I am trying to prove that a Feller continuous Markov kernel on a compact metric space has an invariant distribution.

Let $S\ $be a compact metric space. Let $\mathcal{B}(S)$ be the Borel $\sigma $-algebra on $S$. A Markov kernel $P$ on the measurable space $\left (S ,\mathcal{B}(S)\right )$ is a function $P :S \times \mathcal{B}(S) \rightarrow [0 ,1]$ such that for all $s \in S$, $P\left (s , \cdot \right )$ is a probability measure on $S$ and for each $B \in \mathcal{B}(S)$, $P\left ( \cdot ,B\right )$ is a $\mathcal{B}(S)$-measurable function.

The Markov kernel $P$ is Feller continuous if for any bounded, continuous function $f :S \rightarrow \mathbb{R}$, the function $\int _{S}f\left (s^{ \prime }\right )P\left (s ,ds^{ \prime }\right )$ is continuous. Define the operator $T :\mathcal{P}(S) \rightarrow \mathcal{P}(S)$ where $\mathcal{P}(S)$ is the set of all probability measures on $S$ by \begin{equation}T\lambda (B) =\int _{S}P(s ,B)\lambda (ds) . \end{equation}Say that $\mu $ is an invariant distribution if $T\mu =\mu $.

Theorem: There exist an invariant distribution.

Proof: Since $S$ is compact $\mathcal{P}(S)$ is compact in the weak topology. $\mathcal{P}(S)$ is clearly convex. Furthermore, if $\lambda _{n} \rightarrow \lambda $ (weakly) then for any continuous and bounded function $f$, $\int f(s)T\lambda _{n}(ds) =\int f(s^{ \prime })P(s ,ds^{ \prime })\lambda _{n}(ds) \rightarrow \int f(s^{ \prime })P(s ,ds)\lambda (ds) =\int f(s)T\lambda (ds)$ since $\int _{S}f\left (s^{ \prime }\right )P\left (s ,ds^{ \prime }\right )$is continuous and $\lambda _{n} \rightarrow \lambda $. So $T$ is continuous.

Schauder fixed-point theorem: https://en.wikipedia.org/wiki/Schauder_fixed-point_theorem imply that $T$ has a fixed point.

Is my proof correct? Do you know about other proofs / counter example?

  • 1
    $\begingroup$ That is indeed the standard proof. The only thing to add in is that the weak topology (probabilists lingo) coincides with the relative topology of the weak*-topology (functional analysts lingo) on the Banach space of finite signed measures with the variation norm. $\endgroup$ – Michael Greinecker Nov 2 '17 at 8:02
  • $\begingroup$ One remark that could be made is that Schauder's fixed point theorem is overkill - for linear operators there is an easier proof. But why would anyone want to use weaker theorems... $\endgroup$ – Vogel Nov 5 '17 at 16:55
  • $\begingroup$ Michael Greinecker: Why do we need that the topologies concides? The Banach space of finite signed measures with the variation norm is reflexive? Cant we just work with the weak topology? $\endgroup$ – Mark Nov 23 '17 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.