Suppose that $\{\xi_i, -\infty<i<\infty\}$ is a doubly infinite sequence of random variables, and $\{a_i, -\infty<i<\infty\}$ is an absolutely summable sequence of real numbers. Let

$ {\mathit{X}_\mathit{k}} = \sum\limits_{\mathit{i} = - \infty }^\infty {{\mathit{a}_{\mathit{i}{\rm{ + }}\mathit{k}}}} {\mathit{\xi }_\mathit{i}}, \;\;\;\mathit{k} \ge 1 $

(1.1)

be the moving average process based on $\{\xi_i, -\infty<i<\infty\}$. So far, there were detailed studies about the asymptotic behavior of the moving average process $\{X_k, k\geq1\}$.

Throughout the paper, let $N$ be the standard normal random variable. We denote by $C$ a positive constant which may vary from place to place, $\xrightarrow{d}$ means convergence in distribution, and $\lfloor x\rfloor=\sup\{m:m\leq x, m\in\mathbb{Z}^+\}$. Also we let $\log x=\ln(x\vee e)$ and $\log\log x=\ln(\ln(x\vee e^e))$.

We begin with a brief review of the definition of $\varphi$-mixing. Let $\mathbb{F}_k^l$ denote the $\sigma$-field generated by $X_k, X_{k+1}, \cdots, X_l$ and define

$ \begin{align*} &\varphi(\mathbb{F}_1^k, \mathbb{F}_{k+n}^\infty) :=\sup\{|\mathbb{P}(B|A)-\mathbb{P}(B)|; A\in\mathbb{F}_1^k, B\in\mathbb{F}_{k+n}^\infty\}, \\ &\varrho(\mathbb{F}_1^k, \mathbb{F}_{k+n}^\infty) :=\sup\{|\mathbb{C}orr(U, V)|; U\in\mathbb{L}^2(\mathbb{F}_1^k), V\in\mathbb{L}^2(\mathbb{F}_{k+n}^\infty)\}, \\ &\varphi(n):=\sup_{k\geq1}\varphi(\mathbb{F}_1^k, \mathbb{F}_{k+n}^\infty), \qquad\varrho(n):=\sup_{k\geq1}\varrho(\mathbb{F}_1^k, \mathbb{F}_{k+n}^\infty). \end{align*} $

A sequence $\{X_n\}_{n\geq1}$ of random variables is said to be $\varphi$-mixing if $\varphi(n)\rightarrow0$ and $\varrho$-mixing if $\varrho(n)\rightarrow0$. It is well known that a $\varphi$-mixing sequence is $\varrho$-mixing, since $\varrho(n)\leq 2\varphi^{1/2}(n)$.

In the sequel, we suppose $\{\xi_i, -\infty<i<\infty\}$ is a sequence of identically distributed and $\varphi$-mixing random variables with zero mean and finite variance with $0<\sigma^2=\mathbb{E}\xi_1^2+2\sum\limits_{k=2}^\infty\mathbb{E}\xi_1\xi_k<\infty$ and $\sum\limits_{m=1}^\infty\varphi^{1/2}(m)<\infty$. For the moving average processes $\{X_k, k\geq1\}$ defined in (1.1), where $\{a_i, -\infty<i<\infty\}$ is a sequence of real numbers with $\sum\limits_{i=-\infty}^\infty|a_i|<\infty$, we set $S_n=\sum\limits_{k=1}^nX_k$ and $\tau=\sigma\cdot\sum\limits_{i=-\infty}^\infty a_i$.

Li and Zhang [1] showed the precise rates in the law of the iterated logarithm of the moving average process defined in (1.1) for $\varphi$-mixing or negatively associated sequences under conditions above. For any $\delta\geq0$, if $\mathbb{E}\xi_1^2(\log\log|\xi_1|)^{\delta-1}<\infty$, they proved that

$ \begin{equation}\label{e1.3} _{\mathit{\varepsilon } \searrow 0}^{{\rm{lim}}}\varepsilon^{2\delta+2}\sum\limits_{n = 1}^\infty {} \frac{(\log\log n)^{\delta}}{n\log n}\mathbb{P}\{|S_n|\geq\varepsilon \tau\sqrt{2n\log\log n}\}=\frac{1}{(\delta+1)\sqrt{\pi}}\Gamma(\delta+3/2), \end{equation} $

(1.2)

where $\Gamma(\cdot)$ is a Gamma function.

In this paper, we consider the general law of complete convergence rates of the moving average process $\{X_k, k\geq1\}$ defined in (1.1) for $\varphi$-mixing sequences, and we have the following results.

Theorem 1.1  Suppose that $g(x)$ is a positive and differentiable function defined on $[n_0, \infty)$, which is strictly increasing to $\infty$. For $b>0$, assume that $\phi(x)=g'(x)g^{b-1}(x)$ is monotone nondecreasing or monotone nonincreasing on $[n_0, \infty)$, and if $\phi(x)$ is monotone nondecreasing, we assume that $\lim\limits_{x\rightarrow\infty}\phi(x+1)/\phi(x)=1$. If $\mathbb{E}|\xi_1|^{2+\delta}<\infty$ for some $\delta\geq0$, then we have

$ \begin{eqnarray}\label{hbue1.1} &\lim\limits_{\varepsilon\searrow0}\varepsilon^{\frac{b}{s}} \sum\limits_{n=n_0}^\infty \phi(n)\mathbb{P}\{|S_n|\geq(\varepsilon+a_n)\sqrt{n}g^{s}(n)\}=\frac{\tau^{\frac{b}{s}}\mathbb{E}|N|^{\frac{b}{s}}}{b}, ~~~0<\frac{b}{s}<2+\delta, \\ \label{bue1.1} \end{eqnarray} $

(1.3)

$ \begin{eqnarray} &\lim\limits_{\varepsilon\searrow0}\varepsilon^{\frac{b}{s}} \sum\limits_{n=n_0}^\infty \frac{\phi(n)}{\sqrt{n}}\mathbb{E}\{|S_n|-(\varepsilon+a_n)\sqrt{n}g^{s}(n)\}_+=\frac{s\tau^{\frac{b}{s}+1}\mathbb{E}|N|^{\frac{b}{s}+1}}{b(b+s)}, ~~~0<\frac{b}{s}<1+\delta, \end{eqnarray} $

(1.4)

where $a_n=o(g^{-s}(n))$ as $n\rightarrow\infty$.

