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
Here and in the sequel, \(\mathcal{N}\)
is a standard normal random variable and
\(\stackrel{d}{\rightarrow}\)
denotes the convergence in distribution.
$$ \biggl(\frac{\prod_{k=1}^{n} S_{k}}{n!\mu^{n}} \biggr)^{\frac{1}{\gamma \sqrt{n}}}\stackrel{d}{\rightarrow} {e}^{\sqrt{2}\mathcal{N}} \quad \textit{as } n\to\infty. $$
(1.1)
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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
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.
$$ \lim_{N \to\infty}\frac{1}{\log N}\sum_{n=1}^{N} \frac {1}{n}I \biggl\{ \biggl(\frac{\prod_{k=1}^{n} S_{k}}{n!\mu^{n}} \biggr)^{\frac {1}{\gamma\sqrt{n}}}\leq x \biggr\} =F(x) \quad \textit{a.s. for any } x \in\mathbb{R}, $$
(1.2)
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
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.
$$ \lim_{N \to\infty}\frac{1}{\log N}\sum_{n=1}^{N} \frac {1}{n}g \biggl\{ \biggl(\frac{\prod_{k=1}^{n} S_{k}}{n!\mu^{n}} \biggr)^{\frac {1}{\gamma\sqrt{n}}} \biggr\} = \int_{0}^{\infty}g(x)\,\mathrm{d}F(x) \quad \textit{a.s. for any } x \in\mathbb{R}, $$
(1.3)
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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
where
\(F(\cdot)\)
is the distribution function of the random variable
\(e^{\sqrt{10/3}\mathcal{N}}\).
$$ \lim_{n \to\infty}\frac{1}{\log n}\sum_{k=1}^{n} \frac {1}{k}I \biggl\{ \biggl(\frac{2^{k}\prod_{j=1}^{k} T_{j}}{k!(k+1)!\mu^{k}} \biggr)^{\frac{1}{\gamma\sqrt{k}}}\leq x \biggr\} =F(x) \quad \textit{a.s. for any } x \in\mathbb{R}, $$
(1.4)
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
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.
$$\begin{aligned}& \lim_{N \to\infty}\frac{1}{\log N}\sum_{n=1}^{N} \frac {1}{n}g \biggl( \biggl(\frac{2^{n}\prod_{k=1}^{n} T_{k}}{n!(n+1)!{\mu }^{n}} \biggr)^{\frac{1}{\gamma\sqrt{n}}} \biggr) \\& \quad = \int_{0}^{\infty }g(x)\,\mathrm{d}F(x) \quad \textit{a.s. for any } x \in\mathbb{R}, \end{aligned}$$
(1.5)
Remark 1
Let \(f(x)=g(e^{\sqrt{10/3} x})\), \(t=e^{\sqrt{10/3} x}\). Then 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
$$\begin{aligned}& x=\sqrt{\frac{3}{10}}\log t, \qquad g(t)=f \biggl(\sqrt {\frac{3}{10}} \log t \biggr), \\& g \biggl( \biggl(\frac{2^{n}\prod_{k=1}^{n} T_{k}}{n!(n+1)!{\mu}^{n}} \biggr)^{\frac{1}{\gamma\sqrt{n}}} \biggr) =f \Biggl( \sqrt{\frac{3}{10}}\log \Biggl(\prod_{k=1}^{n} \frac {2T_{k}}{k(k+1)\mu} \Biggr)^{\frac{1}{\gamma\sqrt{n}}} \Biggr) \\& \hphantom{g \biggl( \biggl(\frac{2^{n}\prod_{k=1}^{n} T_{k}}{n!(n+1)!{\mu}^{n}} \biggr)^{\frac{1}{\gamma\sqrt{n}}} \biggr)}=f \Biggl(\frac{1}{\gamma\sqrt{10n/3}}\sum _{k=1}^{n} \log \frac {T_{k}}{k(k+1)\mu/2} \Biggr). \end{aligned}$$
