定义1.1 设$\{ {X_n};n \ge 1\} $是概率空间$\left( {\Omega, F, P} \right)$的一列随机变量, 集合$\Gamma _n^ - = \sigma ({X_i};$ $1 \le i \le n ), $ $\Gamma _n^ + = \sigma \left( {{X_i};i \ge n} \right).$定义

$ \rho \left( n \right) = \mathop {\sup }\limits_{k \ge 1} \mathop {\sup }\limits_{X \in {L_2}\left( {\Gamma _k^ - } \right), Y \in {L_2}\left( {\Gamma _{k + n}^ + } \right)} \frac{{\left| {EXY - EXEY} \right|}}{{\sqrt {E{{\left( {X - EX} \right)}^2}{{\left( {Y - EY} \right)}^2}} }}, $

如果$n \to \infty $时, $\rho \left( n \right) \to 0$, 则称序列$\{ {X_n};n \ge 1\} $是$\rho $ -混合的.

$\rho $ -混合序列首先被Kolmogorov和Rozanov提出, 随后, 很多学者对其极限性质做了很多研究, 已取得许多成果.文献[1]得到了$\rho $ -混合序列部分和最大值不等式; 文献[2]得到了Baum-Katz大数定律的精确渐近性; 文献[3]得到了$\rho$ -混合序列完全矩收敛的精确率等.最近, 文献[4]得到了关于独立同分布随机变量序列完全收敛精确渐近性的一般结果, 即

定理1.1 设$\{ X, {X_n}, n \ge 1\} $为独立同分布的随机变量序列, $EX = 0, $ $E{X^2} = {\sigma ^2}, $ $E{X^2}{\log ^ + }\left| X \right| < \infty .$进一步假设$g(x)$为$[{n_0}, \infty )$上具有非负导数$g'\left( x \right)$的正函数, 且满足$g\left( x \right) \uparrow \infty, $ $x \to \infty .$ $g'\left( x \right)$在$[{n_0}, \infty )$单调, 当$g'\left( x \right)$单调非降时, 满足$\mathop {\lim }\limits_{n \to \infty } \frac{{g'\left( {n + 1} \right)}}{{g'\left( n \right)}} = 1.$ $\phi \left( x \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}}$在$[{n_0}, \infty )$单调, 当$\phi \left( x \right)$单调非降时, 满足$\mathop {\lim }\limits_{n \to \infty } \frac{{\phi \left( {n + 1} \right)}}{{\phi \left( n \right)}} = 1.$且满足$\mathop {\lim \sup }\limits_{n \to \infty } n\phi \left( n \right) < \infty .$则有

$ \mathop {\lim }\limits_{\varepsilon \downarrow 0}- \frac{1}{{\log \varepsilon }}{\sum\limits_{n = {n_0}}^\infty {\phi \left( n \right)E\left| {\frac{{{S_n}}}{{\sqrt n }}} \right|} ^2}I\left( {\left| {{S_n}} \right| \ge \varepsilon \sqrt {ng\left( n \right)} } \right) = 2{\sigma ^2}. $

对独立同分布的随机变量序列很容易得到定理1.1的结论成立, 我们自然希望类似的结论对$\rho $-混合序列也成立, 这正是本文要解决的问题.本文的目的就是证明在一定条件下得到类似定理1.1的结果.

在整篇文章中, 设$\{ X, {X_n};n \ge 1\} $是均值为零的严平稳$\rho $-混合随机变量序列,

$ E{X_1} = 0, ~~ {S_n} = \sum\limits_{i = 1}^n {{X_i}}, ~~ {\sigma ^2} = EX_1^2 + 2\sum\limits_{n = 2}^\infty {E{X_1}{X_n}} > 0, $

$ \left\lfloor x \right\rfloor = \sup \left\{ {m:m \le x, m \in Z} \right\}, ~~ \left\| {{X_1}I\left( {\left| {{X_1}} \right| \le x} \right)} \right\|_2^q = {\left( {EX_1^2I\left( {\left| {{X_1}} \right| \le x} \right)} \right)^{\frac{q}{2}}}, $

$C$在不同的位置表示不同的常数.本文的主要结果如下.

定理1.2  设$g(x)$为$[{n_0}, \infty )$上具有非负导数$g'\left( x \right)$的正函数, 且满足$g\left( x \right) \uparrow \infty, $ $x \to \infty .$ $g'\left( x \right)$在$[{n_0}, \infty )$单调, 当$g'\left( x \right)$单调非降时, 满足$\mathop {\lim }\limits_{n \to \infty } \frac{{g'\left( {n + 1} \right)}}{{g'\left( n \right)}} = 1.$ $\psi \left( x \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}}$在$[{n_0}, \infty )$单调, 当$\psi \left( x \right)$单调非降时, 满足$\mathop {\lim }\limits_{n \to \infty } \frac{{\psi \left( {n + 1} \right)}}{{\psi \left( n \right)}} = 1.$且满足

