在经典马氏链及马氏过程的理论中, 强遍历性曾被很多学者所研究.特别是离散时间马氏链强遍历性的研究, 成果比较丰富.如Paz^[1]讨论了状态可数的非齐次马氏链的强遍历性, 并给出了若干判定条件. Madsen^[2]研究了可由随机核序列所表示的马氏过程的强遍历性, 得到了马氏过程是强遍历的一些充分条件, 同时也研究了强遍历条件下马氏过程的性质.国内的一些学者, 如杨卫国^{[6, 7]}引入了非齐次马氏链绝对平均强遍历性的概念, 并给出了马氏链满足绝对平均强遍历性的条件, 并利用此种强遍历性解决了熵率的存在性问题以及讨论了其在马氏决策过程和信息论中的一些应用.但是有关随机环境中的马氏链强遍历性这方面的研究工作及相关的文献并不多, 这是由于受到环境因素的影响, 相关问题的研究无论从概念或是方法上相比较经典情形而言都要复杂一些.其实在这些问题的研究中, 我们往往并不是直接处理单链$\overrightarrow{X}$的遍历性, 对单链遍历性的研究往往是从$\overrightarrow{\theta}$链开始的.在对$\overrightarrow{\theta}$链相关的问题研究清楚之后, 我们进而再研究单链.在这一方面Cogburn^{[3, 4]}做的工作比较早, 同时他也获得了一些很有用的结果.另外李应求^{[8, 9]}引入了初始时间在任意点的强遍历性及一致强遍历性等概念, 研究了$\overrightarrow{\theta}$链是强遍历及一致强遍历的条件.考虑到$\overrightarrow{\theta}$链本身就是一种非齐次的马氏链, 因而相关的研究往往可以借鉴经典马氏链的研究方法.

本文拟在前人工作的基础上对这方面问题作一些探讨, 首先引入单链强遍历与$\chi$-一致强遍历的定义, 围绕这些定义讨论了相应的充分条件, 同时也得到了与定义等价的一些形式.并利用鞅这一强有力的工具刻画了单链强遍历条件下尾的结构, 最后对强遍历、弱遍历、尾三者之间的关系做了必要的概括.

设$Z$表示整数集, $Z_{+}$表示非负整数集, $\chi$是至多可数集, $\mathcal{A}$是定义在其上的$\sigma$-代数, $(\Theta,\mathcal{B})$是一可测空间, $\{P(\theta)\}_{\theta\in\Theta}$是转移概率族, 若存在概率空间$(\Omega,\mathcal{F},P)$及定义在其上且分别取值于$\chi$和$\Theta$的随机变量序列$\{X_{n}\}_{n\in Z_{+}}$及$\{\xi_{n}\}_{n\in Z}$, 满足

$P(X_{0}\in A|\overrightarrow{\xi})=P(X_{0}\in A|\overrightarrow{\xi}^{0}_{-\infty})\ \ {\hbox{a.s.}} \ \ \forall A\in\mathcal{A},$

(2.1)

$P(X_{n+1}\in A|\overrightarrow{X}^{n}_{0};\overrightarrow{\xi})=P(\xi_{n};X_{n},A)\ \ {\hbox{a.s.}}\ \ \forall A\in \mathcal{A},$

(2.2)

则称$\{X_{n}\}_{n\in Z_{+}}$为随机环境$\{\xi_{n}\}_{n\in Z}$中的马氏链.给定转移概率族$\{P(\theta)\}_{\theta\in \Theta},$必存在概率空间$(\Omega,\mathcal{F},P)$及其上满足 (2.1) 及 (2.2) 式的随机变量序列$\{X_{n}\}_{n\in Z_{+}}$及$\{\xi_{n}\}_{n\in Z}$.事实上, 对于$(\chi,\mathcal{A})$上的分布$\nu$, 设$C(A_{0}\times\cdots\times A_{n})$ $(A_{i}\in \mathcal{A},i=1,2,\cdots,n)$是以$A_{0}\times\cdots\times A_{n}$为底的柱集,

$P_{\nu}^{\overrightarrow{\theta}}(C(A_{0}\times\cdots\times A_{n}))=\int_{A_{0}}\nu(dx_{0})\int_{A_{1}}P(\theta_{0};x_{0},dx_{1})\cdots\int_{A_{n}}P(\theta_{n-1};x_{n-1},dx_{n}).$

