / J. Math. Anal. Appl. 339 (2008) 553–565

If the mixing rate result.

∞

n=1 ρ

2/q (2n ) < ∞

is replaced by

∞

Theorem 1.3. Suppose that limn→∞ ESn2 /n = σ 2 > 0, and  δ EX1 = 0 and EX12 log |X1 | < ∞

n=1 ρ(n) < ∞,

555

one can immediately obtain the following

∞

n=1 ρ(n) < ∞.

Then, for any 0 < δ  1, (1.8)

if and only if lim  2δ ↓0

∞  (log n)δ−1 n=2

n2

∞    2E|N |2δ+2  (−1)n EMn2 I Mn  σ n log n = . δ (2n + 1)2δ+2

(1.9)

n=0

Without loss of generality, in the sequel, we will suppose that σ 2 = 1. 2. Proof of Theorem 1.1 We begin this section by introducing two lemmas, which are helpful in proving the theorem. stationary ρ-mixing random Lemma 2.1. (See Ibragimov .) Assume that {Xn : n  1} is a sequence  of strictly n variables with EX1 = 0 and EX12 < ∞, σn2 = ESn2 → ∞ as n → ∞ and ∞ n=1 ρ(2 ) < ∞. Then D Sn /σn − → N (0, 1)

as n → ∞.

(2.1)

Lemma 2.2. (See Shao .) Let {Xn : n  1} be a sequence of ρ-mixing random variables. Assume that EXn = 0, put Sk (n) = k+n i=k+1 Xi , k  0. Then, for any q  2, there exists K = K(q, ρ(·)) depending only on q and ρ(·) such that

[log n] k+n            q −q q/2 i   P |Xi |  y + Kx n exp K ρ 2 max Xi I |Xi |  y P max Sk (j )  x  1j n

i=k+1

i=1

+ Kx −q n exp K

[log n]



kik+n

2

    max E|Xi |q I |Xi |  y , ρ 2/q 2i kik+n

i=1

for any x > 0 and y > 0 with 2n maxkik+n E|Xi |I (|Xi |  y)  x. In fact, one can easily get ∞  (log n)δ−1 n=2

n2

∞       (log n)δ  P |Sn |   n log n ESn2 I |Sn |   n log n =  2 n n=2

+

∞  (log n)δ−1 n=2

n2

  2xP |Sn |  x dx

√  n log n

=: I1 + I2 . Consequently, to verify (1.6), we only need to study I1 and I2 , respectively. The rest of this section is to present several propositions which are needed in the proof. Proposition 2.1. Suppose that N is a standard normal random variable. Then, for any 0 < δ  1, lim  2δ+2 ↓0

∞    E|N |2δ+2 (log n)δ  P |N |   log n = . n δ+1 n=2

Proof. For the proof see [5, Theorem 1.3].

2

(2.2)

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Y. Zhao / J. Math. Anal. Appl. 339 (2008) 553–565

Proposition 2.2. Suppose that limn→∞ ESn2 /n = σ 2 > 0, and EX12 (log |X1 |)δ < ∞ for any 0 < δ  1. Then lim  2δ+2 ↓0

∞

n=1 ρ

2/q (2n )

< ∞, where q > 2δ + 2,

∞       (log n)δ   P |Sn |   n log n − P |N |   log n  = 0. n

(2.3)

n=1

Proof. Using the standard method, set H () = [exp(M/ 2 )], where M > 4, 0 <  < 1/4. It is easy to get ∞       (log n)δ   P |Sn |   n log n − P |N |   log n  n n=1

 (log n)δ        P |Sn |   n log n − P |N |   log n  n

=

nH ()

+

 (log n)δ        P |Sn |   n log n − P |N |   log n  n

n>H ()

=: Σ1 + Σ2 .

