Let $X = Unif(2, 4)$ and $Y=Poisson(X)$. My goal is to find $P(Y=n)$, but I always seem to get stuck on some nasty integral. Here's what I've tried:
$P(Y=n) =\int_2^4P(Y=n|X=x)*P(X=x)dx = \int_2^4 e^{-x}x^{n}/n!*1/2 dx$
I can see that the $n!$ is a constant, so this is just $1/(2*n!)*\int_2^4 e^{-x}x^{n}dx$
However, I've got no clue how to make progress on that integral, because $n$ is arbitrary. I'd accept an answer that either helps me to find a closed form formula for that integral or another route of tackling this problem that finds a closed form answer for $P(Y=n)$.
It was probably proven in class or it is given in the text that if $Z\sim\text{Gamma}(r, \lambda)$, then the cdf is $$F_Z(z) = 1-\sum_{k=0}^{r-1}e^{-\lambda z}\frac{(\lambda z)^k}{k!}.$$ Then notice that \begin{align*} \frac{1}{2\cdot n!}\int_2^4 e^{-x}x^{n}dx &=\frac{\Gamma(n+1)}{2\cdot n!}\int_2^4\frac{1}{\Gamma(n+1)}1^{n+1}x^{(n+1)-1}e^{-x}\,dx \tag 1\\ &=\frac{n!}{2\cdot n!}\left[F_Z(4)-F_Z(2)\right]\\ &=\frac{1}{2} \left[1-\sum_{k=0}^{n}e^{-1\cdot 4}\frac{(1\cdot 4)^k}{k!}-\left(1-\sum_{k=0}^{n}e^{-1\cdot 2}\frac{(1\cdot 2)^k}{k!}\right)\right], \end{align*} where I recognize $(1)$ as the density of a $\text{Gamma}(r = n+1,\lambda=1)$.