另外假设$\pi$是$(\Theta^{Z},\mathcal{B}^{Z})$上的分布, $P_{\nu}(A\times F)=\int_{F}P_{\nu}^{\overrightarrow{\theta}}(A)\pi(d\overrightarrow{\theta}),$ $\{\widetilde{X}_{n}\}_{n\in Z_{+}},\{\widetilde{\xi}_{n}\}_{n\in Z}$分别是定义在$(\chi^{Z_{+}},\mathcal{A}^{Z_{+}})$及$(\Theta^{Z},\mathcal{B}^{Z})$且取值于$(\chi,\mathcal{A})$、$(\Theta,\mathcal{B})$上的坐标过程, 令$\Omega=\chi^{Z_{+}}\times \Theta^{Z},\mathcal{F}=\mathcal{A}^{Z_{+}}\times\mathcal{B}^{Z},$ $\forall\omega=(\overrightarrow{x},\overrightarrow{\theta})\in \Omega$, $X_{n}(\omega)=\widetilde{X}_{n}(\overrightarrow{x}),\xi_{n}(\omega)=\widetilde{\xi}_{n}(\overrightarrow{\theta}),$不难验证$\{X_{n}\}_{n\in Z_{+}}$为随机环境$\{\xi_{n}\}_{n\in Z}$中的马氏链, 验证的方法类似于文献^[5], 且$\{X_{n}\}_{n\in Z_{+}}$的初始分布就是$\nu.$记$P_{\nu,\overrightarrow{\theta}}(A\times \Theta^{Z})=P_{\nu}^{\overrightarrow{\theta}}(A)$.事实上, $P_{\nu,\overrightarrow{\theta}}(A\times \Theta^{Z})=P_{\nu}(A\times \Theta^{Z}|\overrightarrow{\xi}=\overrightarrow{\theta}),$由 (2.2) 式, 在给定$\overrightarrow{\xi}$的一个现实$\overrightarrow{\theta}$的条件下, $\{X_{n}\}_{n\in Z_{+}}$是非齐次的马氏链, 一般称之为$\overrightarrow{\theta}$链.需要指出的是虽然我们的目的是对单链$\overrightarrow{X}$相关的问题进行研究, 但在讨论问题时却往往可以先从$\overrightarrow{\theta}$链入手.令$P(\theta_{0},\cdots,\theta_{n-1})=P(\theta_{0})\cdots P(\theta_{n-1})$.

定义2.1 若存在分布$q(y),$使得$\forall m\in Z_{+}$及$\forall x \in \chi$, 都有

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{y\in \chi}|P(\theta_{m},\cdots,\theta_{m+n-1};x,y)-q(y)|=0,$

则称$\overrightarrow{\theta}$链是强遍历的.若

$\pi\{\overrightarrow{\theta}\in\Theta^{Z}:\lim\limits_{n\rightarrow\infty}\sum\limits_{y\in \chi}|P(\theta_{m},\cdots,\theta_{m+n-1};x,y)-q(y)|=0\}=1,$

则称单链$\overrightarrow{X}$是强遍历的.

定义2.2 若存在分布$q(y),$使得$\forall m\in Z_{+}$有

$\lim\limits_{n\rightarrow\infty}\sup\limits_{x}\sum\limits_{y\in \chi}|P(\theta_{m},\cdots,\theta_{m+n-1};x,y)-q(y)|=0,$

则称$\overrightarrow{\theta}$链是$\chi$一致强遍历的.若

$\pi\{\overrightarrow{\theta}\in\Theta^{Z}:\lim\limits_{n\rightarrow\infty}\sup\limits_{x}\sum\limits_{y\in \chi}|P(\theta_{m},\cdots,\theta_{m+n-1};x,y)-q(y)|=0\}=1,$

则称单链$\overrightarrow{X}$是$\chi$一致强遍历的.

称行向量相同的随机矩阵为常数随机矩阵, 显然$\overrightarrow{\theta}$链是$\chi$一致强遍历的的一个等价的形式是存在常数随机矩阵$Q$, 使得$\lim\limits_{n\rightarrow \infty}\|P(\theta_{m},\cdots,\theta_{m+n-1})-Q\|=0$.

定义2.3 若$\forall m\in Z_{+}$及$\forall x,y\in \chi$, 都有

$\pi\{\overrightarrow{\theta}:\lim\limits_{n\rightarrow\infty}\sum\limits_{z}|P(\theta_{m},\cdots, \theta_{m+n-1};x,z)-P(\theta_{m},\cdots,\theta_{m+n-1};y,z)|=0\}=1, $

则称单链$\overrightarrow{X}$是弱遍历的.

显然, 若单链$\overrightarrow{X}$是强遍历的, 则其必是弱遍历的.但反之不真.

例 设$\chi=Z,\Theta=\{0,1\},\eta$是$(\Theta,\mathcal{B})$上的分布, $\eta(\{0\})=\eta(\{1\})=\frac{1}{2}.\pi=\eta^{Z},$则$\pi$是$(\Theta^{Z},\mathcal{B}^{Z})$上的平稳分布.令$P,Q$是常数随机矩阵, 且$P\neq Q,P(0)=P,P(1)=Q$, 则单链$\overrightarrow{X}$是弱遍历的, 但其不是强遍历的.事实上, 令

$\tau_{0}^{0}(\overrightarrow{\theta})=\inf\{k\geq 0:\theta_{k}=0\},\tau_{n}^{0}(\overrightarrow{\theta})=\inf\{k> \tau_{n-1}^{0}(\overrightarrow{\theta}):\theta_{k}=0\},\\ \tau_{0}^{1}(\overrightarrow{\theta})=\inf\{k\geq 0:\theta_{k}=1\},\tau_{n}^{1}(\overrightarrow{\theta})=\inf\{k> \tau_{n-1}^{1}(\overrightarrow{\theta}):\theta_{k}=1\},\\ B=\{\overrightarrow{\theta}:\forall n\in Z_{+},\tau_{n}^{0}(\overrightarrow{\theta})<\infty,\tau_{n}^{1}(\overrightarrow{\theta})<\infty\}.$

