I was wondering if anyone could give me a tip on how to proceed with the following question?
Suppose X~Poisson(N), where N~Poisson($\lambda$). What is the PGF of X + N? (Where $\lambda$ is a number)
I know that because the two distributions are not independent, you can't use the normal rules of PGF and multiply the two together. Thanks!
According to the definition in Wikipedia , the probability generating function of a random variable $Y$ is given by $$G(z)=\mathbb{E}[z^{Y}].$$
First, if $Y\sim Poisson(\mu)$ then $\mathbb{E}[z^{Y}]=\sum_{k=0}^{\infty}z^{k}\frac{e^{-\mu}\mu^{k}}{k!}=e^{\mu(z-1)}$. Using this result repeatedly and conditional expectation we can write
\begin{eqnarray} \mathbb{E}[z^{X+N}]&=&\mathbb{E}[\mathbb{E}[z^{X+N}|N]]\\ &=&\mathbb{E}[z^{N}e^{N(z-1)}]\\ &=& \mathbb{E}[e^{N(z+\log z-1)}]\\ &=&\exp[\lambda(e^{z+\log z-1}-1)]=\exp[\lambda(ze^{z-1}-1)]. \end{eqnarray}