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.
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.
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.