WOD随机变量序列的完全收敛性和矩完全收敛性

WOD随机变量序列的完全收敛性和矩完全收敛性

章茜^,, 蔡光辉^,

Complete Convergence and Complete Moment Convergence for WOD Random Variables Sequences

Zhang Qian^,, Cai Guanghui^,

通讯作者: 蔡光辉, E-mail: cghzju@163.com

收稿日期: 2018-06-7

基金资助:

浙江省自然科学基金. LY17A010003
浙江省一流学科A类. 浙江工商大学统计学

Received: 2018-06-7

Fund supported:

the Natural Science Foundation of Zhejiang Province. LY17A010003
the First Class Discipline of Zhejiang-A. 浙江工商大学统计学

作者简介 About authors

章茜,E-mail:qiwa_007@163.com , E-mail：qiwa_007@163.com

摘要

该文采用五段截尾法，将Chen和Sung（2014）^[5]的定理2.1以及Qiu和Chen（2014）^[6]中的定理2.1推广至WOD随机变量序列情形，证明方法较已有的证明方法有所不同.

关键词： WOD随机变量序列 ; 完全收敛性 ; 矩完全收敛性

Abstract

In this paper, we use a new method to improve the corresponding result of Chen and Sung (2014)^[5] and Qiu and Chen (2014)^[6] by truncating the WOD random variables into five parts.

Keywords： WOD random variables sequences ; Complete convergence ; Complete moment convergence

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本文引用格式

章茜, 蔡光辉. WOD随机变量序列的完全收敛性和矩完全收敛性. 数学物理学报[J], 2019, 39(5): 1183-1191 doi:

Zhang Qian, Cai Guanghui. Complete Convergence and Complete Moment Convergence for WOD Random Variables Sequences. Acta Mathematica Scientia[J], 2019, 39(5): 1183-1191 doi:

1 引言

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

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

(1.1)

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

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

(1.2)

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

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

(1.3)

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

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

(1.4)

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 $a_{n}\leq C b_{n}$ .

2 定理的证明

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

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

(2.1)

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

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

(2.2)

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

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

(2.3)

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

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

(2.4)

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

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

(2.5)

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

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

(2.6)

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

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

(2.7)

同理可证

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

(2.8)

对于 $I_{4}$ ,在引理2.4中将 $Y$ 替换成 $4X$ 可得

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

(2.9)

同理可证

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

(2.10)

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

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

(2.11)

$\forall 1\leq k\leq n$ , $x\geq n$ ,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

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

(2.12)

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

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

(2.13)

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

(2.14)

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

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

(2.15)

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

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

(2.16)

同理可证

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

(2.17)

在引理2.4中将 $Y$ 替换成 $4X$ ,可得

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

(2.18)

同理可证

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

(2.19)

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.

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DOI:10.1023/A:1007849609234 [本文引用: 1]

Complete convergence and the law of large numbers

1947

... 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. ...

On the rate of moment convergence of sample sums and extremes

1988

Uniform asymptotics for the finite-time ruin probability of a new dependent risk model with a constant interest rate

2013

... 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 ...

... 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. ...

An intermediate Baum-Katz theorem

2011

... 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. ...

... [4]的基础上,讨论了矩完全收敛性,结果如下. ...

A Baum-Katz theorem for i.i.d. random variables with higher order moments

2014

... 该文采用五段截尾法，将Chen和Sung（2014）^[5]的定理2.1以及Qiu和Chen（2014）^[6]中的定理2.1推广至WOD随机变量序列情形，证明方法较已有的证明方法有所不同. ...

... In this paper, we use a new method to improve the corresponding result of Chen and Sung (2014)^[5] and Qiu and Chen (2014)^[6] by truncating the WOD random variables into five parts. ...

... 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 ...

... 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. ...

... [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. ...

... [5]中的定理2.1推广至WOD的情形. ...

Complete moment convergence for i.i.d. random variables

2014

... In this paper, we use a new method to improve the corresponding result of Chen and Sung (2014)^[5] and Qiu and Chen (2014)^[6] by truncating the WOD random variables into five parts. ...

... 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 ...

... 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). ...

... 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. ...

... [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. ...

... [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. ...

Complete convergence for weighted wums and arrays of rowwise extended negatively dependent random variables

2013

Complete convergence for weighted sums of NSD random variables and its application in the EV regression model

2015

Complete convergence and complete moment convergence for negatively associated sequences random variables

2014

END序列加权和的完全收敛性

2016

END序列加权和的完全收敛性

2016

Complete convergence for negatively orthant dependent random variables

2014

... 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(

$g(n)\equiv 1$

END随机变量序列加权和的矩完全收敛性

2017

... 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(

$g(n)\equiv 1$

END随机变量序列加权和的矩完全收敛性

2017

... 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(

$g(n)\equiv 1$

NOD随机变量序列加权和的矩完全收敛性的等价条件

2013

... 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(

$g(n)\equiv 1$

NOD随机变量序列加权和的矩完全收敛性的等价条件

2013

... 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(

$g(n)\equiv 1$

On complete convergence for widely orthant-dependent random variables and its applications in nonparametric regression models

2014

... 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(

$g(n)\equiv 1$

Complete and complete moment convergence for weighted sums of widely orthant dependent random variables

2014

... 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(

$g(n)\equiv 1$

WOD随机变量加权和的完全收敛性

2015

... 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(

$g(n)\equiv 1$

WOD随机变量加权和的完全收敛性

2015

... 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(

$g(n)\equiv 1$

Complete moment convergence of widely orthant dependent random variables

2017

... 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(

$g(n)\equiv 1$

不同分布WOD随机变量序列的重对数律

2015

... 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(

$g(n)\equiv 1$

不同分布WOD随机变量序列的重对数律

2015

... 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(

$g(n)\equiv 1$

A comparison theorem on moment inequalities between Negatively associated and independent random variables

2000

... 引理2.3^[19] 对所有的

$x>0$

,均有 ...

〈

〉