The joint probability of the random variables $X$ and $Y$ is given by $$P(X=n, Y=m) =\frac{\alpha ^{m}\beta ^{n-m}}{m!(n-m)!}e^{-(\alpha +\beta )} $$ $$m=0,1,2...$$ $$n=m, m+1, m+2, ...$$
a) Evaluate $P(X=n)$ and $P(Y=m)$.
b) Are $X$ and $Y$ independent?
My solution so far:
a) $$f_{XY}(n,m) = P(X=n, Y=m)$$ $$P(X=n) = f_{X}(n) = \sum_{m}f_{XY}(n,m) = \sum_{m=0}^{n}\frac{\alpha ^{m}\beta ^{n-m}}{m!(n-m)!}e^{-(\alpha +\beta )}$$ $$=e^{-(\alpha +\beta )}\frac{1}{n!} \sum_{m=0}^{n}\frac{n!}{m!(n-m)!}\beta^{n-m}\alpha^{m}$$ $$=\frac{(\beta+\alpha)^{n}}{e^{(\alpha+\beta)}n!}.$$
So far, am I doing this correctly?
For $P(Y=m)$:
$$P(Y=m) = f_{Y}(n) = \sum_{n}f_{XY}(n,m) = \sum_{n=m}^{\infty}\frac{\alpha ^{m}\beta ^{n-m}}{m!(n-m)!}e^{-(\alpha +\beta )}$$ $$=e^{-(\alpha +\beta )}\frac{1}{n!} \sum_{n=m}^{\infty}\frac{n!}{m!(n-m)!}\beta^{n-m}\alpha^{m}$$ $$\mbox{ from here-> }=\frac{e^{\beta }\alpha ^{m}}{m!}e^{-(\alpha +\beta )}=\frac{e^{-\alpha }\alpha ^{m}}{m!}=\frac{\alpha ^{m}}{e^{\alpha }m!}.$$ Here, I have no idea how to the get the summation "from here". I used an online summation calculator because I don't know how to get there. Any help as to how to get there would be appreciated!!
b)
$X$ and $Y$ are independent if $f_{X,Y} = f_{X}(x)f_{Y}(y)$.
From a, $$f_{X}(x)f_{Y}(y) = \frac{(\beta+\alpha)^{n}}{e^{(\alpha+\beta)}n!}\frac{\alpha ^{m}}{e^{\alpha }m!}. $$
From given, $$f_{X,Y}= \frac{\alpha ^{m}\beta ^{n-m}}{m!(n-m)!}e^{-(\alpha +\beta )}.$$
So, since $f_{X,Y} \neq f_{X}(x)f_{Y}(y)$, $X$ and $Y$ are NOT independent.
Here, I'm not sure if $f_{X}(x)f_{Y}(y)$ actually reduces to $f_{X,Y}$ somehow.
Thanks for answering!
Your solution to (a) appears correct.
Regarding (b), we simplify and obtain the series expansion for $e^\beta$ as highlighted in blue below:- $$\begin{align}P(Y=m)&=e^{-(\alpha +\beta )}\frac{1}{n!} \sum_{n=m}^{\infty}\frac{n!}{m!(n-m)!}\beta^{n-m}\alpha^{m}\\&=\frac{e^{-(\alpha+\beta)}\alpha^m}{m!}\sum_{n=m}^{\infty}\frac{1}{(n-m)!}\beta^{n-m}&\\&=\frac{e^{-(\alpha+\beta)}\alpha^m}{m!}\color{blue}{\sum_{k=0}^{\infty}\frac{1}{k!}\beta^k}\\&=\frac{e^{-(\alpha+\beta)}\alpha^m}{m!}\color{blue}{e^\beta}\\&=\frac{e^{-\alpha}\alpha^m}{m!}\end{align}$$
Finally, from the expressions obtained from (a) and (b), $X$ and $Y$ are indeed NOT independent.