# Explicit formula for the one-dimensional distributions of a time-homogeneous Markov chain subordinated by a Poisson process

Let $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space, $$(E,\mathcal E)$$ be a measurable space, $$(X_n)_{n\in\mathbb N_0}$$ be a $$(E,\mathcal E)$$-valued time-homogeneous Markov chain on $$(\Omega,\mathcal A,\operatorname P)$$, $$(N_t)_{t\ge0}$$ be a Poisson process on $$(\Omega,\mathcal A,\operatorname P)$$ with intensity $$\lambda\ge0$$ and $$Y_t:=X_{N_t}\;\;\;\text{for }t\ge0.$$

Let $$t\ge0$$. Is there an explicit formula for the distribution of $$Y_t$$?

In particular, if $$X$$ is stationary, will $$Y_t$$ be distributed according to the law of $$X_0$$?

I'll assume that the processes $$X$$ and $$N$$ are independent. Then for each $$A \in \mathcal{E}$$ we have \begin{aligned} \operatorname P(Y_t \in A) &= \sum_{k=0}^\infty \operatorname P(X_k \in A, N_t = k) \\ &= \sum_{k=0}^\infty \operatorname P(X_k \in A) \operatorname P(N_t = k) \\ &= \sum_{k=0}^\infty \operatorname P(X_k \in A) e^{-\lambda t}\frac{\left(\lambda t\right)^k}{k!}. \end{aligned} I don't know if this can be made more explicit in the general case, but we already see that if $$X$$ is stationary, then $$\operatorname P(X_k \in A) = \operatorname P(X_0 \in A)$$ for all $$k$$, so $$\operatorname P(Y_t \in A) = \sum_{k=0}^\infty \operatorname P(X_0 \in A) e^{-\lambda t}\frac{\left(\lambda t\right)^k}{k!} = \operatorname P(X_0 \in A),$$ so yes, in this case the distribution of $$Y_t$$ is that of $$X_0$$.
• I was just going to write down an answer by myself. As you say, assuming independence, we see that $Y$ is a time-homogeneous Markov process with generator $A:=\lambda(\kappa-\operatorname{id})$ and transition semigroup $(e^{tA})_{t\ge0}$. Apr 17, 2019 at 18:11