Suppose $X_0, X_1, X_2, \ldots$ is a Markov chain on a compact metric space $(\Omega, \mathcal{B}, \mathbb{P})$. Call its Markov generator $R f(x) = \mathbb{E}[f(X_1)\; | \; X_0=x]$ ($f$ bounded measurable on $\Omega$).

Let $0 = \tau_0 < \tau_1 < \tau_2 < \cdots$ be a sequence of stopping times which are almost surely finite, and call the Markov chain at these stopping times $Y_k = X_{\tau_k}$. Call its generator $L$.

My question is this. If there exists an $R$-invariant measure $\mathbb{Q}$ on $\Omega$, so that $\int Rf d\mathbb{Q} = \int f d\mathbb{Q}$, is $\mathbb{Q}$ necessarily $L$-invariant?

I'm having trouble proving this. Thanks for any help!

  • $\begingroup$ Consider a finite state space. Take the times so that L is actually constant. Then only in very trivial situations will the two measures agree. $\endgroup$ – Ian Oct 27 '15 at 20:37

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