设$(X_n, {\cal F}_n, n\geq0)$是定义在$(\Omega, {\cal F}, P)$上的一可积随机适应序列, 即${\cal F}_n$是一非降$\sigma$域$({\cal F}_n\subset{\cal F}_{n+1})$, $X_n$可积的且是${\cal F}_n$可测的.

下面给出万成高在整个空间上的的收敛定理, 参见文献[1].

定理A  设$(X_n, {\cal F}_n, n\geq0)$是任意可积随机适应序列, $(a_n, n\geq0)$是非降的正常数列, 若下列条件成立:

$ \sum\limits_{n=1}^\infty {E[\frac{|X_n|^p}{{a_n}^p+a_n|X_n|^{p-1}}|{\cal F}_{n-1}]}<\infty, 1\leq{p}\leq2, $

(1)

则有

$ \sum\limits_{n=1}^\infty {\frac{X_n-E[X_n|{\cal F}_{n-1}]}{a_n}}\ \ \ \ \mbox{a.e. 收敛}. $

类似定理A我们给出一个集合上的强收敛定理.

定理B  设$(X_n, {\cal F}_n, n\geq0)$是任意可积随机适应序列, $(a_n, n\geq0)$是非降的正常数列.

设

$ A=\{\omega:\sum\limits_{n=1}^\infty {E[\frac{|X_n|^p}{{a_n}^p+a_n|X_n|^{p-1}}|{\cal F}_{n-1}]}<\infty\}, 1\leq{p}\leq2, $

(2)

则有

$ \sum\limits_{n=1}^\infty {\frac{X_n-E[X_n|{\cal F}_{n-1}]}{a_n}}\ \ \ \ \mbox{在$A$ 中a.e. 收敛.} $

(3)

定理A在条件(1) 下成立, 定理B在条件(2) 下成立.定理B更优, 因为它推广了Chow的鞅差序列的强大数定理.本文是利用文献[2]的方法给出定理B的证明.

定理B的证明  设$n\geq0$, $X_n^*={X_nI_{(|X_n|\leq a_n)}}$.记$Z_n={\frac{|X_n|^p}{a_n^p+a_n|X_n|^{p-1}}}$, 设$k$为正整数,

$ A_k=\{\omega:\sum\limits_{n=1}^\infty {E[Z_n|{\cal F}_{n-1}]}\leq{k}\}, $

(4)

$ \tau_k=\inf\{n:n\geq1, \sum\limits_{i=1}^{n+1} {E[Z_i|{\cal F}_{i-1}]}>k\}, $

(5)

当(5) 式的右边集为空集时, 令$\tau_k=\infty $, 这样$\sum\limits_{m=1}^{{\tau_k}\Lambda{n}}{Z_m}=\sum\limits_{m=1}^{n}{I_{(\tau_k\geq m)}Z_m} $.由于$I_{(\tau_k\geq m)}$是${\cal F}_{m-1}$可测的, 由$Z_n$的非负性, $\forall {n}$, 有

$ \begin{aligned} E(\sum\limits_{m=1}^{{\tau_k}\Lambda{n}}{Z_m})&=E(\sum\limits_{m=1}^{n}{I_{(\tau_k\geq m)}Z_m} )=E\{\sum\limits_{m=1}^{n}{E[I_{(\tau_k\geq m)}Z_m|{\cal F}_{m-1}]} \} \\ &=E\{\sum\limits_{m=1}^{n}{I_{(\tau_k\geq m)}E[Z_m|{\cal F}_{m-1}]} \} =E\{\sum\limits_{m=1}^{{\tau_k}\Lambda{n}}{E[Z_m|{\cal F}_{m-1}]}\}\leq{k}. \end{aligned} $

(6)

由于$A_k=\{\tau_k=\infty\}$, 于是由(6) 式, $\forall{n}$有

$ \begin{aligned}&\sum\limits_{m=1}^{n}{\int_{A_k}Z_m{dP}}=\sum\limits_{m=1}^{n}{E[I_{A_k}Z_m]}=E\{I_{A_k}\sum\limits_{m=1}^{n}{Z_m}\}=E\{I_{(\tau_k=\infty)}\sum\limits_{m=1}^{n}{Z_m}\}\\ =&E\{I_{(\tau_k=\infty)}\sum\limits_{m=1}^{{\tau_k}\Lambda{n}}{Z_m}\}\leq{E(\sum\limits_{m=1}^{{\tau_k}\Lambda{n}}{Z_m})}\leq{k}, \end{aligned} $

