# An example of Markov-Feller chain with some properties

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.

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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
@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

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 $\pi$.