Joint PMF of two random variables in two different ways

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?

\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}
• OK, I got it but I think that the lower limit of the summations should be $Y=y$ or simply $y$ instead of $y=m$ – mgus Oct 16 '15 at 17:11
• @KonstantinosKonstantinidis The support of the (integer) random variables is $0\leq X\leq Y< \infty$. So when $X=m$ then $m\leq Y< \infty$. – Graham Kemp Oct 16 '15 at 17:30