Hus和Robbins (1947)^[1]提出了完全收敛性的概念, Chow (1988)^[2]则提出了矩完全收敛性的概念.矩完全收敛性是完全收敛性的深化,许多学者进行了研究,得到很多深刻的结果,如文献[4-17]等等.特别地,最近Chen和Sung (2014)^[5]利用三段截尾法获得了如下的Baum-Katz型完全收敛性结果.

定理1.1^[5]  设$\{X, X_{n}, n\geq 1\}$是独立同分布的随机变量序列且$EX=0$,记$S_{n}=\sum\limits_{k=1}^{n}X_{k}$,令$\alpha>1$, $E\exp(\ln^{\alpha}|X|) < \infty$,则有

(1.1)$ \begin{eqnarray}\sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2}}P(|S_{n}|>n)<\infty. \end{eqnarray}$

定理1.1推广了Gut和Stadtüller (2011)^[4]中的定理2.1. Qiu和Chen (2014)^[6]在Gut和Stadtüller (2011)^[4]的基础上,讨论了矩完全收敛性,结果如下.

定理1.2^[6]  设$\{X, X_{n}, n\geq 1\}$是独立同分布的随机变量序列且$EX=0$,记$S_{n}=\sum\limits_{k=1}^{n}X_{k}$,令$\alpha>1$, $E\exp(\ln^{\alpha}|X|) < \infty$,则对所有的$\varepsilon >1$及$q>0$有

(1.2)$ \begin{eqnarray}\sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2+q}}E\{|S_{n}|-\varepsilon n\}^{q}_{+}<\infty. \end{eqnarray}$

本论文考虑将定理1.1及定理1.2推广至WOD情形.下面介绍WOD随机变量序列的定义.

引理1.1^[3]  称$\{X_{n}, n\geq 1\}$是带有控制系数$\{g_{U}(n), n\geq 1\}$的WUOD (widely upper orthant dependent)序列,如果存在有限的实数序列$\{g_{U}(n), n\geq 1\}$,任取$n\geq 1$和$x_{i}\in (-\infty, \infty)$, $1\leq i \leq n$满足

$ P(X_{1}> x_{1}, X_{2}> x_{2}, \cdots, X_{n}> x_{n}) \leq g_{U}(n)\prod\limits_{i=1}^{n} P(X_{i}> x_{i}).$

称$\{X_{n}, n\geq 1\}$是带有控制系数$\{g_{L}(n), n\geq 1\}$的WLOD (widely lower orthant dependent)序列,如果存在有限的实数序列$\{g_{L}(n), n\geq 1\}$,任取$n\geq 1$和$x_{i}\in (-\infty, \infty)$, $1\leq i \leq n$满足

$P(X_{1}\leq x_{1}, X_{2}\leq x_{2}, \cdots, X_{n}\leq x_{n}) \leq g_{L}(n)\prod\limits_{i=1}^{n} P(X_{i}\leq x_{i}).$

称$\{X_{n}, n\geq 1\}$是WOD (widely orthant dependent)序列,如果$\{X_{n}, n\geq 1\}$既是WUOD序列,又是WLOD序列,其控制系数记为$g(n)=\max\{g_{U}(n), g_{L}(n)\}$.

WOD这一概念是由学者Wang等(2013)^[3]提出的,以END (extended negatively dependent)为特殊情形($g(n)\equiv C$),当然, NOD (negatively orthant dependent)更是为其特殊情形($g(n)\equiv 1$).如Qiu等(2014)^[11]研究了NOD随机变量序列的完全收敛性.邱德华等(2017)^[12]研究了END随机变量序列加权和的矩完全收敛性.郭明乐和祝东进(2013)^[13]研究了NOD随机变量序列加权和的矩完全收敛性的等价条件. Wang等(2014)^[14]研究了WOD随机变量序列的完全收敛性和在非参数回归模型中的应用. Qiu和Chen (2014)^[15]在适当条件下获得了WOD随机变量序列加权和的完全收敛性和矩完全收敛性.丁洋等(2015)^[16]利用WOD随机变量序列部分和的Rosenthal型不等式获得了WOD随机变量序列加权和的完全收敛性. Liu等(2017)^[17]获得了WOD随机变量序列的矩完全收敛性.蔡光辉和潘雪艳(2015)^[18]在一个独立随机变量序列的重对数律的基础上,获得了不同分布WOD随机变量序列的重对数律等等.