易证$\pi(B)=1,\forall \overrightarrow{\theta}\in B$及$\forall n\in Z_{+},$显然$P(\theta_{m},\cdots,\theta_{m+n-1})=P$或$Q,$因而

$\sum\limits_{z}|P(\theta_{m},\cdots,\theta_{m+n-1};x,z)-P(\theta_{m},\cdots,\theta_{m+n-1};y,z)|=0,$

从而单链$\overrightarrow{X}$是弱遍历的.令$\tau_{n}^{0}(\overrightarrow{\theta})=n_{0},\tau_{n}^{1}(\overrightarrow{\theta})=n_{1},$于是

$P(\theta_{0},\cdots,\theta_{n_{0}})=P,P(\theta_{0},\cdots,\theta_{n_{1}})=Q,$

这说明单链$\overrightarrow{X}$不是强遍历的.

在对$\overrightarrow{\theta}$链是否为强遍历的判定过程中, 我们可以将定义2.1中的条件进行适当的弱化.

定理3.1 设$q(y)$是$(\chi,\mathcal{A})$上的概率分布.

(1) 若存在$m_{0}\in Z_{+}$, 对任意的$x\in \chi$,

$\lim\limits_{n\rightarrow\infty}\sum\limits_{y\in \chi}|P(\theta_{m_{0}},\cdots,\theta_{m_{0}+n-1};x,y)-q(y)|=0,$

则对任意的$ m\leq m_{0},m\in Z_{+}$, 都有

$\lim\limits_{n\rightarrow\infty}\sum\limits_{y\in \chi}|P(\theta_{m},\cdots,\theta_{m+n-1};x,y)-q(y)|=0.$

(2) 存在单调递增数列$\{m_{i}\}$, $m_{i}\rightarrow\infty$, 且对于每个$m_{i}$及任意的

$x\in \chi,\lim\limits_{n\rightarrow\infty}\sum\limits_{y\in \chi}|P(\theta_{m_{i}},\cdots,\theta_{m_{i}+n-1};x,y)-q(y)|=0,$

则$\overrightarrow{\theta}$链是强遍历的.

证 (1) 不妨假设$m< m_{0},m\in Z_{+}$, 则

$\ \ \ \sum\limits_{y}|P(\theta_{m},\cdots,\theta_{m+n-1};x,y)-q(y)|\\ =\sum\limits_{y^{'}}\sum\limits_{y}|P(\theta_{m},\cdots,\theta_{m_{0}-1};x,y^{'})[P(\theta_{m_{0}},\cdots,\theta_{m+n-1};x,y)-q(y)]|\\ \leq\sum\limits_{y}\sum\limits_{y^{'}}P(\theta_{m},\cdots,\theta_{m_{0}-1};x,y^{'})|P(\theta_{m_{0}},\cdots,\theta_{m+n-1};x,y)-q(y)|,$

(3.1)

由控制收敛定理

$\ \ \ \lim\limits_{n\rightarrow\infty}\sum\limits_{y}\sum\limits_{y^{'}}P(\theta_{m},\cdots,\theta_{m_{0}-1};x,y^{'})|P(\theta_{m_{0}},\cdots,\theta_{m+n-1};x,y)-q(y)|\\ =\sum\limits_{y^{'}}P(\theta_{m},\cdots,\theta_{m_{0}-1};x,y^{'})\lim\limits_{n\rightarrow\infty}\sum\limits_{y}|P(\theta_{m_{0}},\cdots,\theta_{m+n-1};x,y)-q(y)|\\ =0,$

(3.2)

则

$\lim\limits_{n\rightarrow\infty}\sum\limits_{y\in \chi}|P(\theta_{m},\cdots,\theta_{m+n-1};x,y)-q(y)|=0.$

(2) 对任意的$m\in Z_{+},$总存在$m_{i},m\leq m_{i},$且

$\lim\limits_{n\rightarrow\infty}\sum\limits_{y\in \chi}|P(\theta_{m_{i}},\cdots,\theta_{m_{i}+n-1};x,y)-q(y)|=0,$

由 (1),

$\lim\limits_{n\rightarrow\infty}\sum\limits_{y\in \chi}|P(\theta_{m},\cdots,\theta_{m+n-1};x,y)-q(y)|=0.$

故$\overrightarrow{\theta}$链是强遍历的.

而对$\overrightarrow{\theta}$链是$\chi$-一致强遍历的判定, 也有和定理3.1类似的结果.

