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 $X_0$.

  • $\begingroup$ 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 $(e^{tA})_{t\ge0}$. $\endgroup$ – 0xbadf00d Apr 17 at 18:11

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