本文首先在一个独立同分布随机变量序列的Baum-Katz型完全收敛性结果的基础上,获得了一个同分布的WOD随机变量序列的Baum-Katz型完全收敛性的类似结果(见如下定理1.3).

定理1.3  设$\{X, X_{n}, n\geq 1\}$是同分布的WOD随机变量序列且$EX=0$,记$S_{n}=\sum\limits_{k=1}^{n}X_{k}$,令$\alpha>1$, $E\exp(\ln^{\alpha}|X|) < \infty$, $g(n)=O(n^{\theta})$, $0 < \theta < 1$,则有

(1.3) $ \begin{eqnarray} \sum\limits_{n=1}^{\infty}\exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2}}P(|S_{n}|>n)<\infty. \end{eqnarray} $

注1.1  若$g(n)\equiv1$时, $\{X, X_{n}, n\geq 1\}$为同分布的NOD随机变量序列,则定理1.3仍旧成立.

注1.2  若$g(n)\equiv C$时, $\{X, X_{n}, n\geq 1\}$为同分布的END随机变量序列,则定理1.3仍旧成立.

注1.3  若仍旧采用Chen和Sung(2014)^[5]的三段截尾法,我们无法获得WOD情形时的类似Chen和Sung (2014)^[5]中的定理2.1(即文中定理1.1).故本文采用五段截尾法,在所用的条件一致的情况下将Chen和Sung (2014)^[5]中的定理2.1推广至WOD的情形.

其次,本文受Qiu和Chen (2014)^[6]中的定理2.1的启发,获得了一个同分布的WOD随机变量序列的矩完全收敛性的类似结果(见如下定理1.4).

定理1.4  设$\{X, X_{n}, n\geq 1\}$是同分布的WOD随机变量序列且$EX=0$,记$S_{n}=\sum\limits_{k=1}^{n}X_{k}$,令$\alpha>1$, $E\exp(\ln^{\alpha}|X|) < \infty$, $g(n)=O(n^{\theta})$, $0 < \theta < 1$.则对所有的$\varepsilon \geq 1$及$q>0$有

(1.4)$ \begin{eqnarray} \sum\limits_{n=1}^{\infty}\exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2+q}}E\{|S_{n}|-\varepsilon n\}^{q}_{+}<\infty. \end{eqnarray}$

注1.4  若$g(n)\equiv1$时, $\{X, X_{n}, n\geq 1\}$为同分布的NOD随机变量序列,则定理1.4仍旧成立.

注1.5  若$g(n)\equiv C$时, $\{X, X_{n}, n\geq 1\}$为同分布的END随机变量序列,则定理1.4仍旧成立.

注1.6  若仍旧采用Qiu和Chen (2014)^[6]的三段截尾法,我们无法获得WOD情形时的类似Qiu和Chen (2014)^[6]中的定理2.1 (即文中定理1.2).故本文采用五段截尾法,将Qiu和Chen (2014)^[6]中的定理2.1推广至WOD的情形,并且我们得到了$\varepsilon = 1$时的定理2.1.

本文中$ C $表示正常数,且在不同的地方可为不同的值. $a_{n}\ll b_{n}$表示$a_{n}=O(b_{n})$. $a_{n}=O(b_{n})$表示$a_{n}\leq C b_{n}$.

为了证明定理,需要如下4个引理.

