The joint probability of the random variables $X$ and $Y$ is given by $$P(X=n, Y=m) =\frac{\alpha ^{m}\beta ^{n-m}}{m!(n-m)!}e^{-(\alpha +\beta )} $$ $$m=0,1,2...$$ $$n=m, m+1, m+2, ...$$
Evaluate $P(X=n)$ and $P(Y=m)$.
My solution so far: $$f_{XY}(n,m) = P(X=n, Y=m)$$ $$P(X=n) = f_{X}(n) = \sum_{m}f_{XY}(n,m) = \sum_{m=0}^{n}\frac{\alpha ^{m}\beta ^{n-m}}{m!(n-m)!}e^{-(\alpha +\beta )}$$ $$=e^{-(\alpha +\beta )}\frac{1}{n!} \sum_{m=0}^{n}\frac{n!}{m!(n-m)!}\beta^{n-m}\alpha^{m}$$ $$=\frac{(\beta+\alpha)^{n}}{e^{(\alpha+\beta)}n!}.$$
Here, I know that the upper limit is not $\infty$, but I'm not sure how to obtain the upper limit. If the upper limit was n, then the format would fall under a common summation formula and result in the last line answer. How do I get the upper limit? Also, am I doing this correctly in general?
For $P(Y=m)$:
$$P(Y=m) = f_{Y}(n) = \sum_{n}f_{XY}(n,m) = \sum_{n=m}^{\infty}\frac{\alpha ^{m}\beta ^{n-m}}{m!(n-m)!}e^{-(\alpha +\beta )}$$ $$=e^{-(\alpha +\beta )}\frac{1}{n!} \sum_{n=m}^{\infty}\frac{n!}{m!(n-m)!}\beta^{n-m}\alpha^{m}$$ $$=?.$$
Thanks for answering!
Yes, you are doing well.
The upper bound is $n$, because the support is $\{(m,n)\in\Bbb N^2:0\leq m\leq n<\infty\}$.
PS: That will also tell you the bounds for the summation in the other marginal pmf.