$$\begin{aligned}& \lim_{N \to\infty}\frac{1}{\log N}\sum_{n=1}^{N} \frac {1}{n}f \Biggl(\frac{1}{\gamma\sqrt{10n/3}}\sum_{k=1}^{n} \log \frac {T_{k}}{k(k+1)\mu/2} \Biggr) \\& \quad = \int_{-\infty}^{\infty}f(x)\phi(x)\,\mathrm {d}x\quad \mbox{a.s. for any } x \in\mathbb{R}. \end{aligned}$$
(1.6)
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 Let Note that
$$c_{i,n}=2(b_{i,n}-d_{i,n}), \qquad \sum _{i=1}^{n} c_{i,n}^{2} \sim \frac{10n}{3}. $$
$$Y_{i}=\frac{1}{\gamma\sqrt{10i/3}}\sum_{k=1}^{i} \log \frac {T_{k}}{k(k+1)\mu/2}. $$
$$\begin{aligned}& \frac{1}{\gamma}\sum_{k=1}^{i} \biggl( \frac{T_{k}}{k(k+1)\mu/2}-1 \biggr) \\& \quad =\frac{1}{\gamma}\sum_{k=1}^{i} \biggl(\frac{2\sum_{j=1}^{k} S_{j}-k(k+1)\mu }{k(k+1)\mu} \biggr) \\& \quad =\frac{1}{\gamma}\sum_{k=1}^{i} \frac{2}{k(k+1)\mu}\sum_{j=1}^{k}\sum _{l=1}^{j}(X_{l}-\mu) \\& \quad =\frac{1}{\gamma}\sum_{k=1}^{i} \frac{2}{k(k+1)\mu}\sum_{l=1}^{k}\sum _{j=l}^{k}(X_{l}-\mu) \\& \quad =\frac{\mu}{\sigma}\sum_{k=1}^{i} \frac{2}{k(k+1)\mu}\sum_{l=1}^{k}(k+1-l) (X_{l}-\mu) \\& \quad =\sum_{k=1}^{i}\sum _{l=1}^{k} \frac{2(k+1-l)}{k(k+1)}\frac{X_{l}-\mu }{\sigma} \\& \quad =\sum_{l=1}^{i} \sum _{k=l}^{i}\frac{2(k+1-l)}{k(k+1)}\widetilde{X}_{l} \\& \quad =\sum_{l=1}^{i} c_{l,i} \widetilde{X}_{l}=S_{i,i}. \end{aligned}$$
By the fact that \(\log (1+x)=x+\frac{\delta}{2}x^{2}\), where \(|x|<1\), \(\delta\in(-1,0)\), thus we have By the fact that \(\mathbb{E}|X_{1}|^{2}<\infty\), using the Marcinkiewicz-Zygmund strong large number law, we have Thus
$$\begin{aligned} Y_{i}&=\frac{1}{\gamma\sqrt{10i/3}}\sum_{k=1}^{i} \log \frac {T_{k}}{k(k+1)\mu/2} \\ &=\frac{1}{\gamma\sqrt{10i/3}}\sum_{k=1}^{i} \biggl( \frac{T_{k}}{k(k+1)\mu /2}-1 \biggr)+\frac{1}{\gamma\sqrt{10i/3}} \sum_{k=1}^{i} \frac{\delta_{k}}{2} \biggl(\frac{T_{k}}{k(k+1)\mu/2}-1 \biggr)^{2} \\ &=\frac{1}{\sqrt{10i/3}}S_{i,i}+\frac{1}{\gamma\sqrt{10i/3}}\sum _{k=1}^{i}\frac{\delta_{k}}{2} \biggl( \frac{T_{k}}{k(k+1)\mu/2}-1 \biggr)^{2} \\ &=:\frac{1}{\sqrt{10i/3}}S_{i,i}+R_{i}. \end{aligned}$$
$$\begin{aligned}& S_{k}-k\mu=o \bigl(k^{1/2} \bigr) \quad \mbox{a.s.}, \\& \biggl\vert \frac{T_{k}}{k(k+1)\mu/2}-1 \biggr\vert = \biggl\vert \frac{2\sum_{j=1}^{k} S_{j}-k(k+1)\mu}{k(k+1)\mu} \biggr\vert \\& \hphantom{\biggl\vert \frac{T_{k}}{k(k+1)\mu/2}-1 \biggr\vert }\leq\frac{2\vert \sum_{j=1}^{k} (S_{j}-j\mu)\vert }{k(k+1)\mu} \\& \hphantom{\biggl\vert \frac{T_{k}}{k(k+1)\mu/2}-1 \biggr\vert }\leq\frac{2\sum_{j=1}^{k} j^{1/2}}{k(k+1)\mu}\ll\frac {k^{3/2}}{k^{2}}=\frac{1}{k^{1/2}}. \end{aligned}$$
$$ |R_{i}|\ll\frac{1}{\sqrt{i}}\sum_{k=1}^{i} \frac{1}{k}\ll\frac{\log i}{\sqrt{i}}\quad \mbox{a.s.} $$
(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
holds for
\(\lambda\sqrt{n}\geq L\).
