0 引言

定义1  对于随机变量X和Y, 若∀x, y∈R, 有

$ P\left( X\le x,Y\le y \right)\le P\left( X\le x \right)P\left( Y\le y \right), $

则称X和Y是NQD(negatively quadrant dependent)的; 若∀i≠j, X_i和X_j是NQD的, 则称随机变量序列{X_n, n≥1}是两两NQD列.

上述定义由统计学家LEHMAMN^[1]于1966年提出.两两NQD列是一类非常广泛的随机变量序列, 例如两两独立的随机变量列以及NA(negative association)^[2]列就是其特例.因此, 对两两NQD列的研究具有重要的理论意义和应用价值.目前, 关于两两NQD列的极限研究已有许多结果, 详见文献[3-7].近年来, GUT等^[8]获得了一类独立同分布且均值为零的随机变量序列的完全收敛性, 吴群英等^[9]将其推广到NA随机变量序列的情形.受上述文献启发, 本文将文献[9]的定理1.3推广到更为一般的行为两两NQD随机变量阵列加权和的情形.

定义2  随机变量阵列{X_ni, i≥1, n≥1}被随机变量X控制是指:如果存在一常数C>0, 使得对任意的n≥1, i≥1, x≥0, 有P(|X_ni|≥x)≤CP(|X|≥x).

本文中, a_n≪b_n表示存在常数c>0, 使得对所有n≥1, 有a_n≤cb_n. a_n≈b_n表示对所有n≥1, 有a_n≪b_n以及b_n≪a_n.I(A)表示集合A的示性函数, #A表示集合A中元素的个数.

1 引理

引理1^[3]  设随机变量X和Y是NQD的, 则

(1)EXY≤EXEY;

(2) 对∀x, y∈R, 有P(X>x, Y>y)≤P(X>x)P(Y>y);

(3) 若f, g同为非降(或非增)函数, 则f(X)与g(Y)仍为NQD的.

引理2^[3]  设{X_n, n≥1}是两两NQD列, 且EX_n² < ∞(n≥1), 记

$ {T_j}\left( k \right) = \sum\limits_{i = j + 1}^{j + k} {\left( {{X_i} - E{X_i}} \right)} ,j \ge 0, $

则有

(1)E(T_j(k)²)≤$\sum\limits_{i=j+1}^{j+k}{EX_{i}^{2}}$;

(2)$E\left[ {\mathop {\max }\limits_{1 \le k \le n} {{\left( {{T_j}\left( k \right)} \right)}^2}} \right] \le \frac{{4{{\ln }^2}n}}{{{{\ln }^2}2}}\sum\limits_{i = j + 1}^{j + k} {EX_i^2} .$

引理3^[7]  令{X_n, n≥1}是两两NQD随机变量列, 那么对于任意的n≥1以及x≥0, 有

$ \begin{array}{l} {\left( {1 - P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {{X_k}} \right| > x} \right)} \right)^2}\sum\limits_{k = 1}^n {P\left( {\left| {{X_k}} \right| > x} \right)} \le \\ \;\;\;\;\;\;2P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {{X_k}} \right| > x} \right). \end{array} $

2 主要结果及证明

定理1   {X_ni, i≥1, n≥1}是行为两两NQD随机变量阵列, 均值为零, 且被随机变量X随机控制, {a_nk, k≥1, n≥1}是一个实数阵列, 令α>1, ${{\sum\limits_{i=1}^{\infty }{\left| {{a}_{ni}} \right|}}^{2}}$=O(exp(-ln^αn)), 记S_ni=$\sum\limits_{k=1}^{i}{{{a}_{nk}}}{{X}_{nk}}$, 则

$ E\left| X \right|\exp \left( {{{\ln }^\alpha }\left| X \right|} \right) < \infty $

(1)

$ \Leftrightarrow \sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^2}}}P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {{S_{nk}}} \right| > n\beta } \right) < \infty \left( {\beta > 1} \right)} . $

(2)

证明  先证式(1)⇒式(2).

