I have an exercise which is regarding conditional pgfs. I know that: $PGF(z) = E[z^Y] = E[E[z^Y|X]]$.
But I'm not sure how to find $E[z^Y|X]$, because I'm confused about what exactly $z^Y$ is. Say $X$ is exponentially distributed with rate 1 and $Y|X=x$ is Poisson distributed. How would I find $E[z^Y|X]$ and then the PGF?
Thanks in advance.
If $Y|X=x$ is non-negative discrete then $$E[z^Y|X=x]=\sum_{y=0}^\infty z^y P(Y=y|X=x)$$ which if $Y|X=x$ is Poisson distributed with mean and variance $X=x$ might be $e^{x(z-1)}$. Then if $X$ is non-negative continuous with density $p(x)$ you have $$E[E[z^Y|X]] = \int_{x=0}^\infty E[z^Y|X=x] \;p(x)\; dx$$ which if $X$ has an exponential distribution with rate $1$ might be $\frac{1}{2-z}$, though I have not checked and you should do the summation and integration yourself.