Theorem 1.2  Suppose that $g(x)$ is a positive and differentiable function defined on $[n_0, \infty)$, which is strictly increasing to $\infty$. Assume that $\phi(x)=g'(x)g^{-1}(x)$ is monotone nondecreasing or monotone nonincreasing on $[n_0, \infty)$, and if $\phi(x)$ is monotone nondecreasing, we assume that $\lim\limits_{x\rightarrow\infty}\phi(x+1)/\phi(x)=1$. Then for $s>0$, we have

$ \begin{eqnarray}\label{hbue1.2} &&\lim\limits_{\varepsilon\searrow0}\frac{1}{-\log\varepsilon} \sum\limits_{n=n_0}^\infty \frac{g'(n)}{g(n)}\mathbb{P}\{|S_n|\geq(\varepsilon+a_n)\sqrt{n}g^{s}(n)\}=\frac{1}{s}, \\ \label{bue1.2} \end{eqnarray} $

(1.5)

$ \begin{eqnarray} &&\lim\limits_{\varepsilon\searrow0}\frac{1}{-\log\varepsilon} \sum\limits_{n=n_0}^\infty \frac{g'(n)}{\sqrt{n}g(n)}\mathbb{E}\{|S_n|-(\varepsilon+a_n)\sqrt{n}g^{s}(n)\}_+=\frac{\tau}{s}\sqrt{\frac{2}{\pi}}, \end{eqnarray} $

(1.6)

where $a_n=o(g^{-s}(n))$ as $n\rightarrow\infty$.

Remark 1.1  Applying Lemma 2.3 of [1], Theorems 1.1 and 1.2 are still true when $\{\xi_i, -\infty<i<\infty\}$ is a sequence of identically distributed negatively associated random variables with $\mathbb{E}\xi_1=0$, $\mathbb{E}\xi^2_1<\infty$ and $0<\sigma^2=\mathbb{E}\xi_1^2+2\sum\limits_{k=2}^\infty\mathbb{E}\xi_1\xi_k<\infty$.

Remark 1.2  Specially, for $k\geq1$, if we let $a_{2k}=1$ and $a_i=0, -\infty<i<\infty$ for $i\neq 2k$, that is to say, $X_k=\xi_k$ with $\mathbb{E}\xi_1=0$, $\mathbb{E}\xi^2_1<\infty$ and $0<\sigma^2=\mathbb{E}\xi_1^2+2\sum\limits_{k=2}^\infty\mathbb{E}\xi_1\xi_k<\infty$, then for $S_n=\sum\limits_{k=1}^n\xi_k$, Theorems 1.1 and 1.2 are still valid for $\tau=\sigma$.

Remark 1.3  The conditions about $\phi(x)$ and $g(x)$ in the above two theorems are mild for many common functions like $g(x)=x^\gamma$, $(\log x)^\gamma$ and $(\log\log x)^\gamma$ with $\gamma>0$, and the corresponding results were obtained by many researchers.

For proving of our main results, we introduce the following lemmas.

Lemma 2.1(see [2])  Let $\sum\limits_{i=-\infty}^\infty a_i$ be an absolutely convergent series of real numbers with $a=\sum\limits_{i=-\infty}^\infty a_i$ and $k\geq1$, then

$ _{\mathit{n} \to \infty }^{\;\;{\rm{lim}}}\frac{1}{n}\sum\limits_{i=-\infty}^\infty\Big|\sum\limits_{\mathit{j} = \mathit{i} + 1}^{\mathit{i} + \mathit{n}} {} a_j\Big|^k=a^k. $

Lemma 2.2  Suppose $\{\xi_i, -\infty<i<\infty\}$ is a sequence of identically distributed and $\varphi$-mixing random variables with $\mathbb{E}\xi_1=0$, $\mathbb{E}\xi^2_1<\infty$ and $\sum\limits_{m=1}^\infty\varphi^{1/2}(m)<\infty$, and suppose $0<\sigma^2=\mathbb{E}\xi_1^2+2\sum\limits_{k=2}^\infty\mathbb{E}\xi_1\xi_k<\infty$. For the moving average processes $\{X_k, k\geq1\}$ defined in (1.1) with $\sum\limits_{i=-\infty}^\infty|a_i|<\infty$, we set $S_n=\sum\limits_{k=1}^nX_k$. Then we have $\frac{S_n}{\tau\sqrt{n}}\xrightarrow{d}N, $ where $\tau=\sigma\cdot\sum\limits_{i=-\infty}^\infty a_i.$

Proof  The proof is similar to that of Theorem 1 in [3], so we omit it.

Lemma 2.3(see [4])  Let $\{\xi_i; i\geq 1\}$ be a $\varphi$-mixing sequence. $Y_n=\sum\limits_{i=1}^n\xi_i$, $n\geq1$. Suppose that there exists a sequence $\{C_n\}$ of positive numbers such that $\rm{ma}{{\rm{x}}_{1\le i\le \mathit{n}}}\mathbb{E}\mathit{Y}_{\mathit{i}}^{2}\le {{\mathit{C}}_{\mathit{n}}}$. Then for any $q\geq 2$, there exists some constant $C=C(q, \varphi(\cdot))$ such that

$ \mathbb{E}(_{_{1 \le \mathit{i} \le \mathit{n}}}^{{\rm{max}}}|Y_i|^q)\leq C\Big(C_n^{q/2}+\mathbb{E}_{_{1 \le \mathit{i} \le \mathit{n}}}^{{\rm{max}}}|\xi_i|^q\Big). $

In this section, for $M>1$ and $0<\varepsilon<1$, we define

$ \begin{equation}\label{e2.1} b_M(\varepsilon)=\lfloor g^{-1}(M\varepsilon^{-1/s})\rfloor. \end{equation} $

(3.1)

Without loss of generality, we assume $\tau=1$. Next we calculate the left hand side of (1.3) and (1.4) by approximation of partial sums about the tail probability of standard normal random variable $N$.