不失一般性，本文假设a_ni>0, ∀β>1，b_n=$\frac{{{\beta }_{n}}}{{{\ln }^{\alpha }}n}$，其中n≥1.

令

$ \begin{array}{l} X_{ni}^{\left( 1 \right)} = - {b_n}I\left( {{a_{ni}}{X_{ni}} < - {b_n}} \right) + \\ \;\;\;\;\;\;\;\;{a_{ni}}{X_{ni}}I\left( {\left| {{a_{ni}}{X_{ni}}} \right| \le {b_n}} \right) + {b_n}I\left( {{a_{ni}}{X_{ni}} > {b_n}} \right), \end{array} $

$ X_{ni}^{\left( 2 \right)} = \left( {{a_{ni}}{X_{ni}} - {b_n}} \right)I\left( {{b_n} < {a_{ni}}{X_{ni}} < n} \right), $

$ X_{ni}^{\left( 3 \right)} = \left( {{a_{ni}}{X_{ni}} + {b_n}} \right)I\left( { - n < {a_{ni}}{X_{ni}} < - {b_n}} \right), $

$ \begin{array}{l} X_{ni}^{\left( 4 \right)} = \left( {{a_{ni}}{X_{ni}} + {b_n}} \right)I\left( {{a_{ni}}{X_{ni}} < - n} \right) + \\ \;\;\;\;\;\;\;\;\;\;\left( {{a_{ni}}{X_{ni}} - {b_n}} \right)I\left( {{a_{ni}}{X_{ni}} > n} \right). \end{array} $

显然有$S_{nk}^{\left( j \right)}=\sum\limits_{i=1}^{k}{X_{ni}^{\left( j \right)}}$，j=1, 2, 3, 4, 1≤k≤n, n≥1,

$ {S_{nk}} = \sum\limits_{j = 1}^4 {S_{nk}^{\left( j \right)}} . $

注意到

$ \left( {\mathop {\max }\limits_{1 \le k \le n} \left| {{S_{nk}}} \right| > n\beta } \right) \subseteq \bigcup\limits_{j = 1}^4 {\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {S_{nk}^{\left( j \right)}} \right| > \frac{{n\beta }}{4}} \right).} $

因此欲证式(2)，只需证明

$ {I_j} = \sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^2}}}P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {S_{nk}^{\left( j \right)}} \right| > \frac{{n\beta }}{4}} \right) < \infty } , $

j=1, 2, 3, 4即可.

由EX_ni=0及Markov不等式，先证

$ {n^{ - 1}}\mathop {\max }\limits_{1 \le k \le n} E\left| {S_{nk}^{\left( j \right)}} \right| \le $

$ \begin{array}{l} {n^{ - 1}}\sum\limits_{i = 1}^n {\left( {E\left| {{a_{ni}}{X_{ni}}} \right|I\left( {\left| {{a_{ni}}{X_{ni}}} \right| > {b_n}} \right) + } \right.} \\ \;\;\;\;\;\;\left. {{b_n}P\left( {\left| {{a_{ni}}{X_{ni}}} \right| > {b_n}} \right)} \right) \le \end{array} $

$ \begin{array}{l} {n^{ - 1}}\sum\limits_{i = 1}^n {\left( {E\left| {{a_{ni}}{X_{ni}}} \right|\frac{{\left| {{a_{ni}}{X_{ni}}} \right|}}{{{b_n}}}I\left( {\left| {{a_{ni}}{X_{ni}}} \right| > {b_n}} \right) + } \right.} \\ \;\;\;\;\;\;\;\left. {b\frac{{E{{\left| {{a_{ni}}{X_{ni}}} \right|}^2}}}{{b_n^2}}} \right) \ll {n^{ - 1}}b_n^{ - 1}\sum\limits_{i = 1}^n {E{{\left| {{a_{ni}}{X_{ni}}} \right|}^2}} \ll \end{array} $