引理2.1  设$\{X_{n}, n\geq 1\}$为WOD随机变量序列,如果$\{f_{n}(\cdot), n\geq 1\}$是均非升(或均非降)的函数，则$\{f_{n}(X_{n}), n\geq 1\}$仍为WOD随机变量序列.

引理2.2  设$\{X_{n}, n\geq 1\}$为WOD随机变量序列, $EX_{n}=0$, $EX_{n}^{2} < \infty$, $B_{n}=\sum\limits_{k=1}^{n}EX_{k}^{2}$, $S_{n}=\sum\limits_{k=1}^{n}X_{k}$.记控制系数$g(n)=\max \{g_{U}(n), g_{L}(n)\}$.则对所有的$x>0$, $y>0$均有

$P(|S_{n}|\geq x)\leq \sum\limits_{i=1}^{n} P(|X_{i}|\geq y) +2g(n)\exp\bigg\{\frac{x}{y}-\frac{x}{y}\ln (1+\frac{xy}{B_{n}})\bigg\}.$

引理2.3^[19]  对所有的$x>0$,均有

$\ln (1+x)\geq \frac{x}{1+x}+\frac{x^{2}}{2(1+x)^{2}}\bigg [1+\frac{2}{3}\ln (1+x)\bigg].$

引理2.4  设$Y$为一个随机变量,当$\alpha>1$时, $E\exp(\ln^{\alpha}|Y|) < \infty$,则我们有

$\sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n) \frac{\ln^{\alpha-1}n}{n}P(|Y|>\frac{n}{\ln^{\alpha}n})<\infty.$

证  对于足够大的$n_{0}$,当$x\in[n_{0}, +\infty)$时,可得$x/\ln^{\alpha}x$及$ \exp(\ln^{\alpha}x)\ln^{\alpha-1}x /x$为增函数.令$d_{n}=n/\ln^{\alpha}n$,且$\ln^{\alpha}(n+1)-\ln^{\alpha}n=\textit{o}(1)$, $\ln^{\alpha}n-\ln^{\alpha}d_{n}=\textit{o}(1)$,我们有

$\exp\{\ln^{\alpha}(n+1)\}\leq C \exp\{\ln^{\alpha}(n)\}, \ \ \ \exp\{\ln^{\alpha}(n)\}\leq C \exp\{\ln^{\alpha}(d_{n})\}.$

则

$ \begin{eqnarray*} \sum\limits_{n=n_{0}}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n}P(|Y|> d_{n}) &=&\sum\limits_{n=n_{0}}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n}\sum\limits_{i=n}^{\infty}P(d_{i}<|Y|\leq d_{i+1})\nonumber\\ &=&\sum\limits_{i=n_{0}}^{\infty}P(d_{i}<|Y|\leq d_{i+1})\sum\limits_{n=n_{0}}^{i}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n}\nonumber\\ &\leq& \sum\limits_{i=n_{0}}^{\infty}P(d_{i}<|Y|\leq d_{i+1})\int_{n_{0}}^{i+1}{\exp (\ln^{\alpha}x)\frac{\ln^{\alpha-1}x}{x}}\, {\rm d}x\nonumber\\ &\leq &\sum\limits_{i=n_{0}}^{\infty}P(d_{i}<|Y|\leq d_{i+1})\frac{1}{\alpha}\exp\{\ln^{\alpha}(i+1)\}\nonumber\\ &\leq &\sum\limits_{i=n_{0}}^{\infty}P(d_{i}<|Y|\leq d_{i+1})\exp\{\ln^{\alpha}(i)\}\nonumber\\ &\leq &\sum\limits_{i=n_{0}}^{\infty}P(d_{i}<|Y|\leq d_{i+1})\exp\{\ln^{\alpha}(d_{i})\}\nonumber\\ &\leq &C E\exp(\ln^{\alpha}|Y|)<\infty . \end{eqnarray*} $

引理2.4证明完毕.