$$\sup_{a}P \biggl(a\leq\frac{S_{n}}{\sqrt{n}}\leq a+\lambda \biggr) \leq c\lambda $$
Proof
See (20) on p.73 of Berkes et al. [5]. □
Let where \(1<\beta<(\alpha-3)/2\).
$$\begin{aligned}& Z_{k}=\sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{i}f(Y_{i}), \\& Z_{k}^{*}=\sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{i}f(Y_{i})I \biggl\{ f(Y_{i})\leq \frac{k}{(\log k)^{\beta}} \biggr\} , \end{aligned}$$
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 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 We complete the proof of Lemma 2.3. □
$$\begin{aligned} \bigl\{ Z_{k}\neq Z_{k}^{*} \bigr\} \subseteq& \bigl\{ |Y_{i}|\geq f^{-1} \bigl(k/(\log k)^{\beta} \bigr) \mbox{ for some } 2^{k}< i\leq2^{k+1} \bigr\} \\ =& \biggl\{ \biggl\vert \frac{1}{\sqrt{10i/3}}S_{i,i}+R_{i} \biggr\vert \geq f^{-1} \bigl(k/(\log k)^{\beta} \bigr)\geq \bigl(2\log k+(\alpha-2\beta)\log \log k \bigr)^{1/2} \\ &\mbox{for some } 2^{k}< i\leq2^{k+1} \biggr\} . \end{aligned}$$
$$\begin{aligned} \mathbb{P} \bigl(Z_{k}\neq Z_{k}^{*}, \mathrm{i.o.} \bigr) \leq& \mathbb{P} \biggl( \biggl\vert \frac {1}{\sqrt{10i/3}}S_{i,i} \biggr\vert \geq \bigl(2\log \log i+(\alpha-2\beta)\log \log \log i-O(1) \bigr)^{1/2}, \mathrm{i.o.} \biggr) \\ =&0. \end{aligned}$$
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 Obviously \(\sigma_{i}\leq1\), \(\lim_{i \to\infty}\sigma_{i}= 1\).
