A mixed discrete-continuous distribution Let $(X,Y)$ have the mixed discrete-continuous pdf given by:
$$f(x,y)= \begin{cases} \frac{y^{a+x-1}e^{-2y}}{\Gamma(a) x!}\ y>0;x=0,1,2,\ldots \\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \  \ \text{elsewhere} \end{cases}$$
Could you please help me show that this nasty pdf integrates/sums to 1 over the support of $(X,Y)$?
I initially tried to integrate out $Y$, by completing the gamma distribution with parameters, shape**$=a+x$ and **scale=$1/2$   and then sum over $X$ but that lead to a mess. 
Is there another way to go here?
 Thank you.
 A: Really fascinating question. What you want to show is that 
$$\int\limits_{0}^{\infty}\sum\limits_{x=0}^{\infty}\dfrac{y^{a+x-1}e^{-2y}}{\Gamma\left(a\right)x!}\text{ d}y = 1$$
Here's how I approached this problem. Let's isolate the $x$ and $y$ parts as much as possible:
$$\int\limits_{0}^{\infty}\dfrac{y^{a-1}e^{-2y}}{\Gamma\left(a\right)}\left(\sum\limits_{x=0}^{\infty}\dfrac{y^{x}}{x!}\right)\text{ d}y$$
I thought to myself, well, $y > 0$ and the $\displaystyle \sum\limits_{x=0}^{\infty}\dfrac{y^{x}}{x!}$ part looks something like the CDF of a Poission random variable. But the probability mass function of a Poission random variable with mean $\lambda = y$ is 
$$ p\left(y\right) = \dfrac{e^{-y}y^{x}}{x!}\text{,} \quad{} y > 0\text{.}$$
So this suggested to me that I should bring one of the $e^{-y}$ terms into the summation. 
$$\int\limits_{0}^{\infty}\dfrac{y^{a-1}e^{-y}}{\Gamma\left(a\right)}\left(\sum\limits_{x=0}^{\infty}\dfrac{e^{-y}y^{x}}{x!}\right)\text{ d}y$$
Since I've already established that the summation is summing over all probabilities for a Poisson-distributed random variable, it follows that $\displaystyle \sum\limits_{x=0}^{\infty}\dfrac{e^{-y}y^{x}}{x!} = 1$.
So all that remains is to show that 
$$\int\limits_{0}^{\infty}\dfrac{y^{a-1}e^{-y}}{\Gamma\left(a\right)}\text{ d}y = 1\text{.}$$
However, this follows immediately since $\displaystyle f(y) = \dfrac{y^{a-1}e^{-y}}{\Gamma\left(a\right)}$ is the PDF of a Gamma distribution.
