A given joint PMF for two random variables is the following: $$P(X=m,Y=n)=\frac{e^{-7}4^m 3^{n-m}}{m!(n-m)!},\ if\ m=0,1,2,...n\ and\ n\in \aleph,\ 0\ otherwise$$
I want to find the joint PMF of $X$ and $Y-X$. One way is the following: $$P(X=m,Y-X=k)=P(X=m, Y=m+k)=\frac{e^{-7}4^m3^k}{m!k!},\ where\ m,k\in \aleph$$
But I would like to do it in a different way such that: \begin{eqnarray*} P(X=m, Y-X=k)&=&P(Y-X=k|X=m)\cdot P(X=m) \\&=&P(Y=k+m|X=m)\cdot P(X=m) \end{eqnarray*}
It can be easily proven by marginalization that $X$ and $Y$ are not independent. Is there any way to proceed further or should I go with the first option?
Well, in order to obtain those factors, you would use:
$\begin{align}\mathsf P(X=m) &=\sum_{y=m}^\infty \mathsf P(X=m, Y=y) \\[1ex] \mathsf P(Y=k+m\mid X=m) & = \dfrac{\mathsf P(X=m, Y=k+m)}{\sum_{y=m}^{\infty} \mathsf P(X=m,Y=y)} \\[2ex] \therefore \mathsf P(Y=k+m\mid X=m)\cdot \mathsf P(X=m) & = \mathsf P(X=m, Y=k+m)\end{align}$
If you already have the joint mass function, using it to find the marginal and conditional mass functions in order to multiply them together to obtain the joint mass function seems ... rather redundant. Don't give yourself an unnecessary work load.