Let the joint mass of $x$ and $y$ have the joint density as:
$P(X=x,Y=y)=\frac{e^{-1}}{(y-x)!x!2^y}$ where $x=0,1,2,....,y$ and $y=0,1,2...$. It is required to obtain the marginal distribution of both the random variables.
My approach
I obtained the marginal distribution of $Y$ as follows:
$P(Y=y)=\sum_{x=0}^{y}\frac{e^{-1}}{(y-x)!x!2^y}=\frac{e^{-1}}{2^y} \sum_{x=0}^{y}\frac{y!}{y!(y-x)!x!}=\frac{e^{-1}}{2^yy!}\sum_{x=0}^{y} yC_x$
Using the formula $2^y$, we get:
$P(Y=y)=\frac{e^{-1}}{y!},$ where $y=0,1,...$
But, I am not able to obtain the marginal density of $X$ beacuse of the unknown summation formula:
$\frac{e^{-1}}{x!}\sum_{y=0}^{\infty}\frac{1}{(y-x)!2^y}$
Any help to proceed with the above sum would be great.
Thanks.
$x=0,1,2,...,y$ and $y=0,1,2,...$ is equivalent to $x=0,1,2,...$ and $y=x,x+1,...$ so we can write$$f_X(x)=\sum_{y=x}^{\infty}\frac{e^{-1}}{(y-x)!x!2^y}$$by substituting $z=y-x$ we have$$f_X(x)=\sum_{z=0}^{\infty}\dfrac{e^{-1}}{z!x!2^x2^z}=\dfrac{e^{-1}}{x!2^x}\sum_{z=0}^{\infty}\dfrac{2^{-z}}{z!}=\dfrac{e^{-1}}{x!2^x}e^{\frac{1}{2}}=\dfrac{1}{\sqrt ex!2^x}$$