Help finding MGF of mixture distribution

Question

Im having trouble finding any good resources or examples for finding the conditional distribution of two variables. I've tried using double expectations but cant get it to work out. Thanks

I found the pmf of X not conditioned on N below

$ P(x=x) = \frac{e^{-\lambda p}(\lambda p)^x}{x!} $

Then used $ E(e^{tx}) $ to find the mgf

$ M_x(t) = e^{\lambda p(e^t-1) } $

Answer 1

1. You want the moment generating function $M_X(t)=\mathbb{E}[e^{tX}]$.
2. Read this to understand why $\mathbb{E}[e^{tX}]=\mathbb{E}[\mathbb{E}[e^{tX}\mid N]]$.
3. The information given in the problem statement gives you $$\mathbb{E}[e^{tX}\mid N]=(1-p+e^tp)^N. \qquad\qquad \text{(why?)}$$
4. Defining $e^u:=1-p+e^tp$, and using the expression of the moment generating function of a random variable with Poisson distribution given in the problem statement, you're done; just express the result as a function of $t$.