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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 $t\to\infty$?

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  • $\begingroup$ 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 $\mu \kappa_t = \mu$. $\endgroup$ – Sayantan Dec 28 '18 at 1:43
  • $\begingroup$ @Sayantan We've got $\mu_t=\mu_0\otimes \kappa_t$ (product of $\mu_0$ and $\kappa_t$). $\endgroup$ – 0xbadf00d Dec 28 '18 at 10:35

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