√ Write Δn = supx |P (|Sn |  x n) − P (|N|  x)|, note that P (|N |  x) is a continuous function for x  0, and this combined with Lemma 2.1 yields limn→∞ Δn = 0 for any x  0. Then, applying Toeplitz’s lemma (see, e.g., [17, Lemma 6.10]), it follows that  (log n)δ Δn n

 2δ+2 Σ1   2δ+2

nH ()

 δ+1 =  2δ+2 log H ()

 (log n)δ 1 Δn → 0 as  ↓ 0. n log(H ())δ+1

(2.4)

nH ()

Obviously, we have, for the second part Σ2 , Σ2 

 (log n)δ   (log n)δ      P |N |   log n + P |Sn |   n log n n n

n>H ()

n>H ()

=: Σ3 + Σ4 . Notice that H () − 1 

√ H () for M > 4 and 0 <  < 1/4, an easy calculation leads to

 2δ+2 Σ3   2δ+2

 (log n)δ    P |N |   log n n

n>H ()

∞ C √

  y 2δ+1 P |N | > y dy → 0 as M → ∞,

(2.5)

M/4

√ √ uniformly with respect to 0 <  < 1/4. For Σ4 , taking x =  n log n, y = 2 n log n in Lemma 2.2. Note that 2 n > H () if and only if log n > M/ , it turns out that √ √ 2n max1in E|Xi |I (|Xi |  2 n log n) E|Xi |2 I (|Xi |  2 n log n) C  C/M → 0 as M → ∞. (2.6) √  2 log n  n log n Furthermore, applying Lemma 2.2, we have  P

[log n]        q/2 −q q/2 i EX12 I |X1 |  y max |Sj |  x  nP |X1 |  y + Kx n exp K ρ 2

1j n

i=1

Y. Zhao / J. Math. Anal. Appl. 339 (2008) 553–565

+ Kx

−q

[log n]

n exp K



ρ

2/q

557

 i   2 E|X1 |q I |X1 |  y

i=1

=: Σ5 + Σ6 + Σ7 . Recalling the moment condition, which in turn implies      n−1 (log n)δ Σ5 = (log n)δ P |X1 |  2 n log n n>H ()

n>H ()



C

(log n)δ

k=n

n>H ()



C

k     P 2 k log k  |X1 | < 2 (k + 1) log(k + 1) (log n)δ

k>H ()



C

∞      P 2 k log k  |X1 | < 2 (k + 1) log(k + 1)

n=1

  k(log k)δ P k  X12 /4M < k + 1

k>H ()

  δ    CEX12 log |X1 | I |X1 |  2 H () log H () < ∞.

(2.7)

Since q > 2δ + 2, it suffices to show that 

n−1 (log n)δ Σ6  C

n>H ()

 (log n)δ  ( n log n )−q nq/2 n

n>H ()

 C



−q

n

−1

δ−q/2

(log n)

 C

n>H ()

 C −q √

 C −q





 C



n−q/2 (log n)δ−q/2

n>H ()

 C −q

x −1 (log x)δ−q/2 dx

H ()−1

x −1 (log x)δ−q/2 dx  C −2(δ+1) M δ−q/2+1 ,

(2.8)

n>H () n

−1 (log n)δ Σ

6

= 0. Finally, focusing attention on Σ7 , It turns out

   n−q/2 (log n)δ−q/2 E|X1 |q I |X1 |  2 n log n

n>H () −q

H ()

consequently, we have limM→∞  2δ+2 that  n−1 (log n)δ Σ7 n>H ()

−q





    E|X1 |q I 2 j log j  |X1 |  2 (j + 1) log(j + 1)

1j n

    j −q/2+1 (log j )δ−q/2 E|X1 |q I 2 j log j  |X1 |  2 (j + 1) log(j + 1)

j >H ()

 C −2



    (log j )δ−1 E|X1 |2 I 2 j log j  |X1 |  2 (j + 1) log(j + 1)

j >H ()

 C

−2

   E|X1 |2 I |X1 |  2 H () log H () .

(2.9)

 Notice, in particular, that n > H (), it follows that  2δ E|X1 |2 I (|X1 |  2 H () log H ()) → 0 as  ↓ 0 and M → ∞. Putting (2.4), (2.5), (2.7)–(2.9) together, we obtain (2.3). 2

558

Y. Zhao / J. Math. Anal. Appl. 339 (2008) 553–565

Proposition 2.3. Suppose that N is a standard normal random variable. Then, for any 0 < δ  1, lim  2δ ↓0

∞  (log n)δ−1

n2

n=2

 √  E|N |2δ+2 2xP |N|  x/ n = . δ(δ + 1)

(2.10)

 n log n

2

Proof. Refer to Liu and Lin .

Proposition 2.4. Under the conditions of Theorem 1.1,  ∞  (log n)δ−1    2δ lim  2xP |Sn |  x dx −  2 √ ↓0 n nH ()

 n log n

∞ √

  √   2xP |N |  x/ n dx  = 0. 