Proposition 3.1  For $b, s>0$, we have

$ \begin{equation}\label{e2.2} _{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \sum\limits_{\mathit{n} = {\mathit{n}_0}}^\infty {} \phi(n)\mathbb{P}\{|N|\geq(\varepsilon+a_n) g^{s}(n)\}=\frac{\mathbb{E}|N|^{b/s}}{b}. \end{equation} $

(3.2)

Proof  Since $\lim\limits_{n\rightarrow\infty}a_ng^{s}(n)=0$, then for any $\tilde{\delta}>0$ there exists a positive integer $N_0>n_0$ such that for any $n\geq N_0$, we have $ -\tilde{\delta} \leq a_ng^{s}(n)\leq \tilde{\delta}$. If $\phi(x)=g'(x)g^{b-1}(x)$ is monotone nonincreasing, we have

$ \begin{align*} _{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \int_{N_0+1}^\infty \phi(x)\mathbb{P}\{|N|\geq \varepsilon g^{s}(x)+\tilde{\delta}\}dx\leq&_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \sum_{n=N_0+1}^\infty \phi(n)\mathbb{P}\{|N|\geq(\varepsilon+a_n)g^{s}(n)\}\\ \leq&_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \int_{N_0}^\infty \phi(x)\mathbb{P}\{|N|\geq \varepsilon g^{s}(x)-\tilde{\delta}\}dx. \end{align*} $

Let $\tilde{\delta}\downarrow0$, we obtain

$ \begin{equation}\label{equ2.1} _{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \sum\limits_{\mathit{n} = {\mathit{n}_0}}^\infty {} \phi(n)\mathbb{P}\{|N|\geq(\varepsilon+a_n) g^{s}(n)\}=_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \int_{n_0}^\infty \phi(x)\mathbb{P}\{|N|\geq \varepsilon g^{s}(x)\}dx. \end{equation} $

(3.3)

If $\phi(x)$ is monotone nondecreasing, by $\lim\limits_{x\rightarrow\infty}\phi(x+1)/\phi(x)=1$, for the $\tilde{\delta}$ mentioned above, there exists a positive integer $N_1$ such that for any $x\geq N_1$, we have $(1-\tilde{\delta})\phi(x+1)\leq\phi(x)\leq(1+\tilde{\delta})\phi(x-1)$. Let $N_2=\max\{N_0, N_1\}$, thus it holds that

$ \begin{align*} &_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s}\sum\limits_{\mathit{n} = {\mathit{n}_0}}^\infty {} \phi(n)\mathbb{P}\{|N|\geq(\varepsilon+a_n)g^{s}(n)\}\\ =&_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s}\sum_{n=N_2+1}^\infty \phi(n)\mathbb{P}\{|N|\geq(\varepsilon+a_n)g^{s}(n)\}\\ \leq&(1+\tilde{\delta})_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s}\sum_{n=N_2+1}^\infty \phi(n-1)\mathbb{P}\{|N|\geq\varepsilon g^{s}(n)-\tilde{\delta}\}\\ \leq&(1+\tilde{\delta})_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s}\int_{N_2}^\infty \phi(x)\mathbb{P}\{|N|\geq \varepsilon g^{s}(x)-\tilde{\delta}\}dx, \end{align*} $

similarly, it holds that

$ \begin{align*} _{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s}\sum\limits_{\mathit{n} = {\mathit{n}_0}}^\infty {} \phi(n)\mathbb{P}\{|N|\geq(\varepsilon+a_n) g^{s}(n)\} \geq&(1-\tilde{\delta})_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s}\int_{N_2}^\infty \phi(x)\mathbb{P}\{|N|\geq \varepsilon g^{s}(x)+\\ &\tilde{\delta}\}dx. \end{align*} $

Let $\tilde{\delta}\downarrow0$, we obtain (3.3).

Let $y=\varepsilon g^{s}(x)$, we have

$ \begin{align*} &_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \int_{n_0}^\infty g'(x)g^{b-1}(x)\mathbb{P}\{|N|\geq\varepsilon g^{s}(x)\}dx\\ =&\frac{1}{b}_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\int_{\varepsilon g^s(n_0)}^\infty\frac{b}{s}y^{\frac{b}{s}-1} \mathbb{P}\{|N|\geq y\}dy\\ =&\frac{1}{b}\int_{0}^\infty\frac{b}{s}y^{\frac{b}{s}-1} \mathbb{P}\{|N|\geq y\}dy=\frac{\mathbb{E}|N|^{b/s}}{b}. \end{align*} $

Thus proof of this proposition is completed.

Proposition 3.2  For $b, s>0$, we have

$ \begin{equation}\label{e2.3} _{_{\mathit{\varepsilon } \searrow 0}}^{{\rm{lim}}}\varepsilon^{b/s} \sum\limits_{\mathit{n} = {\mathit{n}_0}}^{{\mathit{b}_\mathit{M}}(\mathit{\varepsilon })} {} \phi(n)|\mathbb{P}\{|S_n|\geq(\varepsilon+a_n)\sqrt{n} g^{s}(n)\}-\mathbb{P}\{|N|\geq(\varepsilon+a_n)g^{s}(n)\}|=0. \end{equation} $

(3.4)

Proof  Let

$ \begin{equation}\label{defee1} \Delta_n=_{\mathit{x}\in \mathbb{R}}^{\ \sup }|\mathbb{P}(|S_n|\geq\sqrt{n} x)-\mathbb{P}(|N|\geq x)|, \end{equation} $

(3.5)

then $\Delta_n\rightarrow0$ as $n\rightarrow\infty$ from Lemma 2.2. Using the Toeplitz lemma [5], we have

