$X$ given $\Theta$ is $\operatorname{Poisson}(\Theta)$, $\Theta$ is distributed as $\operatorname{Gamma}(a,b)$. What are the Mean and Variance of $X$?

Question

This is what I have so far:

Joint Distribution (product of pmf of $X$ and pdf of $\Theta$) $$f(x,\theta)=\frac{\theta ^xe^{-\theta}}{x!}\frac{b^a}{\Gamma (a)}\theta ^{a-1}e^{-b\theta}$$ and with a little algebra we can get $$f(x,\theta)=\frac{1}{x!}\frac{b^a}{(b+1)^{x+a}}\frac{\Gamma (x+a)}{\Gamma (a)}\cdot\frac{(b+1)^{x+a}}{\Gamma (x+a)} \theta ^{(x+a)-1}e^{-(b+1)\theta}$$ for easy derivation of the pdf of $X$: $$f(x)=\frac{1}{x!}\frac{b^a}{(b+1)^{x+a}}\frac{\Gamma (x+a)}{\Gamma (a)}\int_0^\infty\frac{(b+1)^{x+a}}{\Gamma (x+a)} \theta ^{(x+a)-1}e^{-(b+1)\theta}d\theta$$ leaving us with$$f(x)=\frac{1}{x!}\frac{b^a}{(b+1)^{x+a}}\frac{\Gamma (x+a)}{\Gamma (a)}$$ Which looks kind of similar to some kind of beta distribution? What kind of distribution can this be identified as? If it's not a standard distribution, how can I calculate the raw moments? Is integration or summation appropriate (Changing $x!$ to $\Gamma (x+1)$)? Thanks for your help!

Answer 1

$$ f(x,\theta) \,d\theta = \frac{\theta^x e^{-\theta}}{x!} \cdot \frac 1 {\Gamma(a)} (b\theta)^{a-1} e^{-b\theta} (b\,d\theta) $$

$$ \operatorname{E}(X) = \operatorname{E}(\operatorname{E}(X\mid \Theta)) = \operatorname{E}(\Theta) = \frac a b. $$ \begin{align} \operatorname{var}(X) & = \operatorname{E}(\operatorname{var}(X\mid\Theta)) + \operatorname{var}(\operatorname{E}(X\mid\Theta)) \\[10pt] & = \operatorname{E}(\Theta) + \operatorname{var}(\Theta). \end{align}

The actual distribution of $X$ given $\Theta$ is a negative binomial distribution, but I can't find the various occasions when I've posted about that. I think one of them may have been about three months ago.