Probability generating function and expectation Let $X$ be Poisson random variable with parameter $Y$, where $Y$ is Poisson random variable, with parameter $\mu$. Prove that, $G_{X+Y}(s)=\exp\{\mu (s\exp^{s-1}-1)\}$
I know that, Poisson r.v. generating function is $G(s)=\exp\{\lambda(s-1)\}$. Do I need to calculate joint probability distribution first ($P(Z)$), where $Z=X+Y$?

Hint: As recalled in the post, $E[s^X\mid Y]=\exp\{Y(s-1)\}$ hence $G_{X+Y}(s)=E[s^Xs^Y]$ is also $G_{X+Y}(s)=E[E[s^X\mid Y]s^Y]=E[\exp\{Y(s-1)\}s^Y]=E[t^Y]$ for $t$= $____$. Furthermore, for every $t$, $E[t^Y]=\exp\{\mu(t-1)\}$ hence $G_{X+Y}(s)=$ $______$.
