非时齐马氏链是马氏过程中较难研究的一种类型,如何判断马氏链的收敛性,特别是非时齐马氏链的收敛性一直是国内外许多学者研究的课题.Dobrushin-Isaacson-Madsen定理研究了有限状态空间非时齐马氏链的收敛性^[4]参考文献的标记与引用本文对一般状态空间非时齐马氏链进行了研究,得到非时齐马氏链收敛的一个充分条件即定理1.1.定理A (Dobrushin-Isaacson-Madsen定理)是定理1.1的特例.

下面Dobrushin-Isaacson-Madsen定理给出了有限状态空间非时齐马氏链的收敛性的充分条件.

定理A (Dobrushin-Isaacson-Madsen定理^[4])设$ X=\{X_1, X_2, \cdots , X_n, \cdots \} $是有限状态空间$ S $上的非时齐马氏链,其第$ n $步转移矩阵为$ {P_n} $,如果他们满足

(A1)  ${P_n}$作为时齐的转移矩阵时有平稳分布${\pi_n};$

(A2)  $\sum\limits_{n}\|\pi_n-\pi_{n+1}\|<\infty ;$

(A3)  或者满足Isaacson-Madsen条件:对于$S$上的任意概率分布向量$\mu, \nu$及正整数$j$,恒有

$\|(\mu-\nu)P_j\cdots P_n\|\rightarrow0, n\rightarrow\infty;$

或者满足Dobrushin条件:对于任意$j$,记$P=(P_j\cdots P_n)=(P_{il}).$$P$的收缩系数$C(P)$满足

$C(P)\rightarrow0, n\rightarrow\infty.$

其中

$C(P)=\frac{1}{2}\sup\limits_{i, k}\sum\limits_l|P_{il}-P_{kl}|.$

那么存在$S$上的概率测度$\pi$,使得

(1)  $\|\pi_n-\pi\|\rightarrow0, n\rightarrow\infty;$

(2)  无论什么初始分布${\mu_0}$下,此非时齐的马链$X$在时刻$n$的分布${\mu_n}$恒有极限

$\|\mu_n-\pi\|\rightarrow0, n\rightarrow\infty.$

其中$\|v\|$是$S$上符号测度$v$的全变差范数.

本文的目的就是将定理A从有限状态空间$S$推广到一般波兰空间$(X, \rho, {\cal B}(X))$上去.设$(X, \rho, {\cal B}(X))$是波兰空间.为叙述方便,先引入几个记号.设$g$是$X$上的可测函数, $v$是${\cal B}(X)$上的符号测度, $K$是$(X, {\cal B}(X))$上的可测核.记

$v(g):=\int{g{\rm d}v}, $

$Kg(x):=\int{K(x, {\rm d}y)g(y)}, $

$(vK)g=v(Kg):=\int v({\rm d}x)\int{K(x, {\rm d}y)g(y)}.$

$\|v\|:=\sup\{|v(g)|:|g|\leq 1\}.$

定义1.1  设$\mu_1, \mu_2$是${\cal B}(X)$上的任意概率测度, $\widetilde{\mu}$是${\cal B}(X)\times {\cal B}(X)$上的概率测度,称$\widetilde{\mu}$是$\mu_1$与$\mu_2$的耦合,若满足边缘性

$\widetilde{\mu}(A_1\times X)=\mu_1(A_1), A_1\in {\cal B}(X), \\\;\widetilde{\mu}(X\times A_2)=\mu_2(A_2), A_2\in {\cal B}(X).$

记$K(\mu_1, \mu_2)$为$\mu_1$与$\mu_2$的全体耦合.

定义1.2  设$\mu_1, \mu_2$是波兰空间$(X, \rho, {\cal B}(X))$上的任意概率测度,令

$W(\mu_1, \mu_2)=\inf\bigg\{\bigg[\int\mu({\rm d}x, {\rm d}y)\rho(x, y)\bigg]:\mu\in K(\mu_1, \mu_2)\bigg\}.$

称之为$\mu_1$与$\mu_2$的最小概率距离.若存在$\overline{\mu}\in K(\mu_1, \mu_2), $使

$W(\mu_1, \mu_2)=\int\overline{\mu}({\rm d}x, {\rm d}y)\rho(x, y).$

则称$\overline{\mu}$是$\mu_1$与$\mu_2$的$\rho$最优耦合.$\rho$最优耦合总是存在的^[1].

