# On the Markov property of a limit process

Let $$E$$ be a locally compact separable metric with countable base. We consider a sequence of Hunt processes $$\{X^{(n)}\}_{n \in \mathbb{N}}$$ on $$E$$. That is, each $$X^{(n)}=(\{X_t^{(n)}\}_{t \in [0,\infty]},\{P_x^{(n)}\}_{x \in E})$$ is a cad-lag Markov process on $$E$$ with the strong Markov property (and the quasi-left-continuity). Let $$m$$ be a probability measure on $$E$$. For $$n \in \mathbb{N}$$, we write $$P_{m}^{(n)}$$ for the law of $$X^{(n)}$$ with initial distribution $$m$$.

We now assume the following condition.

$${\rm \bf(A)}$$ The laws $$\{P_{m}^{(n)}\}_{n\in \mathbb{N}}$$ of $$\{X^{(n)}\}_{n \in \mathbb{N}}$$ are tight in $$D([0,\infty),E)$$. Here, $$D([0,\infty),E)$$ is the space of $$E$$-valued right continuous functions on $$[0,\infty)$$ with finite left limits.

Under the condition $${\rm \bf(A)}$$, we have a subsequential limit $$(X,P)$$ of $$\{X^{(n)},P_m^{(n)}\}_{n \in \mathbb{N}})$$. For a bounded measurable function $$f\colon E \to \mathbb{R}$$, we set $$(P_t f)(x)=E[f(X_t) \mid X_0=x]$$ for $$m$$-a.e. $$x$$. Then, each $$P_t$$ is extended to a contraction operator on $$L^\infty(E,m)$$ (the extended operator is denoted by the same symbol).

Can we show that $$P_t(P_sf)=P_{t+s}f$$ for every $$t,s>0$$ and $$f \in L^\infty(E,m)$$?

That is, I would like to know that $$(X,P)$$ is a time-homogeneous Markov process on $$E.$$

• Haven't gone through the details, but what if we work on $E = \mathbb{R}$, and define $X^{(n)}$ with the following transitions: from $0$ it jumps to either $1/n$ or $-1$ at rate 1. The states $1/n$ and $1$ are absorbing. It seems that in the limit, the state $1/n$ merges with $0$, so the process has probability $1/2$ to stay at 0 forever, which is impossible for a Markov process. May 11, 2022 at 15:50
• @NateEldredge Thank you for your comment. The state $0$ is a cemetery point for your limit process? If so, I feel that it does not necessarily contradict the Markov property. Am I making a mistake? May 11, 2022 at 16:04
• Maybe I have to think some more. I guess the problem is that "initial state" becomes inconsistent in the limit. May 11, 2022 at 16:10
• @NateEldredge Thank you very much. If you know sufficient conditions to reach an affirmative conclusion, please let me know. May 11, 2022 at 16:19
• I think this is often studied in terms of convergence of the associated Dirichlet forms. I don't know the exact results, but "Mosco convergence" comes up a lot. doi.org/10.1006/jfan.1994.1093 seems to be the original paper May 11, 2022 at 16:55

If your semigroups $$\{P_t^{(n)}\}$$ and $$\{P_t\}$$ are Feller then Ethier and Kurtz Theorem 4.2.5 give that your limiting process $$X$$ exists and is a Markov process (indeed, a Feller process).
This may seem obvious (since clearly $$X$$ is Markov if its semigroup is Feller) but without the Feller property, it's not always clear that $$X$$ exists. Ethier and Kurtz Corollary 4.2.8 gives an if and only if guarantee: if your semigroup $$\{P_t\}$$ is strongly continuous (and it is already a positive contraction semigroup) then in order for $$X$$ to exist, $$\{P_t\}$$ must also have a conservative generator, and thus be Feller.