Suppose that $Y|X \sim\operatorname{Poisson}(cX)$ where $c>0$ is a constant and $X\sim\operatorname{Exp}(1)$
(a) Find the marginal mean and variance of $Y$.
(b) Find the marginal pmf of $Y$.
For part $A$ here's what I have so far:
$E(Y|X) = E(cX)$ because $Y|X \sim Poisson(cX)$ so
$$E(Y) = E(E(Y|X)) = E(cX) = cE(X) = c$$
Likewise $Var(Y|X) = cX$, $E(X) = 1$, and $Var(X) = 1$ so
$$Var(Y) = E(Var(Y|X)) + Var(E(Y|X))$$
$$Var(Y) = E(cX) + Var(cX)$$
$$Var(Y) = cE(X) + c^2Var(X)$$
$$Var(Y) = c + c^2$$
For part B I'm a bit stuck but here's what I have so far:
$$f_Y(y) = \int_0^\infty f(x,y)dx$$ where
$$f(x,y) = f(y|x)*f(x) = \dfrac{(cx)^ye^{-cx}}{y!}e^{-x}$$
I am struggling with the integral in part B and for part A I'm not sure if I'm using the right theorems and applying them correctly. Any tips or advice?
Where $n\in\Bbb N$ , $r\in{\Bbb R}^+$ we have:
$$\int_0^\infty x^n\,\mathrm e^{-rx}\,\mathrm d x=\dfrac{n!}{r^{n+1}}$$
Because using integration by parts produces a the recursion:
$$\begin{align}\int_0^\infty x^n\,\mathrm e^{-rx}\,\mathrm d x &= \left.-\dfrac{x^n\mathrm e^{-rx}}{r}\right\vert_{x=0}^{x\to\infty}+\dfrac{n}{r}\int_0^\infty x^{n-1}\mathrm e^{-rx}\,\mathrm d x\\[2ex]&=\dfrac nr\int_0^\infty x^{n-1}\,\mathrm e^{-rx}\,\mathrm d x\\[1ex]\text{and}\qquad\\[2ex] \int_0^\infty \mathrm e^{-r x}\,\mathrm d x &=\dfrac 1 r \end{align}$$