I've been doing this question and I was wondering if my workings are correct, if they are not correct, can you please correct them?
The question is as follows:
My workings are:
$\binom{y}{x} \frac{e^{-1}}{2^y y!}$
We can rewrite this:
$\frac{y!}{x!(y-x)!} \times \frac{e^{-1}}{2^y y!}$
$\frac{1}{x!(y-x)!} \times \frac{e^{-1}}{2^y}$
So we have: $f_{x, y}(x,y) = \frac{1}{x!(y-x)!} \times \frac{e^{-1}}{2^y}$
Now using the formula:
$f_{x} (x,y) = \sum_{y}^{\infty} P_{x,y}(x,y)$
and subbing in $P_{x,y}(x,y)$:
$\frac{e^{-1}}{x!} \sum_{y}^{\infty} \frac{1}{(y-x)!}$
= $\frac{e^{-1}}{x!}$
And in a same fashion, for y we get:
$f_{y}(x,y) = \frac{e^{-1}}{y!}$
First we find the marginal pmf $f_Y(y)$ of $Y$. This is given by $$f_Y(y)=\sum_{x=0}^y \binom{y}{x}\frac{e^{-1}}{2^yy!}.$$ But the sum $\sum_{x=0}^y \binom{y}{x}$ of the binomial coefficients is $2^y$.
It follows that $f_Y(y)=\frac{e^{-1}}{y!}$ for $y=0,1,2,\dots$.
For the marginal pmf $f_X(x)$ of $X$, we have $$f_X(x)=\sum_{y=x}^\infty \binom{y}{x}\frac{e^{-1}}{2^yy!}.$$ By the simplification that you made, we have $$f_X(x)=\frac{e^{-1}}{x!}\sum_{y=x}^\infty \frac{1}{(y-x)!2^y}.$$ Using $\frac{1}{2^y}=\frac{1}{2^x}\cdot \frac{1}{2^{y-x}}$, we obtain $$f_X(x)=\frac{e^{-1}}{2^x x!}\sum_{y=x}^\infty \frac{1}{(y-x)!2^{y-x}}.\tag{1}$$ In (i), replace $y-x$ by $k$. Then $$f_X(x)=\frac{e^{-1}}{2^x x!}\sum_{k}^\infty \frac{1}{k!2^{k}}.\tag{2}$$ We recognize the infinite sum in (2) as the series expansion of $e^{1/2}$. Thus $$f_X(x)=\frac{e^{-1/2}}{2^x x!}$$ for $x=0,1,2,\dots$.