A note on the almost sure central limit theorem for the product of some partial sums

Abstract

Let $(X_n)$ be a sequence of i.i.d., positive, square integrable random variables with $E(X_1)=\mu >0$, $Var(X_1)={\sigma }^2$. Denote by $S_{n,k}={\sum }_{i=1}^n{X}_i-X_k$ and by $\gamma =\sigma /\mu$ the coefficient of variation. Our goal is to show the unbounded, measurable functions g, which satisfy the almost sure central limit theorem, i.e.,

$$\underset{N\to \mathrm{\infty }}{lim}\frac{1}{logN}\sum _{n=1}^N\frac{1}{n}g\left({\left(\frac{{\prod }_{k=1}^n{S}_{n,k}}{{(n-1)}^n{\mu }^n}\right)}^{\frac{1}{\gamma \sqrt{n}}}\right)={\int }_0^{\mathrm{\infty }}g(x)dF(x)\text{a.s.},$$

where $F(\cdot )$ is the distribution function of the random variable $e^{\mathcal{N}}$ and is a standard normal random variable.

MSC:60F15, 60F05.

1 Introduction

The almost sure central limit theorem (ASCLT) has been first introduced independently by Schatte [1] and Brosamler [2]. Since then, many studies have been done to prove the ASCLT in different situations, for example, in the case of function-typed almost sure central limit theorem (FASCLT) (see Berkes et al. [3], Ibragimov and Lifshits [4]). The purpose of this paper is to investigate the FASCLT for the product of some partial sums.

Let $(X_n)$ be a sequence of i.i.d. random variables and define the partial sum $S_n={\sum }_{k=1}^n{X}_k$ for $n\ge 1$. In a recent paper of Rempala and Wesolowski [5], it is showed under the assumption $E(X^2)<\mathrm{\infty }$ and $X>0$ that

$${\left(\frac{{\prod }_{k=1}^n{S}_k}{n!{\mu }^n}\right)}^{\frac{1}{\gamma \sqrt{n}}}\stackrel{d}{\to }{e}^{\sqrt{2}\mathcal{N}},$$

(1)

where is a standard normal random variable, $\mu =E(X)$ and $\gamma =\sigma /\mu$ with ${\sigma }^2=var(X)$. For further results in this field, we refer to Qi [6], Lu and Qi [7] and Rempala and Wesolowski [8].

Recently Gonchigdanzan and Rempala [9] obtained the almost sure limit theorem related to (1) as follows.

Theorem A Let $(X_n)$ be a sequence of i.i.d., positive random variables with $E(X_1)=\mu >0$ and $Var(X_1)={\sigma }^2$. Denote by $\gamma =\sigma /\mu$ the coefficient of variation. Then, for any real x,

$$\underset{N\to \mathrm{\infty }}{lim}\frac{1}{logN}\sum _{n=1}^N\frac{1}{n}I({\left(\frac{{\prod }_{k=1}^n{S}_k}{n!{\mu }^n}\right)}^{\frac{1}{\gamma \sqrt{n}}}\le x)=G(x)\mathit{\text{a.s.}},$$

(2)

where $G(x)$ is the distribution function of $e^{\sqrt{2}\mathcal{N}}$, is a standard normal random variable. Some extensions on the above result can be found in Ye and Wu [10]and the reference therein.

A similar result on the product of partial sums was provided by Miao [11], which stated the following.

Theorem B Let $(X_n)$ be a sequence of i.i.d., positive, square integrable random variables with $E(X_1)=\mu >0$ and $Var(X_1)={\sigma }^2$. Denote by $S_{n,k}={\sum }_{i=1}^n{X}_i-X_k$ and $\gamma =\sigma /\mu$ the coefficient of variation. Then

$${\left(\frac{{\prod }_{k=1}^n{S}_{n,k}}{{(n-1)}^n{\mu }^n}\right)}^{\frac{1}{\gamma \sqrt{n}}}\stackrel{d}{\to }{e}^{\mathcal{N}},$$

(3)

and for any real x,

$$\underset{N\to \mathrm{\infty }}{lim}\frac{1}{logN}\sum _{n=1}^N\frac{1}{n}I({\left(\frac{{\prod }_{k=1}^n{S}_{n,k}}{{(n-1)}^n{\mu }^n}\right)}^{\frac{1}{\gamma \sqrt{n}}}\le x)=F(x)\mathit{\text{a.s.}},$$

(4)

where $F(\cdot )$ is the distribution function of the random variable $e^{\mathcal{N}}$ and is a standard normal random variable.