(2.11)

 n log n

√ 2 Proof. √ Set H () = [exp(M/ )] and denote Δn = supx |P (|Sn |  nx) − P (|N |  x)|. Then, assume that x = (y + ) n log n, by integral formula and transformation, it is enough to show that  ∞ 

∞      √   −2 δ−1  n (log n)  2xP |Sn |  x dx − 2xP |N |  x/ n dx  √  √ nH ()

 n log n



C

n−1 (log n)δ

nH ()

 n log n

       2(y + )P |Sn |  (y + ) n log n − P |N |  (y + ) log n  dy

0





C

n

−1

   2(y + )P |N |  (y + ) log n dy

δ

(log n)

√ 1/4 1/ log nΔn

nH () √ 1/4 1/ log

nΔn

       2(y + )P |Sn |  (y + ) n log n − P |N |  (y + ) log n  dy

+ 0

  2(y + )P |Sn |  (y + ) n log n dy 

+



√ 1/4 1/ log nΔn

=: C

 (log n)δ (Λ1 + Λ2 + Λ3 ). n

nH ()

The estimates of Λ1 and √ Λ2 are similar to those√of Proposition 5.2 in , so we omit them. It remains to estimate Λ3 , taking x = (y + ) n log n, y = 2(y + ) n log n in Lemma 2.2, which yields  nH ()

n

−1

(log n) Λ3  C δ



∞ (log n)

√ 1/4 1/ log nΔn

nH ()

+C

   2(y + )P |X1 |  2(y + ) n log n dy

δ



n−1 (log n)δ−q/2

√ 1/4 1/ log nΔn

nH ()

+C

 nH ()

[log n]    (y + )1−q exp ρ 2i dy

n

−q/2

i=1

∞ (y + )

1−q

δ−q/2

(log n)

1/4

1/ log nΔn

[log n]    2/q i 2 exp ρ i=1

   × E|X1 |q I |X1 |  2(y + ) n log n dy =: Λ4 + Λ5 + Λ6 .

Y. Zhao / J. Math. Anal. Appl. 339 (2008) 553–565

Note that (log H ())δ = M δ / 2δ and EX12 I (|X1 |   Λ4   2δ



n) → 0 as n → ∞. By Toeplitz’s lemma, it follows that 



2(y + )I |X1 |/2 n log n  y + 

δ

(log n) E

nH ()

1/



dy

1/4 log nΔn

√ EX12 I (|X1 |  n)(log n)δ−1



  2δ

559

n

nH ()

√   δ   EX12 I (|X1 |  n)(log n)δ−1  M 1/ log H () → 0 as  ↓ 0. n δ

(2.12)

nH ()

(q−2)/4

Observe that Δn

 Λ5   2δ

→ 0 as n → ∞. using Toeplitz’s lemma again, we have



n

−1

∞ (y + )1−q dy

δ−q/2

(log n)

√ 1/4 1/ log nΔn

nH ()

 Δn(q−2)/4 (log n)δ−1 n

  2δ

nH ()

  δ   Δn(q−2)/4 (log n)δ−1 → 0 as  ↓ 0.  M δ 1/ log H () n

(2.13)

nH ()

Finally, let us estimate Λ6 , recalling q > 2δ + 2. It turns out that 

n−q/2 (log n)δ−q/2

   (y + )1−q E|X1 |q I |X1 |  2(y + ) n log n dy

√ 1/4 1/ log nΔn

nH ()



C

n

−q/2

δ−q/2

(log n)

∞    E|X1 | (y + )1−q I |X1 |   n log n dy q

nH ()

+C

0



n

−q/2

δ−q/2

(log n)

∞     E|X1 | (y + )1−q I  n log n < |X1 |  (y + ) n log n dy

nH ()

q

0

=: Λ7 + Λ8 . Several lines of elementary calculation yield 

Λ7  C 2−q

n−q/2 (log n)−q/2+δ



∞     E|X1 |q I  i log i  |X1 |   (i + 1) log(i + 1) n−q/2 (log n)−q/2+δ

iH ()

C



n=i

   δ   (log i)−1 E|X1 |2 log |X1 | I  i log i  |X1 |   (i + 1) log(i + 1)

iH ()

C

    E|X1 |q I  i log i  |X1 |   (i + 1) log(i + 1)

i=1

nH ()

= C 2−q

n 



  δ    E|X1 |2 log |X1 | I  i log i  |X1 |   (i + 1) log(i + 1)

iH ()

 δ  CE|X1 |2 log |X1 | < ∞.