定理A介绍了有限状态空间上非时齐马氏链的收敛性,在实际应用中是不够的.下面我们将用概率距离和耦合方法研究一般状态空间上非时齐马氏链的收敛性问题.

设$\Phi=\{\Phi_0, \Phi_1, \cdots , \Phi_n, \cdots \}$是$(X, \rho, {\cal B}(X))$上的非时齐马氏链,其第$n$步转移概率$P(\Phi_n\in A|\Phi_{n-1}=x\}:=P_n(x, A).$进一步,若$P_1=P_2=\cdots =P_n$,则称$\Phi$为时齐马氏链.设${\cal P}(X)$为$X$上的全体概率测度, $x_{0}$为$X$上某个给定的点(等价地任意给定的点$x_{0}$).记

${\cal M}=\bigg\{\mu\in{\cal P}(X):\int\rho(x_{0}, x)\mu({\rm d}x)<\infty\bigg\}.$

定理1.1  设$(X, \rho, {\cal B}(X))$是波兰空间,设$\Phi=\{\Phi_0, \Phi_1, \cdots , \Phi_n, \cdots \}$是取值$X$上的非时齐马氏链, $\{P_n;n=1, 2, \cdots \}$是$\Phi$相应的转移概率核序列.如果满足

(i)  每个概率核${P_n}$作为一个时齐马氏链有平稳分布$\pi_n$,且$\pi_n\in {\cal M}$,满足

$\sum\limits_{n}W(\pi_n, \pi_{n+1})<\infty;$

(ii)  $\int\rho(u, v)\widetilde{P}_n(x, y;{\rm d}u, {\rm d}v)\leq\rho(x, y)c_n $, $c_n$与$x, y$无关且$0\leq c_n\leq 1$;

(iii)  $\prod\limits_{k=j}^n c_n\rightarrow0, n\rightarrow\infty$;

(iv)  $\forall\mu\in{\cal M}, $有$\mu P_n\in{\cal M}, n=1, 2, \cdots .$

其中$\widetilde{P}_n(x, y;{\rm d}u, {\rm d}v)$为${P_n}(x, {\rm d}u)$与${P_n}(y, {\rm d}v)$的耦合.那么,存在${\cal M}$上的概率测度$\pi$,使得对${\cal M}$上的概率测度${\mu_0}$,当$n\rightarrow\infty$时有

$W({\mu_0P_1P_2\cdots P_n}, \pi)\rightarrow0, n\rightarrow\infty.$

注  定理中的条件(iv)一般说来是易于验证的,特别地$\rho$是有界距离时这个条件自动满足.

引理2.1  设$(X, \rho, {\cal B}(X))$是波兰空间,则$({\cal M}, W)$是完备的度量空间.

证明见文献[1]第175页定理5.4.

引理2.2  设$\{\mu_n:n\geq1\}\subset{\cal M}$,当$\sum\limits_{n=1}^\infty W(\mu_n, \mu_{n+1})<\infty$时, $\{\mu_n:n\geq1\}$是$\{{\cal M}, W\}$上的Cauchy序列.

证  对任意$\varepsilon>0$,因为$\sum\limits_{n=1}^\infty W(\mu_n, \mu_{n+1})<\infty$,故存在$N$,当$n>N$时,有

$\sum\limits_{k=n}^\infty W(\mu_k, \mu_{k+1})<\varepsilon.$

于是对任意正整数m,由三角不等式知

$W(\mu_n, \mu_{n+m})\leq\sum\limits_{k=n}^{n+m}W(\mu_k, \mu_{k+1})\leq\sum\limits_{k=n}^\infty W(\mu_k, \mu_{k+1})<\varepsilon.$

故$\{\mu_n:n\geq1\}$是Cauchy序列.引理证毕.

引理2.3  设$\mu_1, \mu_2$是任意概率测度, $P_{n}$是波兰空间$(X, \rho, {\cal B}(X))$上的概率核,若

(2.1) $\int\rho(u, v)\widetilde{P}_n(x, y;{\rm d}u, {\rm d}v)\leq\rho(x, y)c_n .$

其中$\widetilde{P}_n(x, y;{\rm d}u, {\rm d}v)$为${P_n}(x, {\rm d}u)$与${P_n}(y, {\rm d}v)$的耦合, $c_n$与$x, y$无关且$0\leq c_n\leq 1$.则有