The purpose of this paper is to investigate the validity of (4) for some class of unbounded measurable functions g.

Throughout this article, $(X_n)$ is a sequence of i.i.d. positive, square integrable random variables with $E(X_1)=\mu >0$ and $Var(X_1)={\sigma }^2$. We denote by $S_{n,k}={\sum }_{i=1}^n{X}_i-X_k$ and by $\gamma =\sigma /\mu$ the coefficient of variation. Furthermore, is the standard normal random variable, Φ is the standard normal distribution function, ϕ is its density function and $a\ll b$ stands for ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}|a_n/b_n|<\mathrm{\infty }$.

2 Main result

We state our main result as follows.

Theorem 1 Let $g(x)$ be a real-valued, almost everywhere continuous function on R such that $|g(e^x)\varphi (x)|\le c{(1+|x|)}^{-\alpha }$ with some $c>0$ and $\alpha >5$. Then, for any real x,

$$\underset{N\to \mathrm{\infty }}{lim}\frac{1}{logN}\sum _{n=1}^N\frac{1}{n}g\left({\left(\frac{{\prod }_{k=1}^n{S}_{n,k}}{{(n-1)}^n{\mu }^n}\right)}^{\frac{1}{\gamma \sqrt{n}}}\right)={\int }_0^{\mathrm{\infty }}g(x)dF(x)\mathit{\text{a.s.}},$$

(5)

where $F(\cdot )$ is the distribution function of the random variable $e^{\mathcal{N}}$.

Let $f(x)=g(e^x)$. By a simple calculation, we can get the following result.

Remark 1 Let $f(x)$ be a real-valued, almost everywhere continuous function on R such that $|f(x)\varphi (x)|\le c{(1+|x|)}^{-\alpha }$ with some $c>0$ and $\alpha >5$. Then (5) is equivalent to

$$\underset{N\to \mathrm{\infty }}{lim}\frac{1}{logN}\sum _{n=1}^N\frac{1}{n}f(\frac{1}{\gamma \sqrt{n}}\sum _{k=1}^{n}log\frac{S_{n,k}}{(n-1)\mu })={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)\varphi (x)dx\text{a.s.}$$

(6)

Remark 2 Lu et al. [12] proved the function-typed almost sure central limit theorem for a type of random function, which can include U-statistics, Von-Mises statistics, linear processes and some other types of statistics, but their results cannot imply Theorem 1.

3 Auxiliary results

In this section, we state and prove several auxiliary results, which will be useful in the proof of Theorem 1.

Let ${\tilde{S}}_n={\sum }_{i=1}^n\frac{X_i-\mu }{\sigma }$ and $U_i=\frac{1}{\gamma \sqrt{i}}{\sum }_{k=1}^{i}log\frac{S_{i,k}}{(i-1)\mu }$. Observe that for $|x|<1$ we have

$$log(1+x)=x+\frac{\theta }{2}{x}^2,$$

where $\theta \in (-1,0)$. Thus

$$\begin{array}{rcl}{U}_i& =& \frac{1}{\gamma \sqrt{i}}\sum _{k=1}^{i}log\frac{S_{i,k}}{(i-1)\mu }\\ =& \frac{1}{\gamma \sqrt{i}}\sum _{k=1}^i(\frac{S_{i,k}}{(i-1)\mu }-1)+\frac{1}{\gamma \sqrt{i}}\sum _{k=1}^i\frac{{\theta }_k}{2}{(\frac{S_{i,k}}{(i-1)\mu }-1)}^2\\ =& \frac{1}{\sqrt{i}}\sum _{k=1}^i\left(\frac{{\sum }_{j\ne k,j\le i}(X_j-\mu )}{(i-1)\sigma }\right)+\frac{1}{\gamma \sqrt{i}}\sum _{k=1}^i\frac{{\theta }_k}{2}{(\frac{S_{i,k}}{(i-1)\mu }-1)}^2\\ =& \frac{1}{\sqrt{i}}\sum _{k=1}^i\frac{X_k-\mu }{\sigma }+\frac{1}{\gamma \sqrt{i}}\sum _{k=1}^i\frac{{\theta }_k}{2}{(\frac{S_{i,k}}{(i-1)\mu }-1)}^2\\ =:& \frac{1}{\sqrt{i}}{\tilde{S}}_i+R_i.\end{array}$$

(7)

