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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 http://math.stackexchange.com/q/139189/23993. But here we wanted to calculate expectation. So, I am not sure about zero for odd $q$.2012-06-10
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    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?2012-06-12
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    @MansT: Thank you. But I still don't understand how to calculate the sum. Could you elaborate please.2012-06-12

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