Almost sure central limit theorem for products of sums of partial sums

Abstract

Considering a sequence of i.i.d. positive random variables, for products of sums of partial sums we establish an almost sure central limit theorem, which holds for some class of unbounded measurable functions.

1 Introduction and main results

Let $\{X_{n};n\geq1\}$ be a sequence of random variables and define $S_{n}=\sum_{i=1}^{n} X_{i}$. Some results as regards the limit theorem of products $\prod_{j=1}^{n}S_{j}$ were obtained in recent years. Rempala and Wesolowski [1] obtained the following asymptotics for products of sums for a sequence of i.i.d. random variables.

Theorem A

Let $\{X_{n};n\geq1\}$ be a sequence of i.i.d. positive square integrable random variables with $\mathbb{E}X_{1}=\mu$, the coefficient of variation $\gamma=\sigma/\mu$, where $\sigma ^{2}=\operatorname{Var}(X_{1})$. Then

$$ \biggl(\frac{\prod_{k=1}^{n} S_{k}}{n!\mu^{n}} \biggr)^{\frac{1}{\gamma \sqrt{n}}}\stackrel{d}{\rightarrow} {e}^{\sqrt{2}\mathcal{N}} \quad \textit{as } n\to\infty. $$

(1.1)

Here and in the sequel, $\mathcal{N}$ is a standard normal random variable and $\stackrel{d}{\rightarrow}$ denotes the convergence in distribution.

Gonchigdanzan and Rempala [2] discussed the almost sure central limit theorem (ASCLT) for the products of partial sums and obtained the following result.

Theorem B

Let $\{X_{n};n\geq1\}$ be a sequence of i.i.d. positive random variables with $\mathbb{E}X_{1}=\mu$, $\operatorname {Var}(X_{1})=\sigma^{2}$ the coefficient of variation $\gamma=\sigma/\mu$. Then

$$ \lim_{N \to\infty}\frac{1}{\log N}\sum_{n=1}^{N} \frac {1}{n}I \biggl\{ \biggl(\frac{\prod_{k=1}^{n} S_{k}}{n!\mu^{n}} \biggr)^{\frac {1}{\gamma\sqrt{n}}}\leq x \biggr\} =F(x) \quad \textit{a.s. for any } x \in\mathbb{R}, $$

(1.2)

where F is the distribution function of the random variable $e^{\sqrt {2}\mathcal{N}}$. Here and in the sequel, $I \{\cdot \}$ denotes the indicator function.

Tan and Peng [3] proved the result of Theorem B still holds for some class of unbounded measurable functions and obtained the following result.

Theorem C

Let $\{X_{n};n\geq1\}$ be a sequence of i.i.d. positive random variables with $\mathbb{E}X_{1}=\mu$, $\operatorname {Var}(X_{1})=\sigma^{2}$, $\mathbb{E}|X_{1}|^{3}<\infty$, the coefficient of variation $\gamma =\sigma/\mu$. Let $g(x)$ be a real valued almost everywhere continuous function on $\mathbb{R}$ such that $|g(e^{x})\phi(x)|\leq c(1+|x|)^{-\alpha}$ with some $c>0$ and $\alpha>5$. Then

$$ \lim_{N \to\infty}\frac{1}{\log N}\sum_{n=1}^{N} \frac {1}{n}g \biggl\{ \biggl(\frac{\prod_{k=1}^{n} S_{k}}{n!\mu^{n}} \biggr)^{\frac {1}{\gamma\sqrt{n}}} \biggr\} = \int_{0}^{\infty}g(x)\,\mathrm{d}F(x) \quad \textit{a.s. for any } x \in\mathbb{R}, $$

(1.3)

where $F(\cdot)$ is the distribution function of the random variable $e^{\sqrt {2}\mathcal{N}}$ and $\phi(x)$ is the density function of the standard normal random variable.

Zhang et al. [4] discussed the almost sure central limit theory for products of sums of partial sums and obtained the following result.

Theorem D

Let $\{X,X_{n};n\geq1\}$ be a sequence of i.i.d. positive square integrable random variables with $\mathbb{E}X=\mu$, $\operatorname{Var}(X)=\sigma^{2}<\infty$, the coefficient of variation $\gamma =\sigma/\mu$. Denote $S_{n}=\sum_{i=1}^{n} X_{i}$, $T_{k}=\sum_{i=1}^{k} S_{i}$. Then

$$ \lim_{n \to\infty}\frac{1}{\log n}\sum_{k=1}^{n} \frac {1}{k}I \biggl\{ \biggl(\frac{2^{k}\prod_{j=1}^{k} T_{j}}{k!(k+1)!\mu^{k}} \biggr)^{\frac{1}{\gamma\sqrt{k}}}\leq x \biggr\} =F(x) \quad \textit{a.s. for any } x \in\mathbb{R}, $$

(1.4)

where $F(\cdot)$ is the distribution function of the random variable $e^{\sqrt{10/3}\mathcal{N}}$.

The purpose of this article is to establish that Theorem D holds for some class of unbounded measurable functions.

Our main result is the following theorem.

