Joint and marginal distributions

**yellowcarrotz** · July 25th 2011, 09:08 AM

Hi, i am trying to answer the following question:

Consider the following joint distribution:

P(X=n, Y=m)= (m+n)(e^-2x) x^(m+n)/(n+m)!
n
where m and n takes values in 0,1,2.......

(the first bit should read m+n choose n)

show that X and Y are independent but have the same distribution as each other.

I know I should find the marginal distributions of X and Y, but Im not sure how to do so in this case, in my notes i only have examples where the joint distribution looks geometric. If i were to do it that way, to work out the marginal distribution of X, would I sum the joint distribution from m=0 to infinity and then work on simplifying it?
Thanks for your help

**girdav** · July 25th 2011, 02:11 PM

Originally Posted by yellowcarrotz

Consider the following joint distribution:
$P(X=n, Y=m)=\binom{n+m}{n} \frac{e^{-2x} x^{m+n}}{(n+m)!}$
where $m$ and $n$ takes values in $\mathbb{N}$ .
Show that $X$ and $Y$ are independent but have the same distribution as each other.

Would I sum the joint distribution from $m=0$ to infinity and then work on simplifying it?