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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.