完全收敛性概念自 Hsu 和 Robbins ^[1] 提出以来, 已有许多学者从不同方面进行了推广并取得了比较完美的结果. 如 Baum 和 Katz ^[2] 获得了实值独立同分布随机变量序列完全收敛性的一个经典结果. Sung 等 ^[3] 把文 [1] 中的结果推广为如下的定理A.

定理 A ^[3]  设$\{X_{ni}, 1\leq i\leq k_n,n\geq 1\}$是行独立随机阵列, $\{a_{n}, n\geq 1\}$为一个非负常数列, 若满足下列条件

(i) $\sum\limits_{n=1}^{\infty}a_n\sum\limits_{i=1}^{k_n}P(|X_{ni}|>0;$

(ii) 存在 $J\geq 2, \delta>0$,使得$\sum\limits_{n=1}^{\infty}a_n\big[\sum\limits_{i=1}^{k_n}EX^2_{ni}I(|X_{ni}|\leq \delta)\big]^J<\infty;$

(iii) $\sum\limits_{i=1}^{k_n}EX_{ni}I(|X_{ni}|\leq \delta)\rightarrow 0, n\rightarrow\infty,$

则$\sum\limits_{n=1}^{\infty}a_nP\{|\sum\limits_{i=1}^{k_n}X_{ni}|>0.$

定义 1^[4] 称随机变量是$X_1,X_2,\cdots, X_n,(n\geq 2)$是NA的, 如果对于集合$\{1,2,\cdots,n\}$的任意两个不相交的非空子集$A$与$B$, 都有 $ {\rm Cov}(f_1(X_i,i\in A),f_2(X_j,j\in B))\leq 0$, 其中$f_1,f_2$是任何两个使得协方差存在且对每个变元均非降(或对每个变元均非增)的函数. 称 随机序列$\{X_n,n\in N\}$ 是NA的, 如果对每个$n\geq 2$, $X_1,X_2,\cdots, X_n$都是NA的.

定义 2^[5] 对固定的$m\geq 1$, 称随机变量序列$\{X_k,k\geq 1\}$ 是$m$-NA的, 如果对任意自然数$n\geq 2$, 序列$k_1,k_2,\cdots, k_n$满足$|k_i-k_j|\geq m$, 其中$1\leq i\neq j\leq n, m\leq n$, $\{X_{k_1},X_{k_2},\cdots, X_{k_n}\}$是NA的.

$m$-NA列是NA列的一个自然推广. 这些概念自引入以来, 不少学者研究了其完全收敛性、强收敛性等性质, 获得了很多结果. 如邱德华 ^[6]研究了行为NA随机阵列的完全收敛性, Chen等 ^[7], Sung ^[8]研究了行NA阵列的完全收敛性, 得到如下定理B:

定理 B^[7] 设$\{X_{ni}, 1\leq i\leq k_n, n\geq 1\}$是行NA随机阵列, $\{a_{n}, n\geq 1\}$为一个非负常数列, 若满足定理A中条件(i)及

(Ⅱ) 存在 $J\geq 1, \delta>0$,使得$\sum\limits_{n=1}^{\infty}a_n\big[\sum\limits_{i=1}^{k_n}\mbox{Var}X_{ni}I(|X_{ni}|\leq \delta)\big]^J\infty,$ 则

$\sum\limits_{n=1}^{\infty}a_nP\big\{\max\limits_{1\leq l\leq k_n}\mid\sum\limits_{i=1}^{l}[X_{ni}-EX_{ni} I(|X_{ni}|\leq\delta)]\mid>\epsilon\big\}﹤\infty, \forall\epsilon>0.$

而对行$m$-NA阵列情形Hu和Volodin ^[5]得到如下定理C.

定理 C^[5] 设$\{X_{ni}, 1\leq i\leq k_n,n\geq 1\}$是行为$m$-NA随机阵列, $\{a_{n}, n\geq 1\}$为一个非负常数列, 若满足定理A中条件(i), (ii) 则

$\sum\limits_{n=1}^{\infty}a_nP\big\{\max\limits_{1\leq l\leq k_n}\mid\sum\limits_{i=1}^{l}[X_{ni}-EX_{ni} I(|X_{ni}|\leq\delta)]\mid>\epsilon\big\}﹤\infty, \forall\epsilon>0.$

本文进一步研究行$m$-NA随机阵列的完全收敛性, 并简化了定理C, 即文[5]中定理1的证明. 全文约定$N$为正整数. $I(A)$为$A$的示性函数.