By the law of iterated logarithm, we have for $k\to \mathrm{\infty }$

$$\underset{1\le k\le i}{max}|\frac{S_{i,k}}{(i-1)\mu }-1|=O\left({(loglogi/i)}^{1/2}\right)\text{a.s.}$$

Therefore,

$$|R_i|=\left|\frac{1}{\gamma \sqrt{i}}\sum _{k=1}^i\frac{{\theta }_k}{2}{(\frac{S_{i,k}}{(i-1)\mu }-1)}^2\right|\ll \frac{1}{\sqrt{i}}\sum _{k=1}^i{(\frac{S_{i,k}}{(i-1)\mu }-1)}^2\ll \frac{loglogi}{i^{1/2}}\text{a.s.}$$

(8)

Obviously,

$$\begin{array}{rl}E|R_i|& =E\left|\frac{1}{\gamma \sqrt{i}}\sum _{k=1}^i\frac{{\theta }_k}{2}{(\frac{S_{i,k}}{(i-1)\mu }-1)}^2\right|\\ \ll \frac{1}{\sqrt{i}}\sum _{k=1}^{i}E{(\frac{S_{i,k}}{(i-1)\mu }-1)}^2\ll \frac{1}{\sqrt{i}}\sum _{k=1}^i\frac{1}{i-1}\ll \frac{1}{i^{1/2}}.\end{array}$$

(9)

Our proof mainly relies on decomposition (7). Properties (8) and (9) will be extensively used in the following parts of this section.

Lemma 1 Let X and Y be random variables. We write $F(x)=P(X<x)$, $G(x)=P(X+Y<x)$. Then

$$F(x-\epsilon )-P(|Y|\ge \epsilon )\le G(x)\le F(x+\epsilon )+P(|Y|\ge \epsilon )$$

for every $\epsilon >0$ and x.

Proof It is Lemma 1.3 of Petrov [13]. □

Lemma 2 Let $(X_n)$ be a sequence of i.i.d. random variables. Let $S_n={\sum }_{k\le n}{X}_k$, $F^s$ denote the distribution function obtained from F by symmetrization, and choose $L>0$ so large that ${\int }_{|x|\le L}{x}^{2}d{F}^s\ge 1$. Then, for any $n\ge 1$, $\lambda >0$,

$$\underset{a}{sup}P(a\le \frac{S_n}{\sqrt{n}}\le a+\lambda )\le A\lambda$$

with some absolute constant A, provided $\lambda \sqrt{n}\ge L$.

Proof It can be obtained from Berkes et al. [3]. □

Lemma 3 Assume that (6) is true for all indicator functions of intervals and for a fixed a.e. continuous function $f(x)=f_0(x)$. Then (6) is also true for all a.e. continuous functions f such that $|f(x)|\le |f_0(x)|$, $x\in R$, and, moreover, the exceptional set of probability 0 can be chosen universally for all such f.

Proof See Berkes et al. [3]. □

In view of Lemma 3 and Remark 1, in order to prove Theorem 1, it suffices to prove (6) for the case when $f(x)\varphi (x)={(1+|x|)}^{-\alpha }$, $\alpha >5$. Thus, in the following part, we put $f(x)\varphi (x)={(1+|x|)}^{-\alpha }$, $\alpha >5$ and

$$\begin{array}{c}{\xi }_k=\sum _{i=2^k+1}^{2^{k+1}}\frac{1}{i}f(U_i),\hfill \\ {\xi }_k^{\ast }=\sum _{i=2^k+1}^{2^{k+1}}\frac{1}{i}f(U_i)I\{f(U_i)\le \frac{k}{{(logk)}^{\beta }}\},\hfill \end{array}$$

where $1<\beta <\frac{1}{2}(\alpha -3)$.

Lemma 4 Under the conditions of Theorem  1, we have $P({\xi }_k\ne {\xi }_k^{\ast }\mathit{\text{ i.o.}})=0$.