$ \begin{align*} &_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \sum_{n=n_0}^{b_M(\varepsilon)} \phi(n)|\mathbb{P}\{|S_n|\geq(\varepsilon+a_n)\sqrt{n} g^{s}(n)\}-\mathbb{P}\{|N|\geq(\varepsilon+a_n)g^{s}(n)\}|\\ \leq&_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \sum_{n=n_0}^{b_M(\varepsilon)} g'(n)g^{b-1}(n)\Delta_n \leq _{\mathit{\varepsilon } \searrow 0}^{\;\lim }\frac{CM^b}{g( b_M(\varepsilon))^b} \sum_{n=n_0}^{b_M(\varepsilon)} g'(n)g^{b-1}(n)\Delta_n=0. \end{align*} $

Proposition 3.3  For $b, s>0$, we have

$ _{\mathit{M}\to \infty }^{\ \ \ \lim }\rm{lim}_{\mathit{\varepsilon }\searrow 0}^{\sup }{{\mathit{\varepsilon }}^{\mathit{b}/\mathit{s}}}\sum\limits_{\mathit{n}>{{\mathit{b}}_{\mathit{M}}}(\mathit{\varepsilon })}{\phi }(\mathit{n})\mathbb{P}\{|\mathit{N}|\ge (\mathit{\varepsilon }+{{\mathit{a}}_{\mathit{n}}}){{\mathit{g}}^{\mathit{s}}}(\mathit{n})\}=0. $

(3.6)

Proof  Since $a_n\rightarrow0$ as $n>b_M(\varepsilon)\rightarrow\infty$, it is enough to show

$ \begin{equation*} _{\mathit{M}\to \infty }^{\ \ \ \lim }\rm{lim}_{\mathit{\varepsilon }\searrow 0}^{\sup }\varepsilon^{\mathit{b}/\mathit{s}} \sum\limits_{\mathit{n} > {\mathit{b}_\mathit{M}}(\mathit{\varepsilon })} {} \phi(\mathit{n})\mathbb{P}\Big\{|\mathit{N}|\geq\frac{\varepsilon}{2} \mathit{g}^{\mathit{s}}(\mathit{n})\Big\}=0. \end{equation*} $

Let $y=\varepsilon g^{s}(x)/2$, we have

$ \begin{align*} &\lim_{M\rightarrow\infty}\lim\sup_{\varepsilon\searrow0}\varepsilon^{b/s} \sum_{n>b_M(\varepsilon)} g'(n)g^{b-1}(n)\mathbb{P}\Big\{|N|\geq\frac{\varepsilon}{2} g^{s}(n)\Big\}\\ \leq&C\lim_{M\rightarrow\infty}\lim\sup_{\varepsilon\searrow0}\varepsilon^{b/s} \int_{b_M(\varepsilon)}^\infty g'(x)g^{b-1}(x)\mathbb{P}\Big\{|N|\geq\frac{\varepsilon}{2} g^{s}(x)\Big\}dx\\ \leq&C\lim_{M\rightarrow\infty}\int_{M^s/2}^\infty\frac{b}{s}y^{\frac{b}{s}-1} \mathbb{P}\{|N|\geq y\}dy=0. \end{align*} $

Therefore this proposition is proved.

Proposition 3.4  If $\mathbb{E}|\xi_1|^{2+\delta}<\infty$ for some $\delta\geq0$, then for $0<b<(2+\delta)s$, we have

$ \begin{equation}\label{e2.5} _{\mathit{M}\to \infty }^{\ \ \ \lim }\rm{lim}_{\mathit{\varepsilon }\searrow 0}^{\sup }\varepsilon^{\mathit{b}/\mathit{s}}\sum\limits_{\mathit{n} > {\mathit{b}_\mathit{M}}(\mathit{\varepsilon })} {} \phi(\mathit{n})\mathbb{P}\{|\mathit{S}_\mathit{n}|\geq(\varepsilon+\mathit{a}_\mathit{n})\sqrt{\mathit{n}}\mathit{g}^{\mathit{s}}(\mathit{n})\}=0. \end{equation} $

(3.7)

Proof  It suffices to show that

$ \begin{equation}\label{buequ2.5} _{\mathit{M}\to \infty }^{\ \ \ \lim }\rm{lim}_{\mathit{\varepsilon }\searrow 0}^{\sup }\varepsilon^{\mathit{b}/\mathit{s}}\sum\limits_{\mathit{n} > {\mathit{b}_\mathit{M}}(\mathit{\varepsilon })} {} \mathit{g}'(\mathit{n})\mathit{g}^{\mathit{b}-1}(\mathit{n})\mathbb{P}\Big\{|\mathit{S}_\mathit{n}|\geq\frac{\varepsilon}{2}\sqrt{\mathit{n}}\mathit{g}^{\mathit{s}}(\mathit{n})\Big\}=0. \end{equation} $

(3.8)

Note that $S_n=\sum\limits_{i=-\infty}^\infty\sum\limits_{k=1}^n a_{k+i}\xi_i=\sum\limits_{i=-\infty}^\infty a_{ni}\xi_i$, where $a_{ni}=\sum\limits_{k=1}^n a_{k+i}$. From Lemma 2.1, we can suppose that

$ \begin{equation}\label{ai} \sum\limits_{\mathit{i} = - \infty }^\infty {} |a_{ni}|^p\leq n, \quad p\geq1, \quad \tilde{a}=\sum\limits_{\mathit{i} = - \infty }^\infty {}|a_{i}|\leq1. \end{equation} $

(3.9)

Next, for $x\geq0$, we set

$ \begin{equation}\label{def1} S_n'(x)=\sum\limits_{\mathit{i} = - \infty }^\infty {} a_{ni}\xi_iI\{|a_{ni}\xi_i|\leq (\varepsilon+x)\sqrt{n}g^{s}(n)\}. \end{equation} $

(3.10)

