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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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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?2012-06-14
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    in your main formula in the RHS you have '\?' is it typo or it suppose to be some math symbol?2012-06-14
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    @Aleks.K: It was a typo. :) There are some results ([1](http://www.jstor.org/stable/2235699), [2](http://www.jstor.org/stable/2957598), [3](http://www.casact.org/library/astin/vol24no1/19.pdf)) 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.2012-06-14
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    @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?2012-06-15
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    @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...2012-07-04
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    @MånsT: Is the same lines would hold in instead of absolute value one would have an $R^{2n}$ norm? I.e. How to represent $E\|g\|^q$?2018-04-12