Theorem 1.1

Let $\{X_{n};n\geq1\}$ be a sequence of i.i.d. positive random variables with $\mathbb{E}X_{1}=\mu$, $\operatorname {Var}(X_{1})=\sigma^{2}$, $\mathbb{E}|X_{1}|^{3}<\infty$, the coefficient of variation $\gamma =\sigma/\mu$. Let $g(x)$ be a real valued almost everywhere continuous function on $\mathbb{R}$ such that $|g(e^{\sqrt{10/3} x})\phi(x)|\leq c(1+|x|)^{-\alpha}$ with some $c>0$ and $\alpha>5$. Denote $S_{n}=\sum_{i=1}^{n} X_{i}$, $T_{k}=\sum_{i=1}^{k} S_{i}$. Then

$$\begin{aligned}& \lim_{N \to\infty}\frac{1}{\log N}\sum_{n=1}^{N} \frac {1}{n}g \biggl( \biggl(\frac{2^{n}\prod_{k=1}^{n} T_{k}}{n!(n+1)!{\mu }^{n}} \biggr)^{\frac{1}{\gamma\sqrt{n}}} \biggr) \\& \quad = \int_{0}^{\infty }g(x)\,\mathrm{d}F(x) \quad \textit{a.s. for any } x \in\mathbb{R}, \end{aligned}$$

(1.5)

where $F(\cdot)$ is the distribution function of the random variable $e^{\sqrt{10/3}\mathcal{N}}$. Here and in the sequel, $\phi(x)$ is the density function of the standard normal random variable.

Remark 1

Let $f(x)=g(e^{\sqrt{10/3} x})$, $t=e^{\sqrt{10/3} x}$. Then

$$\begin{aligned}& x=\sqrt{\frac{3}{10}}\log t, \qquad g(t)=f \biggl(\sqrt {\frac{3}{10}} \log t \biggr), \\& g \biggl( \biggl(\frac{2^{n}\prod_{k=1}^{n} T_{k}}{n!(n+1)!{\mu}^{n}} \biggr)^{\frac{1}{\gamma\sqrt{n}}} \biggr) =f \Biggl( \sqrt{\frac{3}{10}}\log \Biggl(\prod_{k=1}^{n} \frac {2T_{k}}{k(k+1)\mu} \Biggr)^{\frac{1}{\gamma\sqrt{n}}} \Biggr) \\& \hphantom{g \biggl( \biggl(\frac{2^{n}\prod_{k=1}^{n} T_{k}}{n!(n+1)!{\mu}^{n}} \biggr)^{\frac{1}{\gamma\sqrt{n}}} \biggr)}=f \Biggl(\frac{1}{\gamma\sqrt{10n/3}}\sum _{k=1}^{n} \log \frac {T_{k}}{k(k+1)\mu/2} \Biggr). \end{aligned}$$

Since $F(x)$ is the distribution function of the random variable $e^{\sqrt{10/3}\mathcal{N}}$, we can get $F(x)=\Phi (\sqrt{\frac{3}{10}}\log x )$, where $\Phi (x)$ is the distribution function of the standard normal random variable. Hence we have the following: Let $f(x)=g(e^{\sqrt{10/3} x})$ and $f(x)$ be a real valued almost everywhere continuous function on $\mathbb{R}$ such that $|f(x)\phi(x)|\leq c(1+|x|)^{-\alpha}$ with some $c>0$ and $\alpha>5$, then (1.5) is equivalent to

$$\begin{aligned}& \lim_{N \to\infty}\frac{1}{\log N}\sum_{n=1}^{N} \frac {1}{n}f \Biggl(\frac{1}{\gamma\sqrt{10n/3}}\sum_{k=1}^{n} \log \frac {T_{k}}{k(k+1)\mu/2} \Biggr) \\& \quad = \int_{-\infty}^{\infty}f(x)\phi(x)\,\mathrm {d}x\quad \mbox{a.s. for any } x \in\mathbb{R}. \end{aligned}$$

(1.6)

Remark 2

By the proof of Theorem 2 of Berkes et al. [5], in order to prove (1.5), it suffices to show (1.6) holds true for $f(x)\phi(x)=(1+|x|)^{-\alpha}$ with $\alpha>5$. Here and in the sequel, $f(x)$ satisfies $f(x)\phi(x)=(1+|x|)^{-\alpha}$ with $\alpha>5$.

2 Preliminaries

In the following, the notation $a_{n}\sim b_{n}$ means that $\lim_{n \to\infty}a_{n}/ b_{n}= 1$ and $a_{n}\ll b_{n}$ means that $\limsup_{n \to\infty}|a_{n}/ b_{n}|<+\infty$. We denote $b_{k,n}=\sum_{j=k}^{n} \frac {1}{j}$, $c_{k,n}=2\sum_{j=k}^{n} \frac{j+1-k}{j(j+1)}$, $d_{k,n}=\frac {n+1-k}{n+1}$, $\widetilde{X}_{i}=\frac{{X}_{i}-\mu}{\sigma}$, $\widetilde{S}_{k}=\sum_{i=1}^{k}\widetilde{X}_{i}$, ${S}_{k,n}=\sum_{i=1}^{k} c_{i,n}\widetilde{X}_{i}$. By Lemma 2.1 of Wu [6], we can get

$$c_{i,n}=2(b_{i,n}-d_{i,n}), \qquad \sum _{i=1}^{n} c_{i,n}^{2} \sim \frac{10n}{3}. $$