(2.2) $W({\mu_1P_n}, {\mu_2P_n})\leq W(\mu_1, \mu_2).$

证  由(2.1)式有

(2.3) $\int\rho(u, v)\widetilde{P}_n(x, y;{\rm d}u, {\rm d}v)\leq\rho(x, y).$

设$\overline{\mu}$为$\mu_1$和$\mu_2$的$\rho$最优耦合,在$(2.3)$式两边对$\overline{\mu}$求积分得

$\int\overline{\mu}({\rm d}x, {\rm d}y)\int\rho(u, v)\widetilde{P}_n(x, y;{\rm d}u, {\rm d}v)\leq\int\overline{\mu}({\rm d}x, {\rm d}y)\rho(x, y).$

即

(2.4) $\int\overline{\mu}\widetilde{P}_n({\rm d}u, {\rm d}v)\rho(u, v)\leq W(\mu_1, \mu_2).$

下面证明

(2.5) $W(\mu_1P_n, \mu_2P_n)\leq\int\overline{\mu}\widetilde{P}_n({\rm d}u, {\rm d}v)\rho(u, v).$

要证明(2.5)式成立,只需证明$\overline{\mu}\widetilde{P}_n$是$\mu_1P_n$与$\mu_2P_n$的耦合.下面验证边缘性条件

$\begin{eqnarray*}\overline{\mu}\widetilde{P}_n(A\times X)&=&\int\overline{\mu}({\rm d}x, {\rm d}y)\widetilde{P}_n(x, y;A\times X)\\&=&\int\overline{\mu}({\rm d}x, {\rm d}y)P_n(x, A)\\&=&\int\mu_1({\rm d}x)P_n(x, A)\\&=&\mu_1P_n(A).\end{eqnarray*}$

同理可证$\overline{\mu}\widetilde{P}_n(X\times B)=\mu_2P_n(B)$.故$\overline{\mu}\widetilde{P}_n$是$\mu_1P_n$与$\mu_2P_n$的耦合.(2.5)式获证.再结合(2.4)式即得(2.2)式.

为了证明定理1.1,下面分三步来证明.

(a)  对$\forall\mu_1, \mu_2\in{\cal M}$,任意正整数$j$有

$W({\mu_1P_jP_{j+1}\cdots P_n}, {\mu_2P_jP_{j+1}\cdots P_n})\rightarrow0, n\rightarrow\infty;$

(b)  存在概率测度$\pi\in{\cal M}$使得

(1) $W({\pi_n}, \pi)\rightarrow0, n\rightarrow\infty;$

(c)  对初始分布${\mu_0}\in{\cal M}$有

$W({\mu_0P_1P_2\cdots P_n}, \pi)\rightarrow0, n\rightarrow\infty.$

定理1.1的证明  (a) 先证明$\overline{\mu}\widetilde{P}_j\widetilde{P}_{j+1}\cdots \widetilde{P}_n$是$\mu_1P_jP_{j+1}\cdots P_n$与$\mu_2P_jP_{j+1}\cdots P_n$的耦合.其中$\overline{\mu}$是$\mu_1$与$\mu_2$的$\rho$最优耦合, $\widetilde{P}_n(x, y;{\rm d}u, {\rm d}v)$为${P_n}(x, {\rm d}u)$与${P_n}(y, {\rm d}v)$的耦合, $\widetilde{P}_n(x, y;\cdot)$是$(X\times X, {\cal B}(X)\times{\cal B}(X))$上的概率核.为简单见,我们只证$n=2$的情形(一般情形可类似地证明).即证$\overline{\mu}\widetilde{P}_1\widetilde{P}_2$是$\mu_1P_1P_2$与$\mu_2P_1P_2$的耦合.由耦合的边缘性知:

$\widetilde{P}_1(x, y;A\times X)=P_1(x, A), \widetilde{P}_1(x, y;X\times A)=P_1(y, A), x, y\in X, A\in {\cal B}(X);\\\;\widetilde{P}_2(x, y;A\times X)=P_2(x, A), \widetilde{P}_2(x, y;X\times A)=P_2(y, A), x, y\in X, A\in {\cal B}(X).$

下面验证边缘性条件

(3.1) $\overline{\mu}\widetilde{P}_1\widetilde{P}_2(A\times X)=\mu_1P_1P_2(A) , x, y\in X, A\in {\cal B}(X);$

(3.2) $\overline{\mu}\widetilde{P}_1\widetilde{P}_2(X\times A)=\mu_2P_1P_2(A) , x, y\in X, A\in {\cal B}(X).$