Proof Let $f^{-1}$ denote an inverse function of f in some interval, and let α, β satisfy $1<\beta <\frac{1}{2}(\alpha -3)$. It is easy to check that

$$\{{\xi }_k\ne {\xi }_k^{\ast }\}\subseteq \{|U_i|\ge f^{-1}(k/{(logk)}^{\beta })\text{ for some }{2}^k<i\le 2^{k+1}\}$$

and

$$\begin{array}{rcl}f\left({(2logk+(\alpha -2\beta )loglogk)}^{1/2}\right)& =& \frac{k}{{(logk)}^{\beta }}\frac{\sqrt{2\pi }{(logk)}^{\alpha /2}}{{\{1+{(2logk+(\alpha -2\beta )loglogk)}^{1/2}\}}^{\alpha }}\\ \le & \frac{k}{{(logk)}^{\beta }}.\end{array}$$

(10)

Note that the function f is even and strictly increasing for $x\ge x_0$. We have

$$f^{-1}(k/{(logk)}^{\beta })\ge {(2logk+(\alpha -2\beta )loglogk)}^{1/2}.$$

(11)

Observing that $2^k<i\le 2^{k+1}$ implies $k\ge \frac{1}{2}logi$, in view of (8) we get

$$\begin{array}{rcl}P({\xi }_k\ne {\xi }_k^{\ast }\mathit{\text{ i.o.}})& \le & P(|U_i|\ge {(2loglogi+(\alpha -2\beta )logloglogi-O(1))}^{1/2}\mathit{\text{ i.o.}})\\ =& P(|\frac{{\tilde{S}}_i}{\sqrt{i}}+R_i|\ge {(2loglogi+(\alpha -2\beta )logloglogi-O(1))}^{1/2}\mathit{\text{ i.o.}})\\ \le & P(\left|\frac{{\tilde{S}}_i}{\sqrt{i}}\right|\ge {(2loglogi+(\alpha -2\beta )logloglogi-O(1))}^{1/2}\mathit{\text{ i.o.}})\\ =& 0,\end{array}$$

where in the last step we use the assumption $\alpha -2\beta >3$ and a version of the Kolmogorov-Erdös-Feller-Petrovski test (see Feller [14], Theorem 2). This completes the proof of Lemma 4. □

Let $a_k=f^{-1}(k/{(logk)}^{\beta })$ and let $G_i$ and $F_i$ denote, respectively, the distribution function of $U_i$ and $\frac{{\tilde{S}}_i}{\sqrt{i}}$. Set

$$\begin{array}{c}{\sigma }_i^2={\int }_{-\sqrt{i}}^{\sqrt{i}}{x}^{2}d{F}_i(x)-{({\int }_{-\sqrt{i}}^{\sqrt{i}}xd{F}_i(x))}^2,\hfill \\ {\eta }_i=\underset{x}{sup}|G_i(x)-\mathrm{\Phi }\left(\frac{x}{{\sigma }_i}\right)|,\hfill \\ {\epsilon }_i=\underset{x}{sup}|F_i(x)-\mathrm{\Phi }\left(\frac{x}{{\sigma }_i}\right)|.\hfill \end{array}$$

Clearly, ${\sigma }_i\le 1$, ${lim}_{i\to \mathrm{\infty }}{\sigma }_i=1$.

Lemma 5 Under the conditions of Theorem  1, we have

$$\sum _{k\le N}E{\left({\xi }_k^{\ast }\right)}^2\ll \frac{N^2}{{(logN)}^{2\beta }}.$$

Proof Observe now that the relation

$$|{\int }_{-a}^a\psi (x)d(G_1(x)-G_2(x))|\le \underset{-a\le x\le a}{sup}|\psi (x)|\cdot \underset{-a\le x\le a}{sup}|G_1(x)-G_2(x)|$$

(12)

is valid for any bounded, measurable functions ψ and distribution functions $G_1$, $G_2$. Let, as previously, $a_k=f^{-1}(k/{(logk)}^{\beta })$. Thus, for any $2^k<i\le 2^{k+1}$, we obtain that

$$\begin{array}{rcl}E{f}^2(U_i)I\{f(U_i)\le \frac{k}{{(logk)}^{\beta }}\}& =& {\int }_{|x|\le a_k}{f}^2(x)d{G}_i(x)\\ \le & {\int }_{|x|\le a_k}{f}^2(x)d\mathrm{\Phi }\left(\frac{x}{{\sigma }_i}\right)+{\eta }_i\frac{k^2}{{(logk)}^{2\beta }}\\ \ll & {\int }_{|x|\le a_k}{f}^2(x)d\mathrm{\Phi }(x)+{\eta }_i\frac{k^2}{{(logk)}^{2\beta }},\end{array}$$

where in the last step, we have used the fact that ${\sigma }_i\le 1$, ${lim}_{i\to \mathrm{\infty }}{\sigma }_i=1$. Hence, by the Cauchy-Schwarz inequality, we have