Let

$$Y_{i}=\frac{1}{\gamma\sqrt{10i/3}}\sum_{k=1}^{i} \log \frac {T_{k}}{k(k+1)\mu/2}. $$

Note that

$$\begin{aligned}& \frac{1}{\gamma}\sum_{k=1}^{i} \biggl( \frac{T_{k}}{k(k+1)\mu/2}-1 \biggr) \\& \quad =\frac{1}{\gamma}\sum_{k=1}^{i} \biggl(\frac{2\sum_{j=1}^{k} S_{j}-k(k+1)\mu }{k(k+1)\mu} \biggr) \\& \quad =\frac{1}{\gamma}\sum_{k=1}^{i} \frac{2}{k(k+1)\mu}\sum_{j=1}^{k}\sum _{l=1}^{j}(X_{l}-\mu) \\& \quad =\frac{1}{\gamma}\sum_{k=1}^{i} \frac{2}{k(k+1)\mu}\sum_{l=1}^{k}\sum _{j=l}^{k}(X_{l}-\mu) \\& \quad =\frac{\mu}{\sigma}\sum_{k=1}^{i} \frac{2}{k(k+1)\mu}\sum_{l=1}^{k}(k+1-l) (X_{l}-\mu) \\& \quad =\sum_{k=1}^{i}\sum _{l=1}^{k} \frac{2(k+1-l)}{k(k+1)}\frac{X_{l}-\mu }{\sigma} \\& \quad =\sum_{l=1}^{i} \sum _{k=l}^{i}\frac{2(k+1-l)}{k(k+1)}\widetilde{X}_{l} \\& \quad =\sum_{l=1}^{i} c_{l,i} \widetilde{X}_{l}=S_{i,i}. \end{aligned}$$

By the fact that $\log (1+x)=x+\frac{\delta}{2}x^{2}$, where $|x|<1$, $\delta\in(-1,0)$, thus we have

$$\begin{aligned} Y_{i}&=\frac{1}{\gamma\sqrt{10i/3}}\sum_{k=1}^{i} \log \frac {T_{k}}{k(k+1)\mu/2} \\ &=\frac{1}{\gamma\sqrt{10i/3}}\sum_{k=1}^{i} \biggl( \frac{T_{k}}{k(k+1)\mu /2}-1 \biggr)+\frac{1}{\gamma\sqrt{10i/3}} \sum_{k=1}^{i} \frac{\delta_{k}}{2} \biggl(\frac{T_{k}}{k(k+1)\mu/2}-1 \biggr)^{2} \\ &=\frac{1}{\sqrt{10i/3}}S_{i,i}+\frac{1}{\gamma\sqrt{10i/3}}\sum _{k=1}^{i}\frac{\delta_{k}}{2} \biggl( \frac{T_{k}}{k(k+1)\mu/2}-1 \biggr)^{2} \\ &=:\frac{1}{\sqrt{10i/3}}S_{i,i}+R_{i}. \end{aligned}$$

By the fact that $\mathbb{E}|X_{1}|^{2}<\infty$, using the Marcinkiewicz-Zygmund strong large number law, we have

$$\begin{aligned}& S_{k}-k\mu=o \bigl(k^{1/2} \bigr) \quad \mbox{a.s.}, \\& \biggl\vert \frac{T_{k}}{k(k+1)\mu/2}-1 \biggr\vert = \biggl\vert \frac{2\sum_{j=1}^{k} S_{j}-k(k+1)\mu}{k(k+1)\mu} \biggr\vert \\& \hphantom{\biggl\vert \frac{T_{k}}{k(k+1)\mu/2}-1 \biggr\vert }\leq\frac{2\vert \sum_{j=1}^{k} (S_{j}-j\mu)\vert }{k(k+1)\mu} \\& \hphantom{\biggl\vert \frac{T_{k}}{k(k+1)\mu/2}-1 \biggr\vert }\leq\frac{2\sum_{j=1}^{k} j^{1/2}}{k(k+1)\mu}\ll\frac {k^{3/2}}{k^{2}}=\frac{1}{k^{1/2}}. \end{aligned}$$

Thus

$$ |R_{i}|\ll\frac{1}{\sqrt{i}}\sum_{k=1}^{i} \frac{1}{k}\ll\frac{\log i}{\sqrt{i}}\quad \mbox{a.s.} $$

(2.1)

In order to prove Theorem 1.1, we introduce the following lemmas.

Lemma 2.1

Let X and Y be random variables. Set $F(x)=P(X< x)$, $G(x)=P(X+Y< x)$, then for any $\varepsilon>0$ and $x\in\mathbb{R}$,

$$ F(x-\varepsilon)-P \bigl(\vert Y\vert \geq\varepsilon \bigr)\leq G(x)\leq F(x+ \varepsilon )+P \bigl(\vert Y\vert \geq\varepsilon \bigr). $$

Proof

See Lemma 3 on p.16 of Petrov [7]. □

Lemma 2.2

Let $\{X_{n};n\geq1\}$ be a sequence of i.i.d. positive random variables. Denote $S_{n}=\sum_{i=1}^{n} X_{i}$, $F^{s}$ denotes the distribution function obtained from F by symmetrization and choose $L >0$ so large that $\int_{|x|\leq L}x^{2}\,\mathrm{d}F^{s}(x)\geq1$. Then, for any $n\geq1$, $\lambda>0$, there exists a $c>0$ such that

$$\sup_{a}P \biggl(a\leq\frac{S_{n}}{\sqrt{n}}\leq a+\lambda \biggr) \leq c\lambda $$

holds for $\lambda\sqrt{n}\geq L$.