$ \begin{equation} \mathop {\lim \sup }\limits_{n \to \infty } \frac{{g'\left( x \right)}}{{g\left( x \right)}} < \infty, ~~~~ \mathop {\lim \sup }\limits_{n \to \infty } \frac{{g\left( n \right)}}{n} < \infty . \end{equation} $

(1.1)

如果$E{X^{{\rm{2}} + \alpha }}{\log ^ + }\left| X \right| < \infty, $ $\alpha > 0, $ $\mathop {\lim }\limits_{n \to \infty } \frac{{ES_n^2}}{n} = {\sigma ^2} > 0, $且存在常量$q > 2, $使得$\sum\limits_{n = 1}^\infty {{\rho ^{\frac{2}{q}}}({2^n})} < \infty .$则有

$ \begin{equation} \mathop {\lim }\limits_{\varepsilon \downarrow 0}- \frac{1}{{\log \varepsilon }}{\sum\limits_{n = {n_0}}^\infty {\psi \left( n \right)E\left| {\frac{{{S_n}}}{{\sqrt n }}} \right|} ^2}I\left( {\left| {{S_n}} \right| \ge \varepsilon \sigma \sqrt {ng\left( n \right)} } \right) = 2. \end{equation} $

(1.2)

注1.1 满足定理1.2中假设条件的$g\left( x \right)$有很多, 比如$g\left( x \right) = {x^\alpha }, $ ${\left( {\log x} \right)^\beta }, $ ${\left( {\log \log x} \right)^\gamma }$, 其中$\alpha > 0, $ $\beta > 0, $ $\gamma > 0$为某些适当的参数.

注1.2 在定理1.2中取$g\left( x \right) = x$可以得到文献[5]的定理1.

为了证明定理1.2, 首先我们介绍下面三个引理.

引理2.1^[6]  设$\left\{ {{X_n};n \ge 1} \right\}$为严平稳的$\rho $ -混合序列, $E{X_1} = 0$, $EX_1^2 < \infty, $ $\sum\limits_{n = 1}^\infty {\rho ({2^n})} < \infty, $如果$\sigma _n^2\hat = ES_n^2 \to \infty $, $n \to \infty $, 则有$S_n / \sigma _n \to N(0, 1)$, $n \to \infty .$

引理2.2^[7]  设$\left\{ {{X_n};n \ge 1} \right\}$为严平稳的$\rho $ -混合序列, $EX = 0, $ $E{\left| X \right|^{\rm{p}}} < \infty \left( {p \ge 2} \right)$, 且$\mathop {\lim }\limits_{n \to \infty } \frac{{ES_n^2}}{n} = {\sigma ^2} > 0, $ $\sum\limits_{n = 1}^\infty {{\rho ^{\frac{2}{q}}}({2^n})} < \infty, $存在$q \ge p.$函数$g\left( x \right)$是具有非负导数$g'\left( x \right)$的正函数, 且满足$g\left( x \right) \uparrow \infty, $ $x \to \infty .$ $g'\left( x \right)$在$[{n_0}, \infty )$单调, 当$g'\left( x \right)$单调非降时, 满足$\mathop {\lim }\limits_{n \to \infty } \frac{{g'\left( {n + 1} \right)}}{{g'\left( n \right)}} = 1.$存在常数$\delta > 0, $使得$\mathop {\lim \sup }\limits_{n \to \infty } \frac{{g\left( n \right)}}{{{n^\delta }}} < \infty, $则对任意的$v > \max \left( {\frac{1}{q}, \frac{1}{p} + \frac{{2- p}}{{2p\delta }}} \right), $都有

$ \mathop {\lim }\limits_{\varepsilon \downarrow 0} {\varepsilon ^{\frac{1}{v}}}\sum\limits_{n = 1}^\infty {g'\left( n \right)P\left( {\left| {{S_n}} \right| \ge \varepsilon \sigma \sqrt {ng\left( n \right)} } \right)} = E{\left| N \right|^{\frac{1}{v}}}. $

引理2.3^[8]  设$\left\{ {{X_n};n \ge 1} \right\}$为严平稳的$\rho $ -混合序列, $E{X_n} = 0, $ ${S_n} = \sum\limits_{i = 1}^n {{X_i}}, $则对任意的$q \ge 2, $存在常数$ K = K(q, \rho ( \cdot )), $使得对任意的$x > 0$和$y > 0$满足

$ 2n\mathop {\max }\limits_{1 \le i \le n} E\left| {{X_i}} \right|I\left( {\left| {{X_i}} \right| \ge y} \right) \le x, $

有

$ \begin{eqnarray*} P\left( {\max \left| {{S_n}} \right| \ge x} \right) &\leq& \sum\limits_{i = 1}^n {P\left( {\left| {{X_i}} \right| \ge y} \right)}\\ &&+ Kx^{ - q} n^{q/2} \exp \bigg( {K\sum\limits_{i = 1}^{[\log n]} {\rho \left( {{2^i}} \right)} } \bigg)\mathop {\max }\limits_{1 \le i \le n} \left\| {{X_i}I\left( {\left| {{X_i}} \right| \le y} \right)} \right\|_2^q\\ &&+ K{x^{ - q}}n\exp \bigg( {K\sum\limits_{i = 1}^{[\log n]} { \rho ^{2/q} \left( {{2^i}} \right)} } \bigg)\mathop {\max }\limits_{1 \le i \le n} E{\left| {{X_i}} \right|^q}I\left( {\left| {{X_i}} \right| \le y} \right). \end{eqnarray*} $

