Generating function of Poisson random variable of random parameter Poisson distributed

Question

Consider the random variable $X \sim poisson(\lambda)$ where $\lambda > 0$ and $Y \sim poisson(X)$.

How can I calculate the generating function of $Y$ which determines the distribution of $Y$ uniquely?

A hint would be much appreciated.

Answer 1

$$R_{X}(t) = E(t^X) = e^{\lambda(t-1)}$$

$$M_{X}(t) = e^{\lambda(e^t-1)}$$

$$R_{Y}(t) = E(t^{Y}) = E(E(t^{Y}|X)) = E(e^{X(t-1)}) = M_X(t-1) = e^{\lambda(e^{t-1}-1)}$$

$R$ is the generating function and $M$ is the moment generating function. Also note the caution mentioned in the comment by @Did.