Proof

See (20) on p.73 of Berkes et al. [5]. □

Let

$$\begin{aligned}& Z_{k}=\sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{i}f(Y_{i}), \\& Z_{k}^{*}=\sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{i}f(Y_{i})I \biggl\{ f(Y_{i})\leq \frac{k}{(\log k)^{\beta}} \biggr\} , \end{aligned}$$

where $1<\beta<(\alpha-3)/2$.

Lemma 2.3

Under the conditions of Theorem  1.1, we get

$$\mathbb{P} \bigl(Z_{k}\neq Z_{k}^{*}, \mathrm{i.o.} \bigr)=0. $$

Proof

It is easy to get

$$\begin{aligned} \bigl\{ Z_{k}\neq Z_{k}^{*} \bigr\} \subseteq& \bigl\{ |Y_{i}|\geq f^{-1} \bigl(k/(\log k)^{\beta} \bigr) \mbox{ for some } 2^{k}< i\leq2^{k+1} \bigr\} \\ =& \biggl\{ \biggl\vert \frac{1}{\sqrt{10i/3}}S_{i,i}+R_{i} \biggr\vert \geq f^{-1} \bigl(k/(\log k)^{\beta} \bigr)\geq \bigl(2\log k+(\alpha-2\beta)\log \log k \bigr)^{1/2} \\ &\mbox{for some } 2^{k}< i\leq2^{k+1} \biggr\} . \end{aligned}$$

Since $|R_{i}|\ll\frac{\log i}{\sqrt{i}}$ a.s.; see (2.1). By the law of iterated logarithm (Feller [8], Theorem 2), we get

$$\begin{aligned} \mathbb{P} \bigl(Z_{k}\neq Z_{k}^{*}, \mathrm{i.o.} \bigr) \leq& \mathbb{P} \biggl( \biggl\vert \frac {1}{\sqrt{10i/3}}S_{i,i} \biggr\vert \geq \bigl(2\log \log i+(\alpha-2\beta)\log \log \log i-O(1) \bigr)^{1/2}, \mathrm{i.o.} \biggr) \\ =&0. \end{aligned}$$

We complete the proof of Lemma 2.3. □

Let $G_{i}$, $F_{i}$, F denote the distribution functions of $Y_{i}$, $\frac {\widetilde{S}_{i}}{\sqrt{i}}$, $\widetilde{X}_{1}$, respectively. Φ denotes the distribution function of the standard normal distribution function. Set

$$\begin{aligned}& \sigma_{i}^{2} = \int_{-\sqrt{i}}^{\sqrt{i}}x^{2}\,\mathrm{d}F(x)- \biggl( \int _{-\sqrt{i}}^{\sqrt{i}}x\,\mathrm{d}F(x) \biggr)^{2}, \\& \varepsilon_{i} =\sup_{x} \biggl\vert F_{i}(x)-\Phi \biggl(\frac{x}{\sigma_{i}} \biggr) \biggr\vert ,\qquad \theta_{i}=\sup_{x} \biggl\vert G_{i}(x)-\Phi \biggl(\frac{x}{\sigma_{i}} \biggr) \biggr\vert . \end{aligned}$$

Obviously $\sigma_{i}\leq1$, $\lim_{i \to\infty}\sigma_{i}= 1$.

Lemma 2.4

Under the conditions of Theorem  1.1, we have

$$\sum_{k=1}^{N}\mathbb{E} \bigl(Z_{k}^{*} \bigr)^{2}\ll\frac{N^{2}}{(\log N)^{2\beta}}. $$

Proof

Note that the estimation

$$ \biggl\vert \int_{-a}^{a}\Psi(x)\,\mathrm{d} \bigl(H_{1}(x)-H_{2}(x) \bigr) \biggr\vert \leq\sup _{-a\leq x \leq a} \bigl\vert \Psi(x) \bigr\vert \cdot\sup _{-a\leq x \leq a} \bigl\vert H_{1}(x)-H_{2}(x) \bigr\vert $$

