# If $(X_t)_{t\ge0}$ is a Markov process with invariant measure $\mu$, does the distribution of $X_t$ weakly converge to $\mu$ as $t\to\infty$?

Let

• $$(\Omega,\mathcal A,\operatorname P)$$ be a complete probability space
• $$(X_t)_{t\ge0}$$ be a time-homogeneous Markov process on $$(\Omega,\mathcal A,\operatorname P)$$ and $$\kappa_t$$ denote a regular version of the conditional distribution of $$X_t$$ given $$X_0$$ for $$t\ge0$$, i.e. $$\kappa_t(x,B)=\operatorname P\left[X_t\in B\mid X_0=x\right]\;\;\;\text{for all }(x,B)\in\mathbb R\times\mathcal B(\mathbb R)\tag1$$
• $$C_0(\mathbb R)$$ denote the space of continuous functions on $$\mathbb R$$ vanishing at infinity
• $$\mu$$ be a probability measure on $$(\mathbb R,\mathcal B(\mathbb R))$$

As usual, let $$\kappa_tf:=\int\kappa_t(x,{\rm d}y)f(y)$$ for bounded Borel measurable $$f:\mathbb R\to\mathbb R$$ and $$t\ge0$$. Assume $$\kappa_tC_0(\mathbb R)\subseteq C_0(\mathbb R)\;\;\;\text{for all }t\ge0.\tag2$$

Let $$\mu_t(B):=\operatorname P\left[X_t\in B\right]\;\;\;\text{for }B\in\mathcal B(\mathbb R)$$ for $$t\ge0$$. It's easy to see that weak convergence of $$\mu_t$$ to $$\mu$$ as $$t\to\infty$$ implies that $$\mu$$ is invariant with respect to $$(\kappa_t)_{t\ge0}$$, i.e. $$\mu\kappa_t=\mu\;\;\;\text{for all }t\ge0,$$ where $$\mu\kappa_t$$ denotes the composition of $$\mu$$ and $$\kappa_t$$.

Does the converse hold as well, i.e. if $$\mu$$ is invariant with respect to $$(\kappa_t)_{t\ge0}$$, are we able to conclude that $$\mu_t$$ converges weakly to $$\mu$$ as $$t\to\infty$$?

• What is the relation between $\mu_t$, $\kappa_t$ and $\mu$? Is it something like $$\mu_t (B) = \int \kappa_t(x,B) \mu(\, d x) = P^{\mu}(X_t \in B) ?$$ Because otherwise $\mu_t \to \mu$ weakly will not imply $\mu \kappa_t = \mu$. – Sayantan Dec 28 '18 at 1:43
• @Sayantan We've got $\mu_t=\mu_0\otimes \kappa_t$ (product of $\mu_0$ and $\kappa_t$). – 0xbadf00d Dec 28 '18 at 10:35