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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?

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    $\begingroup$ Have you ever use conditional expectation? This might help you to formalize what you want to do with $\phi^n$ when $n$ is a random variable. $\endgroup$
    – md5
    Dec 9, 2020 at 16:23
  • $\begingroup$ @md5 Yes, so $\mathbb{E}[e^{iuY}] = \mathbb{E}[\mathbb{E}[e^{iuY} \ | \ N = n]] = \sum_{n = 0}^\infty \phi^n \mathbb{P}(N = n) = $ $\sum_{n = 0}^\infty \phi^n \frac{e^{-\lambda} \lambda^n}{n!} = e^{-\lambda} e^{\lambda \phi} = e^{\lambda \phi - \lambda}$? $\endgroup$
    – user388557
    Dec 9, 2020 at 16:36
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    $\begingroup$ I'm a bit confused by your notations, but the overall approach and the result seem correct. $\endgroup$
    – md5
    Dec 9, 2020 at 17:42
  • $\begingroup$ @md5 Ok, thank you. Maybe you can make an answer out of it so I can mark this question as solved? $\endgroup$
    – user388557
    Dec 9, 2020 at 18:19

1 Answer 1

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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

$$\mathop{\mathsf{E}}e^{itY}=\mathsf{E}\left[\prod_{i=1}^N \mathsf{E}\left[e^{it X_i}\right]\right]=\mathsf{E}\left[\phi(t)^N\right]=\sum_{k\ge 0} \frac{\phi(t)^k \lambda^k}{k!}e^{-\lambda}=e^{\lambda(\phi(t)-1)}.$$

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