$ {n^{ - 1}}b_n^{ - 1}E{\left| X \right|^2}\sum\limits_{i = 1}^n {{{\left| {{a_{ni}}} \right|}^2}} \ll $

$ {n^{ - 1}}\frac{{{{\ln }^\alpha }n}}{n}\exp \left( {{{\ln }^\alpha }n} \right) \to 0,n \to \infty . $

令${{\tilde I}_1} = \sum\limits_{n = 1}^\infty {\exp } \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^2}}}P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {S_{nk}^{\left( 1 \right)} - ES_{nk}^{\left( 1 \right)}} \right| > \frac{{n\beta }}{4}} \right) $, 欲证I₁ < ∞，只需证明${{{\tilde{I}}}_{1}}$ < ∞即可.

E|X|exp(ln^α|X|) < ∞蕴含着$E{{X}^{2}}<\infty \left( E\left| X \right|\exp \left( {{\ln }^{\alpha }}\left| X \right| \right)>E{{X}^{2}},\alpha >1 \right)$、Markov不等式及引理2.

$ \begin{array}{l} {{\tilde I}_1} \ll \sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^4}}}E\left( {\mathop {\max }\limits_{1 \le k \le n} {{\left( {S_{nk}^{\left( 1 \right)} - ES_{nk}^{\left( 1 \right)}} \right)}^2}} \right)} \ll \\ \;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^4}}}\frac{{4{{\ln }^2}n}}{{{{\ln }^2}2}}\left( {\sum\limits_{i = 1}^\infty {E{{\left( {X_{ni}^{\left( 1 \right)}} \right)}^2}} } \right)} \ll \\ \;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha + 1}}n}}{{{n^4}}}} \sum\limits_{i = 1}^n {\left( {E{{\left| {{a_{ni}}{X_{ni}}} \right|}^2} \times } \right.} \\ \;\;\;\;\;\;\;\left. {I\left( {\left| {{a_{ni}}{X_{ni}}} \right| \le {b_n}} \right) + b_n^2P\left( {\left| {{a_{ni}}{X_{ni}}} \right| > {b_n}} \right)} \right) \ll \\ \;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha + 1}}n}}{{{n^4}}}} \sum\limits_{i = 1}^n {E{{\left| {{a_{ni}}{X_{ni}}} \right|}^2} \ll } \\ \;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha + 1}}n}}{{{n^4}}}} \sum\limits_{i = 1}^n {{{\left| {{a_{ni}}} \right|}^2} \ll } \\ \;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {\frac{{{{\ln }^{\alpha + 1}}n}}{{{n^4}}}} < \infty . \end{array} $

对于I₂，根据X_ni⁽²⁾的定义, X_ni⁽²⁾>0,S_nk⁽²⁾>0,且有

$ \begin{array}{l} \left( {\mathop {\max }\limits_{1 \le k \le n} \left| {S_{nk}^{\left( 2 \right)}} \right| > \frac{{n\beta }}{4}} \right) = \left( {\sum\limits_{n = 1}^n {X_{ni}^{\left( 2 \right)}} > \frac{{n\beta }}{4}} \right) = \\ \;\;\;\;\;\;\;\left( {\sum\limits_{i = 1}^n {\left( {{a_{ni}}{X_{ni}} - {b_n}} \right)I\left( {{b_n} < {a_{ni}}{X_{ni}} < n} \right)} > \frac{{n\beta }}{4}} \right) \subseteq \\ \;\;\;\;\;\;\;\left( {至少存在2个\;k,使得\;{a_{nk}}{X_{nk}} > {b_n}} \right). \end{array} $