$$\begin{array}{rcl}E{\left({\xi }_k^{\ast }\right)}^2& \ll & E{\left[{\left(\sum _{i=2^k+1}^{2^{k+1}}{\left(\frac{1}{i}\right)}^2\right)}^{1/2}{(\sum _{i=2^k+1}^{2^{k+1}}{f}^2(U_i)I\{f(U_i)\le \frac{k}{{(logk)}^{\beta }}\})}^{1/2}\right]}^2\\ \ll & \left(\sum _{i=2^k+1}^{2^{k+1}}\frac{1}{i^2}\right)\left(\sum _{i=2^k+1}^{2^{k+1}}({\int }_{|x|\le a_k}{f}^2(x)d\mathrm{\Phi }(x)+{\eta }_i\frac{k^2}{{(logk)}^{2\beta }})\right)\\ \ll & \frac{1}{2^k}(2^k{\int }_{|x|\le a_k}{f}^2(x)d\mathrm{\Phi }(x)+\frac{k^2}{{(logk)}^{2\beta }}\sum _{i=2^k+1}^{2^{k+1}}{\eta }_i)\\ \ll & {\int }_{|x|\le a_k}\frac{e^{x^2/2}}{{(1+|x|)}^{2\alpha }}dx+\frac{k^2}{{(logk)}^{2\beta }}\sum _{i=2^k+1}^{2^{k+1}}\frac{{\eta }_i}{i}.\end{array}$$

(13)

Note that

$${\int }_0^t\frac{e^{x^2/2}}{{(1+|x|)}^{2\alpha }}dx={\int }_0^{t/2}+{\int }_{t/2}^t\ll t{e}^{t^2/8}+\frac{1}{t^{2\alpha +1}}{\int }_{t/2}^{t}x{e}^{x^2/2}dx\ll \frac{e^{t^2/2}}{t^{2\alpha +1}},$$

and thus by (10) and (11), we have

$${\int }_{|x|\le a_k}\frac{e^{x^2/2}}{{(1+|x|)}^{2\alpha }}dx\ll \frac{e^{a_k^2/2}}{a_k^{2\alpha +1}}\ll f(a_k)\frac{1}{a_k^{\alpha +1}}\ll \frac{k}{{(logk)}^{\beta +(\alpha +1)/2}}.$$

(14)

Now we estimate ${\eta }_i$. By Lemma 1, we have that for some $\epsilon >0$,

$$\begin{array}{rcl}{\eta }_i& =& \underset{x}{sup}|G_i(x)-\mathrm{\Phi }\left(\frac{x}{{\sigma }_i}\right)|\\ \le & \underset{x}{sup}|G_i(x)-F_i(x)|+\underset{x}{sup}|F_i(x)-\mathrm{\Phi }\left(\frac{x}{{\sigma }_i}\right)|\\ =& \underset{x}{sup}|P(U_i\le x)-P(\frac{{\tilde{S}}_i}{\sqrt{i}}\le x)|+{\epsilon }_i\\ =& \underset{x}{sup}|P((\frac{{\tilde{S}}_i}{\sqrt{i}}+R_i)\le x)-P(\frac{{\tilde{S}}_i}{\sqrt{i}}\le x)|+{\epsilon }_i\\ \le & P(|R_i|\ge \epsilon )+\underset{x}{sup}\{P(\frac{{\tilde{S}}_i}{\sqrt{i}}\le x+\epsilon )-P(\frac{{\tilde{S}}_i}{\sqrt{i}}\le x)\}+{\epsilon }_i.\end{array}$$

The Markov inequality and (9) imply that

$$P(|R_i|\ge \epsilon )\le \frac{E|R_i|}{\epsilon }\ll \frac{1}{i^{1/2}\epsilon }.$$

In addition, Lemma 2 yields

$$\underset{x}{sup}\{P(\frac{{\tilde{S}}_i}{\sqrt{i}}\le x+\epsilon )-P(\frac{{\tilde{S}}_i}{\sqrt{i}}\le x)\}\ll \epsilon .$$

Setting $\epsilon =i^{-1/3}$, we have

$${\eta }_i\ll \frac{1}{i^{1/6}}+\frac{1}{i^{1/3}}+{\epsilon }_i.$$

Using Theorem 1 of Friedman et al. [15], we get

$$\sum _{i=1}^{\mathrm{\infty }}\frac{{\epsilon }_i}{i}<\mathrm{\infty }.$$

Hence,

$$\sum _{i=1}^{\mathrm{\infty }}\frac{{\eta }_i}{i}\ll \sum _{i=1}^{\mathrm{\infty }}\frac{\frac{1}{i^{1/6}}+{\epsilon }_i}{i}<\mathrm{\infty },$$