Since $\mathbb{E}\xi_i=0$, then by (3.9) and Markov's inequality, we have

$ \begin{align}\label{def2} \nonumber|\mathbb{E}S_n'(x)|=&\Big|\sum_{i=-\infty}^\infty\mathbb{E}a_{ni}\xi_iI\{|a_{ni}\xi_i|> (\varepsilon+x)\sqrt{n}g^{s}(n)\}\Big|\\ \nonumber\leq&\sum_{i=-\infty}^\infty|a_{ni}|\mathbb{E}|\xi_i|I\{|a_{ni}\xi_i|> (\varepsilon+x)\sqrt{n}g^{s}(n)\}\\ \nonumber \leq& \sum_{i=-\infty}^\infty|a_{ni}|(\mathbb{E}\xi_i^2)^{1/2}\sqrt{\mathbb{P}\{|a_{ni}\xi_i|> (\varepsilon+x)\sqrt{n}g^{s}(n)\}}\\ \leq&Cn\sqrt{\mathbb{P}\{|\xi_1|> (\varepsilon+x)\sqrt{n}g^{s}(n)\}} \leq\frac{C\sqrt{n}}{(\varepsilon+x) g^{s}(n)}. \end{align} $

(3.11)

Since

$ \begin{equation}\label{def3} |S_n|\leq|\mathbb{E}S_n'(x)|+|S_n'(x)-\mathbb{E}S_n'(x)|+\Big|\sum\limits_{\mathit{i} = - \infty }^\infty {} a_{ni}\xi_iI\{|a_{ni}\xi_i|>(\varepsilon+x)\sqrt{n}g^{s}(n)\}\Big|, \end{equation} $

(3.12)

and for $M$ large enough, it holds that

$ \begin{equation*} \frac{|\mathbb{E}S_n'(0)|}{\varepsilon\sqrt{n}g^{s}(n)} \leq\frac{C}{\varepsilon^2g^{2s}(n)}\leq CM^{-2s}<\varepsilon, \quad n>b_M(\varepsilon), \end{equation*} $

then we obtain

$ \begin{align}\label{e3.8} &\mathbb{P}\Big\{|S_n|\geq\frac{\varepsilon}{2}\sqrt{n}g^{s}(n)\Big\} \leq\mathbb{P}\{\sup_{i}|a_{ni}\xi_i|>\varepsilon\sqrt{n}g^{s}(n)\}\\ &+\mathbb{P}\Big\{|S_n'(0)-\mathbb{E}S_n'(0)|\geq\frac{\varepsilon}{4}\sqrt{n}g^{s}(n)\Big\}. \end{align} $

(3.13)

Set

$ I_{nj}=\{j\in\mathbb{Z}, 1/(j+1)<|a_{ni}|\leq 1/j, j=1, 2, \cdots\}, $

then $\cup_{j\geq1}I_{nj}=\mathbb{Z}$, where $\mathbb{Z}$ is the set of all integers. Note that (referred by [6])

$ \begin{equation}\label{inj} \sum\limits_{\mathit{j} = 1}^\mathit{k} {} \#I_{nj}\leq n(k+1). \end{equation} $

(3.14)

On the one hand, we have

$ \begin{align}\label{e3.9} \nonumber &\mathbb{P}\{\sup_{i}|a_{ni}\xi_i|>(\varepsilon+x)\sqrt{n}g^{s}(n)\}\leq\sum_{i=-\infty}^\infty\mathbb{P}\{|a_{ni}\xi_i|\geq(\varepsilon+x)\sqrt{n}g^{s}(n)\}\\ \nonumber\leq&\sum_{j=1}^\infty\sum_{i\in I_{nj}}\mathbb{P}\{|\xi_1|\geq j(\varepsilon+x)\sqrt{n}g^{s}(n)\} \leq\sum_{j=1}^\infty(\# I_{nj})\mathbb{P}\{|\xi_1|\geq j(\varepsilon+x)\sqrt{n}g^{s}(n)\}\\ \nonumber\leq&\sum_{j=1}^\infty\sum_{k=j}^\infty(\# I_{nj})\mathbb{P}\bigg\{k\leq\frac{|\xi_1|}{(\varepsilon+x)\sqrt{n}g^{s}(n)}<k+1\bigg\}\\ \nonumber=&\sum_{k=1}^\infty\sum_{j=1}^k(\# I_{nj})\mathbb{P}\bigg\{k\leq\frac{|\xi_1|}{(\varepsilon+x)\sqrt{n}g^{s}(n)}<k+1\bigg\}\\ \nonumber\leq&\sum_{k=1}^\infty n(k+1)\mathbb{P}\bigg\{k\leq\frac{|\xi_1|}{(\varepsilon+x)\sqrt{n}g^{s}(n)}<k+1\bigg\}\\ \nonumber\leq&\frac{2\sqrt{n}\mathbb{E}|\xi_1|I\{|\xi_1|\geq(\varepsilon+x)\sqrt{n}g^{s}(n)\}}{(\varepsilon+x) g^{s}(n)}\\ \nonumber\leq&\frac{2\mathbb{E}|\xi_1|^{2+\delta}I\{|\xi_1|\geq(\varepsilon+x)\sqrt{n}g^{s}(n)\}}{n^{\delta/2}(\varepsilon+x)^{2+\delta}g^{(2+\delta)s}(n)}\\ \leq&\frac{C}{(\varepsilon+x)^{2+\delta}g^{(2+\delta)s}(n)}. \end{align} $

(3.15)

On the other hand, since $\sum\limits_{m=1}^\infty\varphi^{1/2}(m)<\infty$, then we have

$ \begin{align*} \nonumber \mathbb{E}(S_n'(x)-\mathbb{E}S_n'(x))^2\leq& C\sum_{i=-\infty}^\infty\mathbb{E}(a_{ni}\xi_i)^2I\{|a_{ni}\xi_i|\leq(\varepsilon+x)\sqrt{n}g^{s}(n)\}\\ \leq&C\sum_{i=-\infty}^\infty(a_{ni})^2\mathbb{E}\xi_1^2\leq Cn. \end{align*} $

Thus using Markov's inequality and Lemma 2.3, we have

$ \begin{align}\label{e3.15} \nonumber &\mathbb{P}\Big\{|S_n'(x)-\mathbb{E}S_n'(x)|\geq\frac{(x+\varepsilon)}{4}\sqrt{n}g^{s}(n)\Big\}\\ \nonumber \leq&C\mathbb{E}\{|S_n'(x)-\mathbb{E}S_n'(x)|^{q}\}n^{-q/2}(x+\varepsilon)^{-q}g^{-qs}(n)\\ \nonumber\leq&\frac{Cn^{-q/2}}{(x+\varepsilon)^{q}g^{qs}(n)}\!\sum_{i=-\infty}^\infty\!\mathbb{E}|a_{ni}\xi_1|^{q}I\{|a_{ni}\xi_1|\leq (x+\varepsilon)\sqrt{n}g^{s}(n)\}\\ &+C(x+\varepsilon)^{-q}g^{-qs}(n)=:H_{1}(x)+H_{2}(x), \end{align} $

(3.16)

where we take $q=2+\delta$ (actually, the above inequality holds for any $q\geq2$).