不失一般性, 不妨假设${\sigma ^2} = 1$, 记$A\left( \varepsilon \right) = \left\lfloor {{g^{ - 1}}\left( {{\varepsilon ^{ - 2}}} \right)} \right\rfloor, $其中${g^{ - 1}}\left( x \right)$是$g\left( x \right)$的反函数.当$E{\left| X \right|^p} < \infty, $ $0 < p < \infty, $对任意的常量$b > 0, $则有

$ \begin{eqnarray} E{\left| X \right|^p}I\left( {\left| X \right| \ge b} \right) = {b^p}P\left( {\left| X \right| \ge b} \right) + p\int_b^\infty {{x^{p - 1}}P\left( {\left| X \right| \ge x} \right)} {\rm d}x. \end{eqnarray} $

(3.1)

因此, 式(1.2) 中求和部分可以转化为

$ \begin{eqnarray*} &&\sum\limits_{n = {n_0}}^\infty {\psi \left( n \right)E{{\left| {\frac{{{S_n}}}{{\sqrt n }}} \right|}^2}} I\left( {\left| {{S_n}} \right| \ge \varepsilon \sqrt {ng\left( n \right)} } \right)\\ &=&{\varepsilon ^2}\sum\limits_{n = {n_0}}^\infty {g'\left( n \right)} P\left( {\left| {{S_n}} \right| \ge \varepsilon \sqrt {ng\left( n \right)} } \right) +\sum\limits_{n = {n_0}}^\infty {\psi \left( n \right)} \int_{\varepsilon \sqrt {g\left( n \right)} }^\infty {2xP\left( {\left| {{S_n}} \right| \ge \sqrt n x} \right){\rm d}x.} \end{eqnarray*} $

在引理2.2中取$v = \frac{1}{2}$, 则有

$ \begin{eqnarray} \mathop {\lim }\limits_{\varepsilon \downarrow 0} - \frac{{{\varepsilon ^2}}}{{\log \varepsilon }}\sum\limits_{n = {n_0}}^\infty {g'\left( n \right)} P\left( {\left| {{S_n}} \right| \ge \varepsilon \sqrt {ng\left( n \right)} } \right) = 0. \end{eqnarray} $

(3.2)

为了证明式(1.2), 我们只需要证明

$ \begin{eqnarray} \mathop {\lim }\limits_{\varepsilon \downarrow 0} - \frac{1}{{\log \varepsilon }}\sum\limits_{n = {n_0}}^\infty {\psi \left( n \right)} \int_{\varepsilon \sqrt {g\left( n \right)} }^\infty {2xP\left( {\left| {{S_n}} \right| \ge \sqrt n x} \right){\rm d}x = 2.} \end{eqnarray} $

(3.3)

为了证明式(3.3), 我们需要下列命题.

命题3.1 设$N$为标准正态随机变量.有

$ \begin{eqnarray} \mathop {\lim }\limits_{\varepsilon \downarrow 0} - \frac{1}{{\log \varepsilon }}\sum\limits_{n = {n_0}}^\infty {\psi \left( n \right)} \int_{\varepsilon \sqrt {g\left( n \right)} }^\infty {2xP\left( {\left| N \right| \ge x} \right){\rm d}x = 2.} \end{eqnarray} $

(3.4)

证 参见文献[4]的性质4.1.

命题3.2 在定理1.2的条件下, 有

$ \begin{eqnarray} \mathop {\lim }\limits_{\varepsilon \downarrow 0} - \frac{1}{{\log \varepsilon }}\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)} \left| {\int_{\varepsilon \sqrt {g\left( n \right)} }^\infty {2xP\left( {\left| {{S_n}} \right| \ge \sqrt n x} \right){\rm d}x - \int_{\varepsilon \sqrt {g\left( n \right)} }^\infty {2xP\left( {\left| N \right| \ge x} \right){\rm d}x} } } \right| = 0.\label{eq7} \end{eqnarray} $

(3.5)

证  记${\Lambda _n} = \mathop {\sup }\limits_x \left| {P\left( {{S_n} \ge \sqrt n x} \right) - P\left( {\left| N \right| \ge x} \right)} \right|.$对任意的$x \ge 0, $ $P\left( {\left| N \right| \ge x} \right)$是一连续函数.由引理2.1, 则有$\mathop {\lim }\limits_{n \to \infty } {\Lambda _n} = 0.$