(15)

which, coupled with (13), (14) and the fact $\frac{1}{2}(\alpha +1)>\beta$, yields

$$\begin{array}{rcl}\sum _{k\le N}E{\left({\xi }_k^{\ast }\right)}^2& \ll & \sum _{k\le N}\frac{k}{{(logk)}^{\beta +(\alpha +1)/2}}+\sum _{k\le N}\frac{k^2}{{(logk)}^{2\beta }}\sum _{i=2^k+1}^{2^{k+1}}\frac{{\eta }_i}{i}\\ \ll & \frac{N^2}{{(logN)}^{2\beta }},\end{array}$$

which completes the proof. □

Lemma 6 Let ${\xi }_k^{\ast }={\sum }_{i=2^k+1}^{2^{k+1}}\frac{1}{i}f(U_i)I\{f(U_i)\le \frac{k}{{(logk)}^{\beta }}\}$, ${\xi }_l^{\ast }={\sum }_{i=2^l+1}^{2^{l+1}}\frac{1}{i}f(U_i)I\{f(U_i)\le \frac{l}{{(logl)}^{\beta }}\}$. Under the conditions of Theorem  1, we have for $l\ge l_0$

$$|cov({\xi }_k^{\ast },{\xi }_l^{\ast })|\ll \frac{kl}{{(logk)}^{\beta }{(logl)}^{\beta }}{2}^{-(l-k-1)/4}.$$

Proof We first show the following result, for any $1\le i\le \frac{j}{2}$ and real x, y,

$$|P(U_i\le x,U_j\le y)-P(U_i\le x)P(U_j\le y)|\ll {\left(\frac{i}{j}\right)}^{1/4}.$$

(16)

Letting $\rho =\frac{i}{j}$, the Chebyshev inequality yields

$$P(\left|\frac{{\tilde{S}}_i}{\sqrt{j}}\right|\ge {\rho }^{1/4})\le \frac{1}{j}{\rho }^{-1/2}E{|{\tilde{S}}_i|}^2={\rho }^{1/2}.$$

(17)

Using the Markov inequality and (9), we have

$$P(|R_j|\ge {\rho }^{1/4})\le \frac{E|R_j|}{{\rho }^{1/4}}\ll \frac{1}{j^{1/2}{\rho }^{1/4}}=\frac{1}{j^{1/4}{i}^{1/4}}\le {\rho }^{1/4}.$$

(18)

It follows from Lemma 1, Lemma 2, (17), (18) and the positivity and independence of $(X_n)$ that

$$\begin{array}{r}P(U_i\le x,U_j\le y)\\ =P(U_i\le x,\frac{{\tilde{S}}_j}{\sqrt{j}}+R_j\le y)\\ =P(U_i\le x,\frac{{\tilde{S}}_i}{\sqrt{j}}+\sqrt{1-\rho }\frac{{\tilde{S}}_j-{\tilde{S}}_i}{\sqrt{j-i}}+R_j\le y)\\ \ge P(U_i\le x,\sqrt{1-\rho }\frac{{\tilde{S}}_j-{\tilde{S}}_i}{\sqrt{j-i}}\le y)\\ -P(y-2{\rho }^{1/4}\le \sqrt{1-\rho }\frac{{\tilde{S}}_j-{\tilde{S}}_i}{\sqrt{j-i}}\le y)-P(\left|\frac{{\tilde{S}}_i}{\sqrt{j}}\right|\ge {\rho }^{1/4})-P(|R_j|\ge {\rho }^{1/4})\\ \ge P(U_i\le x,\sqrt{1-\rho }\frac{{\tilde{S}}_j-{\tilde{S}}_i}{\sqrt{j-i}}\le y)-(4A+O(1)+1){\rho }^{1/4}\\ =P(U_i\le x)P(\sqrt{1-\rho }\frac{{\tilde{S}}_j-{\tilde{S}}_i}{\sqrt{j-i}}\le y)-(4A+O(1)+1){\rho }^{1/4}.\end{array}$$

(19)