定理1.3的证明  $\forall 1\leq k\leq n$,令

$\begin{eqnarray*} &&X_{k}(1)=-b_{n}I(X_{k}<-b_{n})+X_{k}I(|X_{k}|\leq b_{n})+b_{n}I(X_{k}>b_{n}), \nonumber\\&&X_{k}(2)=(X_{k}-b_{n})I(b_{n}\leq X_{k}<b_{n}+c_{n})+c_{n}I(X_{k}>b_{n}+c_{n}), \nonumber\\&&X_{k}(3)=(X_{k}+b_{n})I(-b_{n}-c_{n}<X_{k}\leq -b_{n})-c_{n}I(X_{k}<-b_{n}-c_{n}), \nonumber\\&&X_{k}(4)=(X_{k}-b_{n}-c_{n})I(X_{k}\geq b_{n}+c_{n}), \nonumber\\&&X_{k}(5)=(X_{k}+b_{n}+c_{n})I(X_{k}\leq- b_{n}-c_{n}), \end{eqnarray*} $

其中$b_{n}=n/4\ln^{2\alpha}n$, $c_{n}=n/4\ln^{\alpha}n$.

则

(2.1) $ \begin{eqnarray} \sum\limits_{n=1}^{\infty} \exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2}}P(|S_{n}|>n) &\leq& \sum\limits_{l=1}^{5}\sum\limits_{n=1}^{\infty} \exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2}}P(|\sum\limits_{k=1}^{n}X_{k}(l)|>n)\nonumber\\ &:=&\sum\limits_{l=1}^{5}I_{l}. \end{eqnarray} $

先证$I_{1} < \infty$.由引理2.1,可得$\{X_{k}(1), 1\leq k \leq n, n\geq1\}$也是WOD的.由$E\exp(\ln^{\alpha}|X|) < \infty$, $\alpha>1$,推出$EX^{2} < \infty$.注意到$EX=0$,我们有

(2.2) $ \begin{eqnarray} \bigg|\sum\limits_{k=1}^{n}EX_{k}(1)\bigg|& \leq& \sum\limits_{k=1}^{n}[E|X_{k}|I(|X_{k}|\leq b_{n})+b_{n}I(|X_{k}|>b_{n})]\nonumber\\ &= &n E|X|I(|X|>b_{n})+n b_{n}EI(|X|>b_{n})\nonumber\\ &\leq& n \frac{EX^{2}I(|X|>b_{n})}{b_{n}}+n b_{n}^{-1}EX^{2}I(|X|>b_{n})\nonumber\\ &=&\frac{2n EX^{2}I(|X|>b_{n})}{b_{n}} \nonumber\\ &=&\textit{o}(\ln^{2\alpha}n), \quad n \rightarrow\infty. \end{eqnarray} $

由(2.2)式可知,对于足够大的$n_{0}$,当$n\geq n_{0}$时,可得$|\sum\limits_{k=1}^{n}EX_{k}(1)|\leq \ln^{2\alpha}n$.由引理2.2,引理2.3, $E|X_{k} (1)|\leq b_{n}$,令$x=n-\ln^{2\alpha}n$, $y=2b_{n}=n/2\ln^{2\alpha}n$,注意到, $B_{n}=\sum\limits_{k=1}^{n}E[X_{k}(1)-EX_{k}(1)]^{2}\leq n EX^{2}$,当$n\geq n_{0}$时,我们可得