则

$ \begin{array}{l} P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {S_{nk}^{\left( 2 \right)}} \right| > \frac{{n\beta }}{4}} \right) = P\left( {\sum\limits_{i = 1}^n {X_{ni}^{\left( 2 \right)}} > \frac{{n\beta }}{4}} \right) \le \\ \;\;\;\;\;\;\;\;\;\;\sum\limits_{1 \le {i_1} < {i_2} \le n} {P\left( {{a_{n{i_1}}}{X_{n{i_1}}} > {b_n},{a_{n{i_2}}}{X_{n{i_2}}} > {b_n}} \right)} \le \\ \;\;\;\;\;\;\;\;\;\;\sum\limits_{1 \le {i_1} < {i_2} \le n} {\prod\limits_{{\rm{j}} = 1}^2 {P\left( {{a_{n{i_j}}}{X_{n{i_j}}} > {b_n}} \right)} } \le \\ \;\;\;\;\;\;\;\;\;\;{\left( {\sum\limits_{i = 1}^n {P\left( {\left| {{a_{ni}}X} \right| > {b_n}} \right)} } \right)^2} \le \\ \;\;\;\;\;\;\;\;\;\;{\left( {\sum\limits_{i = 1}^n {b_n^{ - 2}E{{\left| {{a_{ni}}X} \right|}^2}} } \right)^2} \ll \\ \;\;\;\;\;\;\;\;\;\;{\left( {b_n^{ - 2}\sum\limits_{i = 1}^n {{{\left| {{a_{ni}}} \right|}^2}} } \right)^2} \ll {\left( {b_n^{ - 2}\exp \left( { - {{\ln }^\alpha }n} \right)} \right)^2}. \end{array} $

所以

$ \begin{array}{l} {I_2} = \sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^2}}}P\left( {\sum\limits_{i = 1}^n {X_{ni}^{\left( 2 \right)}} > \frac{{n\beta }}{4}} \right)} \ll \\ \;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^2}}} \cdot b_n^{ - 4} \cdot {{\left( {\exp \left( { - {{\ln }^\alpha }n} \right)} \right)}^2}} \ll \\ \;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {\exp \left( { - {{\ln }^\alpha }n} \right)\frac{{{{\ln }^{3\alpha - 1}}n}}{{{n^6}}} < \infty } . \end{array} $

对于I₃，根据X_ni⁽³⁾的定义, X_ni⁽³⁾ < 0, S_nk⁽³⁾ < 0, 同理可得I₃ < ∞.

最后证I₄ < ∞.根据X_ni⁽⁴⁾的定义,

$ \begin{array}{l} P\left( {\mathop {\max }\limits_{1 \le k \le n} S_{nk}^{\left( 4 \right)} > \frac{{n\beta }}{4}} \right) = P\left( {\sum\limits_{i = 1}^n {X_{ni}^{\left( 4 \right)}} > \frac{{n\beta }}{4}} \right) \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;P\left( {\bigcup\limits_{i = 1}^n {\left( {\left| {{a_{ni}}{X_{ni}}} \right| > n} \right)} } \right) \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{i = 1}^n {P\left( {\left| {{a_{ni}}{X_{ni}}} \right| > n} \right)} . \end{array} $

令I_nj={i∈J; (j+1)^-1 < |a_ni|≤j^-1}, j=1, 2, …, 则有$\mathop \cup \limits_{j \ge 1} {I_{nj}} = J$, 由文献[10]有

