# Discrete random variables with joint pdf

The question asks: Find $c$ for discrete random variables $X$ and $Y$ with joint pdf:

$$\displaystyle f(x,y)=c\frac{2^{(x+y)}}{x!y!}$$

where $x=0,1,2...$ and $y=0,1,2...$ and $0$ otherwise.

If it was continuous I'd integrate but since its discrete I don't know what I'm supposed to do to find c. I've only seen discrete joint problems where actual probabilities have been provided rather than this so I'm not sure where to start. Thanks.

• Well, instead of integrating, you have to sum. And instead of a double integral, it will be a double sum. Start by fixing $x$ and summing over $y$. That will give you the sum over each vertical line in the grid. Then sum those over $x$ to get the total sum. Your answer will depend on $c$, but since you know that the total mass must be equal to $1$, this will give you an equation which you can solve to evaluate $c$. – Unwisdom Mar 2 '14 at 6:17
• $\displaystyle{\large c = \left(\sum_{x = 0}^{\infty}{2^{x} \over x!}\right)^{-2}}$. – Felix Marin Mar 2 '14 at 6:20

HINT:

$$\sum_{x=0}^\infty\sum_{y=0}^\infty\ \left( \dfrac{c\ 2^{(x+y)}}{x!y!}\right)=1\iff c\sum_{x=0}^\infty \dfrac{2^x}{x!}\sum_{y=0}^\infty \dfrac{2^y}{y!}=1$$

• Thanks very much. I found this other problem that is virtually identical to mine (with k instead of c) and the they use a Maclaurin rule $\sum_{x=0}^\infty$ (t^n)/n! =e^t from here so it makes sense now. math.stackexchange.com/questions/582044/… – Helena Shaw Mar 2 '14 at 19:33

Just to add the result from Wolfram Alpha

$$\displaystyle{\large c = \left(\sum_{x = 0}^{\infty}{2^{x} \over x!}\right)^{-2}} = \frac{1}{(e^{2})^{2}} => = c= \frac{1}{(e^{4})}$$

http://www.wolframalpha.com/input/?i=%5Csum_%7Bx%3D0%7D%5E%7B%5Cinfty%7D+%5Cfrac%7B2%5E%7Bx%7D%7D%7Bx%21%7D