(7)

因此可得

$ \sum\limits_{m=1}^{\infty}{\int_{A_k}Z_m{dP}}\leq{k}. $

(8)

由于当$|X_n|>a_n$时, $\frac{1}{Z_n}=(\frac{a_n}{|X_n|})^{p}+\frac{a_n}{|X_n|}\leq{2}\quad(1\leq{p}\leq{2}), $于是有

$ \begin{aligned}&\sum\limits_{n=1}^{\infty}{P(A_k(X_n\neq{X_n^*}))}=\sum\limits_{n=1}^{\infty}{\int_{A_k(X_n\neq{X_n^*})}{dP}}\leq{\sum\limits_{n=1}^{\infty}{\int_{A_k(X_n\neq{X_n^*})}2Z_n{dP}}} \\ \leq&{\sum\limits_{n=1}^{\infty}{\int_{A_k}2Z_n{dP}}}\leq{2k}.\end{aligned} $

(9)

于是由Borel-Cantelli引理知$P(A_k(X_n\neq{X_n^*})\, \, {\rm i.o.})=0$, 于是有

$ \sum\limits_{n=1}^{\infty}{(X_n-X_n^*)/a_n} \ \ \mbox{在$A_k$ 中a.e. 收敛}. $

由于$A=\bigcup\limits_{k}{A_k}$, 所以有

$ \sum\limits_{n=1}^{\infty}{(X_n-X_n^*)/a_n} \ \ \mbox{在$A$ 中a.e. 收敛}. $

(10)

设

$ Y_m=(X_m^*-E[X_m^*|{\cal F}_{m-1}])/a_m. $

(11)

设$\lambda=1, -1$且

$ t_n(\lambda)=\frac{\exp\{\lambda\sum\limits_{m=1}^{n}Y_m\}}{\prod\limits_{m=1}^n{E[\exp(\lambda Y_m)|{\cal F}_{m-1}]}} \, \, \, \ \ \, n\geq{1}, $

(12)

由于$E[t_n|{\cal F}_{n-1}]=t_{n-1}$且$E|t_n(\lambda)|=Et_n(\lambda)=Et_1(\lambda)=1$, 故$\{t_n(\lambda), n\geq{1}\}$为非负鞅, 由Doob鞅收敛定理得

$ \lim\limits_{n\rightarrow{\infty}}t_n(\lambda)=t_\infty(\lambda) \ \ {\rm a.e.}, $

(13)

由不等式

$ 0\leq{\exp(x)-1-x} \leq {x^2\exp(|x|)}\, \, \, \, \, \, \, \, \forall{x}\in{R}, $

(14)

注意到$|Y_n|\leq{2}$, 且$E[Y_n|{\cal F}_{n-1}]=0$ a.e.及

$ \begin{aligned} E[Y_n^2|{\cal}{F_{n-1}}]&=E[((X_n^*-E[X_n^*|{\cal F}_{n-1}])/a_n)^2|{\cal F}_{n-1}] \\ &=(E[(X_n^*)^2|{\cal F}_{n-1}]-(E[X_n^*|{\cal F}_{n-1}])^2)/a_n^2\\ &\leq E[(X_n^*)^2|{\cal F}_{n-1}]/a_n^2\, \, \, \, \, \, \, \, \, {\rm a.e.}, \end{aligned} $

(15)

有

$ \begin{aligned}0&\leq{E[\exp(\lambda{Y_n})|{\cal F}_{n-1}]-1}=E[\exp(\lambda{Y_n})-\lambda{Y_n}-1|{\cal F}_{n-1}]\\ &\leq{E[|\lambda{Y_n}|^2e^{|\lambda{Y_n}|}|{\cal F}_{n-1}]}\leq{e^2E[Y_n^2|{\cal F}_{n-1}]}\leq{e^2E[(X_n^*)^2|{\cal F}_{n-1}]/a_n^2}\, \, \, \, \, \, \, \, \, {\rm a.e.}, \end{aligned} $

(16)

由于$|X_n|\leq{a_n}$时, $|X_n^*|\leq{a_n}$, 并且$f(x)=\frac{|x|^p}{a^p+a|x|^{p-1}}\, \, \, \, \, (a>0)$在$[0, +\infty)$是单调递增的, 有