$$\begin{aligned}& \sigma_{i}^{2} = \int_{-\sqrt{i}}^{\sqrt{i}}x^{2}\,\mathrm{d}F(x)- \biggl( \int _{-\sqrt{i}}^{\sqrt{i}}x\,\mathrm{d}F(x) \biggr)^{2}, \\& \varepsilon_{i} =\sup_{x} \biggl\vert F_{i}(x)-\Phi \biggl(\frac{x}{\sigma_{i}} \biggr) \biggr\vert ,\qquad \theta_{i}=\sup_{x} \biggl\vert G_{i}(x)-\Phi \biggl(\frac{x}{\sigma_{i}} \biggr) \biggr\vert . \end{aligned}$$
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 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 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 By the same methods as that on p.72 of Berkes et al. [5], we get Now we estimate \(\theta_{i}\). By Lemma 2.1, for any \(\varepsilon>0\), we have By the Markov inequality and (2.1), we have By Lemma 2.2, we have By the Berry-Esseen inequality, we have Let \(\varepsilon=i^{-1/3}\), then Therefore, there exists \(\varepsilon_{0}>0\) such that By Theorem 1 of Friedman et al. [9], we have Hence By the fact that \((\alpha+1)/2>\beta\), we have We complete the proof of Lemma 2.4. □
$$ \biggl\vert \int_{-a}^{a}\Psi(x)\,\mathrm{d} \bigl(H_{1}(x)-H_{2}(x) \bigr) \biggr\vert \leq\sup _{-a\leq x \leq a} \bigl\vert \Psi(x) \bigr\vert \cdot\sup _{-a\leq x \leq a} \bigl\vert H_{1}(x)-H_{2}(x) \bigr\vert $$
(2.2)
$$\begin{aligned}& \mathbb{E}f^{2}(Y_{i})I \biggl\{ f(Y_{i})\leq \frac{k}{(\log k)^{\beta }} \biggr\} \\& \quad = \int_{|x|\leq a_{k}}f^{2}(x)\,\mathrm{d}G_{i}(x) \\& \quad \leq \int_{|x|\leq a_{k}}f^{2}(x)\,\mathrm{d}\Phi \biggl( \frac{x}{\sigma _{i}} \biggr)+\theta_{i}\frac{k^{2}}{(\log k)^{2\beta}} \\& \quad \ll \int_{|x|\leq a_{k}}f^{2}(x)\,\mathrm{d}\Phi(x)+ \theta_{i}\frac {k^{2}}{(\log k)^{2\beta}}; \end{aligned}$$
$$\begin{aligned} \mathbb{E} \bigl(Z_{k}^{*} \bigr)^{2} &\ll\mathbb{E} \Biggl( \Biggl(\sum_{i=2^{k}+1}^{2^{k+1}} \biggl( \frac{1}{i} \biggr)^{2} \Biggr)^{1/2} \Biggl(\sum _{i=2^{k}+1}^{2^{k+1}}f^{2}(Y_{i})I \biggl\{ f(Y_{i}) \leq\frac{k}{(\log k)^{\beta}} \biggr\} \Biggr)^{1/2} \Biggr)^{2} \\ &\ll \Biggl(\sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{i^{2}} \Biggr) \Biggl(\sum_{i=2^{k}+1}^{2^{k+1}} \biggl( \int_{|x|\leq a_{k}}f^{2}(x)\,\mathrm{d}\Phi (x)+ \theta_{i}\frac{k^{2}}{(\log k)^{2\beta}} \biggr) \Biggr) \\ &\ll\frac{1}{2^{k}} \Biggl(2^{k} \int_{|x|\leq a_{k}}f^{2}(x)\,\mathrm{d}\Phi (x)+ \frac{k^{2}}{(\log k)^{2\beta}}\sum_{i=2^{k}+1}^{2^{k+1}}\theta _{i} \Biggr) \\ &\ll \int_{|x|\leq a_{k}}\frac{e^{x^{2}/2}}{(1+|x|)^{2\alpha}}\,\mathrm {d}x+\frac{k^{2}}{(\log k)^{2\beta}} \sum_{i=2^{k}+1}^{2^{k+1}}\frac {\theta_{i}}{i}. \end{aligned}$$
$$\int_{|x|\leq a_{k}}\frac{e^{x^{2}/2}}{(1+|x|)^{2\alpha}}\,\mathrm{d}x\ll \frac{k}{(\log k)^{\beta+(\alpha+1)/2}}. $$
