# Markov property implies strong Markov property when state space is countable

In general it cannot be said that a Markov process must satisfy the strong Markov property (see for example A Markov process which is not strong Markov process (follow up 2) ).

But what if the state space is countable/discrete? Is it true that a continuous-time Markov process taking values on $\mathbb{Z}$ must satisfy the strong Markov property?

• If I understand [Kai Lai Chung, "Markov chains with stationary probabilities" (1960), p. 171, p. 174] correctly then if $(X_t)_{t \geq 0}$ is a continuous-time Markov chain with (1) a standard transition matrix $P(t)$ (i.e. $P_{ij}(t) \to \delta_{ij}$ as $t \to 0$ for each two states $i,j$) and (2) right-continuous sample paths $t \mapsto X_t(\omega)$ then $X_t$ has the strong Markov property. Not all Markov chains satsify (1). There are Markov chains that satisfy (1) but not (2), e.g. those that have an instantaneous state. Maybe these can be taken into account for a counterexample. – yadaddy Jul 19 '18 at 10:59