注意到$\mu_1P_1P_2(A)=\int\!\!\!\int\mu_1({\rm d}x)P_1(x, {\rm d}u)P_2(u, A)$,而

$\begin{eqnarray*}\overline{\mu}\widetilde{P}_1\widetilde{P}_2(A\times X)&=&\int\!\!\!\int\overline{\mu}({\rm d}x, {\rm d}y)\widetilde{P}_1(x, y;{\rm d}u, {\rm d}v)\widetilde{P}_2(u, v;A\times X) \\&=&\int\!\!\!\int\overline{\mu}({\rm d}x, {\rm d}y)\widetilde{P}_1(x, y;{\rm d}u, {\rm d}v)P_2(u, A) \\&=&\int\overline{\mu}({\rm d}x, {\rm d}y)\int P_1(x, {\rm d}u)P_2(u, A) \\&=&\int\mu_1({\rm d}x)\int P_1(x, {\rm d}u)P_2(u, A) \\&=&\int\!\!\!\int\mu_1({\rm d}x) P_1(x, {\rm d}u)P_2(u, A) \\&=&\mu_1P_1P_2(A).\end{eqnarray*}$

即(3.1)式成立,类似地,可证(3.2)式成立.

于是有

$\begin{eqnarray*}&&W({\mu_1P_jP_{j+1}\cdots P_n}, {\mu_2P_jP_{j+1}\cdots P_n})\\&=&\inf\bigg\{\bigg[\int\overline{\mu}\widetilde{P}_j\widetilde{P}_{j+1}\cdots \widetilde{P}_n({\rm d}u, {\rm d}v)\rho(u, v)\bigg]:\overline{\mu}\in K(\mu_1, \mu_2)\bigg\}\\&\leq&\int\overline{\mu}\widetilde{P}_j\widetilde{P}_{j+1}\cdots \widetilde{P}_n({\rm d}u, {\rm d}v)\rho(u, v).\end{eqnarray*}$

反复利用条件(ii)得

$\begin{eqnarray*}\int\overline{\mu}\widetilde{P}_j\widetilde{P}_{j+1}\cdots \widetilde{P}_n({\rm d}u, {\rm d}v)\rho(u, v)&\leq&\int\overline{\mu}\widetilde{P}_j\widetilde{P}_{j+1}\cdots \widetilde{P}_{n-1}({\rm d}u, {\rm d}v)\rho(u, v)c_n\\&\leq&\cdots \\&\leq&\int\overline{\mu}({\rm d}x, {\rm d}y)\rho(u, v)\prod _{k=j}^{n}c_n\\&=& W(\mu_1, \mu_2)\prod _{k=j}^{n}c_n.\end{eqnarray*}$

由于$\mu_1, \mu_2\in{\cal M}$,从而有

$\begin{eqnarray*}W(\mu_1, \mu_2)&=&\int\rho(x, y)\overline{\mu}({\rm d}x, {\rm d}y)\\&\leq&\int\rho(x, x_0)\overline{\mu}({\rm d}x, {\rm d}y)+\int\rho(x_0, y)\overline{\mu}({\rm d}x, {\rm d}y)\\&=&\int\rho(x, x_0)\mu_1({\rm d}x)+\int\rho(x_0, y)\mu_2({\rm d}y)\\&<& \infty.\end{eqnarray*}$

结合条件(iii) $\prod\limits_{k=j}^n c_n\rightarrow0, n\rightarrow\infty$证得(a)成立.

(b) 由条件(i)知

$\sum\limits_{n}W(\pi_n, \pi_{n+1})<\infty .$

由引理$(2.1)$知$\{{\cal M}, W\}$是完备的度量空间.又由引理$(2.2)$知$\{\pi_n:n\geq1\}$是$\{{\cal M}, W\}$上的Cauchy序列.于是存在$\pi\in{\cal M}, $使得$W(\pi_n, \pi)\rightarrow0, n\rightarrow\infty.$结论(b)得证.