定理3.2 (1) 若存在$m_{0}\in Z_{+}$及常数矩阵$Q,$使得

$\lim\limits_{n\rightarrow \infty}\|P(\theta_{m_{0}},\cdots,\theta_{m_{0}+n-1})-Q\|=0,$

则对任意的$m\leq m_{0},m\in Z_{+},$都有

$\lim\limits_{n\rightarrow \infty}\|P(\theta_{m},\cdots,\theta_{m+n-1})-Q\|=0.$

(2) 存在单调递增数列$\{m_{i}\}$, $m_{i}\rightarrow\infty$, 且对于每个$m_{i}$, 都有

$\lim\limits_{n\rightarrow \infty}\|P(\theta_{m_{i}},\cdots,\theta_{m_{i}+n-1})-Q\|=0,$

则$\overrightarrow{\theta}$链是$\chi$-一致强遍历的.

证 (1) 不妨假设$m< m_{0},m\in Z_{+}$, $q(y)$是矩阵$Q$位于$y$列上的元素, 于是

$\ \ \ \lim\limits_{n\rightarrow \infty}\|P(\theta_{m},\cdots,\theta_{m+n-1})-Q\|=\lim\limits_{n\rightarrow \infty}\sup\limits_{x}\sum\limits_{y}|P(\theta_{m},\cdots,\theta_{m+n-1};x,y)-q(y)|\\ =\lim\limits_{n\rightarrow \infty}\sup\limits_{x}\sum\limits_{y}|\sum\limits_{y^{'}}P(\theta_{m},\cdots,\theta_{m_{0}-1};x,y^{'})[P(\theta_{m_{0}},\cdots,\theta_{m+n-1};y^{'},y)-q(y)]|\\ \leq\lim\limits_{n\rightarrow \infty} \sup\limits_{x}\sum\limits_{y^{'}}\sum\limits_{y}P(\theta_{m},\cdots,\theta_{m_{0}-1};x,y^{'})|P(\theta_{m_{0}},\cdots,\theta_{m+n-1};y^{'},y)-q(y)|\\ \leq\lim\limits_{n\rightarrow \infty} \sup\limits_{x}\sum\limits_{y^{'}}\sum\limits_{y}P(\theta_{m},\cdots,\theta_{m_{0}-1};x,y^{'})\cdot\sup\limits_{y^{'}}\sum\limits_{y}|P(\theta_{m_{0}},\cdots,\theta_{m+n-1};y^{'},y)-q(y)| \\ =\lim\limits_{n\rightarrow \infty}\sup\limits_{y^{'}}\sum\limits_{y}|P(\theta_{m_{0}},\cdots,\theta_{m+n-1};y^{'},y)-q(y)|.$

(3.3)

注意到

$\sup\limits_{y^{'}}\sum\limits_{y}|P(\theta_{m_{0}},\cdots,\theta_{m+n-1};y^{'},y)-q(y)|=\|P(\theta_{m_{0}},\cdots,\theta_{m_{0}+n-1})-Q\|\rightarrow 0,$

则对任意的$m\leq m_{0},m\in Z_{+},$都有$\lim\limits_{n\rightarrow\infty}\sum\limits_{y\in \chi}|P(\theta_{m},\cdots,\theta_{m+n-1};x,y)-q(y)|=0$.

(2) 对任意的$m_{0}\in Z_{+},$总存在$m_{i}\geq m,$且$\lim\limits_{n\rightarrow \infty}\|P(\theta_{m_{i}},\cdots,\theta_{m_{i}+n-1})-Q\|=0,$由 (1),

$\lim\limits_{n\rightarrow \infty}\|P(\theta_{m},\cdots,\theta_{m+n-1})-Q\|=0,$

故$\overrightarrow{\theta}$链是$\chi$-一致强遍历的.

在定义2.1中, 分布$q(y)$与$m$是无关的, 但在验证单链的强遍历性时, 可以忽略这一信息.事实上, 将$q(y)$换成$q^{(m)}(y),$也不会影响结果.

定理3.3 若对任意的$m\in Z_{+},$都存在$(\chi,\mathcal{A})$上的分布$q^{(m)}(\cdot),$使得对任意的$x\in \chi,$有

$\pi\{\overrightarrow{\theta}:\forall y\in \chi,\lim\limits_{n\rightarrow\infty}P(\theta_{m},\cdots,\theta_{m+n-1};x,y)=q^{(m)}(y)\}=1,$

则$\overrightarrow{X}$是强遍历的.

证 令$B=\bigcap\limits_{m}\bigcap\limits_{x}\{\overrightarrow{\theta}:\forall y\in \chi,\lim\limits_{n\rightarrow\infty}P(\theta_{m},\cdots,\theta_{m+n-1};x,y)=q^{(m)}(y)\},$则$\pi(B)=1.$对任意的$\overrightarrow{\theta}\in B,$由于

$P(\theta_{0},\cdots,\theta_{m+n-1};x,y)=\sum\limits_{z}P(\theta_{0},\cdots,\theta_{m-1};x,z)P(\theta_{m},\cdots,\theta_{m+n-1};z,y),$

令$\mu (\{z\})=P(\theta_{0},\cdots,\theta_{m-1};x,z),$则$\int_{\chi}\mu(dz)=1.$由控制收敛定理