(2.14)

560

Y. Zhao / J. Math. Anal. Appl. 339 (2008) 553–565

In view of the moment condition, it follows that 

Λ8  C

n

−q/2

−q/2+δ

(log n)

   E|X1 | I |X1 | >  n log n q

nH ()



C

0

n

−q/2

−q/2+δ

(log n)

   E|X1 | I |X1 | >  n log n

C

n



−1

−1+δ

(log n)

C

    E|X1 |2 I  i log i < |X1 |   (i + 1) log(i + 1)

i     E|X1 |2 I  i log i < |X1 |   (i + 1) log(i + 1) n−1 (log n)−1+δ n=1

−1

(log i)

iH ()



C

∞  i=n

iH ()



(x + )1−q dx |X | √ 1 2 n log n

nH ()

C

q

nH ()



∞    (x + )1−q I |X1 |  2(x + ) n log n dx

  δ    E|X1 | log |X1 | I  i log i < |X1 |   (i + 1) log(i + 1) 2

  δ    E|X1 |2 log |X1 | I  i log i < |X1 |   (i + 1) log(i + 1)

iH ()

 δ  CE|X1 |2 log |X1 | < ∞.

(2.15)

Combining (2.12)–(2.15), consequently, we obtain (2.11).

2

Proposition 2.5. Under the conditions of Theorem 1.1, lim  ↓0

lim 

 (log n)δ−1 n2

n>H ()

 (log n)δ−1 n2

↓0

n>H ()

 √  2xP |N |  x/ n dx = 0,

(2.16)

  2xP |Sn |  x dx = 0.

(2.17)

√  n log n

∞ √  n log n

Proof. The proof of (2.16) is quite routine, we omit it. Applying Lemma 2.2, the proof of (2.17) are exposed as follows: 

n

−2

∞ δ−1

(log n)

√  n log n

n>H ()

C



n

−1

∞ δ

(log n)

n>H ()

C

  2xP |Sn |  x dx

 n>H ()

0

n

−1

∞ δ

(log n)

   2(x + )P |Sn |  (x + ) n log n dx 

   2(x + ) nP |X1 |  2(x + ) n log n dx

0

[log n]      −q q/2 + (x + ) n log n n exp ρ 2i dx i=1

[log n]     −q      2/q i q 2 E|X1 | I |X1 |  2(x + ) n log n dx n exp ρ + (x + ) n log n i=1

Y. Zhao / J. Math. Anal. Appl. 339 (2008) 553–565



=: C

n

−1

561

∞ 2(x + )(II 1 + II 2 + II 3 ) dx.

δ

(log n)

n>H ()

0

Recalling the moment condition, it suffices to prove that 

n

−1

∞ (log n)

n>H ()



2(x + )II 1 dx  C

δ

δ

(log n)

n>H ()

0





C

δ

(log n) E

n>H ()

∞  CE

   2xP |X1 |  2x n log n dx

X12  x

   2xI |X1 |  2x n log n dx



√  δ−1    log |X1 | − log x  I |X1 |  x I |X1 |  M dx



√  δ    CEX12 log |X1 | − log   I |X1 |  M δ   CEX12 log |X1 | + CEX12 | log |δ < ∞.

(2.18)

In light of q > 2δ + 2, the proof of part II 2 follows immediately. Consequently, turn to the last part II 3 , it leads to 

n

−1

∞ (log n)

n>H ()



2(x + )II 3 dx

δ 0



n

−q/2

−q/2+δ

(log n)

∞    E|X1 | (x + )1−q I |X1 |  2(x + ) n log n dx q

n>H ()

0

∞     −q/2 −q/2+δ q n (log n) E|X1 | (x + )1−q I |X1 |   n log n dx = n>H ()

+

0



n

−q/2

−q/2+δ

(log n)

∞     E|X1 | (x + )1−q I  n log n < |X1 |  2(x + ) n log n dx

n>H ()

q

0

=: II 4 + II 5 . A careful calculation shows that     n−q/2 (log n)−q/2+δ E|X1 |q I |X1 |   n log n II 4  C 2−q n>H ()



 C 2−q

n>H ()



= C 2−q

i>H ()



