Density of discrete random vector Let $(X,Y)$ be a discrete random vector with density
$$
f(x,y)=c\frac{2^{x+y}}{x!y!} I_{(x,y)\in \mathbb{N}_0 \times \mathbb{N}_0}(x,y).
$$
I am supposed to compute the value of the constant $c$.
I know that for $f$ to be a density function, it must hold 
$$
\sum_{(x,y)\in\mathbb{N}_0 \times \mathbb{N}_0} f(x,y) = 1.
$$
By using this, I tried:
\begin{align}
1 &= \sum_{(x,y)\in\mathbb{N}_0 \times \mathbb{N}_0} c\frac{2^{x+y}}{x!y!} \\ 
&= c \sum_{x\in\mathbb{N}_0} \frac{2^x}{x!} \sum_{y\in\mathbb{N}_0} \frac{2^y}{y!} \\
&= c \sum_{n=0}^{\infty} \frac{2^n}{n!}\sum_{n=0}^{\infty} \frac{2^n}{n!} \\
&= c \cdot (e^2)^2 \\ &= ce^4,
\end{align}
which means that $c=e^{-4}$. Am I correct? I'm not completely sure if the second equality sign holds.
 A: Addressing you concern, you can refer to the following general result. Let $\{a_{m,n}\}$ be a collection of numbers indexed by the natural numbers in both entries (so $m\in\mathbb{N_0}$ and $n\in\mathbb{N_0}$).
If $a_{m,n}\geq 0$ for all $m,n\in\mathbb{N_0}$, then
$$\sum_{(m,n)\in\mathbb{N_0}\times\mathbb{N_0}}a_{m,n} = \sum_{m\in\mathbb{N_0}}\sum_{n\in\mathbb{N_0}}a_{m,n}=\sum_{n\in\mathbb{N_0}}\sum_{m\in\mathbb{N_0}}a_{m,n},$$
with the understanding that the common value can be $+\infty$ in case of divergence. This follows from a more general result called Tonelli's theorem.
In your case, we have that $a_{m,n}=\frac{2^{m+n}}{m!n!}$, which are certainly positive terms. Thus
$$
\begin{align*}
\sum_{(m,n)\in\mathbb{N_0}\times\mathbb{N_0}}\frac{2^{m+n}}{m!n!}&=\sum_{m\in\mathbb{N_0}}\sum_{n\in\mathbb{N_0}}\frac{2^{m+n}}{m!n!}\\
&=\sum_{m\in\mathbb{N_0}}\sum_{n\in\mathbb{N_0}}\frac{2^m2^n}{m!n!}\\
&=\sum_{m\in\mathbb{N_0}}\frac{2^m}{m!}\sum_{n\in\mathbb{N_0}}\frac{2^n}{n!},\\
\end{align*}$$
where the first equality follows from the result I mentioned above, and where the last equality follows from $\frac{2^m}{m!}$ not depending on $n$.