(2.2)

holds for any bounded, measurable function $\Psi(x)$ and the distribution functions $H_{1}(x)$, $H_{2}(x)$. Thus for $2^{k}< i\leq 2^{k+1}$, we get

$$\begin{aligned}& \mathbb{E}f^{2}(Y_{i})I \biggl\{ f(Y_{i})\leq \frac{k}{(\log k)^{\beta }} \biggr\} \\& \quad = \int_{|x|\leq a_{k}}f^{2}(x)\,\mathrm{d}G_{i}(x) \\& \quad \leq \int_{|x|\leq a_{k}}f^{2}(x)\,\mathrm{d}\Phi \biggl( \frac{x}{\sigma _{i}} \biggr)+\theta_{i}\frac{k^{2}}{(\log k)^{2\beta}} \\& \quad \ll \int_{|x|\leq a_{k}}f^{2}(x)\,\mathrm{d}\Phi(x)+ \theta_{i}\frac {k^{2}}{(\log k)^{2\beta}}; \end{aligned}$$

here and in the sequel $a_{k}=f^{-1}(\frac{k}{(\log k)^{\beta }})$. Hence, by the Cauchy-Schwarz inequality and the fact that $f(x)\phi(x)=(1+|x|)^{-\alpha}$, we obtain

$$\begin{aligned} \mathbb{E} \bigl(Z_{k}^{*} \bigr)^{2} &\ll\mathbb{E} \Biggl( \Biggl(\sum_{i=2^{k}+1}^{2^{k+1}} \biggl( \frac{1}{i} \biggr)^{2} \Biggr)^{1/2} \Biggl(\sum _{i=2^{k}+1}^{2^{k+1}}f^{2}(Y_{i})I \biggl\{ f(Y_{i}) \leq\frac{k}{(\log k)^{\beta}} \biggr\} \Biggr)^{1/2} \Biggr)^{2} \\ &\ll \Biggl(\sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{i^{2}} \Biggr) \Biggl(\sum_{i=2^{k}+1}^{2^{k+1}} \biggl( \int_{|x|\leq a_{k}}f^{2}(x)\,\mathrm{d}\Phi (x)+ \theta_{i}\frac{k^{2}}{(\log k)^{2\beta}} \biggr) \Biggr) \\ &\ll\frac{1}{2^{k}} \Biggl(2^{k} \int_{|x|\leq a_{k}}f^{2}(x)\,\mathrm{d}\Phi (x)+ \frac{k^{2}}{(\log k)^{2\beta}}\sum_{i=2^{k}+1}^{2^{k+1}}\theta _{i} \Biggr) \\ &\ll \int_{|x|\leq a_{k}}\frac{e^{x^{2}/2}}{(1+|x|)^{2\alpha}}\,\mathrm {d}x+\frac{k^{2}}{(\log k)^{2\beta}} \sum_{i=2^{k}+1}^{2^{k+1}}\frac {\theta_{i}}{i}. \end{aligned}$$

By the same methods as that on p.72 of Berkes et al. [5], we get

$$\int_{|x|\leq a_{k}}\frac{e^{x^{2}/2}}{(1+|x|)^{2\alpha}}\,\mathrm{d}x\ll \frac{k}{(\log k)^{\beta+(\alpha+1)/2}}. $$

Now we estimate $\theta_{i}$. By Lemma 2.1, for any $\varepsilon>0$, we have

$$\begin{aligned} \theta_{i} =&\sup_{x} \biggl\vert G_{i}(x)-\Phi \biggl(\frac{x}{\sigma_{i}} \biggr) \biggr\vert \\ \leq&\sup_{x} \bigl\vert G_{i}(x)-F_{i}(x) \bigr\vert +\sup_{x} \biggl\vert F_{i}(x)-\Phi \biggl(\frac{x}{\sigma_{i}} \biggr) \biggr\vert \\ =&\sup_{x} \biggl\vert P(Y_{i}\leq x)-P \biggl( \frac{\widetilde{S}_{i}}{\sqrt {i}}\leq x \biggr) \biggr\vert +\varepsilon_{i} \\ \leq&\sup_{x} \biggl\vert P(Y_{i}\leq x)-P \biggl(\frac{S_{i,i}}{\sqrt {10i/3}}\leq x \biggr) \biggr\vert + \sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x \biggr)-P \biggl( \frac{\widetilde{S}_{i}}{\sqrt{i}}\leq x \biggr) \biggr\vert +\varepsilon _{i} \\ \leq&\sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}+R_{i} \leq x \biggr)-P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x+\varepsilon \biggr) \biggr\vert \\ &{} +\sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x+\varepsilon \biggr)-P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x \biggr) \biggr\vert \\ &{} +\sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x \biggr)-P \biggl(\frac{\widetilde{S}_{i}}{\sqrt{i}}\leq x \biggr) \biggr\vert + \varepsilon_{i} \\ \leq& P \bigl(\vert R_{i}\vert \geq\varepsilon \bigr)+\sup _{x} \biggl\vert P \biggl(\frac {S_{i,i}}{\sqrt{10i/3}}\leq x+ \varepsilon \biggr)-P \biggl(\frac {S_{i,i}}{\sqrt{10i/3}}\leq x \biggr) \biggr\vert \\ &{}+ \sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x \biggr)-P \biggl(\frac{\widetilde{S}_{i}}{\sqrt{i}}\leq x \biggr) \biggr\vert + \varepsilon_{i}. \end{aligned}$$

By the Markov inequality and (2.1), we have

$$P \bigl(\vert R_{i}\vert \geq\varepsilon \bigr)\leq \frac{\mathbb{E}|R_{i}|}{\varepsilon}\ll \frac{\log i}{\sqrt{i}\varepsilon}. $$

By Lemma 2.2, we have

$$\sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x+ \varepsilon \biggr)-P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x \biggr) \biggr\vert \ll \varepsilon. $$

