Expectation from a hypergeometric distribution

Hi,

I am solving an expectation problem, in which I arrived at the following sum expression to evaluate:

$\displaystyle E\left[\frac{X}{2m-X}\right] = \sum_{x_i=0}^{m} \frac{x_i}{2m-x_i} \cdot \frac{\binom{m}{x_i} \binom{n-m}{m-x_i}}{\binom{n}{m}}$

I'm kinda stuck, could someone give me a hint what to do, or where to look for, to compute the expected value here?

Thanks!

Re: Expectation from a hypergeometric distribution

Hey herrfz.

Have you tried simplifying all the factorial terms? (Remember nCr = n!/(r!*(n-r)!)).

Re: Expectation from a hypergeometric distribution

Could you give me a specific hint which part of the factorials that can be simplified? I tried expanding the factorial terms (see below), but couldn't spot any simplifications, not the ones that (I think) will be helpful anyways... thanks..

$\displaystyle \frac{\binom{m}{x_i}\binom{n-m}{m-x_i}}{\binom{n}{m}} = \frac{\frac{m!}{(m-x_i)!x_i!}\frac{(n-m)!}{(n-2m+x_i)!(m-x_i)!}}{\frac{n!}{(n-m)!m!}}$

===

EDIT: there was a mistake on the factorial expansion

Re: Expectation from a hypergeometric distribution

That isn't quite right since (n-m)C(m-xi) = (n-m)!/[(m-xi)! * [n - m - [m - xi]]! = (n-m)!/[(m-xi)! * [n + xi]!]

Note that (n-m)!/(n-xi)! = (n-m)(n-m-1)...(n-m+xi) if m < xi and then use the denominator if it's the other way around. You have similar situations with the others.

This doesn't look easy although a hint from wikipedia says:

Hypergeometric distribution - Wikipedia, the free encyclopedia

which is another piece of information that I think may be useful.

Re: Expectation from a hypergeometric distribution

Ah yes, indeed there's a mistake..

It is indeed pretty tricky; the identity 1) from Wikipedia (sum of probability) is hard to use in the expectation expression, and identity 2) won't change the expansion, because the swapped variables are the same ($\displaystyle m$ in the first binomial term is the same as the $\displaystyle m$ in $\displaystyle m-x_i$ in the second binomial term of the nominator).

Re: Expectation from a hypergeometric distribution

One idea is to transform it so that [X + 2m - 2m]/[2m - X] = -1 + 2m/[X-2m].