$ (\frac{X_n^*}{a_n})^2\leq{(\frac{|X_n^*|}{a_n})^p}=\frac{|X_n^*|^p}{a_n^p+a_n|X_n^*|^{p-1}}\cdot\frac{a_n^p+a_n|X_n^*|^{p-1}}{|a_n^*|^p}\leq{2Z_n}\, \, \, \, \, \, \, \, \, (1\leq{p}\leq{2}), $

(17)

由(16) 与(17) 式有

$ 0\leq{E[\exp(\lambda{Y_n})|{\cal F}_{n-1}]-1}\leq{2e^2E[Z_n|{\cal F}_{n-1}]}\, \, \, \, \, \, \, \, \, {\rm a.e.}, $

(18)

由(18) 与(2) 式是有$\sum\limits_{n=1}^{\infty}{(E[\exp(\lambda{Y_n})|{\cal F}_{n-1}]-1])} $在$A$ 中a.e. 收敛, 或等价地

$ \prod\limits_{n=1}^{\infty}{E[\exp(\lambda{Y_n})|{\cal F}_{n-1}]} \ \ \mbox{在$A$ 中a.e. 收敛}. $

(19)

由(12), (13) 和(19) 式有

$ \lim\limits_{n\rightarrow{\infty}}\, \exp\{\lambda\sum\limits_{m=1}^{n}Y_m\} \ \ \mbox{在$A$ 中a.e. 收敛}. $

(20)

因为上式对$\lambda=1, -1$成立, 所以有

$ \sum\limits_{n=1}^{\infty}{Y_n}=\sum\limits_{n=1}^{\infty}{(X_n^*-E[X_n^*|{\cal F}_{n-1}])/a_n} \ \ \mbox{在$A$ 中a.e. 收敛}. $

(21)

当$|X_n|>a_n$时有

$ (\frac{|X_n|}{a_n})=\frac{|X_n|^p}{a_n^p+a_n|X_n|^{p-1}}\cdot\frac{a_n^{p-1}+a_n|X_n|^{p-1}}{|X_n|^{p-1}}\leq{2Z_n}\, \, \, \, \, \, \, \, \, (1\leq{p}\leq{2}), $

(22)

所以有

$ \begin{aligned} &|E[X_n|{\cal F}_{n-1}]-E[X_n^*|{\cal F}_{n-1}])/a_n| \leq E[|X_n-X_n^*|/a_n|{\cal F}_{n-1}]\\ =& E[|X_n|/a_nI_{(|X_n|>a_n)}|{\cal F}_{n-1}] \leq E[2Z_nI_{(|X_n|>a_n)}|{\cal F}_{n-1}] \leq 2E[2Z_n|{\cal F}_{n-1}]\, \, \, \, \, \, \, \, \, {\rm a.e.}. \end{aligned} $

(23)

由(23) 和(2) 式有

$ \sum\limits_{n=1}^{\infty}{(E[X_n|{\cal F}_{n-1}-E[X_n^*|{\cal F}_{n-1}])/a_n} \ \ \mbox{在$A$ 中~a.e. 收敛}. $

(24)

由(10), (21) 及(24) 知(3) 式成立.

注  在定理B中令$P(A)=1$, 可得定理A.

推论1  设$(X_n, {\cal F}_n, n\geq0)$是鞅差序列, $\{a_n, n\geq{0}\}$是非降的正数列.设$A$由(2) 定义, 则有

$ \sum\limits_{n=1}^{\infty}{\frac{X_n}{a_n}} \ \ \mbox{在$A$ 中a.e. 收敛}. $

(25)

证  注意到$E[X_n|{\cal F}_{n-1}]=0$, 于是从定理B可得本推论.

推论2(Chow, 参见文献[3, 第249页练习8]  设$(X_n, {\cal F}_n, n\geq0)$是鞅差序列, $\{a_n, n\geq{0}\}$是非降的正数列.设

$ B=\{\omega:\sum\limits_{n=1}^\infty {a_n^{-p}E[|X_n|^p|{\cal F}_{n-1}]}<\infty\}, 1\leq{p}\leq2, $

(26)

则有

$ \sum\limits_{n=1}^{\infty}{\frac{X_n}{a_n}} \ \ \mbox{在$B$ 中~a.e. 收敛}. $

(27)

证  令$A$如定理B定义, 因为$\frac{|X_n|^p}{a_n^p+a_n|X_n|^{p-1}}\leq\frac{|X_n|^p}{a_n^p}$, 则有$B\subseteq{A}$, 由推论1可得本推论.