$\lim\limits_{n\rightarrow\infty}P(\theta_{0},\cdots, \theta_{m-1};x,y)=\lim\limits_{n\rightarrow\infty}\int_{\chi}P(\theta_{m}, \cdots,\theta_{m+n-1};z,y)\mu(dz)\\ =\int_{\chi}\lim\limits_{n\rightarrow\infty}P(\theta_{m}, \cdots,\theta_{m+n-1};z,y)\mu(dz)=q^{(m)}(y).$

(3.4)

注意到$\lim\limits_{n\rightarrow\infty}P(\theta_{0},\cdots,\theta_{m+n-1};x,y)=q^{(0)}(y),$则$q^{(m)}(y)=q^{(0)}(y).$令$q(y)=q^{(0)}(y),$则

$\lim\limits_{n\rightarrow\infty}\sum\limits_{z}P(\theta_{m},\cdots,\theta_{m+n-1};x,z)=1=\sum\limits_{z}q(z)=\sum\limits_{z}\lim\limits_{n\rightarrow\infty}P(\theta_{m},\cdots,\theta_{m+n-1};x,z), $

于是由Fatou引理, 有

$\lim\limits_{n\rightarrow\infty}\sum\limits_{z}[P(\theta_{m},\cdots,\theta_{m+n-1};x,z)+q(z)] -\mathop{\overline\lim}\limits_{n\rightarrow\infty}\sum\limits_{z}|P(\theta_{m},\cdots,\theta_{m+n-1};x,z)-q(z)| \\ =\mathop{\underline\lim}\limits_{n\rightarrow\infty}\sum\limits_{z}[P(\theta_{m},\cdots,\theta_{m+n-1};x,z)+q(z)-|P(\theta_{m},\cdots,\theta_{m+n-1}; x,z)-q(z)|]\\ \geq\sum\limits_{z}\mathop{\underline\lim}\limits_{n\rightarrow\infty}[P(\theta_{m},\cdots,\theta_{m+n-1};x,z)+q(z)-|P(\theta_{m},\cdots, \theta_{m+n-1};x,z)-q(z)|]\\ =\sum\limits_{z}\lim\limits_{n\rightarrow\infty}[P(\theta_{m},\cdots,\theta_{m+n-1};x,z)+q(z)] -\mathop{\overline\lim}\limits_{n\rightarrow\infty}\sum\limits_{z}|P(\theta_{m},\cdots,\theta_{m+n-1};x,z)-q(z)|,$

(3.5)

故

$\mathop{\overline\lim}\limits_{n\rightarrow\infty}\sum\limits_{z}|P(\theta_{m},\cdots,\theta_{m+n-1};x,z)-q(z)| \leq\sum\limits_{z}\mathop{\overline\lim}\limits_{n\rightarrow\infty}|P(\theta_{m},\cdots,\theta_{m+n-1};x,z)-q(z)|=0, $

则$\overrightarrow{X}$是强遍历的.

在单链强遍历的定义中, 可以理解为$m$时刻单链从某固定点出发, 因而单链的初始分布是退化的.事实上, 下面的定理表明, 将初始的退化分布换成一般的分布对单链的强遍历性并无影响.

定理3.4 以下条件是等价的

(1) $\overrightarrow{X}$是强遍历的.

(2) 存在$(\chi,\mathcal{A})$上的分布$q(y),$使得对任意的$m\in Z_{+},x\in \chi,$都有

$\pi\{\overrightarrow{\theta}:\forall y\in \chi,\lim\limits_{n\rightarrow\infty}P(\theta_{m},\cdots,\theta_{m+n-1}; x,y)=q(y)\}=1.$

(3) 存在$(\chi,\mathcal{A})$上的分布$q(y),$使得对$(\chi,\mathcal{A})$上任意的分布$\nu$及任意的$m\in Z_{+},$都有

$\pi\{\overrightarrow{\theta}:\forall y\in \chi,\lim\limits_{n\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(X_{n}=y)=q(y)\}=1.$

(4) 存在$(\chi,\mathcal{A})$上的分布$q(y),$使得对$(\chi,\mathcal{A})$上任意的分布$\nu$及任意的$m\in Z_{+},$都有

$\pi\{\overrightarrow{\theta}:\forall y\in \chi,\lim\limits_{n\rightarrow\infty}\sum\limits_{y}|P_{\nu,\overrightarrow{\theta}}(X_{n}=y)-q(y)|=0\}=1.$

证 (1)、(2) 的等价性由定理3.3可知 (3)$\Rightarrow$(2) 是显然的.下面证明 (2)$\Rightarrow$(3), 由控制收敛定理,

$P_{\nu,\overrightarrow{\theta}}(X_{n}=y)=\int_{\chi}P_{x,\overrightarrow{\theta}}(X_{n}=y)\nu (dx)\\ \rightarrow \int_{\chi}\lim\limits_{n\rightarrow\infty}P_{x,\overrightarrow{\theta}}(X_{n}=y)\nu (dx)=\int_{\chi}\lim\limits_{n\rightarrow\infty}P(\theta_{m},\cdots,\theta_{m+n-1}; x,y)\nu (dx)=q(y),$

(3.6)

故 (2)$\Rightarrow$(3).

