# Compute the characteristic function of $Z_n=\sum\limits_{k=1}^{\xi_n}X_k$

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?

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.

• Just a question: what if $\phi_X(t)=0$ for some $t$? Is it still meaningful to write $\log (\phi_X(t))$? Jul 4, 2020 at 20:58
• @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. Jul 4, 2020 at 21:00
• @SangchulLee Beautiful! Thank you very much. Jul 4, 2020 at 21:26