$$\begin{aligned} \theta_{i} =&\sup_{x} \biggl\vert G_{i}(x)-\Phi \biggl(\frac{x}{\sigma_{i}} \biggr) \biggr\vert \\ \leq&\sup_{x} \bigl\vert G_{i}(x)-F_{i}(x) \bigr\vert +\sup_{x} \biggl\vert F_{i}(x)-\Phi \biggl(\frac{x}{\sigma_{i}} \biggr) \biggr\vert \\ =&\sup_{x} \biggl\vert P(Y_{i}\leq x)-P \biggl( \frac{\widetilde{S}_{i}}{\sqrt {i}}\leq x \biggr) \biggr\vert +\varepsilon_{i} \\ \leq&\sup_{x} \biggl\vert P(Y_{i}\leq x)-P \biggl(\frac{S_{i,i}}{\sqrt {10i/3}}\leq x \biggr) \biggr\vert + \sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x \biggr)-P \biggl( \frac{\widetilde{S}_{i}}{\sqrt{i}}\leq x \biggr) \biggr\vert +\varepsilon _{i} \\ \leq&\sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}+R_{i} \leq x \biggr)-P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x+\varepsilon \biggr) \biggr\vert \\ &{} +\sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x+\varepsilon \biggr)-P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x \biggr) \biggr\vert \\ &{} +\sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x \biggr)-P \biggl(\frac{\widetilde{S}_{i}}{\sqrt{i}}\leq x \biggr) \biggr\vert + \varepsilon_{i} \\ \leq& P \bigl(\vert R_{i}\vert \geq\varepsilon \bigr)+\sup _{x} \biggl\vert P \biggl(\frac {S_{i,i}}{\sqrt{10i/3}}\leq x+ \varepsilon \biggr)-P \biggl(\frac {S_{i,i}}{\sqrt{10i/3}}\leq x \biggr) \biggr\vert \\ &{}+ \sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x \biggr)-P \biggl(\frac{\widetilde{S}_{i}}{\sqrt{i}}\leq x \biggr) \biggr\vert + \varepsilon_{i}. \end{aligned}$$
$$P \bigl(\vert R_{i}\vert \geq\varepsilon \bigr)\leq \frac{\mathbb{E}|R_{i}|}{\varepsilon}\ll \frac{\log i}{\sqrt{i}\varepsilon}. $$
$$\sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x+ \varepsilon \biggr)-P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x \biggr) \biggr\vert \ll \varepsilon. $$
$$\begin{aligned}& \sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}}\leq x \biggr)-P \biggl(\frac{\widetilde{S}_{i}}{\sqrt{i}}\leq x \biggr) \biggr\vert \\& \quad \leq\sup_{x} \biggl\vert P \biggl(\frac{S_{i,i}}{\sqrt{10i/3}} \leq x \biggr)-\Phi(x) \biggr\vert +\sup_{x} \biggl\vert P \biggl(\frac{\widetilde{S}_{i}}{\sqrt {i}}\leq x \biggr)-\Phi(x) \biggr\vert \\& \quad \ll\frac{1}{i^{1/2}}+\frac{1}{i^{1/2}}. \end{aligned}$$
$$\theta_{i}\ll\frac{\log i}{i^{1/6}}+ \frac{1}{i^{1/3}}+ \frac {1}{i^{1/2}}+\varepsilon_{i}. $$
$$\theta_{i}\ll\frac{1}{i^{\varepsilon_{0}}}+\varepsilon_{i}. $$
$$\sum_{i=1}^{\infty}\frac{\varepsilon_{i}}{i}< \infty. $$
$$\sum_{i=1}^{\infty}\frac{\theta_{i}}{i}\ll\sum _{i=1}^{\infty}\frac{\frac {1}{i^{\varepsilon_{0}}}+\varepsilon_{i}}{i}< \infty. $$
$$\sum_{k=1}^{N}\mathbb{E} \bigl(Z_{k}^{*} \bigr)^{2}\ll\sum_{k=1}^{N} \frac{k}{(\log k)^{\beta+(\alpha+1)/2}} +\sum_{k=1}^{N} \frac{k^{2}}{(\log k)^{2\beta}}\sum_{i=2^{k}+1}^{2^{k+1}} \frac{\theta_{i}}{i}\ll\frac{N^{2}}{(\log N)^{2\beta}}. $$
Lemma 2.5
Under the conditions of Theorem
1.1, for
\(l\geq l_{0}\), we have
where
τ
is a constant
\(0<\tau\leq1/8\).
