probability mass function of $Y-X$ (two Poisson random variables) It is question 55 on page 86 from Ross's book(Introduction to Probability Theory)
Suppose that the joint probability mass function of $X$ and $Y$ is
$$ P(X = i, Y = j) = {j\choose i} e^{-2\lambda} \frac{\lambda^j}{j!}, \quad 0\le i\le j$$
(a) Find the probability mass function of $Y$ 
(b) Find the probability mass function of $X$
(c) Find the probability mass function of $Y - X$
My answers: 
(b): $f_Y(j) = e^{-2\lambda} {(2\lambda)}^j/j!$
(a): $f_X(j) = e^{-\lambda} \lambda^j/j!$
for (c): to calculate $P\{Y - X = n\}$
suppose $x = k,$ $y = n + k,$
then
\begin{align}
P\{Y - X = n\} & = P\{x = k, Y = n + k\} = P(X = k) P(Y = n + k) \\[10pt]
& = \sum_{k=0}^n \left(e^{-\lambda} \frac{\lambda^k}{k!} \cdot e^{-2\lambda}  \frac{(2\lambda)^{n+k}}{(n+k)!}\right)
\end{align}
then I was stuck here.
Thanks in advance 
Update 1: follow the suggestions by @Michael Hardy


*

*P( Y-X|Y ) = P(X|Y): Calculate this first(Why???)
Intuitively, Y - X and X complement each other when Y is given.
$$ P(X = i |Y = j) = \sum_{j=0} {j\choose i} e^{-2\lambda} \frac{\lambda^j}{j!}$$  $$ =e^{-2\lambda}\frac{\lambda^j}{j!} \sum_{j=0} {j\choose i} $$
$$ P(Y-X=j-i |Y=j) = \sum_{j=0} {j\choose j-i} e^{-2\lambda} \frac{\lambda^j}{j!}$$  $$ =e^{-2\lambda}\frac{\lambda^j}{j!} \sum_{j=0} {j\choose j-i} $$


*Calculate P(Y-X).


suppose Y - X = n, x = i
P(Y-X = n) = $$ \sum_{i=0}^n {i+n\choose i} e^{-2\lambda} \frac{\lambda^{i+n}}{(i+n)!} $$
= $$e^{-2\lambda}\frac{\lambda^{n}}{n!} \sum_{i=0}^n\frac{\lambda^i}{i!} $$
= $$e^{-2\lambda}\frac{\lambda^{n}}{n!} e^{\lambda} $$
= $$e^{-\lambda}\frac{\lambda^{n}}{n!} $$
it is the same as @Mohit 's result.
 A: $$\Pr(X = i, Y = j) = {j\choose i} e^{-2\lambda} \frac{\lambda^j}{j!}, \quad 0\le i\le j$$
\begin{align}
\Pr(Y=j) = {} & \sum_{i=0}^j \Pr(X=i\ \&\ Y=j) \\[10pt]
= {} & \sum_{i=0}^j \binom j i e^{-2\lambda} \frac{\lambda^j}{j!}. \\
& \text{In this sum, everything to the right of $\dbinom j i$ does} \\
& \text{not change as $i$ goes from $0$ to $j$. Therefore it can} \\
& \text{be pulled out:} \\[10pt]
= {} & e^{-2\lambda} \frac{\lambda^j}{j!} \sum_{i=0}^j \binom j i \\[10pt]
= {} & e^{-2\lambda} \frac{\lambda^j}{j!} \cdot 2^j \text{ by the binomial theorem} \\[10pt]
= {} & e^{-2\lambda} \frac{(2\lambda)^j}{j!}. \\[15pt]
\text{Therefore } Y \sim {} & \operatorname{Poisson}(2\lambda).
\end{align}
\begin{align}
\Pr(X=i) = {} & \sum_{j=i}^\infty \binom j i e^{-2\lambda} \frac{\lambda^j}{j!} \\
& \text{Notice that this sum starts with $j=i,$ not with} \\
& \text{$j=0$ or $j={}$something else, since we're told at} \\
& \text{the outset that $i\le j$.} \\[10pt]
& = e^{-2\lambda} \sum_{j=i}^\infty \frac{\lambda^{j-i}}{(j-i)!} \cdot\frac{\lambda^i}{i!} \\
& \ldots\text{ and now $\lambda^i/(i!)$ does not change as $j$ goes from} \\& \text{$i$ to $\infty,$ so it can be pulled out:} \\[10pt]
= {} & e^{-2\lambda} \frac{\lambda^i}{i!} \sum_{j=i}^\infty \frac{\lambda^{j-i}}{(j-i)!} \\[10pt]
= {} & e^{-2\lambda} \frac{\lambda^i}{i!} \sum_{k=0}^\infty \frac{\lambda^k}{k!} \text{ where } k = j-i \\[10pt]
= {} & e^{-2\lambda} \frac{\lambda^i}{i!} \cdot e^{\lambda} \\[10pt]
= {} & e^{-\lambda} \frac{\lambda^i}{i!}. \\[15pt]
\text{Therefore } X \sim {} & \operatorname{Poisson}(\lambda).
\end{align}
To show that $Y-X\sim\operatorname{Poisson}(\lambda),$ first show that the conditional distribution of $Y-X$ given $Y$ is the same as the conditional distribution of $X$ given $Y.$ Then that conclusion follows.
It seems regrettable that the exercise didn't have a part $(d)$ in which you show that $X$ and $Y-X$ are actually independent.
A: Careful, $X, Y$ aren't necessarily independent. Your idea was right though:
$$
P(Y-X = n) = \sum_{k=0}^\infty P(x = k, Y = n + k) = \sum_{k=0}^\infty {n+k\choose k} e^{-2\lambda}  \lambda^{n+k}/(n+k)! \\
=e^{-2\lambda}\lambda^{n}/n! \sum_{k=0}^\infty   \lambda^{k}/k!  = e^{-\lambda}\lambda^{n}/n!\\
$$
But this is just the Poission distribution with parameter $\lambda$. 