 C 2−q

n−q/2 (log n)−q/2+δ

n 

    E|X1 |q I  i log i  |X1 |   (i + 1) log(i + 1)

i=1 ∞     E|X1 |q I  i log i  |X1 |   (i + 1) log(i + 1) n−q/2 (log n)−q/2+δ n=i

    i −q/2+1 (log i)−q/2+δ E|X1 |q I  i log i  |X1 |   (i + 1) log(i + 1)

i>H ()

C



  δ    (log i)−1 E|X1 |2 log |X1 | I  i log i  |X1 |   (i + 1) log(i + 1)

i>H ()

C



i>H ()

  δ    E|X1 |2 log |X1 | I  i log i  |X1 |   (i + 1) log(i + 1)

562

Y. Zhao / J. Math. Anal. Appl. 339 (2008) 553–565

  δ    CE|X1 |2 log |X1 | I |X1 |   H () log H () < ∞.

(2.19)

Finally, we have 

II 5  C

n

−q/2

−q/2+δ

(log n)

   E|X1 | I |X1 | >  n log n q

n>H ()



C

0

n−1 (log n)−1+δ



C

    E|X1 |2 I  i log i < |X1 |   (i + 1) log(i + 1)

i     E|X1 |2 I  i log i < |X1 |   (i + 1) log(i + 1) n−1 (log n)−1+δ n=1

i>H ()



∞  i=n

n>H ()

C

∞    (x + )1−q I |X1 |  2(x + ) n log n dx

−1

(log i)

  δ    E|X1 | log |X1 | I  i log i < |X1 |   (i + 1) log(i + 1) 2

i>H ()

  δ    CE|X1 |2 log |X1 | I |X1 |   H () log H () < ∞.

(2.20) 2

With the aid of (2.18)–(2.20), one can complete the proof of (2.17).

Proof of Theorem 1.1. The proof follows immediately from the above five propositions.

2

3. Proofs of Theorems 1.2 and 1.3 Note that (1.7) is the maximal version of (1.6), if we make some modification of the proof of (1.6), Theorem 1.2 will follow from Lemma 2.2 together with the following lemma. Lemma 3.1. (See Shao .) Let {X be a strictly stationary sequences of ρ-mixing random variables n : n  1} n ) < ∞. If σ 2 = ES 2 → ∞, n → ∞, then lim 2 2 ρ(2 satisfying EX1 = 0, EX12 < ∞ and ∞ n→∞ σn /n = σ . Furn n n=1 √ thermore, if σ > 0, then Wn ⇒ W , where Wn (t) = S[nt] /σ n, 0  t  1, “⇒” means weak convergence in D[0, 1] with Skorohod topology, in particular,   √ D (3.1) → sup W (t). Mn /σ n − 0t1

Indeed, it suffices to show that ∞  (log n)δ−1 n=2

n2

∞       (log n)δ  P Mn   n log n EMn2 I Mn   n log n =  2 n n=2

+

∞  (log n)δ−1 n=2

n2

2xP (Mn  x) dx √

 n log n

=: I3 + I4 . To pave the way for the proofs of I3 and I4 , several propositions will be given as follows. Proposition 3.1. Suppose that {W (t): t  0} is a standard Wiener process (Brownian motion). Then, for any 0 < δ  1, lim  2δ+2 ↓0

∞ ∞ 2E|N |2δ+2      (log n)δ  (−1)n P sup W (s)   2 log n = . n δ+1 (2n + 1)2δ+2 0s1 n=1

Proof. Refer to Huang et al. .

n=0

2

(3.2)

Y. Zhao / J. Math. Anal. Appl. 339 (2008) 553–565

563

Proposition 3.2. For any 0 < δ  1, we have lim 

↓0

∞  (log n)δ−1

n2

n=2

 2xP

∞   √ (−1)n 2E|N |2δ+2  sup W (t)  x/ n dx = . δ(δ + 1) (2n + 1)2δ+2 0t1

(3.3)

n=0

 n log n

Proof. By Fubini’s theorem together with Proposition 3.1, it turns out that lim 

↓0

∞ 

n

−2

∞ y

↓0

−1

∞ δ−1

(log y)