(4)$\Rightarrow$(3) 是显然的.下面证明 (1)$\Rightarrow$(4).由控制收敛定理, 有

$\lim\limits_{n\rightarrow\infty}\sum\limits_{y}|P_{\nu,\overrightarrow{\theta}}(X_{n}=y)-q(y)| =\lim\limits_{n\rightarrow\infty}\sum\limits_{y}|\int_{\chi}P_{x,\overrightarrow{\theta}}(X_{n}=y)\nu(dx)-q(y)|\\ \leq\lim\limits_{n\rightarrow\infty}\int_{\chi}\sum\limits_{y}|P_{x,\overrightarrow{\theta}}(X_{n}=y)-q(y)|\nu(dx) \\ =\int_{\chi}\lim\limits_{n\rightarrow\infty}\sum\limits_{y}|P_{x,\overrightarrow{\theta}}(X_{n}=y)-q(y)|\nu(dx)=0.$

(3.7)

上述的四个条件中, (4) 最强, (2) 最弱, 在判断单链的强遍历性时, 用 (2) 最方便, 而在处理强遍历性的性质时, 用 (4) 则最好.

单链若是强遍历的则会直接影响到尾的结构, 下面的结论则刻画了这一事实.

定理3.5 设$\overrightarrow{X}$是强遍历的, $\mathcal{F}^{\infty}=\bigcap\limits_{n=1}^{\infty}\sigma(X_{n},X_{n+1},\cdots),$ $\nu$是$(\chi,\mathcal{A})$上的分布, 则$\forall M\in \mathcal{F}^{\infty},$ $P_{\nu,\overrightarrow{\theta}}(M)=0$或$1$.

证 由于$\overrightarrow{X}$是强遍历的, 则存在$B\in \mathcal{B}^{Z}$及$(\chi,\mathcal{A})$上的分布$q(y),$满足$\pi(B)=1,$且$\forall\overrightarrow{\theta}\in B,$ $P_{\nu,\overrightarrow{\theta}}(X_{n}=y)\rightarrow q(y).$令$A=\{X_{n}=y$ i.o.$\}$, 则

$P_{\nu,\overrightarrow{\theta}}(A)=\lim\limits_{n\rightarrow\infty}P_{\nu,\overrightarrow{\theta}} (\bigcup\limits_{m=n}^{\infty}\{X_{m}=y\})\geq \lim\limits_{n\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(X_{n}=y)= q(y),$

(3.8)

若$q(y)>0,$则$P_{\nu,\overrightarrow{\theta}}(A)>0.$设$A^{'}\subseteq A,A^{'}\in \mathcal{F}^{\infty},P_{\nu,\overrightarrow{\theta}}(A^{'})>0,\mathcal{F}_{\leq n}=\sigma(X_{0},\cdots,X_{n}),\mathcal{F}_{\infty}=\mathop{\vee}\limits_{n=0}^{\infty}\mathcal{F}_{\leq n}.$由鞅收敛定理, 有

$P_{\nu,\overrightarrow{\theta}}(A^{'}|\mathcal{F}_{\leq n})\rightarrow P_{\nu,\overrightarrow{\theta}}(A^{'}|\mathcal{F}_{\infty})=I_{A^{'}}\quad {\hbox{a.s.}}, $

对任意的$\gamma\in (0,1),$令$E^{(n)}=\{z:P_{\nu,\overrightarrow{\theta}}(A^{'}|X_{n}=z)>\gamma\}.$由于

$P_{\nu,\overrightarrow{\theta}}(A^{'}|\mathcal{F}_{\leq n})=P_{\nu,\overrightarrow{\theta}}(A^{'}|X_{n})=\sum\limits_{z}P_{\nu,\overrightarrow{\theta}}(A^{'}|X_{n}=z)\cdot I_{\{X_{n}=z\}}\rightarrow I_{A^{'}}, $

(3.9)

因而$\forall\omega\in A^{'},$存在正整数$N,$当$n\geq N$时, 若$X_{n}(\omega)=z,$就有$P_{\nu,\overrightarrow{\theta}}(A^{'}|X_{n}=z)>\gamma.$则$\omega\in \mathop{\underline{\lim}}\limits_{n\rightarrow\infty}\{X_{n}\in E^{(n)}\}, $从而$A^{'}\subseteq \mathop{\underline{\lim}}\limits_{n\rightarrow\infty}\{X_{n}\in E^{(n)}\}.$同样地, $\forall\omega\in (A^{'})^{c},$存在正整数$N,$当$n\geq N$时, 若$X_{n}(\omega)=z,$就有$P_{\nu,\overrightarrow{\theta}}(A^{'}|X_{n}=z)\leq\gamma.$则$\omega\in \mathop{\underline{\lim}}\limits_{n\rightarrow\infty}\{X_{n}\notin E^{(n)}\}, $从而$\mathop{\overline{\lim}}\limits_{n\rightarrow\infty}\{X_{n}\in E^{(n)}\}\subseteq A^{'}.$于是便有

