Suppose $X$ and $Y$ are discrete random variable with joint p.f. Find the marginal p.f. of $X$ and $Y$ The joint probability function of x and y is:
$$f(x,y)=\frac{e^{-2}}{x!(y-x)!}$$ where $$x = 0, 1, ..., y; y= 0, 1, ...$$
My solution to this problem:
The marginal p.f. of x is:
$$f_x(x)=\sum_{y=0}^\infty \frac{e^{-2}}{x!(y-x)!}$$
Then $$f_x(x)=\frac{e^{-2}}{x!}\sum_{y=0}^\infty\frac{1}{(y-x)!}$$
and this is where I got stuck. 
In the solution, it writes $$f_x(x)=\sum_{\color{red}{y=x}}^\infty \frac{e^{-2}}{x!(y-x)!}$$
$$=\frac{e^{-2}}{x!}\sum_{t=0}^\infty\frac{1}{t!}$$
$$=\frac{e^{-2}}{x!}e$$
$$=\frac{e^{-1}}{x!}$$
I understand that $y=x$ as the initial condition since when $y=0$, $x=0$. However, when $y$ starts to increase, the series will not always conform the series of Euler's number $e$. For example when $y=2$, $x$ can also be 2 then the third entry in the series would be $\frac{1}{(2-2)!}$ which is 1 again like the first two entries in the series. 
How to interpret this solution?
 A: The joint probability only applies when $y \ge x$ (e.g., $(y-x)!$, apart from using the Gamma function, doesn't make sense when $y \lt x \implies y - x \lt 0$). This is why the summation starts at $y = x$ as it's given in the solution. Note the function $f_x(x)$ depends only on $x$, with $y$ being a summation variable. Thus, as $y$ increases on the RHS, the value of $x$ is not changing. So your statement of

For example when $y=2$, $x$ can also be 2 then the third entry in the series would be $\frac{1}{(2-2)!}$ which is 1 again like the first two entries in the series.

is not correct. If $x = 2$, then the first entry in the series would be for $y = 2$ since $y$ starts at $x = 2$, i.e., it's not the third entry.
Also, note that in the summation expression of
$$\sum_{y=x}^\infty \frac{1}{(y-x)!} \tag{1}\label{eq1A}$$
you have for some value of $x$, starting $y$ at $x$ and incrementing it by $1$ each time in the summation up to infinity. If you let $t = y - x$, then when $y = x$, you have $t = 0$, when $y = x + 1$, you have $t = 1$, etc. Thus, you get
$$\sum_{y=x}^\infty \frac{1}{(y-x)!} = \sum_{t=0}^\infty \frac{1}{t!} \tag{2}\label{eq2A}$$
Since $\frac{e^{-2}}{x!}$ doesn't involve $y$ at all, it's a constant factor in all of the terms so you can move it outside of the summation. You thus get
$$\begin{equation}\begin{aligned}
\sum_{y=x}^\infty \frac{e^{-2}}{x!(y-x)!} & = \frac{e^{-2}}{x!}\sum_{y=x}^\infty \frac{1}{(y-x)!} \\
& = \frac{e^{-2}}{x!}\sum_{t=0}^\infty \frac{1}{t!}
\end{aligned}\end{equation}\tag{1}\label{eq3A}$$
A: You've written $f_x(x)$ where you should have $f_X(x).$ For example, if $x=4$ then $f_X(x) = f_X(4) = \Pr(X=4).$ The number $4$ replaces $x$ but it doesn't replace $X.$ Without this distinction one cannot even understand something like $\Pr(X\le x).$
You wrote $x=0,1,\ldots,y,$ so the range of values of $x$ depends on $y.$ You always have $x\le y.$ Thus, for example, when $x=4$ then $y \ge 4,$ i.e. $y\in\{4,5,6,\ldots\}.$
Then you have
\begin{align}
f_X(4) = {} & \sum_{y\,=\,4}^\infty \frac{e^{-2}}{x!(y-x)!} \\[8pt]
= {} & \frac {e^{-2}} {x!} \sum_{y\,=\,4}^\infty \frac 1 {(y-4)!} \\
& \text{since as $y$ goes from 4 to $\infty,$} \\
& \text{$e^{-2}/x!$ does not change,} \\[10pt]
= {} & \frac{e^{-2}}{x!} \sum_{w\,=\,0}^\infty \frac 1 {w!} \\
& \text{where $w=y-4.$} \\
& \text{When $y=4,$ then $w=0.$} \\[10pt]
= {} & \frac{e^{-2}}{x!} \cdot e = \frac{e^{-1}}{x!}.
\end{align}
