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Let $X$ be $\rm{Hypergeometric}(2n,\ell,n)$ and $E(X)=\frac{1}{2} \ell=:\mu$. Is it possible and how to approximate the $q$-th central moment $E(X-\mu)^q$ of the hypergeometric distribution by the moments of binomial distribution?

Thank you.

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Assuming $n \geqslant \ell$: $$ \mathbb{P}(X=k) = \frac{\binom{2n-\ell}{n-k}}{\binom{2n}{n}} \binom{\ell}{k} I\left(0 \leqslant k \leqslant \ell\right) $$ In the large $n$ limit, using Stirling's approximation: $$ \frac{\binom{2n -\ell}{n-k}}{\binom{2n}{n}} =\underbrace{\frac{(2n-\ell)!}{(2n)!}}_{ \approx n^{-\ell} 2^{-\ell}} \underbrace{\frac{n!}{(n-k)!}}_{\approx n^{k}} \underbrace{\frac{n!}{(n-\ell +k)!}}_{\approx n^{\ell-k}} \to 2^{-\ell} $$ Hence: $$ \lim_{n\to \infty} \mathbb{P}(X=k) = 2^{-\ell} \binom{\ell}{k} I\left(0 \leqslant k \leqslant \ell\right) $$ meaning that in the limit of large $n$, $X$ converges in distribution to central binomial random variable.

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Does it mean that for large $n$ I can simply say that $E(X-\mu)^q=E(Y-\mu)^q=\sigma^q(q-1)!!$ for even q and $0$ for the odd $q$? Here by $Y$ I've denoted the binomial rndom variable and $\sigma$ as in here en.wikipedia.org/wiki/Normal_distribution#Moments –  Aleks.K Jun 19 '12 at 22:31
    
No, it does not. As $n$ grows larger $X$ converges in distribution to a symmetric binomial random variable $Y$. Hence we can say $\lim_{n \to \infty} \mathbb{E}\left(X- \mu\right)^q = \mathbb{E}\left(Y- \mu\right)^q$. By the central moment of the symmetric binomial random variables do not equal $\sigma^q (q-1)!!$, that expression is the value of central moment of the normal random variable. I am not aware of a closed form expression for the central moment of the symmetric binomial random variable, only that its moment generating function is $\cos^\ell\left(\frac{t}{2}\right)$. –  Sasha Jun 19 '12 at 23:07
    
Thank you. Is there a general formula for the $q$-th central moment of hypergeometric random variable? –  Aleks.K Jun 20 '12 at 0:26
    
No, there is no general formula for the $q$-th central moment of the hypergeometric random variable, as far as I know. However, there exists a closed form expression for $\mathbb{E}\left( \binom{X}{r} \right) = \frac{ \binom{-\ell}{r} \binom{-n}{r}}{\binom{-2n}{r}} (-1)^r$. –  Sasha Jun 20 '12 at 2:43
    
@Sasha: Since there is no general formula for the $q$=th central moment of the hypergeometric random variable, is it possible to get an asymptotic upper bound for E(X-\mu)^q? I guess one should use a concentration? –  Michael Jul 4 '12 at 14:22
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