(2.3)$\begin{eqnarray} P(|\sum\limits_{k=1}^{n}X_{k}(1)|>n)&\leq &P(|\sum\limits_{k=1}^{n}(X_{k}(1)-EX_{k}(1))|>n- \ln^{2\alpha}n)\nonumber\\ & \leq& \sum\limits_{k=1}^{n} P(|X_{k}(1)-EX_{k}(1)|\geq y)+2 g(n)\exp\bigg\{\frac{x}{y}-\frac{x}{y}\ln(1+\frac{xy}{B_{n}})\bigg\}\nonumber\\& \leq &\sum\limits_{k=1}^{n} P(|X_{k}(1)|\geq y-b_{n})+2 g(n)\exp\bigg\{\frac{x}{y}-\frac{x}{y}\ln(1+\frac{xy}{B_{n}})\bigg\}\nonumber\\ &=&2g(n)\exp\bigg\{\frac{x}{y}(1-\ln(1+\frac{xy}{B_{n}}))\bigg\}\nonumber\\ &\leq& 2g(n)\exp\bigg\{\frac{x}{y}(\frac{1}{1+\frac{xy}{B_{n}}} -\frac{(\frac{xy}{B_{n}})^{2}}{2(1+\frac{xy}{B_{n}})^{2}}(1+\frac{2}{3}\ln(1+\frac{xy}{B_{n}})))\bigg\}\nonumber\\ &\leq&2 Cg(n)\exp\bigg\{\frac{x}{y}\frac{B_{n}}{B_{n}+xy}\bigg\}\exp \bigg\{-\frac{x}{y}\frac{(xy)^{2}}{2(B_{n}+xy)^{2}}\bigg\}\nonumber\\ &\leq&2 C g(n) \exp\bigg\{\frac{B_{n}}{y^{2}}\}\exp\{-\frac{x^{3}y}{2(B_{n}+xy)^{2}}\bigg\}\nonumber\\ & =&2 C g(n)\exp\bigg\{\frac{nEX^{2}}{(\frac{n}{2\ln^{2\alpha}n})^{2}}\bigg\} \exp\bigg \{-\frac{(1-\frac{\ln^{2\alpha}n}{n})^{3} \ln^{2\alpha}n}{4[\frac{\ln^{2\alpha}nEX^{2}}{n}+\frac{1}{2}- \frac{\ln^{2\alpha}n}{2n}]^{2}}\bigg\}\\ &\ll& 2g(n)\exp\{-\ln^{2\alpha}n\}. \end{eqnarray}$

将(2.3)式代入到(2.1)式中,由$g(n)=O(n^{\theta})$, $0 < \theta < 1$,我们可得

(2.4)$\begin{eqnarray} I_{1}& \ll& \sum\limits_{n=1}^{\infty} \exp\{-\ln^{2\alpha}n+ \ln^{\alpha}n\}\ln^{\alpha-1}n\frac{1}{n^{2}}g(n)\nonumber\\ &\leq& C\sum\limits_{n=1}^{\infty} \exp\{-\ln^{2\alpha}n+\ln^{\alpha}n\}\ln^{\alpha-1}n\frac{1}{n^{2-\theta}}\nonumber\\ &<&\infty. \end{eqnarray}$

对于$I_{2}$,由引理2.1,可得$\{X_{k}(2), 1\leq k \leq n, n\geq1\}$也是WOD的.由$E\exp(\ln^{\alpha}|X|) < \infty$, $\alpha > 1$,推出$EX^{2} < \infty$.可得

(2.5)$ \begin{eqnarray} \bigg|\sum\limits_{k=1}^{n}EX_{k}(2)\bigg|& =& \bigg|\sum\limits_{k=1}^{n}E\{(X_{k}-b_{n})I(b_{n}\leq X_{k}< b_{n}+c_{n})+c_{n}I(X_{k}>b_{n}+c_{n})\}\bigg|\nonumber\\ &\leq &n E|X|I(|X|>b_{n})+\frac{n^{2}}{4\ln^{\alpha}n}EI(|X|>b_{n})\nonumber\\ &\leq &n \frac{EX^{2}I(|X|>b_{n})}{b_{n}}+\frac{n^{2}}{4\ln^{\alpha}n}\frac{EX^{2}I(|X|>b_{n})}{b_{n}^{2}}\nonumber\\ &=&\textit{o}(\ln^{3\alpha}n), \quad n \rightarrow\infty. \end{eqnarray} $