$$\bigl\vert \operatorname{Cov} \bigl(Z_{k}^{*},Z_{l}^{*} \bigr) \bigr\vert \ll\frac{kl}{(\log k)^{\beta }(\log l)^{\beta}}2^{-(l-k)\tau}, $$
Proof
For \(1\leq i \leq j/2\), \(j\geq j_{0}\) and any x, y, we first prove Let \(\rho=\frac{i}{j}\). By the Chebyshev inequality, we have where \(\tau_{1}\) is a constant \(0<\tau_{1}\leq1/8\).
$$ \bigl\vert P(Y_{i}\leq x,Y_{j}\leq y)-P(Y_{i} \leq x)P(Y_{j}\leq y) \bigr\vert \ll \biggl(\frac {i}{j} \biggr)^{\tau}. $$
(2.3)
$$P \biggl( \biggl\vert \frac{S_{i,i}}{\sqrt{10j/3}} \biggr\vert \geq\rho^{1/8} \biggr) =P \biggl( \biggl\vert \frac{S_{i,i}}{\sqrt{10i/3}} \biggr\vert \geq\sqrt{ \frac {j}{i}} {\rho}^{1/8} \biggr)\leq\frac{i}{j} \rho^{-1/4}\leq{\rho }^{1/8}\leq\rho^{\tau_{1}}, $$
By the Markov inequality and (2.1), for \(j\geq j_{0}\), we have where \(\tau_{2}\) is a constant, \(0<\tau_{2}\leq1/8\).
$$P \bigl(\vert R_{j}\vert \geq\rho^{1/8} \bigr)\leq \frac{\mathbb{E}|R_{j}|}{\rho^{1/8}}\ll \frac{\log j}{j^{1/2}\rho^{1/8}} =\rho^{1/8}\frac{\log j}{j^{1/4}i^{1/4}}\ll \rho^{\tau_{2}}, $$
By the Markov inequality, we have where \(\tau_{3}\) is a constant that satisfies \(0<\tau_{3}\leq1/8\).
$$\begin{aligned}& P \biggl( \biggl\vert \sqrt{1-\rho}\frac{c_{i+1,j}\widetilde{S}_{i}}{\sqrt {10(j-i)/3}} \biggr\vert \geq \rho^{1/8} \biggr) \\& \quad =P \biggl( \biggl\vert \frac{\widetilde{S}_{i}}{\sqrt{i}} \biggr\vert \geq\sqrt{ \frac {10/3j}{i}}\frac{1}{c_{i+1,j}}{\rho}^{1/8} \biggr) \\& \quad \leq\frac{3}{10}{\rho}^{3/4}(c_{i+1,j})^{2}= \frac{3}{10}{\rho }^{3/4}(2b_{i+1,j}-2d_{i+1,j})^{2} \\& \quad =\frac{3}{10}{\rho}^{3/4} \Biggl[ \Biggl(2\sum _{k=i+1}^{j}\frac{1}{k} \Biggr)^{2}+4 \biggl(\frac {j+1-i-1}{j+1} \biggr)^{2}-8 \Biggl(\sum _{k=i+1}^{j}\frac{1}{k} \Biggr) \frac {j+1-i-1}{j+1} \Biggr] \\& \quad \ll{\rho}^{3/4} \biggl[ \biggl(\log \frac{j}{i} \biggr)^{2}+ \biggl(\frac {j-i}{j+1} \biggr)^{2}- \frac{j-i}{j+1}\log \frac{j}{i} \biggr] \\& \quad \ll\rho^{\tau_{3}}, \end{aligned}$$
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 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 A similar argument, holds for some constant \(M'\). Thus we prove that (2.3) holds.