$\mathop{\overline{\lim}}\limits_{n\rightarrow\infty}\{X_{n}\in E^{(n)}\}=\mathop{\underline{\lim}}\limits_{n\rightarrow\infty}\{X_{n}\in E^{(n)}\}=A^{'},$

(3.10)

因而$\lim\limits_{n\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(X_{n}\in E^{(n)})=P_{\nu,\overrightarrow{\theta}}(A^{'}).$由于$A^{'}\subseteq \{X_{n}=y\quad {\hbox{i.o.}}\},$ $\forall\omega\in A^{'},$由 (3.10) 式, $\omega\in \mathop{\underline{\lim}}\limits_{n\rightarrow\infty}\{X_{n}\in E^{(n)}\},$且存在$\{n_{k}\},$使得$X_{n_{k}}(\omega)=y,X_{n_{k}}(\omega)\in E^{(n_{k})},$故$y\in E^{(n_{k})}.$这样便有$\{X_{n_{k}}=y\}\subseteq\{X_{n_{k}}\in E^{(n_{k})}\}.$由于

$\lim\limits_{k\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(X_{n_{k}}\in E^{(n_{k})})=\lim\limits_{n\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(X_{n}\in E^{(n)})=P_{\nu,\overrightarrow{\theta}}(A^{'}), $

(3.11)

$\lim\limits_{k\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(X_{n_{k}}=y)=\lim\limits_{n\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(X_{n}=y) =q(y),$

(3.12)

则$q(y)\leq P_{\nu,\overrightarrow{\theta}}(A^{'}),$这说明$A$中必含原子, 即存在$A_{0}\subseteq A,A_{0}\in \mathcal{F}^{\infty},A_{0}$是原子.

设$m\leq n,$由于$\{P_{\nu,\overrightarrow{\theta}}(X_{m}=y|\mathcal{F}_{\geq n})\}_{n\geq m}$是反鞅, 则由反鞅收敛定理, 有

$P_{\nu,\overrightarrow{\theta}}(X_{m}=y|X_{n})=P_{\nu,\overrightarrow{\theta}}(X_{m}=y|\mathcal{F}_{\geq n})\rightarrow P_{\nu,\overrightarrow{\theta}}(X_{m}=y|\mathcal{F}^{\infty})\quad {\hbox{a.s.}},$

(3.13)

$\forall \omega\in A_{0},$有$P_{\nu,\overrightarrow{\theta}}(X_{m}=y|\mathcal{F}^{\infty})(\omega)=P_{\nu,\overrightarrow{\theta}}(X_{m}=y|A_{0}),$同时注意到$A_{0}\subseteq A,$则存在正整数数列$\{n_{k}\},X_{n_{k}}(\omega)=y.$于是

$P_{\nu,\overrightarrow{\theta}}(X_{m}=y|X_{n_{k}})(\omega)=P_{\nu,\overrightarrow{\theta}}(X_{m}=y|X_{n_{k}}=y),$

(3.14)

故

$\lim\limits_{k\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(X_{m}=y|X_{n_{k}})(\omega)\\ = \lim\limits_{k\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(X_{m}=y|X_{n})(\omega) = P_{\nu,\overrightarrow{\theta}}(X_{m}=y|\mathcal{F}^{\infty})(\omega)=P_{\nu,\overrightarrow{\theta}}(X_{m}=y|A_{0}),$

这样便有

$\lim\limits_{k\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(X_{m}=y|X_{n_{k}}=y)=P_{\nu,\overrightarrow{\theta}}(X_{m} =y|A_{0}).$

(3.15)

注意到

$\lim\limits_{k\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(X_{m}=y|X_{n_{k}}=y)\\ = \lim\limits_{k\rightarrow\infty}[P_{\nu,\overrightarrow{\theta}}(X_{m}=y)\frac{P_{\nu,\overrightarrow{\theta}}(X_{n_{k}}=y|X_{m}=y)} {P_{\nu,\overrightarrow{\theta}}(X_{n_{k}}=y)}]=P_{\nu,\overrightarrow{\theta}}(X_{m}=y), \\ P_{\nu,\overrightarrow{\theta}}(X_{m}=y|A_{0})=P_{\nu,\overrightarrow{\theta}}(X_{m}=y)\cdot\frac{P_{\nu,\overrightarrow{\theta}}(A_{0}|X_{m}=y)}{P_{\nu,\overrightarrow{\theta}}(A_{0})},$

则$P_{\nu,\overrightarrow{\theta}}(A_{0}|X_{m}=y)=P_{\nu,\overrightarrow{\theta}}(A_{0}).$由于

$\lim\limits_{n\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(A_{0}|X_{m})=\lim\limits_{n\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(A_{0}|\mathcal{F}_{\leq m})=I_{A_{0}}\quad {\hbox{a.s.}}$

(3.16)

故

$\lim\limits_{k\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(A_{0}|X_{n_{k}}) (\omega)=\lim\limits_{m\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(A_{0}|X_{m})(\omega)=I_{A_{0}}(\omega),$

(3.17)

因而$\lim\limits_{k\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(A_{0}|X_{n_{k}}=y)=1,$而$\lim\limits_{k\rightarrow\infty}P_{\nu,\overrightarrow{\theta}}(A_{0}|X_{n_{k}}=y)=P_{\nu,\overrightarrow{\theta}}(A_{0}),$于是就有$P_{\nu,\overrightarrow{\theta}}(A_{0})=1,$从而$P_{\nu,\overrightarrow{\theta}}(A)=1.$这说明$A$是$\mathcal{F}^{\infty}$中的原子.