However, from (3.14), it holds that

$ \begin{align*} \nonumber &\sum_{i=-\infty}^\infty\mathbb{E}|a_{ni}\xi_1|^{2+\delta}I\{|a_{ni}\xi_1|\leq (x+\varepsilon)\sqrt{n}g^{s}(n)\}\\ \nonumber \leq&\sum_{j=1}^\infty\sum_{i\in I_{nj}}j^{-(2+\delta)}\mathbb{E}|\xi_1|^{2+\delta}I\{|\xi_1|<(j+1)(x+\varepsilon)\sqrt{n}g^{s}(n)\}\\ \nonumber =&\sum_{j=1}^\infty(\# I_{nj})j^{-(2+\delta)}\mathbb{E}|\xi_1|^{2+\delta}I\{|\xi_1|<(j+1)(x+\varepsilon)\sqrt{n}g^{s}(n)\}\\ \nonumber =&\sum_{j=1}^\infty\sum_{k=0}^j(\# I_{nj})j^{-(2+\delta)}\mathbb{E}|\xi_1|^{2+\delta}I\bigg\{k\leq\frac{|\xi_1|}{(x+\varepsilon)\sqrt{n}g^{s}(n)}<k+1\bigg\}\\ \nonumber =&\sum_{j=1}^\infty(\# I_{nj})j^{-(2+\delta)}\mathbb{E}|\xi_1|^{2+\delta}I\bigg\{0\leq\frac{|\xi_1|}{(x+\varepsilon)\sqrt{n}g^{s}(n)}<1\bigg\}\\ &+\sum_{k=1}^\infty\sum_{j=k}^\infty(\# I_{nj})j^{-(2+\delta)}\mathbb{E}|\xi_1|^{2+\delta}I\bigg\{k\leq\frac{|\xi_1|}{(x+\varepsilon)\sqrt{n}g^{s}(n)}<k+1\bigg\}\\ \nonumber =:&L_1(x)+L_2(x). \end{align*} $

Since

$ \sum\limits_{\mathit{j} = \mathit{k}}^\infty {} \frac{\#I_{nj}}{(j+1)^{2+\delta}}(k+1)^{1+\delta}\leq\sum\limits_{\mathit{j} = \text{1}}^\infty {} \frac{\#I_{nj}}{j+1} \leq\sum\limits_{\mathit{i} = -\infty }^\infty {} |a_{ni}|=\sum\limits_{\mathit{j} = \text{1}}^\infty {} \sum\limits_{\mathit{i} \in {\mathit{I}_{\mathit{nj}}}} {} |a_{ni}|\leq n, $

then we have

$ \begin{equation}\label{knj} \sum\limits_{\mathit{j} = \mathit{k}}^\infty {} (\#I_{nj})j^{-(2+\delta)}\leq Cnk^{-(1+\delta)}. \end{equation} $

(3.17)

Therefore, using (3.17), we have

$ \begin{align*} L_1(x)\leq& Cn\mathbb{E}|\xi_1|^{2+\delta}I\{|\xi_1|<(x+\varepsilon)\sqrt{n}g^{s}(n)\}\leq Cn, \\ L_2(x)\leq&Cn\sum_{k=1}^\infty k^{-(1+\delta)}\mathbb{E}|\xi_1|^{2+\delta} I\bigg\{k\leq\frac{|\xi_1|}{(x+\varepsilon)\sqrt{n}g^{s}(n)}<k+1\bigg\}\\ \leq&Cn\sum_{k=1}^\infty\mathbb{E}|\xi_1|^{2+\delta} I\bigg\{k\leq\frac{|\xi_1|}{(x+\varepsilon)\sqrt{n}g^{s}(n)}<k+1\bigg\}\\ \leq&Cn\mathbb{E}|\xi_1|^{2+\delta} I\{|\xi_1|\geq(x+\varepsilon)\sqrt{n}g^{s}(n)\}\leq Cn. \end{align*} $

Thus we have

$ \begin{equation}\label{e3.10} H_1(x)\leq\frac{Cn^{-\delta/2}}{(\varepsilon+x)^{2+\delta}g^{(2+\delta)s}(n)}\leq\frac{C}{(\varepsilon+x)^{2+\delta}g^{(2+\delta)s}(n)}, \end{equation} $

(3.18)

then using (3.13), (3.15), (3.16) and (3.18) with $x=0$, it holds that

$ \begin{align*} &\lim_{M\rightarrow\infty}\lim\sup_{\varepsilon\searrow0}\varepsilon^{b/s}\sum_{n>b_M(\varepsilon)} g'(n)g^{b-1}(n)\mathbb{P}\Big\{|S_n|\geq\frac{\varepsilon}{2}\sqrt{n}g^{s}(n)\Big\}\\ \leq&\lim_{M\rightarrow\infty}\lim\sup_{\varepsilon\searrow0}C\varepsilon^{\frac{b}{s}-(2+\delta)}\sum_{n>b_M(\varepsilon)} g'(n)[g(n)]^{b-(2+\delta)s-1}\\ \leq&\lim_{M\rightarrow\infty}\lim\sup_{\varepsilon\searrow0}C\varepsilon^{\frac{b}{s}-(2+\delta)}[g(b_M(\varepsilon))]^{b-(2+\delta)s}\\ \leq&\lim_{M\rightarrow\infty}\lim\sup_{\varepsilon\searrow0}CM^{b-(2+\delta)s}=0. \end{align*} $

Therefore we complete the proof of (3.8).

