Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a Polish space and $C(X)$ denote the space of all bounded and continuous functions on $X$. We consider a Markov chain $(\xi_n)_{n\geq 0}$ with transition probability $P:X\times \mathcal{B}_X \rightarrow \left[0,1\right]$ and assume that the related Markov semigroup $(P^n)_{n\geq1}$ satisfies the Chapman–Kolmogorov equation: $$P^{n+1}(x,A)=\int\limits_{X}P^n(y,A)\,P(x,dy), \mbox { for } x\in X,\, A\in \mathcal{B}_X.$$ Routinely, we define a dual operator $U^n$ related to $P^n$ as follows: $$U^n f(x)=\int\limits_{X} f(y)P^n(x,dy), \mbox{ for any } f\in C(X)$$ and Markov operator: $$P^n \mu (A)=\int\limits_{X} P^n(x,A)\, \mu(dx), \mbox{ for any } A\in \mathcal{B}_X$$ We also assume that $P$ has the Feller property, that means, $Uf \in C(X)$, for any $f\in C(X)$.

Recently, I've proved the following theorem:

If the chain $(\xi_n)_{n\geq 0}$ has the Feller property and the following conditions are satisfied:

(i) for all $f\in C(X)$ with bounded support the family $\{U^n f: n \in \mathbb{N}\}$ is equicontinuous in each point of $X$,

(ii) the family of measures $\{P^n(x, \cdot): n\in\mathbb{N}\}$ is tight, for any $x\in X$,

(iii) there is a point $z\in X$ such that $\bigwedge\limits_{\delta>0}\, \bigvee\limits_{N\in\mathbb{N}}\,\bigwedge\limits_{x\in X}\,P^N(x,B(z,\delta))>0$,

then, there is an unique probabilistic invariant measure for $(\xi_n)_{n\geq 0}$ and $(P^n(x,\cdot))_{n\geq1}$ converges to $\pi$ weakly and uniformly in $x$ from compact subsets of X.

I have problems with two things:

  1. Does the sequence $(P^n \mu)_{n\geq 1}$ also converge weakly to $\pi$, for any probabilistic measure $\mu$ on $X$? (I know, that the set of point measures is dense in space of all probabilistic measures with weak topology, but I cant use it without the nonexpansivity $P$ in the total variation norm).

  2. Could anyone give an example of Markov - Feller chain, which satisfied conditions (i) - (iii)?

I will be very grateful especially for answer to my second question.

share|improve this question
1  
Perhaps this is too trivial, but an irreducible Markov chain on a finite state space satisfies your properties. (Actually, it doesn't quite have to be irreducible; there can be some transient states also.) I think that any positive recurrent irreducible chain on a countable discrete state space works, too. And I didn't check it, but I would bet that classical examples of recurrent processes on $\mathbb{R}^n$ would work too. For instance, Brownian motion on an open set with reflecting boundary, or an Ornstein-Uhlenbeck process. –  Nate Eldredge Aug 24 '11 at 1:29
1  
@dawid: to add to the answer by Nate, in $\mathbb R^n$ you can consider kernels having continuous densities. A trivial example can be when the density is strictly positive and does not depend on $x$. –  Ilya Aug 24 '11 at 8:41
    
Thanks for comments, I checked such chains on countable discrete state space and I think it should work, but actually I would be satisfied with an example of a proper chain on state space, which is Polish but non locally compact, becouse exactly I generalized this theorem on such spaces... And what's about my first question? What do you think about that? –  dawid Aug 24 '11 at 17:54

1 Answer 1

up vote 1 down vote accepted

For 1., I think that this follows from convergence of $P^n(x,\cdot)$. Let $f\in C(X)$. Then $F_n(x) := \int f \;\;\mathrm{d}P^n(x, \cdot)$ is bounded uniformly in $n$ (by the same bound as $f$) and converges point-wise to $\int f\;\; \mathrm{d}\pi$ by your result. Dominated convergence yields $$ \int f \;\;\mathrm{d}P^n\mu = \int F(x) \;\;\mu(\mathrm{d}x) \to \int f\;\; \mathrm{d}\pi. $$ Hence $P^n\mu$ converges weakly to $\pi$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.