The question reads:

Suppose $Y \sim gamma(\alpha= 2; \beta= 1)$ and the conditional distribution of $X$ given Y=y is Poisson with mean = y. Find the marginal probability mass function of $X$.

So far I've done:

$f(x,y) = f(X|Y=y)f_Y(y)$

$= \frac{y^xe^{-y}}{x!}ye^{-y}$

$f_X(x) = \int_o^\infty \frac{y^{x+1}e^{-2y}}{x!} dy$

but then after that I'm not sure if there's a special identity related to the Gamma distribution or something that will help with integration. The answer is supposed to be $\frac{x+1}{2^{x+2}}$

Thanks in advance


Note: Integrate by Parts to show for all natural numbers, $n$: $$\int_0^\infty y^{n+1} e^{-2y}\mathsf d y = \tfrac 12(n+1)\int_0^\infty y^ne^{-2y}\mathsf d y $$

Also $\int\limits_0^\infty e^{-2y}\mathsf d y=\tfrac 12$.

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