$ \begin{eqnarray*} && \sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)} \left| {\int_{\varepsilon \sqrt {g\left( n \right)} }^\infty {2xP\left( {\left| {{S_n}} \right| \ge \sqrt n x} \right){\rm d}x- \int_{\varepsilon \sqrt {g\left( n \right)} }^\infty {2xP\left( {\left| N \right| \ge x} \right){\rm d}x} } }\right|\\ &=&\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)} \left| {\int_0^{\Lambda _n^{ - \frac{1}{4}}} {2\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)P\left( {\left| {{S_n}} \right| \ge \sqrt n \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)} \right){\rm d}x}} \right.\\ &&-\left. \int_0^{\Lambda _n^{ - \frac{1}{4}}} {2\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)P\left( {\left| N \right| \ge \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)} \right){\rm d}x} \right|\\ &&+\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)} \left| {\int_{\Lambda _n^{ - \frac{1}{4}}}^\infty {2\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)P\left( {\left| N \right| \ge \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)} \right){\rm d}x} } \right|\\ &&+\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)} \left| {\int_{\Lambda _n^{ - \frac{1}{4}}}^\infty {2\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)P\left( {\left| {{S_n}} \right| \ge \sqrt n \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)} \right){\rm d}x} } \right|\\ &=&\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)} \left( {{B_1} + {B_2} + {B_3}} \right). \end{eqnarray*} $

现在, 我们估计${B_1}$和${B_2}$, 由Markov不等式及Toeplitz's引理, 则有

$ \begin{array}{l} - \frac{1}{{\log \varepsilon }}\mathop \sum \limits_{n = {n_0}}^{A\left( \varepsilon \right)} \psi \left( n \right){B_1} \le - \frac{1}{{\log \varepsilon }}\mathop \sum \limits_{n = {n_0}}^{A\left( \varepsilon \right)} \psi \left( n \right)\int_0^{\Lambda _n^{ - \frac{1}{4}}} {2\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right){\Lambda _n}} {\rm{d}}x\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \le - \frac{C}{{\log \varepsilon }}\mathop \sum \limits_{n = {n_0}}^{A\left( \varepsilon \right)} \psi \left( n \right){\Lambda _n}{\left( {\Lambda _n^{ - \frac{1}{4}} + \varepsilon \sqrt {g\left( n \right)} } \right)^2}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \le - \frac{C}{{\log \varepsilon }}\mathop \sum \limits_{n = {n_0}}^{A\left( \varepsilon \right)} \psi \left( n \right){\left( {\Lambda _n^{1/4} + \Lambda _n^{1/2}} \right)^2} \to 0,\varepsilon \downarrow 0. \end{array} $

(3.6)

$ \begin{array}{l} - \frac{1}{{\log \varepsilon }}\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right){B_2}} \le - \frac{C}{{\log \varepsilon }}\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)\int_{\Lambda _n^{ - \frac{1}{4}}}^\infty {\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)\frac{1}{{{{\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)}^4}}}} {\rm{d}}x} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \le - \frac{C}{{\log \varepsilon }}\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)} \Lambda _n^{1/2} \to 0,\varepsilon \downarrow 0. \end{array} $

(3.7)

在引理2.3中取$x = \sqrt n \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right), $ $y = 2\sqrt n \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)$, 当$x \to \infty, $则有

$ \begin{eqnarray} &&\frac{{2n\mathop {\max }\limits_{1 \le i \le n} E\left| {{X_i}} \right|I\left( {\left| {{X_i}} \right| \ge {\rm{2}}\sqrt n \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)} \right)}}{{\sqrt n \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)}}\\ &\leq&C\frac{{E{{\left| {{X_{\rm{1}}}} \right|}^{\rm{2}}}I\left( {\left| {{X_{\rm{1}}}} \right| \ge {\rm{2}}\sqrt n \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)} \right)}}{{{{\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)}^{\rm{2}}}}} \le \frac{{\rm{C}}}{{{x^2}}} \to 0. \end{eqnarray} $

由引理2.3, 我们将${B_3}$分成三部分, 则有

$ \begin{eqnarray*} &&\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)} {B_3} \\ &\leq& C\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {n\psi \left( n \right)} \int_{\Lambda _n^{ - \frac{1}{4}}}^\infty {\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)P\left( {\left| {{X_1}} \right| \ge 2\sqrt n \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)} \right)} {\rm d}x\\ &&+C\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)} {\int_{\Lambda _n^{ - \frac{1}{4}}}^\infty {{{\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)}^{1 - q}}\left( {EX_1^2I\left( {\left| {{X_1}} \right| \le 2\sqrt n \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)} \right)} \right)} ^{^{\frac{q}{2}}}}{\rm d}x\\ &&+C\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\frac{{\psi \left( n \right)}}{{{n^{\frac{q}{2}{\rm{ - 1}}}}}}} {\int_{\Lambda _n^{ - \frac{1}{4}}}^\infty {{{\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)}^{1 - q}}E\left| {{X_1}} \right|} ^q}I\left( {\left| {{X_1}} \right| \le 2\sqrt n \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)} \right){\rm d}x\\ &=& {B_{31}} + {B_{32}} + {B_{33}}. \end{eqnarray*} $