From Propositions 3.1-3.4, applying the triangle inequality, we complete the proof of (1.3). Next we show (1.4). For simplicity, we let $a_n=0$ and omit the discussion of $\phi(x)$, but the process is similar to that of Proposition 3.1.

Proposition 3.5  For $b, s>0$, one has that

$ \begin{equation}\label{hbue2.1} _{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \sum\limits_{\mathit{n} = {\mathit{n}_0}}^\infty {} \phi(n)\mathbb{E}\{|N|-\varepsilon g^{s}(n)\}_+=\frac{s\mathbb{E}|N|^{b/s+1}}{b(b+s)}. \end{equation} $

(3.19)

Proof  We calculate that

$ \begin{align*} &_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \sum\limits_{\mathit{n} = {\mathit{n}_0}}^\infty {} g'(n)g^{b-1}(n)\mathbb{E}\{|N|- \varepsilon g^{s}(n)\}_+\\ =&_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \int_{n_0}^\infty g'(y)g^{b-1}(y)\mathbb{E}\{|N|- \varepsilon g^{s}(y)\}_+dy\\ =&_{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s} \int_{n_0}^\infty g'(y)g^{b-1}(y)dy\int_{\varepsilon g^{s}(y)}^\infty\mathbb{P}\{|N|\geq x\}dx \quad\text{(double integrals)}\\ =&\frac{1}{ s}_{\mathit{\varepsilon } \searrow 0}^{\;\lim } \int_{\varepsilon g^{s}(n_0)}^\infty t^{b/s-1}dt\int_t^\infty\mathbb{P}\{|N|\geq x\}dx\qquad(t=\varepsilon g^{s}(y))\\ =&\frac{1}{ s}_{\mathit{\varepsilon } \searrow 0}^{\;\lim } \int_{\varepsilon g^{s}(n_0)}^\infty\mathbb{P}\{|N|\geq x\}dx\int_{\varepsilon g^{s}(n_0)}^x t^{b/s-1}dt\\ =&\frac{1}{ b}_{\mathit{\varepsilon } \searrow 0}^{\;\lim } \int_{\varepsilon g^{s}(n_0)}^\infty x^{b/s}\mathbb{P}\{|N|\geq x\}dx=\frac{s\mathbb{E}|N|^{b/s+1}}{b(b+s)}. \end{align*} $

Thus proof of this proposition is completed.

Proposition 3.6  For $b, s>0$, one has that

$ \begin{equation}\label{hbue2.2} _{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s}\sum\limits_{\mathit{n} = {\mathit{n}_0}}^{{\mathit{b}_\mathit{M}}(\mathit{\varepsilon })} {} \phi(n)\bigg|\mathbb{E}\{|N|-\varepsilon g^{s}(n)\}_+-\mathbb{E}\bigg\{\frac{|S_n|}{\sqrt{n}}-\varepsilon g^{s}(n)\bigg\}_+\bigg|=0. \end{equation} $

(3.20)

Proof  It holds that

$ \begin{align*} &\sum_{n=n_0}^{b_M(\varepsilon)} \phi(n)\bigg|\mathbb{E}\{|N|-\varepsilon g^{s}(n)\}_+-\mathbb{E}\bigg\{\frac{|S_n|}{\sqrt{n}}-\varepsilon g^{s}(n)\bigg\}_+\bigg|\\ =&\sum_{n=n_0}^{b_M(\varepsilon)} \phi(n)\bigg|\int_0^{\infty}\mathbb{P}\big(|N|\geq \varepsilon g^{s}(n)+x\big)-\mathbb{P}\bigg(\frac{|S_n|}{\sqrt{n}}\geq \varepsilon g^{s}(n)+x\bigg)dx\bigg|\\ \leq&\sum_{n=n_0}^{b_M(\varepsilon)} \phi(n)g^{s}(n)\int_0^{\infty}\bigg|\mathbb{P}\big(|N|\geq (x+\varepsilon)g^{s}(n)\big)-\mathbb{P}\bigg(\frac{|S_n|}{\sqrt{n}}\geq (x+\varepsilon)g^{s}(n)\bigg)\bigg|dx\\ \leq&\sum_{n=n_0}^{b_M(\varepsilon)} \phi(n)g^{s}(n)\bigg[\int_0^{l(n)}\bigg|\mathbb{P}\big(|N|\geq (x+\varepsilon)g^{s}(n)\big)-\mathbb{P}\bigg(\frac{|S_n|}{\sqrt{n}}\geq (x+\varepsilon)g^{s}(n)\bigg)\bigg|dx\\ &+\int_{l(n)}^\infty\mathbb{P}\big(|N|\geq (x+\varepsilon)g^{s}(n)\big)dx+\int_{l(n)}^\infty\mathbb{P}\big(|S_n|\geq (x+\varepsilon)\sqrt{n}g^{s}(n)\big)dx\bigg]\\ =:&\sum_{n=n_0}^{b_M(\varepsilon)} \phi(n)g^{s}(n)\big[J_1+J_2+J_3\big], \end{align*} $

where $l(n)=g^{-s}(n)\Delta_n^{-1/2}$ and $\Delta_n$ is defined in (3.5). It is easy to see that

$ \begin{eqnarray}\label{e6.1} &&J_1\leq \frac{\Delta_n^{1/2}}{g^{s}(n)}, \nonumber\\ &&J_2\leq\int_{l(n)}^\infty\frac{\mathbb{E}N^2}{(x+\varepsilon)^2g^{2s}(n)}dx\leq \frac{C}{l(n)g^{2s}(n)}\leq \frac{C\Delta_n^{1/2}}{g^{s}(n)}. \end{eqnarray} $

(3.21)

Next for $J_3$, from (3.10)-(3.12) and the fact that for $x\geq l(n)$ and $n$ large enough, we have