By the Berry-Esseen inequality, we have

$$\begin{aligned}& \sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x \biggr)-P \biggl(\frac{\widetilde{S}_{i}}{\sqrt{i}}\leq x \biggr) \biggr\vert \\& \quad \leq\sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}} \leq x \biggr)-\Phi(x) \biggr\vert +\sup_{x} \biggl\vert P \biggl(\frac{\widetilde{S}_{i}}{\sqrt {i}}\leq x \biggr)-\Phi(x) \biggr\vert \\& \quad \ll\frac{1}{i^{1/2}}+\frac{1}{i^{1/2}}. \end{aligned}$$

Let $\varepsilon=i^{-1/3}$, then

$$\theta_{i}\ll\frac{\log i}{i^{1/6}}+ \frac{1}{i^{1/3}}+ \frac {1}{i^{1/2}}+\varepsilon_{i}. $$

Therefore, there exists $\varepsilon_{0}>0$ such that

$$\theta_{i}\ll\frac{1}{i^{\varepsilon_{0}}}+\varepsilon_{i}. $$

By Theorem 1 of Friedman et al. [9], we have

$$\sum_{i=1}^{\infty}\frac{\varepsilon_{i}}{i}< \infty. $$

Hence

$$\sum_{i=1}^{\infty}\frac{\theta_{i}}{i}\ll\sum _{i=1}^{\infty}\frac{\frac {1}{i^{\varepsilon_{0}}}+\varepsilon_{i}}{i}< \infty. $$

By the fact that $(\alpha+1)/2>\beta$, we have

$$\sum_{k=1}^{N}\mathbb{E} \bigl(Z_{k}^{*} \bigr)^{2}\ll\sum_{k=1}^{N} \frac{k}{(\log k)^{\beta+(\alpha+1)/2}} +\sum_{k=1}^{N} \frac{k^{2}}{(\log k)^{2\beta}}\sum_{i=2^{k}+1}^{2^{k+1}} \frac{\theta_{i}}{i}\ll\frac{N^{2}}{(\log N)^{2\beta}}. $$

We complete the proof of Lemma 2.4. □

Lemma 2.5

Under the conditions of Theorem  1.1, for $l\geq l_{0}$, we have

$$\bigl\vert \operatorname{Cov} \bigl(Z_{k}^{*},Z_{l}^{*} \bigr) \bigr\vert \ll\frac{kl}{(\log k)^{\beta }(\log l)^{\beta}}2^{-(l-k)\tau}, $$

where τ is a constant $0<\tau\leq1/8$.

Proof

For $1\leq i \leq j/2$, $j\geq j_{0}$ and any x, y, we first prove

$$ \bigl\vert P(Y_{i}\leq x,Y_{j}\leq y)-P(Y_{i} \leq x)P(Y_{j}\leq y) \bigr\vert \ll \biggl(\frac {i}{j} \biggr)^{\tau}. $$

(2.3)

Let $\rho=\frac{i}{j}$. By the Chebyshev inequality, we have

$$P \biggl( \biggl\vert \frac{S_{i,i}}{\sqrt{10j/3}} \biggr\vert \geq\rho^{1/8} \biggr) =P \biggl( \biggl\vert \frac{S_{i,i}}{\sqrt{10i/3}} \biggr\vert \geq\sqrt{ \frac {j}{i}} {\rho}^{1/8} \biggr)\leq\frac{i}{j} \rho^{-1/4}\leq{\rho }^{1/8}\leq\rho^{\tau_{1}}, $$

where $\tau_{1}$ is a constant $0<\tau_{1}\leq1/8$.

By the Markov inequality and (2.1), for $j\geq j_{0}$, we have

$$P \bigl(\vert R_{j}\vert \geq\rho^{1/8} \bigr)\leq \frac{\mathbb{E}|R_{j}|}{\rho^{1/8}}\ll \frac{\log j}{j^{1/2}\rho^{1/8}} =\rho^{1/8}\frac{\log j}{j^{1/4}i^{1/4}}\ll \rho^{\tau_{2}}, $$

where $\tau_{2}$ is a constant, $0<\tau_{2}\leq1/8$.

By the Markov inequality, we have

$$\begin{aligned}& P \biggl( \biggl\vert \sqrt{1-\rho}\frac{c_{i+1,j}\widetilde{S}_{i}}{\sqrt {10(j-i)/3}} \biggr\vert \geq \rho^{1/8} \biggr) \\& \quad =P \biggl( \biggl\vert \frac{\widetilde{S}_{i}}{\sqrt{i}} \biggr\vert \geq\sqrt{ \frac {10/3j}{i}}\frac{1}{c_{i+1,j}}{\rho}^{1/8} \biggr) \\& \quad \leq\frac{3}{10}{\rho}^{3/4}(c_{i+1,j})^{2}= \frac{3}{10}{\rho }^{3/4}(2b_{i+1,j}-2d_{i+1,j})^{2} \\& \quad =\frac{3}{10}{\rho}^{3/4} \Biggl[ \Biggl(2\sum _{k=i+1}^{j}\frac{1}{k} \Biggr)^{2}+4 \biggl(\frac {j+1-i-1}{j+1} \biggr)^{2}-8 \Biggl(\sum _{k=i+1}^{j}\frac{1}{k} \Biggr) \frac {j+1-i-1}{j+1} \Biggr] \\& \quad \ll{\rho}^{3/4} \biggl[ \biggl(\log \frac{j}{i} \biggr)^{2}+ \biggl(\frac {j-i}{j+1} \biggr)^{2}- \frac{j-i}{j+1}\log \frac{j}{i} \biggr] \\& \quad \ll\rho^{\tau_{3}}, \end{aligned}$$

where $\tau_{3}$ is a constant that satisfies $0<\tau_{3}\leq1/8$.