我们来分别处理${B_{31}}$和${B_{32}}$, 由Markov不等式、Toeplitz's引理及Fubini定理, $0 \le {\Lambda _n} \le 1, $则有

$ \begin{eqnarray} - \frac{1}{{\log \varepsilon }}{B_{31}} &\leq& - \frac{C}{{\log \varepsilon }}\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {n\psi \left( n \right)E\left( {\int_{\Lambda _n^{ - \frac{1}{4}}}^\infty {\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)} I\left( {\frac{{\left| {{X_1}} \right|}}{{2\sqrt n }} \ge x + \varepsilon \sqrt {g\left( n \right)} } \right){\rm d}x} \right)}\\ &\leq&- \frac{C}{{\log \varepsilon }}\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {n\psi \left( n \right)} \int_{\rm{0}}^\infty {\frac{{EI\left( {\left| {{X_1}} \right| \ge 2\sqrt n } \right)}}{{{{\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)}^{ - 1}}}}} I\left( {x + \varepsilon \sqrt {g\left( n \right)} \le \frac{{\left| {{X_1}} \right|}}{{{\rm{2}}\sqrt n }}} \right){\rm d}x\\ &\leq& - \frac{C}{{\log \varepsilon }}\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {n\psi \left( n \right)EI\left( {\left| {{X_1}} \right| \ge {\rm{2}}\sqrt {\rm{n}} } \right)} \int_{\rm{0}}^{\frac{{\left| {{{\rm{X}}_{\rm{1}}}} \right|}}{{{\rm{2}}\sqrt {\rm{n}} }}{\rm{ - }}\varepsilon \sqrt {g\left( n \right)} } {\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)} {\rm d}x\\ &\leq& - \frac{C}{{\log \varepsilon }}\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)EX_1^2I\left( {\left| {{X_1}} \right| \ge 2\sqrt n } \right)} \to 0, \varepsilon \downarrow 0. \end{eqnarray} $

(3.8)

$ \begin{equation} - \frac{1}{{\log \varepsilon }}{B_{32}}\le - \frac{C}{{\log \varepsilon }}\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)} \Lambda _n^{\frac{{q - 2}}{4}} \to 0, \varepsilon \downarrow 0. \label{eq11} \end{equation} $

(3.9)

为估计${B_{3{\rm{3}}}}$, 我们将${B_{3{\rm{3}}}}$分成两部分

$ \begin{eqnarray} {B_{33}} &\leq& C\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {{n^{1 - \frac{q}{2}}}\psi \left( n \right)} {\int_0^\infty {{{\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)}^{1 - q}}E\left| {{X_1}} \right|} ^q}I\left( {\left| {{X_1}} \right| \le 2\sqrt n \left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)} \right){\rm d}x\\ &\leq&C\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {{n^{1 - \frac{q}{2}}}\psi \left( n \right)} {\int_0^\infty {{{\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)}^{1 - q}}E\left| {{X_1}} \right|} ^q}I\left( {\left| {{X_1}} \right| \le 2\varepsilon \sqrt {ng\left( n \right)} } \right){\rm d}x\\ &&+C\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\frac{{\psi \left( n \right)}}{{{n^{\frac{q}{2} - 1}}}}} \int_0^\infty {\frac{{E{{\left| {{X_1}} \right|}^q}}}{{{{\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)}^{q - 1}}}}} I\left( {0 < \frac{{\left| {{X_1}} \right|}}{{{\rm{2}}\sqrt n }} - \varepsilon \sqrt {g\left( n \right)} \le x} \right){\rm d}x\\ &=& {B_{331}} + {B_{332}}. \end{eqnarray} $

再分别处理${B_{{\rm{3}}31}}$和${B_{{\rm{3}}32}}$

$ \begin{eqnarray} {B_{331}} &\leq& C{\varepsilon ^{2 - q}}\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {{n^{1 - \frac{q}{2}}}g'\left( n \right){g^{ - \frac{q}{2}}}\left( n \right)E{{\left| {{X_1}} \right|}^q}I\left( {\left| {{X_1}} \right| \le 2\varepsilon \sqrt {ng\left( n \right)} } \right)} \\ &\leq&C{\varepsilon ^{2 - q}}\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\frac{{{n^{1 - \frac{q}{2}}}g'\left( n \right)}}{{{g^{\frac{q}{2}}}\left( n \right)}}\sum\limits_{i = 1}^n {E{{\left| {{X_1}} \right|}^q}I\left( {2\varepsilon \sqrt {ig\left( i \right)} \le \left| {{X_1}} \right| \le 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)} } \\ &\leq&C{\varepsilon ^{2 - q}}\sum\limits_{i = {n_0}}^{A\left( \varepsilon \right)} {E{{\left| {{X_1}} \right|}^q}I\left( {2\varepsilon \sqrt {ig\left( i \right)} \le \left| {{X_1}} \right| \le 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)\sum\limits_{n = i}^\infty {\frac{{{n^{1 - \frac{q}{2}}}g'\left( n \right)}}{{{g^{\frac{q}{2}}}\left( n \right)}}} } \\ &\leq& C{\varepsilon ^{2 - q}}\sum\limits_{i = {n_0}}^{A\left( \varepsilon \right)} {{i^{1 - \frac{q}{2}}}{g^{1 - \frac{q}{2}}}\left( i \right)E{{\left| {{X_1}} \right|}^q}I\left( {2\varepsilon \sqrt {ig\left( i \right)} \le \left| {{X_1}} \right| \le 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)}\\ &\leq& C\sum\limits_{i = {n_0}}^{A\left( \varepsilon \right)} {E{{\left| {{X_1}} \right|}^2}I\left( {2\varepsilon \sqrt {ig\left( i \right)} \le \left| {{X_1}} \right| \le 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)}\\ &\leq&CEX_1^2 < \infty . \end{eqnarray} $