由于$q(y)$是$(\chi,\mathcal{A})$上的分布, 则必存在$y\in \chi,$使得$q(y)>0,$因而$P_{\nu,\overrightarrow{\theta}}(X_{n}=y\quad {\hbox{i.o.}})=1,$且$\{X_{n}=y\quad {\hbox{i.o.}}\}$是$\mathcal{F}^{\infty}$中的原子.设$M\in \mathcal{F}^{\infty},$则$P_{\nu,\overrightarrow{\theta}}(M\cap \{X_{n}=y\quad {\hbox{i.o.}}\})=0$或$P_{\nu,\overrightarrow{\theta}}(M\cap \{X_{n}=y\quad {\hbox{i.o.}}\})=1,$故$P_{\nu,\overrightarrow{\theta}}(M)=0$或$1.$

最后顺便指出, 虽然平凡尾未必能导出单链是强遍历的, 但却可以推出单链的弱遍历性, 这需要用到下面的引理.

引理^[4] 设$\nu$为$(\chi,\mathcal{A})$上的任意分布, 若对任意的$A\in \mathcal{F}^{\infty},P_{\nu,\overrightarrow{\theta}}(A)=0$或 $1.$则$\forall m\in Z_{+},y\in \chi,P_{\nu,\overrightarrow{\theta}}(X_{m}=y)>0$, 有

$\lim\limits_{n\rightarrow\infty}\sum\limits_{z}|P_{\nu,\overrightarrow{\theta}}(X_{m+n}=z|X_{m}=y)-P_{\nu,\overrightarrow{\theta}}(X_{m+n}=z)| =0.$

(3.18)

在$\pi$是平稳的前提下, 利用上面的引理, 我们不难得出强遍历、弱遍历、平凡尾三者的关系.

命题 $\overrightarrow{X}$强遍历的$\Rightarrow$对$(\chi,\mathcal{A})$上的任意分布$\nu,$及$\pi$-${\hbox{a.e.}}\overrightarrow{\theta},$$\mathcal{F}^{\infty}$关于$P_{\nu,\overrightarrow{\theta}}$是平凡的$\Rightarrow$$\overrightarrow{X}$弱遍历的.

命题的第一个推断即是定理3.5, 下面证明第二个推断.事实上, 若对$(\chi,\mathcal{A})$上的任意分布$\nu,$及$\pi$-${\hbox{a.e.}}\overrightarrow{\theta},$$\mathcal{F}^{\infty}$关于$P_{\nu,\overrightarrow{\theta}}$是平凡的.则$\forall x,y\in \chi,$只需令$\nu=\frac{1}{2}(\delta_{x}+\delta_{y}),B=\{\overrightarrow{\theta}:\mathcal{F}^{\infty}$关于$P_{\nu,\overrightarrow{\theta}}$是平凡的$\}, \forall\overrightarrow{\theta}\in B,$由引理, 便有

$\lim\limits_{n\rightarrow\infty}\sum\limits_{z} |P_{\nu,\overrightarrow{\theta}}(X_{n}= z|X_{0}=x)-P_{\nu,\overrightarrow{\theta}}(X_{n}=z)|=0,$

(3.19)

$\lim\limits_{n\rightarrow\infty}\sum\limits_{z} |P_{\nu,\overrightarrow{\theta}}(X_{n}= z|X_{0}=y)-P_{\nu,\overrightarrow{\theta}}(X_{n}=z)|=0$

(3.20)

于是

$\lim\limits_{n\rightarrow\infty}\sum\limits_{z\in \chi}|P(\theta_{0},\cdots,\theta_{n-1};x,z)-P(\theta_{0},\cdots,\theta_{n-1};y,z)|\\ \leq\lim\limits_{n\rightarrow\infty}\sum\limits_{z} (|P_{\nu,\overrightarrow{\theta}}(X_{n}= z|X_{0}=x)-P_{\nu,\overrightarrow{\theta}}(X_{n}=z)|\\ + |P_{\nu,\overrightarrow{\theta}}(X_{n}= z|X_{0}=y)-P_{\nu,\overrightarrow{\theta}}(X_{n}=z)|)\\ =0,$

从而

$\pi\{\overrightarrow{\theta}\in \Theta^{Z}:\lim\limits_{n\rightarrow\infty}\sum\limits_{z\in \chi}|P(\theta_{0},\cdots,\theta_{n-1};x,z)-P(\theta_{0},\cdots,\theta_{n-1};y,z)|=0\}=1,$

由于$\pi$是平稳分布, 则$\overrightarrow{X}$是弱遍历的.