∞ 2δ−1

=

 2sP

2 δ

  sup W (t)  s ds 0t1

0

=



s 2δ+1 P

s 2u2δ−1 du 0

  sup W (t)  s ds 0t1

0

=

0t1

u

0

  sup W (t)  s ds

 2sP

du

  sup W (t)  s ds 0t1

√  log y

∞ 2u

 2sP

dy

2

=

0t1

√  n log n

  √ sup W (t)  x/ n dx

2xP

(log n)

n=2

= lim 



δ−1

∞ 2E|N |2δ+2  (−1)n . δ(δ + 1) (2n + 1)2δ+2

2

(3.4)

n=0

Proof of Theorem 1.2. Along the same lines as that of the proof of Theorem 1.1, together with Lemma 3.1 and Propositions 3.1 and 3.2, one can easily complete the proof. 2 To prove Theorem 1.3, the definition and the proposition are necessary, which reads as follows. ψ(n) := sup sup

k1 A∈F − k

sup

+ B∈Fk+n P (A)P (B)>0

|P (AB) − P (A)P (B)| , (P (A)P (B))1/2

the sequence {Xn : n  1} is said to be ψ -mixing, if ψ(n) → 0 as n → ∞. It is well known that ψ(n)  ρ(n). Proposition 3.3. Suppose that limn→∞ ESn2 /n = σ 2 > 0, and     P (Mn   n log n )  λnP |X1 |  2 n log n ,

∞

n=1 ρ(n) < ∞,

if (1.9) holds. Then, for any  > 0, (3.5)

where λ is some positive constant. Proof. We first show that    nP |X1 |  2 n log n → 0 as n → ∞.

(3.6)

According to (1.9), we have, for any  > 0, ∞    (log n)δ  P Mn   n log n < ∞. n n=2

Observe that |Xn | = |Sn − Sn−1 |  |Sn | + |Sn−1 |. Then, from (3.7), we have

(3.7)

564

Y. Zhao / J. Math. Anal. Appl. 339 (2008) 553–565 ∞   (log n)δ  P max |Xk |  2 n log n < ∞, n 1kn

 > 0,

(3.8)

n=2

and 

 δ  log 2j P max |Xk |   2j log 2j 1k2j

2j +1

C

 (log n)δ   P max |Xk |   n log n/2/2 → 0, n 1kn j

j → ∞.

(3.9)

n=2

Taking a = (log n)δ , notice that P (AB) = P (A)P (B) + {P (A)P (B) − P (AB)} and the relation between ρ-mixing and ψ-mixing. One can get   P max |Xk |   n log n 1kn

[n/a]    |Xia |   n log n P i=1

=

[n/a] 

   P max |Xj a | <  n log n, |Xia |   n log n j
i=1



[n/a] 

     P max |Xj a | <  n log n P |Xia |   n log n j
i=1

     − ρ(a)P 1/2 max |Xj a |   n log n P 1/2 |Xia |   n log n j
(3.10)

1j
  in turn imUsing the fact that P (max1j <2i |Xj | <  2i log 2i ) → 1 as i → ∞ and ∞ n=1 ρ(n) < ∞, which  ∞ plies (3.6). Finally, note that ∞ ρ(n) < ∞, so one can choose a positive integer m such that n=1 i=1 ρ(mi) < 1. It follows that   P max |Xk |   n log n 1kn



[n/m] 



   P |Xi |   n log n −

    P |Xim |   n log n, |Xj m |   n log n

1i
i=1

[n/m]       2  − [n/m]P |X1 |   n log n ρ(mi)  [n/m]P |X1 |   n log n − [n/m]P |X1 |   n log n





   [n/m]P |X1 | >  n log n





   1 − [n/m]P |X1 |   n log n −

[n/m] 

ρ(mi) .

i=1

(3.11)

i=1

Taking λ = [n/m]/n, thus (3.5) follows from (3.6) and (3.11).

2

Proof of Theorem 1.3. According to Theorem 1.2, the sufficient part is obvious. The necessary part follows from Propositions 3.3 together with the following result: ∞>

∞  n=2

n

−1

∞ δ

(log n)

0

   2(x + 1)P Mn  (x + 1) n log n dx

Y. Zhao / J. Math. Anal. Appl. 339 (2008) 553–565

C

∞ 

n

−1

∞    (log n) (x + 1)nP |X1 |  2(x + 1) n log n dx δ

n=N

0



∞ ∞   δ (log n) I n log n   CE (x + 1) 0

 CEX12

565

n=N

 δ log |X1 | .

   |X1 | |X1 | I x +1 dx 2(x + 1) 2

2

Acknowledgment Many thanks are due to the referee for a careful reading of the manuscript and for valuable suggestions.

References