(3.10)

$ \begin{eqnarray} {B_{332}} &\leq& C\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {{n^{1 - \frac{q}{2}}}\psi \left( n \right)E{{\left| {{X_1}} \right|}^q}I\left( {\left| {{X_1}} \right| > 2\varepsilon \sqrt {ng\left( n \right)} } \right)} \int_{\frac{{\left| {{X_1}} \right|}}{{2\sqrt n }} - \varepsilon \sqrt {g\left( n \right)} }^\infty {{{\left( {x + \varepsilon \sqrt {g\left( n \right)} } \right)}^{1 - q}}} {\rm d}x\\ &\leq&C\sum\limits_{n = {n_0}}^{A\left( \varepsilon \right)} {\psi \left( n \right)\sum\limits_{i = n}^\infty {E{{\left| {{X_1}} \right|}^2}I\left( {2\varepsilon \sqrt {ig\left( i \right)} < \left| {{X_1}} \right| \le 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)} } \\ &\leq&C\sum\limits_{i = {n_0}}^{A\left( \varepsilon \right)} {E{{\left| {{X_1}} \right|}^2}I\left( {2\varepsilon \sqrt {ig\left( i \right)} < \left| {{X_1}} \right| \le 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)\sum\limits_{n = {n_0}}^i {\frac{{g'\left( n \right)}}{{g\left( n \right)}}} } \\ &\leq&C\sum\limits_{i = {n_0}}^{A\left( \varepsilon \right)} {\log g\left( i \right)E{{\left| {{X_1}} \right|}^2}I\left( {2\varepsilon \sqrt {ig\left( i \right)} < \left| {{X_1}} \right| \le 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)} \\ &\leq&CE{\left| {{X_1}} \right|^2}{\log ^ + }\left| {{X_1}} \right| < \infty . \end{eqnarray} $

(3.11)

由式(3.10) 和(3.11), 则有$\mathop {\lim }\limits_{\varepsilon \downarrow 0} - \frac{1}{{\log \varepsilon }}{B_{33}} = 0.$因此, 由式(3.6)--(3.11), 知式(3.5) 成立.

命题3.3 在定理1.2的条件下, 有

$ \begin{equation} \mathop {\lim }\limits_{\varepsilon \downarrow 0} - \frac{1}{{\log \varepsilon }}\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\psi \left( n \right)} \int_{\varepsilon \sqrt {g\left( n \right)} }^\infty {2xP\left( {\left| N \right| \ge x} \right){\rm d}x = 0.} \end{equation} $

(3.12)

$ \begin{equation} \mathop {\lim }\limits_{\varepsilon \downarrow 0} - \frac{1}{{\log \varepsilon }}\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\psi \left( n \right)} \int_{\varepsilon \sqrt {g\left( n \right)} }^\infty {2xP\left( {\left| {{S_n}} \right| \ge \sqrt n x} \right){\rm d}x = 0.} \end{equation} $

(3.13)

证 式(3.12) 的证明很容易得到, 此处略.在引理2.3中取$x = \sqrt n x, $ $y = 2\sqrt n x$, 当$x \to \infty, $则有

$ \begin{eqnarray} \frac{{2n\mathop {\max }\limits_{1 \le i \le n} E\left| {{X_i}} \right|I\left( {\left| {{X_i}} \right| \ge {\rm{2}}\sqrt n x} \right)}}{{\sqrt n x}} \le C\frac{{E{{\left| {{X_{\rm{1}}}} \right|}^{\rm{2}}}I\left( {\left| {{X_{\rm{1}}}} \right| \ge {\rm{2}}\sqrt n x} \right)}}{{{x^{\rm{2}}}}} \le \frac{{\rm{C}}}{{{x^2}}} \to 0. \end{eqnarray} $