定理 1 设$\{X_{ni}, 1\leq i\leq k_n, n\geq 1\}$是一个均值为零的行$m$-NA随机阵列, 且$EX^2_{ni}\infty, 1\leq i\leq k_n,n\geq 1,$ $\{a_{n}, n\geq 1\}$为一个非负常数列, 若满足定理A中条件(i)及

(Ⅱ)'   存在 $J\geq 1$, 使得

$\sum\limits_{n=1}^{\infty}a_n\big(\sum\limits_{i=1}^{k_n}EX^2_{ni}\big)^J\infty,$

则

$\sum\limits_{n=1}^{\infty}a_nP\big\{\max\limits_{1\leq l\leq k_n}\mid\sum\limits_{i=1}^{l}X_{ni}\mid>\epsilon\big\}﹤\infty, \forall\epsilon>0.$

注   该定理把定理C中条件(ii)的$J\geq 2$ 改进为$J\geq 1$, 由该定理可以推导出定理C.

在证明定理之前先引述文[9]一个引理.

引理 1^[9] 设$\{X_{i}, 1\leq i\leq n\}$是一个均值为零的NA列, 且$EX^2_{i}\infty, 1\leq i\leq n,n\geq 2.$ 令$B_n=\sum\limits_{i=1}^nEX^2_i, S_k=\sum\limits_{i=1}^kX_i$, 则$\forall x>0, a>0,$

$ P\big\{\max\limits_{1\leq k\leq n}S_k>x\big\}\leq P\big\{\max\limits_{1\leq i\leq n} X_{i}>a\big\}+2\exp\{-\frac{x^2}{8B_n}\}+2[\frac{B_n}{4(ax+B_n)}]^{x/(12a)}, \\ P\big\{\max\limits_{1\leq k\leq n}\mid S_k\mid>x\big\}\leq 2P\big\{\max\limits_{1\leq i\leq n}\mid X_{i}\mid>a\big\}+4\exp\{-\frac{x^2}{8B_n}\}+4[\frac{B_n}{4(ax+B_n)}]^{x/(12a)}.$

引理 2   $\{X_{i}, i\geq 1\}$是一个均值为零的$m$-NA列, 且$EX^2_{i}\infty, i\geq 1.$ 令$B_n=\sum\limits_{i=1}^nEX^2_i$, 则对$n\geq m, x>0, a>0,$

$\begin{eqnarray} P\big\{\max\limits_{1\leq k\leq n}S_k>x\big\}\leq m[P\big\{\max\limits_{1\leq i\leq n} X_{i}>a\big\}+2\exp\{-\frac{x^2}{8m^2B_n}\}\nonumber\\ +2(\frac{mB_n}{4(ax+mB_n)})^{x/(12ma)}]. \end{eqnarray}$

(2.1)

$\begin{eqnarray} P\big\{\max\limits_{1\leq k\leq n}\mid S_k\mid>x\big\}\leq 2m[P\big\{\max\limits_{1\leq i\leq n}\mid X_{i}\mid>a\big\}+2\exp\{-\frac{x^2}{8m^2B_n}\}\nonumber\\ +2(\frac{mB_n}{4(ax+mB_n)})^{x/(12ma)}].\end{eqnarray}$

(2.2)

$\begin{eqnarray} P\big\{\max\limits_{1\leq k\leq n}\mid S_k\mid﹥x\big\}\leq 2m[P\big\{\max\limits_{1\leq i\leq n}\mid X_{i}\mid﹥a\big\}+4(\frac{2mB_n}{3ax})^{x/(12ma)}]. \end{eqnarray}$

(2.3)

证   对给定的$1\leq k\leq n$, 取$r=[\frac{n}{m}]$(表示取整), 令

$Y_i= \left \{ \begin{array}{ll} X_i, 1\leq i \leq n ;\\ 0, i>n, \\ \end{array} \right.\\ T_{mk+j}=\sum\limits_{i=0}^kY_{mi+j},1\leq j\leq m. $

显然对$1\leq j\leq m,m\leq n,$ $\{Y_{mi+j},i=0,1,\cdots, r\}$是NA的. 由

$\{\max\limits_{1\leq k\leq n}S_k\geq x\}\subset\{\max\limits_{0\leq k\leq r}T_{m k+1}\geq \frac{x}{m}\}\cup\cdots \cup\{\max\limits_{0\leq k\leq r}T_{mk+m}\geq\frac{x}{m}\}.$