$$\begin{aligned}& P(Y_{i}\leq x,Y_{j}\leq y) \\& \quad =P \biggl(Y_{i}\leq x,\frac{S_{j,j}}{\sqrt{10j/3}}+R_{j}\leq y \biggr) \\& \quad =P \biggl(Y_{i}\leq x,\frac{S_{i,i}}{\sqrt{10j/3}}+\sqrt{1-\rho} \frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} +\sqrt{1-\rho}\frac{c_{i+1,j}\widetilde{S}_{i}}{\sqrt {10(j-i)/3}}+R_{j}\leq y \biggr) \\& \quad \geq P \biggl(Y_{i}\leq x,\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} \leq y \biggr) \\& \qquad {} -P \biggl(y-3{\rho}^{1/8}\leq\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} \leq y \biggr)-P \biggl( \biggl\vert \frac{S_{i,i}}{\sqrt{10j/3}} \biggr\vert \geq\rho ^{1/8} \biggr) \\& \qquad {} -P \biggl( \biggl\vert \sqrt{1-\rho}\frac{c_{i+1,j}\widetilde{S}_{i}}{\sqrt {10(j-i)/3}} \biggr\vert \geq\rho^{1/8} \biggr) -P \bigl(\vert R_{j}\vert \geq \rho^{1/8} \bigr) \\& \quad \geq P \biggl(Y_{i}\leq x,\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} \leq y \biggr)-\rho^{\tau} \\& \quad = P(Y_{i}\leq x)P \biggl(\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} \leq y \biggr)-\rho^{\tau}. \end{aligned}$$
$$P(Y_{i}\leq x,Y_{j}\leq y)=P(Y_{i}\leq x)P \biggl(\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} \leq y \biggr)+M\rho^{\tau}. $$
$$P(Y_{i}\leq x)P(Y_{j}\leq y)=p(Y_{i}\leq x)P \biggl(\sqrt{1-\rho}\frac {S_{j,j}-S_{i,i}-c_{i+1,j}\widetilde{S}_{i}}{\sqrt{10(j-i)/3}} \leq y \biggr)+M' \rho^{\tau}, $$
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 Thus we have We complete the proof of Lemma 2.5. □
$$\begin{aligned}& \biggl\vert \operatorname{Cov} \biggl(f(Y_{i})I \biggl\{ f(Y_{i})\leq\frac{k}{(\log k)^{\beta}} \biggr\} ,f(Y_{j})I \biggl\{ f(Y_{j})\leq\frac{l}{(\log l)^{\beta}} \biggr\} \biggr) \biggr\vert \\& \quad = \biggl\vert \int_{|x|\leq a_{k}} \int_{|y|\leq a_{l}}f(x)f(y)\,\mathrm {d} \bigl(G_{i,j}(x,y)-G_{i}(x)G_{j}(y) \bigr) \biggr\vert \\& \quad \ll\frac{kl}{(\log k)^{\beta}(\log l)^{\beta}} \biggl(\frac {i}{j} \biggr)^{\tau} \ll \frac{kl}{(\log k)^{\beta}(\log l)^{\beta }}2^{-(l-k-1)\tau}. \end{aligned}$$
$$\bigl\vert \operatorname{Cov} \bigl(Z_{k}^{*},Z_{l}^{*} \bigr) \bigr\vert \ll\frac{kl}{(\log k)^{\beta }(\log l)^{\beta}}2^{-(l-k)\tau}. $$
Lemma 2.6
Under the conditions of Theorem
1.1, denoting
\(\eta _{k}=Z_{k}^{*}-\mathbb{E}Z_{k}^{*}\), we have
$$\mathbb{E} \Biggl(\sum_{k=1}^{N} \eta_{k} \Biggr)^{2}=O \biggl(\frac{N^{2}}{(\log N)^{2\beta-1}} \biggr). $$
3 Proof of theorem
By Lemma 2.6, we have Letting \(N_{k}=[e^{k\lambda}]\), \((2\beta-1)^{-1}<\lambda<1\), we get which implies Note that for \(2^{k}< i\leq2^{k+1}\), 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 Then by (3.2), (3.3), and (2.2) we get Thus Using \(\sum_{i=1}^{L}1/i=\log L+O(1)\) and \(\sum_{i=1}^{\infty }\frac{\theta_{i}}{i}<\infty\), we get Thus by (3.1), we get Then by Lemma 2.3, we have 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
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.