推论3  $\{X_n, n\geq{1}\}$为独立的随机变量序列, $\{a_n, n\geq{1}\}$如前定义.如果

$ \sum\limits_{n = 1}^\infty {E\left[{\frac{{\left| {X_n } \right|^p }}{{a_n ^p + a_n \left| {X_n } \right|^{p-1} }}} \right]} < \infty \quad 1 \le p \le 2. $

(28)

则$ \sum\limits_{n = 1}^\infty {\frac{{X_n -EX_n }}{{a_n }}} \quad \, {\rm a.e.}\mbox{收敛}. $

证  在定理B中取${\cal F}_n = \sigma \left({X_1, X_2, \cdots, X_n }\right)$, 由于$\{X_n, n\geq{1}\}$为独立的随机变量序列, 故有

$ E\left[{\frac{{|X_n| ^p }}{{a_n ^p + a_n \left| {X_n } \right|^{p-1}}}\left| {{\cal F}_{n-1} } \right.} \right] = E\left[{\frac{{|X_n| ^p }}{{a_n ^p + a_n \left| {X_n } \right|^{p-1}}}} \right]\quad {\rm a.e.} $

及$ E\left[{X_n \left| {{\cal F}_{n-1} } \right.} \right] = EX_n \quad {\rm a.e.}. $由定理A可得本推论.

由推论3易证以下两推论.

推论4   (参见文献[4, p.387])设$\{X_n, n\geq{1}\}$为独立的随机变量序列, 且$EX_n=0$, 如果

$ \sum\limits_{n = 1}^\infty {E\frac{{X_n ^2 }}{{1 + \left| {X_n } \right|}} < \infty }, $

则$ \sum\limits_{n = 1}^\infty {X_n } \, {\rm a.e.} \mbox{收敛}. $

推论5   (Kolmogorov, 参见文献[4, p.389])设$\{X_n, n\geq{1}\}$为独立的随机变量序列, $\{a_n, n\geq{1}\}$如前定义.如果

$ \sum\limits_{n = 1}^\infty {\frac{{EX_n ^2 }}{{a_n ^2 }} < \infty }, $

(29)

则$ \sum\limits_{n = 1}^\infty {\frac{{X_n -EX_n }}{{a_n }}} \, {\rm a.e.} \mbox{收敛}. $

显然推论3是推论5的推广, 这是因为当$p=2$时条件(28) 比条件(29) 弱.下面的例子证明了当$p=2$时(28) 比(29)“真弱”.

例1  设$\{X_n, n\geq{1}\}$为独立的随机变量序列,$X_n$的分布密度为

$ f_n (x) = \left\{ \begin{array}{l} \frac{2}{{n^2 x^3 }}\quad x \ge 1/n, \\ \quad 0\quad \;x < 1/n. \\ \end{array} \right. $

要证明

$ \sum\limits_{n = 1}^\infty {\frac{{EX_n ^2 }}{{n^2 }}} = + \infty, \sum\limits_{n = 1}^\infty {E\left[{\frac{{X_n ^2 }}{{n^2 + n\left| X _n\right| }}} \right] < \infty }. $

事实上, 由于$\displaystyle EX_n ^2 = \int_{1/n}^\infty {\frac{2}{{n^2 x}}} dx = + \infty, $所以$\displaystyle \sum\limits_{n = 1}^\infty {\frac{{EX_n ^2 }}{{n^2 }}} = + \infty, $又因为

$ E\left[{\frac{{X_n ^2 }}{{n^2 + n\left| {X_n } \right|}}} \right] = \int_{1/n}^\infty {\frac{2}{{n^2 x^3 }}} \frac{{x^2 }}{{n^2 + nx}}dx = \int_{1/n}^\infty {\frac{{2n}}{{n^3 }}} \left( {\frac{1}{{n^2 x}} - \frac{1}{{n^2 x + n^3 }}} \right)dx = \frac{2}{{n^{\rm 4} }}\ln \left( {{\rm 1 + }n^{\rm 2} } \right), $

由于

$ \sum\limits_{n = 1}^\infty {\frac{2}{{n^{\rm 4} }}\ln \left( {{\rm 1 + }n^{\rm 2} } \right) < \infty }. $

故

$ \sum\limits_{n = 1}^\infty {E\left[{\frac{{X_n ^2 }}{{n^2 + n \left| {X_n } \right|}}} \right] < \infty } . $

所以当$a_n=n, p=2$时, $\{X_n, n\geq{1}\}$满足(28) 式, 但不满足(29) 式.