根据引理1可得

$ P\{\max\limits_{1\leq k\leq n}S_k\geq x\}\leq\sum\limits_{j=1}^mP\{\max\limits_{0\leq k\leq r}T_{mk+j}\geq \frac{x}{m}\}\\ \leq\sum\limits_{j=1}^mP\{\max\limits_{0\leq i\leq r}Y_{mi+j}> a\}+2\sum\limits_{j=1}^m\exp\{-\frac{x^2}{8m^2\sum\limits_{i=0}^rEY^2_{mi+j}}\} +2\sum\limits_{j=1}^m[\frac{m\sum\limits_{i=0}^rEY^2_{mi+j}}{4(ax+m\sum\limits_{i=0}^rEY^2_{mi+j})}]^{\frac{x}{12ma}}\\ \leq m[P\{\max\limits_{1\leq i\leq n}X_i\geq a\}+2\exp\{-\frac{x^2}{8m^2B_n}\}+2(\frac{mB_n}{4(ax+mB_n)})^{\frac{x}{12ma}}].$

下证(2.3)式,由$e^{-x}﹤\frac{1}{x},x﹥0$, 有

$\exp\{-\frac{x^2}{8m^2B_n}\}=(\exp\{-\frac{3ax}{2mB_n}\})^{x/(12ma)}(\frac{2mB_n}{3ax})^{x/(12ma)},$

又有$(\frac{mB_n}{4(ax+mB_n)})^{x/(12ma)}(\frac{2mB_n}{3ax})^{x/(12ma)}$, 从而由(2.2)式可得(2.3)式.

定理 1证明 令$x=\epsilon, a=\frac{\epsilon}{12mJ}$, 根据(2.3)式有

$\begin{aligned} P\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^lX_{ni}|>\epsilon\} \leq 2m[P\big\{\max\limits_{1\leq i\leq k_n}\mid X_{ni}\mid>\epsilon/(12mJ)\big\}+4(\frac{8m^2J}{\epsilon^2})^{J}B_n^J]\\ \leq 2m\sum\limits_{i=1}^{k_n}P\big\{\mid X_{ni}\mid>\epsilon/(12mJ)\big\}+8^{J+1}m^{2J+1}J^J\epsilon^{-2J}B_n^J. \end{aligned} $

由定理条件(i)和(Ⅱ)'知结论成立.

下证由定理1推定理C:

定理 C证明   令$Y_{ni}=X_{ni}I(|X_{ni}|\leq \delta)+\delta I(X_{ni}>\delta)-\delta I(X_{ni}﹤-\delta), \delta﹥0,1\leq k\leq k_n,n\geq 1,$ 则 $\{Y_{ni},1\leq i\leq k_n,n\geq 1\}$ 也为$m$-NA列^[5] .

$ P\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^l[X_{ni}-EX_{ni}I(|X_{ni}|\leq \delta)]|>\epsilon\}\\ \leq \sum\limits_{i=1}^{k_n}P\big\{\mid X_{ni}\mid>\delta\big\} +P\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^l[X_{ni}I(|X_{ni}|\leq \delta)-EX_{ni}I(|X_{ni}|\leq \delta)]|>\epsilon\}\\ \leq \sum\limits_{i=1}^{k_n}P\big\{\mid X_{ni}\mid>\delta\big\} +P\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^l[Y_{ni}-EY_{ni}]|>\frac{\epsilon}{2}\}\\ +P\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^l[\delta I(X_{ni}>\delta)-\delta I(X_{ni}＜-\delta)- \delta P(X_{ni}>\delta) +\delta P(X_{ni}＜-\delta)]|>\frac{\epsilon}{2}\}\\ \leq P\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^l[Y_{ni}-EY_{ni}]|>\frac{\epsilon}{2}\} +\frac{2\delta}{\epsilon}E\big\{\max\limits_{1\leq l\leq k_n}\big( \sum\limits_{i=1}^l[I(|X_{ni}|>\delta)+P(|X_{ni}|>\delta)]\big)\big\}\\ +\sum\limits_{i=1}^{k_n}P\big\{\mid X_{ni}\mid>\delta\big\}\\ \leq P\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^l[Y_{ni}-EY_{ni}]|>\frac{\epsilon}{2}\}+(1+\frac{4\delta}{\epsilon})\sum\limits_{i=1}^{k_n} P\big\{\mid X_{ni}\mid>\delta\big\}. $

由条件(i)只需证明$\sum\limits_{n=1}^{\infty}a_nP\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^l[Y_{ni}-EY_{ni}]|>\epsilon\}\infty$即可.

