Let $M$ be a Poisson distributed random variable with mean 4, and let $X_{1},X_{2},...$ be independent identically distributed copies of a Bernoulli random variable $X$ with $\mathbb{P}(X=1)=\frac{3}{4}=1-\mathbb{P}(X=0)$. What is the Generating function $\mathbb{E}\{z^{S}\}$ of $S=2\sum_{i=1}^{M}X_{i}$.
So I know that using the property of independence I may apply that: \begin{equation} \mathbb{E}\{z^{2\sum_{i=1}^{M}X_{i}}\}=\mathbb{E}\{z^{2X_{1}}\}\mathbb{E}\{z^{2X_{2}}\}\cdot\cdot\cdot\mathbb{E}\{z^{2X_{M}}\} \end{equation} What I do not understand is how the $M$ Poisson distributed random variable works here in combination with the Bernoulli random variable. Could someone explain that to me?
You have to assume that $M$ is independent of $(X_i)$.
$Ez^{2\sum\limits_{i=1}^{M}X_i} =\sum\limits_{n=1}^{\infty}Ez^{2\sum\limits_{i=1}^{n}X_i} e^{-4}\frac {4^{n}} {n!}$. Since $Ez^{2\sum\limits_{i=1}^{n}X_i}=(Ez^{2X_1})^{n}=(\frac 1 4 +\frac 3 4z^{2})^{n}$ we get $Ez^{2\sum\limits_{i=1}^{M}X_i}=\sum\limits_{n=1}^{\infty}e^{-4}\frac {4^{n}} {n!} (\frac 1 4 +\frac 3 4z^{2})^{n}$. The sum of this series is $e^{-4} e^{1+3z^{2}}$.