Computing Joint Moment Generating Function Good morning, Stack Exchange. My problem is that I have a joint distrubtion of discrete random variables with the following PMF and support:
$f(x,y) = \frac{2^{x-y}  e^{-3}}{x!(y-x)!}$ when $x = 0, 1, 2, ... y$ and $y = 0, 1, 2, ... \infty $ and $f(x,y) = 0$ otherwise
So, my problem here is that I know that to get the MGF,  $M(t_1, t_2)$, we must evaluate the sum:
$$\sum_{y=0}^{\infty} \sum_{x=0}^{y} \frac{e^{x_1t_1 + yt_2} *  e^{-3} * 2^{y-x}}{x!{y-x!}}$$
My strategy for evaluating the inner sum was to move all the terms that contain a $y$ to the outermost sum  and to move the constant $e^{-3}$ out of the sum entirely, hence solve the inner sum as $$\sum_{x=0}^{y} \frac{e^{xt_1} * 2^{-x}}{x!{y-x!}}$$
The hint from the textbook tells me that this sum is easy to evaluate if the term $2^{-x}$ weren't there, but I am not sure how to deal with the sum once this term is present. Am I on the right track here, and how do I solve this innermost sum and help simplifying the sum in general would be much appreciated. Thanks for reading.
 A: You have PMF:
$$
f(x,y) =
\begin{cases}
 \frac{2^{x-y}  e^{-\frac{3}{2}}}{x!(y-x)!},& x = 0, 1, 2, \ldots y\text{ and } y = 0, 1, 2,\ldots\\
0& \text{otherwise}
\end{cases}
$$
And the MGF is:
\begin{align}
M_{X,Y}(t_1,t_2)
&=
E(e^{t_1x+t_2y})\\
&=
\sum\limits_{y=0}^\infty\sum\limits_{x=0}^y
e^{t_1x+t_2y}\frac{2^{x-y}  e^{-\frac{3}{2}}}{x!(y-x)!}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty 2^{-y}e^{t_2y}\sum\limits_{x=0}^y
e^{t_1x}\frac{2^{x}  }{x!(y-x)!}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty 2^{-y}e^{t_2y}\sum\limits_{x=0}^y
\frac{\left(2e^{t_1}\right)^{x} }{x!(y-x)!}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}2^{-y}e^{t_2y}\sum\limits_{x=0}^y
\frac{y! \left(2e^{t_1}\right)^{x} }{x!(y-x)!}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}2^{-y}e^{t_2y}\sum\limits_{x=0}^y
\begin{pmatrix}y\\x\end{pmatrix}\left(2e^{t_1}\right)^{x} \\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}2^{-y}e^{t_2y}\sum\limits_{x=0}^y
\begin{pmatrix}y\\x\end{pmatrix}\left(2e^{t_1}\right)^{x} 1^{y-x}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}2^{-y}e^{t_2y}(2e^{t_1}+1)^y\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}\left(2^{-1}e^{t_2}(2e^{t_1}+1)\right)^y\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}\left(e^{t_1+t_2}+\dfrac{1}{2}e^{t_2}\right)^y\\
&=
e^{-\frac{3}{2}} e^{e^{t_1+t_2}+\frac{1}{2}e^{t_2}}\\
&=
e^{e^{t_1+t_2}+\frac{1}{2}e^{t_2}-\frac{3}{2}}.
\end{align}
Note
I use the fact:
(1) Binomial theorem
$$\sum\limits_{x=0}^n \begin{pmatrix}n\\x\end{pmatrix}a^x b^{n-x} = (a+b)^n$$
(2) The Taylor series about $x=0$ of $e^x$,
$$\sum\limits_{n=0}^\infty \dfrac{x^n}{n!}=e^x.$$
