0
$\begingroup$

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 $X_0$?

$\endgroup$
1
$\begingroup$

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 $X_0$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.