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I'm revising for an exam and have no idea how to approach this question:

Let $\{N_t\}_{t\geq 0}$ be a homogeneous Poisson process of parameter $\lambda > 0$. Let $\{X_k\}_{k\geq 0}$ be a sequence of random variables, identically independently distributed, with $E[X_k] = 0$ and $Var[X_k] = \sigma^2 < +\infty$. Is the process defined by $Y_t = X_{N_t}$ strictly stationary?

I'm guessing the fact that the Poisson process has independent increments and we have:

$P(N_{t_2} - N_{t_1} = k) = \dfrac{\lambda (t_2 - t_1))^k}{k!} e^{-\lambda (t_2 - t_1)}$ is important, but not sure how it relates, any hints appreciated!

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1 Answer 1

If the process $(N_t)$ and the sequence $(X_k)$ are independent, then indeed the identity $Y_t=X_{N_t}$ defines a strictly stationary process $(Y_t)$.

To show this, consider $Y^s_t=Y_{s+t}$, then, for every nonnegative time $s$, conditionally on $N_s=m$, the process $(Y^s_t)_{t\geqslant0}$ is such that, for every $t\geqslant0$, $$ Y^s_t=X_{m+N'_{t}},\qquad N'_t=N_{t+s}-N_s. $$ The process $N'$ is independent on $N_s$ hence $Y^s$ is independent on $N_s$. Furthermore, for every $m$, $(X_{m+k})_{k\geqslant0}$ is distributed like $(X_k)_{k\geqslant0}$ hence $(X_{m+N'_t})_{t\geqslant0}$ is distributed like $(X_{N_t})_{t\geqslant0}$ and the result follows.

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