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Let vector $a\in 2n $ is such that first $l$ of its coordinates are $1$ and the rest are $0$ ($a=(1,\ldots, 1,0, \ldots, 0)$). Let $\pi$ be $k$-th permutation of set $\{1, \ldots, 2n\}$. Define $$g=\left|\sum_{i=1}^n a_{\pi(i)}-\sum_{i=n+1}^{2n}a_{\pi(i)}\right|.$$

Using Hypergeometric distribution calculate /approximate the $q$-th moment $E|g|^q,$ for any $q\ge 2$.

I've got that the $q$-th moment is $$ E|g|^q=\sum_{k=0}^l\frac{{l \choose k}{2n-l \choose n-k}(2k-l)^q}{{2n\choose n}}. $$ But now I am stuck...

Thank you for your help.

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I've also interested in that question some time ago. In fact, using Stirling's approximation formula, you'll get the same sum as in But here we wanted to calculate expectation. So, I am not sure about zero for odd $q$. – David Jun 10 '12 at 13:38
By comparing the last expression to the probability function of the hypergeometric distribution, you see that $E|g|^q=E(2X-l)^q$, where $X$ is $Hypergeometric(2n, l, n)$. Does that help? – MånsT Jun 12 '12 at 19:08
@MansT: Thank you. But I still don't understand how to calculate the sum. Could you elaborate please. – Aleks.K Jun 12 '12 at 22:38

By comparing the last expression to the probability function of the hypergeometric distribution, you see that $E|g|^q=E(2X−l)^q$, where $X$ is $\rm{Hypergeometric}(2n,l,n).$

Therefore $E(X)=\frac{nl}{2n}=l/2=:\mu$. Thus $$E|g|^q=E(2X−l)^q={2}^qE(X-l/2)^q=2^qE(X-\mu)^q.$$

Expressed in words, $E|g|^q$ is $2^q$ times the $q$:th central moment of $X$.

The central moments of the hypergeometric distribution are known and can be computed (preferably not by hand...).

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Thank you. Is it possible to find a general formula for the q-th central moment of the hypergeometric distribution? Or the upper bound of it? – Aleks.K Jun 14 '12 at 16:01
in your main formula in the RHS you have '\?' is it typo or it suppose to be some math symbol? – Aleks.K Jun 14 '12 at 16:02
@Aleks.K: It was a typo. :) There are some results (1, 2, 3) regarding how to compute hypergeometric moments. They probably go beyond what can be considered homework problems for most courses though... As $n\rightarrow\infty$ you can approximate the hypergeometric distribution with the binomial, but I'm not sure if that helps here. – MånsT Jun 14 '12 at 17:28
@MansT:You said that the hypergeometric distribution can be approximated with the binomial. Does it mean that $E(X-\mu)^q\sim E(Y-E(Y))^q,$ where $Y$ is Binomial? – David Jun 15 '12 at 17:12
@MansT: Is it some possibilities to get an asymptotic upper bound for your $2^qE(X-\mu)^q$? I think one can use consentration for that... – Michael Jul 4 '12 at 14:15

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