We can obtain an analogous upper estimate for the first probability in (19) by the same way. Thus

$$P(U_i\le x,U_j\le y)=P(U_i\le x)P(\sqrt{1-\rho }\frac{{\tilde{S}}_j-{\tilde{S}}_i}{\sqrt{j-i}}\le y)-\theta (4A+O(1)+1){\rho }^{1/4},$$

where $|\theta |\le 1$. A similar argument yields

$$P(U_i\le x)P(U_j\le y)=P(U_i\le x)P(\sqrt{1-\rho }\frac{{\tilde{S}}_j-{\tilde{S}}_i}{\sqrt{j-i}}\le y)-{\theta }^{\prime }(4A+O(1)+1){\rho }^{1/4},$$

where $|{\theta }^{\prime }|\le 1$, and (16) follows. Letting $G_{i,j}(x,y)$ denote the joint distribution function of $U_i$ and $U_j$, in view of (12), (16), we get for $l\ge l_0$

$$\begin{array}{r}|cov(f(U_i)I\{f(U_i)\le \frac{k}{{(logk)}^{\beta }}\},f(U_j)I\{f(U_j)\le \frac{l}{{(logl)}^{\beta }}\})|\\ =|{\int }_{|x|\le a_k}{\int }_{|y|\le a_l}f(x)f(y)d(G_{i,j}(x,y)-G_i(x)G_j(y))|\\ \ll \frac{kl}{{(logk)}^{\beta }{(logl)}^{\beta }}{2}^{-(l-k-1)/4},\end{array}$$

where the last relation follows from the facts that: f is strictly increasing for $x\ge x_0$, $f(a_i)=\frac{i}{{(logi)}^{\beta }}$ and $2^k<i\le 2^{k+1}$, $2^l<j\le 2^{l+1}$. Thus

$$|cov({\xi }_k^{\ast },{\xi }_l^{\ast })|\ll \frac{kl}{{(logk)}^{\beta }{(logl)}^{\beta }}{2}^{-(l-k-1)/4}.$$

 □

Lemma 7 Under the conditions of Theorem  1, letting ${\zeta }_k={\xi }_k^{\ast }-E{\xi }_k^{\ast }$, we have

$$E{({\zeta }_1+\cdots +{\zeta }_N)}^2=O\left(\frac{N^2}{{(logN)}^{2\beta -1}}\right),N\to \mathrm{\infty }.$$

Proof By Lemma 6, we have

$$|\underset{l-k>40logN}{\sum _{1\le k\le l\le N}}E({\zeta }_k{\zeta }_l)|\ll \frac{N^2}{{(logN)}^{2\beta }}{N}^2{2}^{-10logN}=o(1).$$

On the other hand, letting $\parallel \cdot \parallel$ denote the $L_2$ norm, Lemma 5 and the Cauchy-Schwarz inequality imply

$$\begin{array}{rcl}|\underset{l-k\le 40logN}{\sum _{1\le k\le l\le N}}E({\zeta }_k{\zeta }_l)|& \le & \underset{l-k\le 40logN}{\sum _{1\le k\le l\le N}}\parallel {\zeta }_k\parallel \parallel {\zeta }_l\parallel \\ \le & \underset{l-k\le 40logN}{\sum _{1\le k\le l\le N}}\parallel {\xi }_k^{\ast }\parallel \parallel {\xi }_l^{\ast }\parallel \\ =& \sum _{0\le j\le 40logN}\sum _{k=1}^{N-j}\parallel {\xi }_k^{\ast }\parallel \parallel {\xi }_{k+j}^{\ast }\parallel \\ \le & {\left(\sum _{k=1}^N{\parallel {\xi }_k^{\ast }\parallel }^2\right)}^{1/2}{\left(\sum _{l=1}^N{\parallel {\xi }_l^{\ast }\parallel }^2\right)}^{1/2}40logN\\ =& O\left(\frac{N^2}{{(logN)}^{2\beta -1}}\right),\end{array}$$

and Lemma 7 is proved. □

4 Proof of the main result

We only prove the property in (6), since, in view of Remark 1, it is sufficient for the proof of Theorem 1.

