Let $X_1, X_2, …$ be iid random variables with characteristic function $\phi$ and let $N$ be Poisson distributed with parameter $\lambda$. Define $Y = \sum_{n = 1}^N X_n$. Express the characteristic function $\psi$ of $Y$ in terms of $\lambda$ and $\phi$.
I am having some trouble with this exercise. We know that the characteristic function of $X_1 + … + X_n$ is $\phi^n$, but the amount of terms in $Y$ is not fixed, so we cannot use this directly. By definition we have
$$\psi(u) = \sum_{k = 0}^\infty\left( e^{iuk} \mathbb{P}\left(\sum_{n = 1}^\infty X_n 1_{\{n \leq N\}} = k)\right)\right)$$
but I do not know how to compute the probability in there. Any hints?
Best Answer
This is going to be a mere reformulation of your comment, possibly with more rigourous notations. First
$$\mathop{\mathsf{E}}e^{itY}=\mathsf{E}\left[\mathsf{E}\left[e^{it\sum_{n=1}^N X_n}\mid N\right] \right]=\mathsf{E}\left[\prod_{i=1}^N \mathsf{E}\left[e^{it X_i}\mid N\right]\right]$$
where we used the independence of $(X_i)$ and $N$ (this is not mentioned in your statement, but I assume this is the case). For the same reasons
$$\mathop{\mathsf{E}}e^{itY}=\mathsf{E}\left[\prod_{i=1}^N \mathsf{E}\left[e^{it X_i}\right]\right]=\mathsf{E}\left[\phi(t)^N\right]=\sum_{k\ge 0} \frac{\phi(t)^k \lambda^k}{k!}e^{-\lambda}=e^{\lambda(\phi(t)-1)}.$$