由引理2.3, 我们将式(3.13) 分成三部分, 则有

$ \begin{eqnarray} &&\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\psi \left( n \right)} \int_{\varepsilon \sqrt {g\left( n \right)} }^\infty {2xP\left( {\left| {{S_n}} \right| \ge \sqrt n x} \right){\rm d}x}\\ &\leq&C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {ng'\left( n \right)} \int_0^\infty {\left( {x + \varepsilon } \right)P\left( {\left| {{X_1}} \right| \ge 2\sqrt {ng\left( n \right)} \left( {x + \varepsilon } \right)} \right){\rm d}x} \\ &&+C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {g'\left( n \right){g^{ - \frac{q}{2}}}\left( n \right)} {\int_0^\infty {{{\left( {x + \varepsilon } \right)}^{1 - q}}\left( {EX_1^2I\left( {\left| {{X_1}} \right| \le 2\sqrt {ng\left( n \right)} \left( {x + \varepsilon } \right)} \right)} \right)} ^{^{\frac{q}{2}}}}{\rm d}x\\ &&+C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\frac{{{n^{1 - \frac{q}{2}}}g'\left( n \right)}}{{{g^{\frac{q}{2}}}\left( n \right)}}} {\int_0^\infty {{{\left( {x + \varepsilon } \right)}^{1 - q}}E\left| {{X_1}} \right|} ^q}I\left( {\left| {{X_1}} \right| \le 2\sqrt {ng\left( n \right)} \left( {x + \varepsilon } \right)} \right){\rm d}x\\ &=&{D_1} + {D_2} + {D_3}. \end{eqnarray} $

我们来分别处理${D_1}$和${D_2}$, 存在充分大的数$M > 0, $则

$ \begin{eqnarray} {D_1} &\leq& C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {ng'\left( n \right)} E\left( {\int_\varepsilon ^\infty {xI\left( {\left| {{X_1}} \right| \ge 2\sqrt {ng\left( n \right)} x} \right){\rm d}x} } \right)\\ &\leq& C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\frac{{g'\left( n \right)}}{{{n^{\frac{\alpha }{2}}}{g^{\frac{{{\rm{2}} + \alpha }}{2}}}\left( n \right)}}E\left( {\int_\varepsilon ^\infty {\frac{{X_1^{{\rm{2}} + \alpha }}}{{{x^{{\rm{1}} + \alpha }}}}I\left( {\left| {{X_1}} \right| \ge x} \right)I\left( {\left| {{X_1}} \right| \ge \sqrt M } \right){\rm d}x} } \right)} \\ &\leq& C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\frac{{g'\left( n \right)}}{{{g^{\frac{{{\rm{2}} + \alpha }}{2}}}\left( n \right)}}E\left( {\int_\varepsilon ^\infty {\frac{{X_1^{{\rm{2}} + \alpha }}}{x}I\left( {\left| {{X_1}} \right| \ge x} \right)I\left( {\left| {{X_1}} \right| \ge \sqrt M } \right){\rm d}x} } \right)}\\ &\leq&CEX_1^{{\rm{2}} + \alpha }\left| {\log \left| {{X_1}} \right| - \log \varepsilon } \right|I\left( {\left| {{X_1}} \right| \ge \sqrt M } \right)\\ &\leq&CEX_1^{{\rm{2}} + \alpha }\log \left| {{X_1}} \right|I\left( {\left| {{X_1}} \right| \ge \sqrt M } \right) + CEX_1^{{\rm{2}} + \alpha }\log \varepsilon I\left( {\left| {{X_1}} \right| \ge \sqrt M } \right)\\ &\lessdot& \infty . \end{eqnarray} $

(3.14)

$ \begin{equation}{D_2}\le C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\frac{{g'\left( n \right)}}{{g\left( n \right)}}} \frac{1}{{{{\left( {\varepsilon \sqrt {g\left( n \right)} } \right)}^{q - 2}}}} < \infty .\label{eq17} \end{equation} $

(3.15)

为估计${D_3}, $我们将${D_3}$分成两部分

$ \begin{eqnarray} {D_3} &\leq&C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\frac{{{n^{1 - \frac{q}{2}}}g'\left( n \right)}}{{{g^{\frac{q}{2}}}\left( n \right)}}} {\int_0^\infty {{{\left( {x + \varepsilon } \right)}^{1 - q}}E\left| {{X_1}} \right|} ^q}I\left( {\left| {{X_1}} \right| \le 2\sqrt {ng\left( n \right)} \left( {x + \varepsilon } \right)} \right){\rm d}x\\ &\leq&C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\frac{{{n^{1 - \frac{q}{2}}}g'\left( n \right)}}{{{g^{\frac{q}{2}}}\left( n \right)}}} {\int_0^\infty {{{\left( {x + \varepsilon } \right)}^{1 - q}}E\left| {{X_1}} \right|} ^q}I\left( {\left| {{X_1}} \right| \le 2\varepsilon \sqrt {ng\left( n \right)} } \right){\rm d}x\\ &&+C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\frac{{{n^{1 - \frac{q}{2}}}g'\left( n \right)}}{{{g^{\frac{q}{2}}}\left( n \right)}}\int_0^\infty {\frac{{E{{\left| {{X_1}} \right|}^q}}}{{{{\left( {x + \varepsilon } \right)}^{q - 1}}}}} } I\left( {\varepsilon < \frac{{\left| {{X_1}} \right|}}{{2\sqrt {ng\left( n \right)} }} \le \left( {x + \varepsilon } \right)} \right){\rm d}x\\ &=&{D_{31}} + {D_{32}}. \end{eqnarray} $

