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Let $\{X_k\}_{k\ge1}$ be an i.i.d. sequence and let $\{\xi_n\}_{n\ge1}$ be a sequence of Poisson random variables with $E\xi_n=n\,\,(n=1,2,...)$. Assume independence between $\{X_k\}_{k\ge1}$ and$\{\xi_n\}_{n\ge1}$. Compute the characteristic function of the variable: \begin{align*} Z_n=\sum_{k=1}^{\xi_n}X_k \end{align*} (More precisely, represent the characteristic function of $Z_n$ in terms of the characteristic function of $X_1$).

Here is what I have so far:

$E\xi_n=n\implies \xi_n\in\text{Poi}(n)$ and \begin{align*} \phi_{Z_n}=\phi_{\sum\limits_{k=1}^{\xi_n}X_k}=\prod\limits_{k=1}^{\xi_n}\phi_{X_k}=\big[\phi_{X_1}\big]^{\xi_n} \end{align*}

but I am not exactly sure how to deal with that $\xi_n$ in the exponent, is there a way that I can break this expression down further?

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Both $X_k$'s and $\xi_n$ are random, so you cannot compute the characteristic function that way. In order to compute it correctly, we proceed using the law of iterated expectation:

$$\phi_{Z_n}(t)=\mathbb{E}[e^{itZ_n}]=\mathbb{E}[\mathbb{E}[e^{itZ_n}\mid\xi_n]]$$

Then by the independence, the inner conditional expectation is computed by

$$\mathbb{E}[e^{itZ_n}\mid\xi_n]=\phi_{X_1}(t)^{\xi_n}.$$

Plugging this back,

$$ \phi_{Z_n}(t) = \mathbb{E}[\phi_{X_1}(t)^{\xi_n}] = \sum_{j=0}^{\infty} \frac{(\phi_{X_1}(t) n)^j}{j!}e^{-n} = e^{n(\phi_{X_1}(t)-1)}. $$

For more details about $Z_n$, the keyword compound poisson distribution might be helpful.

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  • $\begingroup$ Just a question: what if $\phi_X(t)=0$ for some $t$? Is it still meaningful to write $\log (\phi_X(t))$? $\endgroup$ Jul 4, 2020 at 20:58
  • $\begingroup$ @FormulaWriter, My original intention was to use the known formula for the moment generating function of $\xi_n$, but on a second thought, it might be easier and more transparent to directly derive them, which does not suffer the log-of-zero issue. Let me reformulate my answer. $\endgroup$ Jul 4, 2020 at 21:00
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    $\begingroup$ @SangchulLee Beautiful! Thank you very much. $\endgroup$ Jul 4, 2020 at 21:26

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