Find the moment generating function of sum of dependent Poisson distributions

Question

Suppose that $X \sim Poi(\mu)$ and independently $Y \sim Poi(\lambda)$.

Show that $X + Y ∼ Poi(\mu + \lambda)$.

Suppose that $X\mid Y \sim Poi(Y)$ and $Y ∼ Poi(\lambda)$.

Find the moment generating function of $X + Y$.

I can prove $X + Y \sim Poi(\mu + \lambda)$ by combining the probability generating functions of $X$ and $Y$, but I am lost on the second part to find the MGF with dependence. Does $X\mid Y \sim Poi(Y)$ imply that $X + Y \sim Poi(Poi(\lambda) + \lambda)$? How do I handle a Poisson of a Poisson?

Answer 1

Hint:$$E[e^{t(X+Y)}] = E[E[e^{t(X+Y)} \mid Y]] = E[e^{tY} E[e^{tX} \mid Y]]$$ The inner expectation is the conditional MGF of $X$ given $Y$ evaluated at $t$; its value will depend on $t$ and $Y$. Then take the outer expectation with respect to $Y$.