由(2.5)式可知,对于足够大的$n_{0}$,当$n\geq n_{0}$时,可得$|\sum\limits_{k=1}^{n}EX_{k}(2)|\leq \ln^{3\alpha}n$.由引理2.2,引理2.3, $E|X_{k}(2)|\leq c_{n}$,令$x=n-\ln^{3\alpha}n$, $y=2 c_{n}=n/2\ln^{\alpha}n$,注意到$B_{n}=\sum\limits_{k=1}^{n}E[X_{k}(2)-EX_{k}(2)]^{2}\leq n EX^{2}$,当$n\geq n_{0}$时,类似于(2.3)的证明方法,我们可得

(2.6)$ \begin{eqnarray} P(|\sum\limits_{k=1}^{n}X_{k}(2)|>n)& \leq& P(|\sum\limits_{k=1}^{n}(X_{k}(2)-EX_{k}(2))|>n- \ln^{3\alpha}n)\nonumber\\ &\leq& P(|\sum\limits_{k=1}^{n}(X_{k}(2)-EX_{k}(2))|\geq y)+2 g(n)\exp\bigg\{\frac{x}{y}-\frac{x}{y} \ln(1+\frac{xy}{B_{n}})\bigg\}\nonumber\\ & \leq& \sum\limits_{k=1}^{n} P(|X_{k}(2)|\geq y-c_{n})+2 g(n)\exp\bigg\{\frac{x}{y}-\frac{x}{y} \ln(1+\frac{xy}{B_{n}})\bigg\}\nonumber\\ & \leq &2 C g(n) \exp\bigg\{\frac{B_{n}}{y^{2}}\bigg\}\exp\bigg\{-\frac{x^{3}y}{2(B_{n}+xy)^{2}}\bigg\}\nonumber\\ &=&2 C g(n) \exp\bigg\{\frac{nEX^{2}}{(\frac{n}{2\ln^{\alpha}n})^{2}}\bigg\} \exp\bigg \{-\frac{(1-\frac{\ln^{3\alpha}n}{n})^{3} \ln^{\alpha}n}{4[\frac{\ln^{\alpha}nEX^{2}}{n}+\frac{1}{2}- \frac{\ln^{3\alpha}n}{2n}]^{2}}\bigg\}\nonumber\\ &\ll& 2 g(n)\exp\{-\ln^{\alpha}n\}. \end{eqnarray} $

将(2.6)式代入到(2.1)式中,由$g(n)=O(n^{\theta})$, $0 < \theta < 1$,我们可得

(2.7) $ \begin{eqnarray} I_{2}\ll \sum\limits_{n=1}^{\infty} \ln^{\alpha-1}n\frac{1}{n^{2}}g(n)\leq C\sum\limits_{n=1}^{\infty} \ln^{\alpha-1}n\frac{1}{n^{2-\theta}}<\infty.\end{eqnarray} $

同理可证

(2.8)$\begin{equation} I_{3}\ll\sum\limits_{n=1}^{\infty} \ln^{\alpha-1}n\frac{1}{n^{2-\theta}}<\infty.\end{equation}$

对于$I_{4}$,在引理2.4中将$Y$替换成$4X$可得

(2.9)$ \begin{eqnarray}I_{4}& \ll &\sum\limits_{n=1}^{\infty}\exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n}P(|X|>b_{n}+c_{n})\nonumber\\ &\leq& C \sum\limits_{n=1}^{\infty} \exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n}P(|X|>\frac{n}{4\ln^{\alpha}n})\nonumber\\ &<&\infty. \end{eqnarray} $

同理可证

(2.10) $ \begin{equation} I_{5}\ll\sum\limits_{n=1}^{\infty} \exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n}P(|X|>\frac{n}{4\ln^{\alpha}n})<\infty. \end{equation} $

由(2.4)-(2.10)式,我们可得(1.3)式成立.至此定理1.3的证明完毕.