令$A=\{n:\sum\limits_{i=1}^{k_n}P(|X_{ni}|>\delta)\leq 1\}, B=\{n:\sum\limits_{i=1}^{k_n}P(|X_{ni}|>\delta)> 1\}$,则

$\sum\limits_{n=1}^{\infty}a_nP\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^l[Y_{ni}-EY_{ni}]|﹥\epsilon\}\\ =\sum\limits_{n\in A}a_nP\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^l[Y_{ni}-EY_{ni}]|﹥\epsilon\} +\sum\limits_{n\in B}a_nP\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^l[Y_{ni}-EY_{ni}]|﹥\epsilon\}\\ \leq\sum\limits_{n\in A}a_nP\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^l[Y_{ni}-EY_{ni}]|﹥\epsilon\}+\sum\limits_{n\in B}a_n.$

由条件(i)及集合B的定义知$\sum\limits_{n\in B}a_n \infty$, 故只需证明

$\sum\limits_{n\in A}a_nP\{\max\limits_{1\leq l\leq k_n}|\sum\limits_{i=1}^l[Y_{ni}-EY_{ni}]|>\epsilon\}\infty.$

(2.4)

注意到 $\{Y_{ni}-EY_{ni},1\leq i\leq k_n,n\geq 1\}$ 是一个均值为零且二阶矩有限的行$m$-NA随机阵列. 对$Y_{ni}-EY_{ni}$要利用定理1, 只需证其在集合A上满足 定理1的条件. 由$C_r$ -不等式有

$\sum\limits_{n\in A}a_n(\sum\limits_{i=1}^{k_n}E(Y_{ni}-EY_{ni})^2)^J \\ \leq 2^{2J-1}\sum\limits_{n\in A}a_n(\sum\limits_{i=1}^{k_n} \mbox{Var}(X_{ni}I(|X_{ni}|\leq \delta)))^J \\ +2^{2J-1}\sum\limits_{n\in A}a_n(\sum\limits_{i=1}^{k_n} \mbox{Var}(\delta I(X_{ni}﹥\delta)-\delta I(X_{ni}＜-\delta)))^J\\ \leq 2^{2J-1}\sum\limits_{n\in A}a_n(\sum\limits_{i=1}^{k_n} \mbox{Var}(X_{ni}I(|X_{ni}|\leq \delta)))^J +2^{2J-1}\delta^{2J}\sum\limits_{n\in A}a_n(\sum\limits_{i=1}^{k_n}P(|X_{ni}|﹥\delta))^J \\ \leq 2^{2J-1}\sum\limits_{n\in A}a_n(\sum\limits_{i=1}^{k_n} \mbox{Var}(X_{ni}I(|X_{ni}|\leq \delta)))^J +2^{2J-1}\delta^{2J}\sum\limits_{n\in A}a_n\sum\limits_{i=1}^{k_n}P(|X_{ni}|﹥\delta).$

由此可得对$Y_{ni}-EY_{ni}$定理1的条件(Ⅱ)'满足.

下证对$Y_{ni}-EY_{ni}$满足定理1的条件(i)也成立.对给定的$\epsilon>0,$ 不妨设$0 ﹤ \epsilon ﹤ 4\delta $, 令$Y'_{ni}=X_{ni}I(|X_{ni}|\leq \frac{\epsilon}{4}), Y''_{ni}=Y_{ni}-Y'_{ni}$, 于是

$\begin{aligned} \sum\limits_{i=1}^{k_n}P(|Y_{ni}-EY_{ni}|>\epsilon)\leq \sum\limits_{i=1}^{k_n}P(|Y'_{ni}-EY'_{ni}|>\frac{\epsilon}{2})+\sum\limits_{i=1}^{k_n}P(|Y''_{ni}-EY''_{ni}|>\frac{\epsilon}{2})\\ =\sum\limits_{i=1}^{k_n}P(|Y''_{ni}-E Y''_{ni}|>\frac{\epsilon}{2})\leq \frac{4}{\epsilon}\sum\limits_{i=1}^{k_n}E|X''_{ni}|\\ \leq \frac{4\delta}{\epsilon}\sum\limits_{i=1}^{k_n}P(|X_{ni}|>\delta) + \frac{4\delta}{\epsilon}\sum\limits_{i=1}^{k_n}P(|X_{ni}|>\frac{\epsilon}{4}). \end{aligned} $

根据条件(i)可得$\sum\limits_{n \in A} {{a_n}} \sum\limits_{i = 1}^{{k_n}} P (|{Y_{ni}} - E{Y_{ni}}| > ) ﹤ \infty .$ 由定理1可得(2.4)式成立.