$ \begin{equation*} \frac{|\mathbb{E}S_n'(x)|}{(x+\varepsilon)\sqrt{n}g^{s}(n)}\leq\frac{C}{(x+\varepsilon)^2g^{2s}(n)} \leq\frac{C}{l^2(n)g^{2s}(n)}\leq C\Delta_n<\varepsilon, \end{equation*} $

then we obtain

$ \begin{align}\label{e6.5} \nonumber \mathbb{P}\Big\{|S_n|\geq(x+\varepsilon)\sqrt{n}g^{s}(n)\Big\} \leq&\mathbb{P}\{\sup_{i}|a_{ni}\xi_i|>(x+\varepsilon)\sqrt{n}g^{s}(n)\}\\ &+\mathbb{P}\Big\{|S_n'(x)-\mathbb{E}S_n'(x)|\geq\frac{(x+\varepsilon)}{2}\sqrt{n}g^{s}(n)\Big\}. \end{align} $

(3.22)

Thus using (3.15), (3.16) and (3.18) with $\delta=0$, we have

$ J_3\leq\int_{l(n)}^\infty\frac{C}{(x+\varepsilon)^2g^{2s}(n)}dx\leq \frac{C}{l(n)g^{2s}(n)}\leq \frac{C\Delta_n^{1/2}}{g^{s}(n)}. $

Therefore using (3.21) and the Toeplitz lemma, we have

$ _{\mathit{\varepsilon } \searrow 0}^{\;\lim }\varepsilon^{b/s}\sum\limits_{\mathit{n} = {\mathit{n}_0}}^{{\mathit{b}_\mathit{M}}(\mathit{\varepsilon })} {} \phi(n)g^{s}(n)\big[J_1+J_2+J_3\big]\leq_{\mathit{\varepsilon } \searrow 0}^{\;\lim }C\varepsilon^{b/s}\sum\limits_{\mathit{n} = {\mathit{n}_0}}^{{\mathit{b}_\mathit{M}}(\mathit{\varepsilon })} {} g'(n)g^{b-1}(n)\Delta_n^{1/2}=0. $

Proposition 3.7  For $b, s>0$, one has that

$ _{\mathit{M}\to \infty }^{\ \ \lim }\rm{lim}_{\mathit{\varepsilon }\searrow 0}^{\sup }{{\mathit{\varepsilon }}^{\mathit{b}/\mathit{s}}}\sum\limits_{\mathit{n}>{{\mathit{b}}_{\mathit{M}}}(\mathit{\varepsilon })}{\phi }(\mathit{n})\mathbb{E}{{\{|\mathit{N}|-\mathit{\varepsilon }{{\mathit{g}}^{\mathit{s}}}(\mathit{n})\}}_{+}}=0. $

(3.23)

Proof  It is easy from the proof of Proposition 3.5.

Proposition 3.8  For $0<b<(1+\delta)s$, one has that

$ _{\mathit{M}\to \infty }^{\ \ \lim }\rm{lim}_{\mathit{\varepsilon }\searrow 0}^{\sup }{{\mathit{\varepsilon }}^{\mathit{b}/\mathit{s}}}\sum\limits_{\mathit{n}>{{\mathit{b}}_{\mathit{M}}}(\mathit{\varepsilon })}{\frac{\phi (\mathit{n})}{\sqrt{\mathit{n}}}}\mathbb{E}{{\{|{{\mathit{S}}_{\mathit{n}}}|-\mathit{\varepsilon }\sqrt{\mathit{n}}{{\mathit{g}}^{\mathit{s}}}(\mathit{n})\}}_{+}}=0. $

(3.24)

Proof  For $M$ large enough and any $x\geq0$, it holds that

$ \begin{equation*} \frac{|\mathbb{E}S_n'(x)|}{(\varepsilon+x)\sqrt{n}g^{s}(n)} \leq\frac{C}{(\varepsilon+x)^2g^{2s}(n)}\leq\frac{C}{\varepsilon^2g^{2s}(n)}\leq CM^{-2s}<\varepsilon. \end{equation*} $

Then using (3.22), (3.15), (3.16) and (3.18), we have

$ \mathbb{P}\Big\{|S_n|\geq(x+\varepsilon)\sqrt{n}g^{s}(n)\Big\}\leq\frac{C}{(\varepsilon+x)^{2+\delta}g^{(2+\delta)s}(n)}. $

Hence it holds that

$ \begin{align*} &\lim_{M\rightarrow\infty}\limsup_{\varepsilon\searrow0}\varepsilon^{b/s} \sum_{n>b_M(\varepsilon)}\frac{g'(n)g^{b-1}(n)}{\sqrt{n}}\mathbb{E}\{|S_n|- \varepsilon\sqrt{n}g^s(n)\}_+\\ =&\lim_{M\rightarrow\infty}\limsup_{\varepsilon\searrow0}\varepsilon^{b/s} \sum_{n>b_M(\varepsilon)}g'(n)g^{b+s-1}(n)\int_0^\infty\mathbb{P}\{|S_n|\geq (\varepsilon+x)\sqrt{n}g^s(n)\}dx\\ \leq&\lim_{M\rightarrow\infty}\limsup_{\varepsilon\searrow0}\varepsilon^{b/s} \sum_{n>b_M(\varepsilon)}g'(n)g^{b+s-1}(n)\int_0^\infty\frac{C} {(x+\varepsilon)^{2+\delta}g^{(2+\delta)s}(n)}dx\\ \leq&\lim_{M\rightarrow\infty}\limsup_{\varepsilon\searrow0}C\varepsilon^{\frac{b}{s}-1-\delta} \sum_{n>b_M(\varepsilon)}g'(n)[g(n)]^{b-(1+\delta)s-1}\\ \leq&\lim_{M\rightarrow\infty}\limsup_{\varepsilon\searrow0}C\varepsilon^{\frac{b}{s}-1-\delta} [g(b_M(\varepsilon))]^{b-(1+\delta)s}\\ \leq&\lim_{M\rightarrow\infty}CM^{b-(1+\delta)s}=0, \end{align*} $

where $b<(1+\delta)s$.

Finally, the proof of (1.4) is completed by combining Propositions 3.5-3.8 together and using the triangle inequality. We omit the proof of Theorem 1.2 since the idea is similar, and we only need to replace $\varepsilon^{b/s}$ by $1/(-\log\varepsilon)$ with $b=0$.