定理1.4的证明

$ \begin{eqnarray}& &\sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2+q}}E\{|S_{n}|-\varepsilon n\}^{q}_{+} \nonumber\\ &=&\varepsilon^{q} \sum\limits_{n=1}^{\infty}\exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2+q}} \int_{0}^{n}{qx^{q-1}P(|S_{n}|-\varepsilon n>\varepsilon x)}\, {\rm d}x \nonumber\\ & &+\varepsilon^{q}\sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2+q}} \int_{n}^{\infty}{qx^{q-1}P(|S_{n}|-\varepsilon n>\varepsilon x)}\, {\rm d}x\nonumber\\ &\ll&\sum\limits_{n=1}^{\infty}\exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2}}P(|S_{n}|>\varepsilon n)\nonumber\\ && +\sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2+q}} \int_{n}^{\infty}{x^{q-1}P(|S_{n}|>\varepsilon x)}\, {\rm d}x.\nonumber \end{eqnarray} $

因为$\varepsilon \geq 1$,由定理1.3可知$\sum\limits_{n=1}^{\infty}\exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2}}P(|S_{n}| > \varepsilon n) < \infty$.因此,为了证明(1.4)式成立,我们只需证明

(2.11) $ \begin{eqnarray} \sum\limits_{n=1}^{\infty}\exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2+q}}\int_{n}^{\infty}{x^{q-1}P(|S_{n}|>\varepsilon x)}\, {\rm d}x<\infty.\end{eqnarray} $

$\forall 1\leq k\leq n$, $x\geq n$,令

$\begin{eqnarray*}&&Y_{k}(1)=-b_{x}I(X_{k}<-b_{x})+X_{k}I(|X_{k}|\leq b_{x})+b_{x}I(X_{k}>b_{x}), \nonumber\\&&Y_{k}(2)=(X_{k}-b_{x})I(b_{x}\leq X_{k}<b_{x}+c_{x})+c_{x}I(X_{k}>b_{x}+c_{x}), \nonumber\\&&Y_{k}(3)=(X_{k}+b_{x})I(-b_{x}-c_{x}<X_{k}\leq -b_{x})-c_{x}I(X_{k}<-b_{x}-c_{x}), \nonumber\\&&Y_{k}(4)=(X_{k}-b_{x}-c_{x})I(X_{k}\geq b_{x}+c_{x}), \nonumber\\&&Y_{k}(5)=(X_{k}+b_{x}+c_{x})I(X_{k}\leq- b_{x}-c_{x}), \end{eqnarray*}$

其中$b_{x}=x/4\ln^{2\alpha}x$, $c_{x}=x/4\ln^{\alpha}x$.

所以

(2.12)$\begin{eqnarray}& &\sum\limits_{n=1}^{\infty}\exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2+q}}\int_{n}^{\infty}{x^{q-1}P(|S_{n}|>\varepsilon x)}\, {\rm d}x\nonumber\\ &\leq& \sum\limits_{J=1}^{5} \sum\limits_{n=1}^{\infty}\exp(\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2+q}}\int_{n}^{\infty}{x^{q-1} P(|\sum\limits_{k=1}^{n}Y_{k}(J)|>\varepsilon x)}\, {\rm d}x\nonumber\\ &:=&\sum\limits_{J=1}^{5}I'_{J}. \end{eqnarray}$

类似地运用证明(2.3)和(2.6)式的方法,当$x\geq n \geq n_{0}$时,可得

(2.13)$\begin{equation}P(|\sum\limits_{k=1}^{n}Y_{k}(1)|>\varepsilon x)\ll 2g(n)\exp\{-\ln^{2\alpha}x\}, \end{equation}$

(2.14)$\begin{equation}P(|\sum\limits_{k=1}^{n}Y_{k}(2)|>\varepsilon x)\ll 2g(n)\exp\{-\ln^{\alpha}x\}.\end{equation}$