$$\mathbb{E} \Biggl(\frac{1}{N}\sum_{k=1}^{N} \eta_{k} \Biggr)^{2}=O \bigl((\log N)^{1-2\beta} \bigr). $$
$$\mathbb{E} \Biggl(\frac{1}{N_{k}}\sum_{k=1}^{N_{k}} \eta_{k} \Biggr)^{2}< \infty, $$
$$ \lim_{k \to\infty}\frac{1}{N_{k}}\sum_{k=1}^{N_{k}}{ \eta}_{k}=0\quad \mbox{a.s.} $$
(3.1)
$$\begin{aligned}& \mathbb{E}f(Y_{i})I \biggl\{ f(Y_{i})\leq\frac{k}{(\log k)^{\beta }} \biggr\} \\& \quad = \int_{|x|\leq a_{k}}f(x)\,\mathrm{d}G_{i}(x) = \int_{|x|\leq a_{k}}f(x)\,\mathrm{d}\Phi \biggl(\frac{x}{\sigma_{i}} \biggr)+ \int _{|x|\leq a_{k}}f(x)\,\mathrm{d} \biggl(G_{i}(x)-\Phi \biggl(\frac{x}{\sigma_{i}} \biggr) \biggr). \end{aligned}$$
(3.2)
$$ \lim_{k \to\infty}\sup_{2^{k}< i\leq2^{k+1}} \biggl\vert \int_{|x|\leq a_{k}}f(x)\,\mathrm{d}\Phi \biggl(\frac{x}{\sigma_{i}} \biggr)-a \biggr\vert =0. $$
(3.3)
$$\begin{aligned}& \biggl\vert \mathbb{E}f(Y_{i})I \biggl\{ f(Y_{i})\leq \frac{k}{(\log k)^{\beta}} \biggr\} -a \biggr\vert \\& \quad \leq \biggl\vert \int_{|x|\leq a_{k}}f(x)\,\mathrm{d}\Phi \biggl(\frac{x}{\sigma_{i}} \biggr)-a \biggr\vert + \biggl\vert \int _{|x|\leq a_{k}}f(x)\,\mathrm{d} \biggl(G_{i}(x)-\Phi \biggl(\frac{x}{\sigma _{i}} \biggr) \biggr) \biggr\vert \\& \quad \leq o_{k}(1)+\frac{k\theta_{i}}{(\log k)^{\beta}}. \end{aligned}$$
$$\mathbb{E}Z_{k}^{*}=a\sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{i}+\zeta_{k}\frac {k}{(\log k)^{\beta}}\sum _{i=2^{k}+1}^{2^{k+1}} \frac{\theta_{i}}{i}+o_{k}(1), \quad |\zeta_{k}|\leq1. $$
$$\begin{aligned} \biggl\vert \frac{\mathbb{E}(\sum_{k=1}^{N}Z_{k}^{*})}{\log 2^{N+1}}-a \biggr\vert &\ll\frac{1}{N}\sum _{k=1}^{N}\frac{k}{(\log k)^{\beta}}\sum _{i=2^{k}+1}^{2^{k+1}} \frac{\theta_{i}}{i}+o_{N}(1) \\ &=O \bigl((\log N)^{-\beta} \bigr)+o_{N}(1) \\ &=o_{N}(1). \end{aligned}$$
$$\lim_{k \to\infty}\frac{\sum_{k=1}^{N_{k}}Z_{k}^{*}}{\log 2^{N_{k}+1}}=a\quad \mbox{a.s.} $$
$$ \lim_{k \to\infty}\frac{\sum_{k=1}^{N_{k}}Z_{k}}{\log 2^{N_{k}+1}}=a\quad \mbox{a.s.} $$
(3.4)
$$\lim_{N \to\infty}\frac{\sum_{k=1}^{N}Z_{k}}{\log 2^{N+1}}=a\quad \mbox{a.s.}, $$
Acknowledgements
The authors would like to thank the editor and the referees for their very valuable comments by which the quality of the paper has been improved. This research is supported by the National Natural Science Foundation of China (71271042) and the Guangxi China Science Foundation (2013GXNSFAA278003). It is also supported by the Research Project of Guangxi High Institution (YB2014150).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
FF conceived of the study and drafted and completed the manuscript. DW participated in the discussion of the manuscript. FF and DW read and approved the final manuscript.