 Solution

For the finite state space $S$, the generator or Q-matrix $G=(g_{ij})$ is defined byIts off-diagonal entries are nonnegative jump rates and $g_{ii}=-{\sum }_{j\ne i}{g}_{ij}$. The transition semigroup of a continuous-time Markov chain is $P(t)=e^{tG}$.

An invariant distribution of a continuous-time Markov chain is a probability row vector $\pi$ satisfying $\pi P(t)=\pi$ for every $t\ge 0$. Differentiation at zero gives $\pi G=0$; conversely, $\pi G=0$ implies $\pi e^{tG}=\pi$. ThusInvariant distributions need not be unique: each closed communicating class supports one, and their convex combinations are invariant. A finite irreducible chain has a unique invariant distribution.

Let $0=T_0<T_1<\cdots$ be the arrival times of the independent rate-$\lambda$ Poisson process. The increments $T_{n+1}-T_n$ are independent exponential variables of rate $\lambda$. Conditional on $Y_n=X_{T_n}=i$, the Markov property, time homogeneity, and independence of the next increment giveHence $(Y_n)$ is a discrete-time Markov chain. Its transition matrix is the Poisson-sampled continuous-time Markov chain resolvent

If $\pi G=0$, then $\pi e^{tG}=\pi$ and integration gives $\pi P_{\lambda }=\pi$. Conversely, if a probability row vector $\nu$ satisfies $\nu P_{\lambda }=\nu$, multiplyingon the right by $\lambda I-G$ yields $\nu G=0$. Therefore the sampled chain and the original chain have exactly the same invariant distributions.