事实上, $\forall a>0$, $b>0$, $\alpha>1$,有$(a+b)^{\alpha}\geq a^{\alpha}+b^{\alpha}$成立.将(2.13)式代入(2.12)式,由$g(n)=O(n^{\theta})$, $0 < \theta < 1$,我们有

(2.15) $ \begin{eqnarray}I'_{1}& \ll&\sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2+q}} \int_{n}^{\infty}{x^{q-1}2g(n) \exp(-\ln^{2\alpha}x)}\, {\rm d}x\nonumber\\ &\leq& \sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2-\theta+q}}n^{q}\exp(-\ln^{2\alpha}n) \int_{1}^{\infty}{t^{q-1}\exp(-\ln^{2\alpha}t)}{\rm d}t\nonumber\\ & \leq& C\sum\limits_{n=1}^{\infty} \exp\{-\ln^{2\alpha}n+\ln^{\alpha}n\}\ln^{\alpha-1}n\frac{1}{n^{2-\theta}}\nonumber\\ &<&\infty. \end{eqnarray} $

将(2.14)式代入(2.12)式,由$g(n)=O(n^{\theta})$, $0 < \theta < 1$,我们有

(2.16) $ \begin{eqnarray}I'_{2}& \ll&\sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2+q}} \int_{n}^{\infty}{x^{q-1}2g(n) \exp(-\ln^{\alpha}x)}\, {\rm d}x\nonumber\\ &\leq& \sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{2-\theta+q}}n^{q}\exp(-\ln^{\alpha}n) \int_{1}^{\infty}{t^{q-1}\exp(-\ln^{\alpha}t)}{\rm d}t\nonumber\\ &\ll&\sum\limits_{n=1}^{\infty} \frac{\ln^{\alpha-1}n}{n^{2-\theta}}\nonumber\\ &<&\infty. \end{eqnarray} $

同理可证

(2.17) $ \begin{equation} I'_{3} \ll\sum\limits_{n=1}^{\infty}\frac{ \ln^{\alpha-1}n}{n^{2-\theta}} <\infty. \end{equation} $

在引理2.4中将$Y$替换成$4X$,可得

(2.18)$ \begin{eqnarray}I'_{4}& \ll&\sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{1+q}}\int_{n}^{\infty}{x^{q-1} P(|X|>\frac{x}{4\ln^{\alpha}x})}\, {\rm d}x\nonumber\\&=&\sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{1+q}}\sum\limits_{j=n}^{\infty}\int_{j}^{j+1}{x^{q-1} P(|X|>\frac{x}{4\ln^{\alpha}x})}\, {\rm d}x\nonumber\\&\ll& \sum\limits_{n=1}^{\infty}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{1+q}}\sum\limits_{j=n}^{\infty}P(|X|>\frac{j}{4\ln^{\alpha}j})j^{q-1}\nonumber\\&=&\sum\limits_{j=1}^{\infty}P(|X|>\frac{j}{4\ln^{\alpha}j})j^{q-1}\sum\limits_{n=1}^{j}\exp (\ln^{\alpha}n)\frac{\ln^{\alpha-1}n}{n^{1+q}}\nonumber\\ &\ll& \sum\limits_{j=1}^{\infty}\exp (\ln^{\alpha}j)\frac{\ln^{\alpha-1}j}{j}P(|X|>\frac{j}{4\ln^{\alpha}j})\nonumber\\&<&\infty.\end{eqnarray} $

同理可证

(2.19)$\begin{equation}I'_{5}\ll \sum\limits_{j=1}^{\infty}\exp (\ln^{\alpha}j)\frac{\ln^{\alpha-1}j}{j}P(|X|>\frac{j}{4\ln^{\alpha}j})<\infty.\end{equation}$

由(2.15)-(2.19)式,我们可得(2.12)式成立,即(1.4)式得证.至此定理1.4的证明完毕.