By Lemma 2.2 and the fact that $\rho=\frac{i}{j}$, $1\leq i \leq j/2$, we have

$$P \biggl(y-3{\rho}^{1/8}\leq\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}}\leq y \biggr)\ll\frac{{\rho}^{1/8}}{\sqrt{1-\rho}}\ll\rho^{1/8}. $$

Set $\tau=\min\{\tau_{1},\tau_{2},\tau_{3},1/8\}$, we get

$$\begin{aligned}& P(Y_{i}\leq x,Y_{j}\leq y) \\& \quad =P \biggl(Y_{i}\leq x,\frac{S_{j,j}}{\sqrt{10j/3}}+R_{j}\leq y \biggr) \\& \quad =P \biggl(Y_{i}\leq x,\frac{S_{i,i}}{\sqrt{10j/3}}+\sqrt{1-\rho} \frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} +\sqrt{1-\rho}\frac{c_{i+1,j}\widetilde{S}_{i}}{\sqrt {10(j-i)/3}}+R_{j}\leq y \biggr) \\& \quad \geq P \biggl(Y_{i}\leq x,\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} \leq y \biggr) \\& \qquad {} -P \biggl(y-3{\rho}^{1/8}\leq\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} \leq y \biggr)-P \biggl( \biggl\vert \frac{S_{i,i}}{\sqrt{10j/3}} \biggr\vert \geq\rho ^{1/8} \biggr) \\& \qquad {} -P \biggl( \biggl\vert \sqrt{1-\rho}\frac{c_{i+1,j}\widetilde{S}_{i}}{\sqrt {10(j-i)/3}} \biggr\vert \geq\rho^{1/8} \biggr) -P \bigl(\vert R_{j}\vert \geq \rho^{1/8} \bigr) \\& \quad \geq P \biggl(Y_{i}\leq x,\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} \leq y \biggr)-\rho^{\tau} \\& \quad = P(Y_{i}\leq x)P \biggl(\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} \leq y \biggr)-\rho^{\tau}. \end{aligned}$$

We can get a similar upper estimate for $P(Y_{i}\leq x,Y_{j}\leq y)$ in the same way. Thus there exists some constant M such that

$$P(Y_{i}\leq x,Y_{j}\leq y)=P(Y_{i}\leq x)P \biggl(\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} \leq y \biggr)+M\rho^{\tau}. $$

A similar argument,

$$P(Y_{i}\leq x)P(Y_{j}\leq y)=p(Y_{i}\leq x)P \biggl(\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} \leq y \biggr)+M' \rho^{\tau}, $$

holds for some constant $M'$. Thus we prove that (2.3) holds.

Let $G_{i,j}(x,y)$ be the joint distribution function of $Y_{i}$ and $Y_{j}$. By (2.2) and (2.3), for $2^{k}< i\leq2^{k+1}$, $2^{l}< j\leq 2^{l+1}$, $l-k\geq3$, $l\geq l_{0}$, we can get

$$\begin{aligned}& \biggl\vert \operatorname{Cov} \biggl(f(Y_{i})I \biggl\{ f(Y_{i})\leq\frac{k}{(\log k)^{\beta}} \biggr\} ,f(Y_{j})I \biggl\{ f(Y_{j})\leq\frac{l}{(\log l)^{\beta}} \biggr\} \biggr) \biggr\vert \\& \quad = \biggl\vert \int_{|x|\leq a_{k}} \int_{|y|\leq a_{l}}f(x)f(y)\,\mathrm {d} \bigl(G_{i,j}(x,y)-G_{i}(x)G_{j}(y) \bigr) \biggr\vert \\& \quad \ll\frac{kl}{(\log k)^{\beta}(\log l)^{\beta}} \biggl(\frac {i}{j} \biggr)^{\tau} \ll \frac{kl}{(\log k)^{\beta}(\log l)^{\beta }}2^{-(l-k-1)\tau}. \end{aligned}$$

Thus we have

$$\bigl\vert \operatorname{Cov} \bigl(Z_{k}^{*},Z_{l}^{*} \bigr) \bigr\vert \ll\frac{kl}{(\log k)^{\beta }(\log l)^{\beta}}2^{-(l-k)\tau}. $$

We complete the proof of Lemma 2.5. □

Lemma 2.6

Under the conditions of Theorem  1.1, denoting $\eta _{k}=Z_{k}^{*}-\mathbb{E}Z_{k}^{*}$, we have

$$\mathbb{E} \Biggl(\sum_{k=1}^{N} \eta_{k} \Biggr)^{2}=O \biggl(\frac{N^{2}}{(\log N)^{2\beta-1}} \biggr). $$