再分别估计${D_{31}}$和${D_{32}}$

$ \begin{eqnarray} {D_{31}}&\leq&C{\varepsilon ^{2 - q}}\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {{n^{1 - \frac{q}{2}}}g'\left( n \right){g^{ - \frac{q}{2}}}\left( n \right)E{{\left| {{X_1}} \right|}^q}I\left( {\left| {{X_1}} \right| \le 2\varepsilon \sqrt {ng\left( n \right)} } \right)}\\ &\leq&C{\varepsilon ^{2 - q}}\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\frac{{{n^{1 - \frac{q}{2}}}g'\left( n \right)}}{{{g^{\frac{q}{2}}}\left( n \right)}}\sum\limits_{i = 1}^n {E{{\left| {{X_1}} \right|}^q}I\left( {2\varepsilon \sqrt {ig\left( i \right)} \le \left| {{X_1}} \right| < 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)} }\\ &\leq&C{\varepsilon ^{2 - q}}\sum\limits_{i = A\left( \varepsilon \right) + 1}^\infty {E{{\left| {{X_1}} \right|}^q}I\left( {2\varepsilon \sqrt {ig\left( i \right)} \le \left| {{X_1}} \right| < 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)\sum\limits_{n = i}^\infty {\frac{{{n^{1 - \frac{q}{2}}}g'\left( n \right)}}{{{g^{\frac{q}{2}}}\left( n \right)}}} }\\ &\leq&C{\varepsilon ^{2 - q}}\sum\limits_{i = A\left( \varepsilon \right) + 1}^\infty {{i^{1 - \frac{q}{2}}}{g^{1 - \frac{q}{2}}}\left( i \right)E{{\left| {{X_1}} \right|}^q}I\left( {2\varepsilon \sqrt {ig\left( i \right)} \le \left| {{X_1}} \right| < 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)}\\ &\leq&CE{\left| {{X_1}} \right|^2}I\left( {\left| {{X_1}} \right| \ge 2\varepsilon \sqrt {A\left( \varepsilon \right)g\left( {A\left( \varepsilon \right)} \right)} } \right) < \infty .\label{eq18} \end{eqnarray} $

(3.16)

$ \begin{eqnarray} {D_{32}} &\leq&C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {{n^{1 - \frac{q}{2}}}g'\left( n \right)} {g^{ - \frac{q}{2}}}\left( n \right)E{\left| {{X_1}} \right|^q}I\left( {\left| {{X_1}} \right| > 2\varepsilon \sqrt {ng\left( n \right)} } \right)\int_{\frac{{\left| {{X_1}} \right|}}{{2\sqrt {ng\left( n \right)} }} - \varepsilon }^\infty {{{\left( {x + \varepsilon } \right)}^{1 - q}}} {\rm d}x\\ &\leq& C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\frac{{g'\left( n \right)}}{{g\left( n \right)}}} E{\left| {{X_1}} \right|^2}I\left( {\left| {{X_1}} \right| > 2\varepsilon \sqrt {ng\left( n \right)} } \right)\\ &\leq&C\sum\limits_{n = A\left( \varepsilon \right) + 1}^\infty {\frac{{g'\left( n \right)}}{{g\left( n \right)}}} \sum\limits_{i = n}^\infty {E{{\left| {{X_1}} \right|}^2}I\left( {2\varepsilon \sqrt {ig\left( i \right)} < \left| {{X_1}} \right| \le 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)}\\ &\leq&C\sum\limits_{i = A\left( \varepsilon \right) + 1}^\infty {E{{\left| {{X_1}} \right|}^2}I\left( {2\varepsilon \sqrt {ig\left( i \right)} < \left| {{X_1}} \right| \le 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)} \sum\limits_{n = 1}^i {\frac{{g'\left( n \right)}}{{g\left( n \right)}}}\\ &\leq&C\sum\limits_{i = A\left( \varepsilon \right) + 1}^\infty {\log g\left( i \right)E{{\left| {{X_1}} \right|}^2}I\left( {2\varepsilon \sqrt {ig\left( i \right)} < \left| {{X_1}} \right| \le 2\varepsilon \sqrt {\left( {i + 1} \right)g\left( {i + 1} \right)} } \right)}\\ &\leq&CE{\left| {{X_1}} \right|^2}{\log ^ + }\left| {{X_1}} \right|I\left( {\left| {{X_1}} \right| > 2\varepsilon \sqrt {A\left( \varepsilon \right)g\left( {A\left( \varepsilon \right)} \right)} } \right) < \infty, \label{eq19} \end{eqnarray} $

(3.17)

由式(3.14)、(3.15)、(3.16) 和(3.17), 则有

$ \mathop {\lim }\limits_{\varepsilon \downarrow 0} - \frac{1}{{\log \varepsilon }}\left( {{D_2} + {D_3}} \right) = 0, $

因此, 式(3.13) 成立.

由命题3.1及三角不等式可知式(3.3) 成立, 即定理1.2成立.