# Marginal PMF of $X$ given joint PMF of $(X,Y)$ is $k \cdot \frac {2^{x+y}}{x!y!}$

Let $X$ and $Y$ be random variables with joint probability mass function $f(x,y) = k \cdot \dfrac {2^{x+y}}{x!y!}$, for $x, y \in \{ 0, 1, 2, \cdots \}$ and for a positive constant $k$.

How can I derive the marginal probability mass function of $X$? How do I evaluate $k$? Are $X$ and $Y$ independent? Furthermore, how could I derive the probability mass function of $T=X+Y$?

• For the marginal pmf of X I know that px(x) = sum of y from o to inifinity of p(x,y) which leads to sum y from o to infinity of k[2^(x+y)/(x!y!)] but I'm not sure where to go from here. When I solve for k, I know the sum of x from 0 to infinity and y from 0 to infinity of p(x,y) equals 1. Commented Nov 26, 2013 at 16:15
• @nbro Please no \dfrac in titles unless one must (which is not the case here).
– Did
Commented Mar 19, 2018 at 22:07

For the marginal distribution of $X$, note that $$f_X(x)=\Pr(X=x)=k\frac{2^x}{x!} \sum_{y=0}^\infty \frac{2^y}{y!}.\tag{1}$$ Recall that $e^t$ has Maclaurin series $\sum_{0}^\infty \frac{t^n}{n!}$. Thus we recognize the sum in (1) as $e^2$. So $$f_X(x)=ke^2\frac{2^x}{x!}.$$ Summing again over all $x$, we get $ke^4$. It follows that $k=e^{-4}$ and $$f_X(x)=e^{-2}\frac{2^x}{x!}.$$

The same argument shows that $f_Y(y)=\Pr(Y=y)=e^{-2}\frac{2^y}{y!}$.

If follows that $$f_{X,Y}(x,y)=f_X(x)f_Y(y),$$ and therefore $X$ and $Y$ are independent.

Side comment: The independence follows more simply from the fact that the joint distribution function $f_{X,Y}(x,y)$ factors as a function of $x$ times a function of $y$.

Note that $X$ and $Y$ each have Poisson distribution with parameter $2$.

For the distribution of $X+Y$, we want the distribution of a sum of two independent Poisson random variables with parameter $2$. This sum has Poisson distribution with parameter $4$.

Or else we can compute. Let $W=X+Y$. We want to compute $\Pr(W=w)$. The sum $X+Y$ can be $w$ in various ways: $X=0$, $Y=w$; $X=1$, $Y=w-1$; and so on up to $X=w$, $Y=0$. Thus $$\Pr(W=w)=\sum_{k=0}^w e^{-4} \frac{2^{w}}{k!(w-k)!}.\tag{2}$$ Multiply top and bottom by $w!$, and note that $\frac{w!}{k!(w-k)!}=\binom{w}{k}$. The expression (2) is therefore equal to $$e^{-4}\frac{2^w}{w!}\sum_{k=0}^w \binom{w}{k}.$$ The sum of the binomial coefficients is $2^w$. It follows that $$\Pr(W=w)=e^{-4} \frac{2^{2w}}{w!}.$$ Things look nicer if instead of $2^{2w}$ we write $4^w$.

• I recognize a patient and good math-teacher here, who tries to step down to the level of the student to show him from there the way(s) that lead upstairs. As a side-effect you give guys like me tips how to become such a math-teacher too.(+1) Commented Nov 27, 2013 at 11:03

Hints:

$e^{z}=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}$

and as a pmf:

$\sum_{x=0}^{\infty}\sum_{y=0}^{\infty}f\left(x,y\right)=1$

Notice that $\sum_{x=0}^{\infty}\sum_{y=0}^{\infty}\frac{2^{x+y}}{x!y!}=\sum_{x=0}^{\infty}\left(\frac{2^{x}}{x!}\sum_{y=0}^{\infty}\frac{2^{y}}{y!}\right)=\sum_{y=0}^{\infty}\frac{2^{y}}{y!}\times\sum_{x=0}^{\infty}\frac{2^{x}}{x!}=...$

You should know $k$ by now, and:

$p_{X}\left(x\right)=\sum_{y=0}^{\infty}f\left(x,y\right)=k\sum_{y=0}^{\infty}\frac{2^{x+y}}{x!y!}=k$$\frac{2^{x}}{x!}\sum_{y=0}^{\infty}\frac{2^{y}}{y!}=...$