Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having trouble to compute the following sum: $$ \sum_{k=0}^n(n-2k)^p \frac{{n \choose k}{2m-n \choose m-k}}{{2m \choose m}} $$ Here $p\geq 2$.

To simplify the question, we can even assume that $n/2-C\sqrt n\leq k \leq n/2+C\sqrt n$. Any help or sources will be very helpful.

Thank you.

share|improve this question
    
So you asymptotics or exact value? –  Norbert Apr 11 '12 at 23:52
    
I would like to get exact value, but I don't know even how to get asymptotic. I will apreciate any explanation. –  Michael Apr 12 '12 at 0:58
    
Even Mathematica doesn't know the answer... –  Norbert Apr 12 '12 at 9:41
    
I guess it's implied that $n\le p$. Besides, the first factor can be negative (if p is odd). Nothing wrong with that, but feels a little strange. Are you sure you got it right? –  leonbloy Apr 12 '12 at 15:07
    
Yes. Its right. What if to make substitution $2k-n=m$? –  Michael Apr 12 '12 at 15:35
add comment

2 Answers

up vote 0 down vote accepted
+50

The search term "moment generating function of hypergeometric distribution" may produce some information. One form of the MGF is listed at the Wikipedia page on the hypergeometric distribution, but probably you know that and want a form where the symmetry around $n/2$ is used. Posting to stats.stackexchange.com or adding a (statistics) tag could be useful.

share|improve this answer
add comment

The expression you gave can be simplified to

$$ S = \sum_{k=0}^n(n-2k)^p \frac{{m \choose {n-k}}{m \choose k}}{{2m \choose n}} $$

This is same as

$$ S = \sum_{k=0}^n(2k-n)^p \frac{{m \choose {n-k}}{m \choose k}}{{2m \choose n}} $$

by substituting n-k as k.

So at least in the case where p is odd, this shows that S = -S, so S = 0; not sure what to do when p is even.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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