Proof

It follows from Lemma 2.4 and Lemma 2.5 that Lemma 2.6 also holds true. The proof is similar to that of Lemma 4 of Berkes et al. [5]. So we omit it here. □

3 Proof of theorem

By Lemma 2.6, we have

$$\mathbb{E} \Biggl(\frac{1}{N}\sum_{k=1}^{N} \eta_{k} \Biggr)^{2}=O \bigl((\log N)^{1-2\beta} \bigr). $$

Letting $N_{k}=[e^{k\lambda}]$, $(2\beta-1)^{-1}<\lambda<1$, we get

$$\mathbb{E} \Biggl(\frac{1}{N_{k}}\sum_{k=1}^{N_{k}} \eta_{k} \Biggr)^{2}< \infty, $$

which implies

$$ \lim_{k \to\infty}\frac{1}{N_{k}}\sum_{k=1}^{N_{k}}{ \eta}_{k}=0\quad \mbox{a.s.} $$

(3.1)

Note that for $2^{k}< i\leq2^{k+1}$,

$$\begin{aligned}& \mathbb{E}f(Y_{i})I \biggl\{ f(Y_{i})\leq\frac{k}{(\log k)^{\beta }} \biggr\} \\& \quad = \int_{|x|\leq a_{k}}f(x)\,\mathrm{d}G_{i}(x) = \int_{|x|\leq a_{k}}f(x)\,\mathrm{d}\Phi \biggl(\frac{x}{\sigma_{i}} \biggr)+ \int _{|x|\leq a_{k}}f(x)\,\mathrm{d} \biggl(G_{i}(x)-\Phi \biggl(\frac{x}{\sigma_{i}} \biggr) \biggr). \end{aligned}$$

(3.2)

Set $a=\int_{-\infty}^{\infty}f(x)\,\mathrm{d}\Phi(x)$. Noting that $\sigma_{i}\leq1$, $\lim_{i \to\infty}\sigma_{i}=1$, we have

$$ \lim_{k \to\infty}\sup_{2^{k}< i\leq2^{k+1}} \biggl\vert \int_{|x|\leq a_{k}}f(x)\,\mathrm{d}\Phi \biggl(\frac{x}{\sigma_{i}} \biggr)-a \biggr\vert =0. $$

(3.3)

Then by (3.2), (3.3), and (2.2) we get

$$\begin{aligned}& \biggl\vert \mathbb{E}f(Y_{i})I \biggl\{ f(Y_{i})\leq \frac{k}{(\log k)^{\beta}} \biggr\} -a \biggr\vert \\& \quad \leq \biggl\vert \int_{|x|\leq a_{k}}f(x)\,\mathrm{d}\Phi \biggl(\frac{x}{\sigma_{i}} \biggr)-a \biggr\vert + \biggl\vert \int _{|x|\leq a_{k}}f(x)\,\mathrm{d} \biggl(G_{i}(x)-\Phi \biggl(\frac{x}{\sigma _{i}} \biggr) \biggr) \biggr\vert \\& \quad \leq o_{k}(1)+\frac{k\theta_{i}}{(\log k)^{\beta}}. \end{aligned}$$

Thus

$$\mathbb{E}Z_{k}^{*}=a\sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{i}+\zeta_{k}\frac {k}{(\log k)^{\beta}}\sum _{i=2^{k}+1}^{2^{k+1}} \frac{\theta_{i}}{i}+o_{k}(1), \quad |\zeta_{k}|\leq1. $$

Using $\sum_{i=1}^{L}1/i=\log L+O(1)$ and $\sum_{i=1}^{\infty }\frac{\theta_{i}}{i}<\infty$, we get

$$\begin{aligned} \biggl\vert \frac{\mathbb{E}(\sum_{k=1}^{N}Z_{k}^{*})}{\log 2^{N+1}}-a \biggr\vert &\ll\frac{1}{N}\sum _{k=1}^{N}\frac{k}{(\log k)^{\beta}}\sum _{i=2^{k}+1}^{2^{k+1}} \frac{\theta_{i}}{i}+o_{N}(1) \\ &=O \bigl((\log N)^{-\beta} \bigr)+o_{N}(1) \\ &=o_{N}(1). \end{aligned}$$

Thus by (3.1), we get

$$\lim_{k \to\infty}\frac{\sum_{k=1}^{N_{k}}Z_{k}^{*}}{\log 2^{N_{k}+1}}=a\quad \mbox{a.s.} $$

Then by Lemma 2.3, we have

$$ \lim_{k \to\infty}\frac{\sum_{k=1}^{N_{k}}Z_{k}}{\log 2^{N_{k}+1}}=a\quad \mbox{a.s.} $$

(3.4)

The relation $\lambda<1$ implies $\lim_{k \to\infty}N_{k+1}/N_{k}=1$, thus (3.4) and the positivity of the $Z_{k}$ yield

$$\lim_{N \to\infty}\frac{\sum_{k=1}^{N}Z_{k}}{\log 2^{N+1}}=a\quad \mbox{a.s.}, $$

i.e. (1.6) holds for the subsequence $N=2^{k}$. Using again the positivity of the terms, we get (1.6). We complete the proof of Theorem 1.1.