(c)  为证(c),即要证对$\forall\varepsilon>0, \exists N\in Z^{+}, $ s.t.当$n>N$时有

(3.3) $W({\mu_0P_1P_2\cdots P_n}, \pi)<\varepsilon.$

对任意的正整数$j\leq n, $利用三角不等式有

(3.4) $W({\mu_0P_1P_2\cdots P_n}, \pi)<W({\mu_0P_1P_2\cdots P_n}, {\pi P_jP_{j+1}\cdots P_n})+W({\pi P_jP_{j+1}\cdots P_n}, \pi).$

对上式第二项反复利用三角不等式并注意条件$\pi_mP_m=\pi_m$推得

$\begin{eqnarray*}&&W({\pi P_jP_{j+1}\cdots P_n}, \pi)\\&\leq& W({\pi P_jP_{j+1}\cdots P_n}, {\pi_j P_jP_{j+1}\cdots P_n})+W({\pi_j P_jP_{j+1}\cdots P_n}, \pi)\\&\leq &W(\pi, \pi_j)+W({\pi_j P_jP_{j+1}\cdots P_n}, \pi)\\&=&W(\pi, \pi_j)+W({\pi_j P_{j+1}\cdots P_n}, \pi) \qquad (\pi_j P_j=\pi_j)\\&\leq& W(\pi, \pi_j)+W({\pi_j P_{j+1}\cdots P_n}, {\pi_{j+1} P_{j+1}\cdots P_n})+W({\pi_{j+1}P_{j+1}\cdots P_n}, \pi) \\&\leq &W(\pi, \pi_j)+W({\pi_j}, {\pi_{j+1}})+W({\pi_{j+1}P_{j+2}\cdots P_n}, \pi) \\&\leq&\cdots \\&\leq& W(\pi, \pi_j)+\sum\limits_{k=1}^{n-j} W({\pi_{j+k-1}}, {\pi_{j+k}})+W({\pi_n}, \pi).\end{eqnarray*}$

从而有

(3.5) $W({\pi P_jP_{j+1}\cdots P_n}, \pi)\leq W(\pi, \pi_j)+\sum\limits_{k=1}^{n-j} W({\pi_{j+k-1}}, {\pi_{j+k}})+W({\pi_n}, \pi).$

由(b)知,当$j\rightarrow\infty$时, $W(\pi, \pi_j)\rightarrow0, $所以$\exists M_1\in Z^{+}, $ s.t.当$j>M_1$时,有

(3.6) $W(\pi, \pi_j)<\frac{\varepsilon}{4}.$

同理,当$n\rightarrow\infty$时$W({\pi_n}, \pi)\rightarrow0, $所以$\exists N_1\in Z^{+}, $ s.t.当$n>N_1$时,有

(3.7) $W({\pi_n}, \pi)<\frac{\varepsilon}{4}.$

由条件$\sum\limits_{n}W(\pi_n, \pi_{n+1})<\infty $知,当$j\rightarrow\infty$时, $\sum\limits_{n=j}^{\infty}W(\pi_n, \pi_{n+1})\rightarrow0, $即$\exists M_2\in Z^{+}, $ s.t.当$j>M_2$时,有

(3.8) $\sum\limits_{k=1}^{n-j} W({\pi_{j+k-1}}, {\pi_{j+k}})\leq\sum\limits_{n=j}^{\infty}W(\pi_n, \pi_{n+1})<\frac{\varepsilon}{4}.$

由${\mu_0}\in{\cal M}$和条件(iv)易证得$\mu_0P_1P_2\cdots P_{j-1}\in{\cal M}, $由(a)知

$\begin{eqnarray*}&&W({\mu_0P_1P_2\cdots P_n}, {\pi P_jP_{j+1}\cdots P_n})\\&=&W((\mu_0P_1P_2\cdots P_{j-1}){P_jP_{j+1}\cdots P_n}, {\pi P_jP_{j+1}\cdots P_n})\rightarrow0, \;\;\;\; n\rightarrow\infty. \end{eqnarray*}$

所以$\exists N_2\in Z^{+}, $ s.t.当$n>N_2$时,有

(3.9) $W({\mu_0P_1P_2\cdots P_n}, {\pi P_jP_{j+1}\cdot \cdot \cdot P_n})<\frac{\varepsilon}{4}.$

最后,取$N=\max\{N_1, N_2\}, M=\max\{M_1, M_2\}, $在$(3.6)$和$(3.8)$式中取$j=M, $则当$n>N$时, (3.6)-(3.9)都成立.把(3.6)-(3.8)式代入$(3.5)$式得

(3.10) $W({\pi_{M}P_{M}\cdots P_{n}}, \pi)<\frac{3\varepsilon}{4}.$

再把(3.9)和(3.10)式代入(3.4)式得到(3.3)式成立.结论(c)得证.定理证毕.