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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…

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Thank you for your help.

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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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