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 $T=X+Y$?
 A: 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 $4^w$. 
A: 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!}=...$
