# Joint and marginal distributions

• Jul 25th 2011, 09:08 AM
yellowcarrotz
Joint and marginal distributions
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?
• Jul 25th 2011, 02:11 PM
girdav
Re: Joint and marginal distributions
Quote:

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

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

Yes, by this way you will find $\displaystyle P(X=n)$, and by symmetry $\displaystyle P(Y=m)$.
• Jul 26th 2011, 12:00 AM
yellowcarrotz
Re: Joint and marginal distributions
ok thanks for your help :) i'm not quite show what to do about the (n+m) bit?
n
• Jul 26th 2011, 01:19 AM
girdav
Re: Joint and marginal distributions
Write $\displaystyle \binom{n+m}n\frac 1{(m+n)!}$ as $\displaystyle \frac 1{n!m!}$.
• Jul 26th 2011, 08:52 AM
yellowcarrotz
Re: Joint and marginal distributions
Got it thanks :)