Proof of Theorem 1 By Lemma 7 we have

$$E{\left(\frac{{\zeta }_1+\cdots +{\zeta }_N}{N}\right)}^2=O\left({(logN)}^{1-2\beta }\right),$$

and thus setting $N_k=[exp(k^{\lambda })]$ with ${(2\beta -1)}^{-1}<\lambda <1$, we get

$$\sum _{k=1}^{\mathrm{\infty }}E{\left(\frac{{\zeta }_1+\cdots +{\zeta }_{N_k}}{N_k}\right)}^2<\mathrm{\infty },$$

and therefore

$$\underset{k\to \mathrm{\infty }}{lim}\frac{{\zeta }_1+\cdots +{\zeta }_{N_k}}{N_k}=0\text{a.s.}$$

(20)

Observe now that for $2^k<i\le 2^{k+1}$ we have

$$\begin{array}{rcl}Ef(U_i)I\{f(U_i)\le \frac{k}{{(logk)}^{\beta }}\}& =& {\int }_{|x|\le a_k}f(x)d{G}_i(x)\\ =& {\int }_{|x|\le a_k}f(x)d\mathrm{\Phi }\left(\frac{x}{{\sigma }_i}\right)+{\int }_{|x|\le a_k}f(x)d(G_i(x)-\mathrm{\Phi }\left(\frac{x}{{\sigma }_i}\right)).\end{array}$$

Put $m={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)d\mathrm{\Phi }(x)$. Since ${\sigma }_i\le 1$, ${lim}_{i\to \mathrm{\infty }}{\sigma }_i=1$ and $a_k\to \mathrm{\infty }$ as $k\to \mathrm{\infty }$, we have

$$\underset{k\to \mathrm{\infty }}{lim}\underset{2^k<i\le 2^{k+1}}{sup}|{\int }_{|x|\le a_k}f(x)d\mathrm{\Phi }\left(\frac{x}{{\sigma }_i}\right)-m|=0,$$

and thus, using (12), we get

$$|Ef(U_i)I\{f(U_i)\le \frac{k}{{(logk)}^{\beta }}\}-m|\le \frac{k{\eta }_i}{{(logk)}^{\beta }}+o_k(1).$$

Thus we have

$$E{\xi }_k^{\ast }=m\sum _{i=2^k+1}^{2^{k+1}}\frac{1}{i}+{\vartheta }_k\frac{k}{{(logk)}^{\beta }}\sum _{i=2^k+1}^{2^{k+1}}\frac{{\eta }_i}{i}+o_k(1),|{\vartheta }_k|\le 1.$$

Consequently, using the relation ${\sum }_{i\le L}1/i=logL+O(1)$ and (15), we conclude

$$\begin{array}{rcl}|\frac{E({\xi }_1^{\ast }+\cdots +{\xi }_N^{\ast })}{log{2}^{N+1}}-m|& \ll & \frac{1}{N}\sum _{k\le N}\frac{k}{{(logk)}^{\beta }}\sum _{i=2^k+1}^{2^{k+1}}\frac{{\eta }_i}{i}+o_N(1)\\ =& O\left({(logN)}^{-\beta }\right)+o_N(1)=o_N(1),\end{array}$$

and thus (20) gives

$$\underset{k\to \mathrm{\infty }}{lim}\frac{{\xi }_1^{\ast }+\cdots +{\xi }_{N_k}^{\ast }}{log{2}^{N_k+1}}=m\text{a.s.}$$

By Lemma 4 this implies

$$\underset{k\to \mathrm{\infty }}{lim}\frac{{\xi }_1+\cdots +{\xi }_{N_k}}{log{2}^{N_k+1}}=m\text{a.s.}$$

(21)

The relation $\lambda <1$ implies ${lim}_{k\to \mathrm{\infty }}{N}_{k+1}/N_k=1$, and thus (21) and the positivity of ${\xi }_k$ yield

$$\underset{N\to \mathrm{\infty }}{lim}\frac{{\xi }_1+\cdots +{\xi }_N}{log{2}^{N+1}}=m\text{a.s.},$$

(22)

i.e., (6) holds for the subsequence $\{2^{N+1}\}$. Now, for each $N\ge 4$, there exists n, depending on N, such that $2^{n+1}\le N\le 2^{n+2}$. Then

$$\frac{{\xi }_1+{\xi }_2+\cdots +{\xi }_n}{log{2}^{n+1}}\le \frac{{\sum }_{i=1}^N\frac{1}{i}f(U_i)}{logN}\frac{logN}{log{2}^{n+1}}\le \frac{{\xi }_1+{\xi }_2+\cdots +{\xi }_{n+2}}{log{2}^{n+2}}\frac{log{2}^{n+2}}{log{2}^{n+1}}$$

(23)

by the positivity of each term of $({\xi }_k)$. Noting that $(n+1)log2\sim logN\sim (n+2)log2$ as $N\to \mathrm{\infty }$, we get (6) by (22) and (23). □