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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.

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  • $\begingroup$ I see a $z$ in your nasty pdf! $\endgroup$ – Stat May 30 '14 at 16:24
  • $\begingroup$ @Stat It's an $x$, sorry about that. $\endgroup$ – JohnK May 30 '14 at 16:25
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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.

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    $\begingroup$ Very good, thank you. Also $e^x=\sum_{i=0}^{\infty} \frac{x^n}{n!}$ ;) $\endgroup$ – JohnK May 30 '14 at 19:00
  • $\begingroup$ I was thinking about that, but for some reason, I thought it was $\displaystyle e^{x} = \sum\limits_{i = 1}^{\infty}\dfrac{n^{x}}{n!}$ when I wrote it at the time! Of course, that's not correct. :P $\endgroup$ – Clarinetist May 30 '14 at 19:02

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