$ \sum\limits_{j = 1}^\infty {\# {I_{nj}}} \le n\left( {k + 1} \right). $

则

$ \begin{array}{l} \sum\limits_{i = 1}^n {P\left( {\left| {{a_{ni}}{X_{ni}}} \right| > n} \right)} \ll \sum\limits_{j = 1}^\infty {\sum\limits_{i \in {I_{nj}}} {P\left( {\left| X \right| > jn} \right)} } = \\ \;\;\;\;\;\;\sum\limits_{j = 1}^\infty {\left( {\# {I_{nj}}} \right)\sum\limits_{k \ge jn} {P\left( {k < \left| X \right| \le k + 1} \right)} } = \\ \;\;\;\;\;\;\sum\limits_{k = n}^\infty {\sum\limits_{j = 1}^{\left[ {k/n} \right]} {\left( {\# {I_{nj}}} \right)P\left( {k < \left| X \right| \le k + 1} \right)} } \ll \\ \;\;\;\;\;\;\sum\limits_{k = n}^\infty {n\left( {\frac{k}{n} + 1} \right)P\left( {k < \left| X \right| \le k + 1} \right)} = \\ \;\;\;\;\;\;\sum\limits_{k = n}^\infty {\left( {k + n} \right)P\left( {k < \left| X \right| \le k + 1} \right)} . \end{array} $

所以,

$ \begin{array}{l} {I_4} \ll \sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^2}}}} \sum\limits_{k = n}^\infty {\left( {k + n} \right)P\left( {k < \left| X \right| \le k + 1} \right)} \ll \\ \;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^2}}}} \sum\limits_{k = n}^\infty {kP\left( {k < \left| X \right| \le k + 1} \right)} = \\ \;\;\;\;\;\;\;\sum\limits_{k = 1}^\infty {kP\left( {k < \left| X \right| \le k + 1} \right)} \sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^2}}}} \approx \\ \;\;\;\;\;\;\;\sum\limits_{k = 1}^\infty {k\exp \left( {{{\ln }^\alpha }k} \right)EI\left( {k < \left| X \right| \le k + 1} \right)} \approx \\ \;\;\;\;\;\;\;\sum\limits_{k = 1}^\infty {E\left| X \right|\exp \left( {{{\ln }^\alpha }\left| X \right|} \right)I\left( {k < \left| X \right| \le k + 1} \right)} \approx \\ \;\;\;\;\;\;\;E\left| X \right|\exp \left( {{{\ln }^\alpha }\left| X \right|} \right) < \infty . \end{array} $

现在, 证明式(2)⇒式(1).显然, 式(2) 蕴含着

$ \sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^2}}}} P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {{a_{nk}}{X_{nk}}} \right| > n} \right) < \infty . $

(3)

则有

$ P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {{a_{nk}}{X_{nk}}} \right| > n} \right) \to 0,n \to \infty . $

因此, 对于足够大的n，有

$ P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {{a_{nk}}{X_{nk}}} \right| > n} \right) < \frac{1}{2}. $

由引理1得{a_niX_ni}仍然为两两NQD阵列.由引理3, 有

$ \sum\limits_{k = 1}^n {P\left( {\left| {{a_{nk}}{X_{nk}}} \right| > n} \right)} \le 8P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {{a_{nk}}{X_{nk}}} \right| > n} \right). $

(4)

由式(3) 及式(4), 有

$ \sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^2}}}} \sum\limits_{k = 1}^n {P\left( {\left| {{a_{nk}}{X_{nk}}} \right| > n} \right) < \infty } . $

根据I₄ < ∞的证明过程,

$ \begin{array}{l} E\left| X \right|\exp \left( {{{\ln }^\alpha }\left| X \right|} \right) \approx \\ \;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^2}}}} \sum\limits_{k = 1}^n {P\left( {\left| {{a_{nk}}{X_{nk}}} \right| > n} \right)} \le \\ \;\;\;\;\;\;\;8\sum\limits_{n = 1}^\infty {\exp \left( {{{\ln }^\alpha }n} \right)\frac{{{{\ln }^{\alpha - 1}}n}}{{{n^2}}}} \sum\limits_{k = 1}^n {P\left( {\left| {{a_{nk}}{X_{nk}}} \right| > n} \right) < \infty } . \end{array} $

证毕.

注  文献[9]定理1.3描述的是一类NA随机变量序列的完全收敛性定理，本文将此定理推广到更一般的行为两两NQD随机变量阵列加权和的情形，在增加权条件的基础上，通过采用不同的截尾方法，亦得到了类似的结论.