proof of variance of the hypergeometric distribution

We will first prove a useful property of binomial coefficients. We know

{n\\choose k}=\\frac{n!}{k!(n-k)!}.

This can be transformed to

{n\\choose k}=\\frac{n}{k}\\frac{(n-1)!}{(k-1)!(n-1-(k-1))!}=\\frac{n}{k}{n-1%\\choose k-1}.

(1)

The variance $\\operatorname{Var}[X]$ of $X$ is given by:

\\operatorname{Var}[X]=\\sum_{x=0}^{n}\\left(x-\\frac{nK}{M}\\right)^{2}\\frac{{K%\\choose x}{M-K\\choose n-x}}{{M\\choose n}}.

We expand the right hand side:

$\\displaystyle\\operatorname{Var}[X]$	$\\displaystyle=$	$\\displaystyle\\sum_{x=0}^{n}\\frac{x^{2}{K\\choose x}{M-K\\choose n-x}}{{M\\choose n}}$
		$\\displaystyle-\\frac{2nK}{M}\\sum_{x=0}^{n}\\frac{x{K\\choose x}{M-K\\choose n-x}}{%{M\\choose n}}$
		$\\displaystyle+\\frac{n^{2}K^{2}}{M^{2}}\\sum_{x=0}^{n}\\frac{{K\\choose x}{M-K%\\choose n-x}}{{M\\choose n}}.$

The second of these sums is the expected value of the hypergeometric distribution, the third sum is $1$ as it sums up all probabilities in the distribution. So we have:

\\operatorname{Var}[X]=-\\frac{n^{2}K^{2}}{M^{2}}+\\sum_{x=0}^{n}\\frac{x^{2}{K%\\choose x}{M-K\\choose n-x}}{{M\\choose n}}.

In the last sum for $x=0$ we add nothing so we can write:

\\operatorname{Var}[X]=-\\frac{n^{2}K^{2}}{M^{2}}+\\sum_{x=1}^{n}\\frac{x^{2}{K%\\choose x}{M-K\\choose n-x}}{{M\\choose n}}.

Applying equation (1) and $x=(x-1)+1$ we get:

\\operatorname{Var}[X]=-\\frac{n^{2}K^{2}}{M^{2}}+\\frac{nK}{M}\\sum_{x=1}^{n}%\\frac{(x-1){K-1\\choose x-1}{M-K\\choose n-x}}{{M-1\\choose n-1}}+\\frac{nK}{M}%\\sum_{x=1}^{n}\\frac{{K-1\\choose x-1}{M-K\\choose n-x}}{{M-1\\choose n-1}}.

Setting $l:=x-1$ the first sum is the expected value of a hypergeometric distribution and is therefore given as $\\frac{(n-1)(K-1)}{M-1}$ . The second sum is the sum over all the probabilities of a hypergeometric distribution and is therefore equal to $1$ . So we get:

$\\displaystyle\\operatorname{Var}[X]$	$\\displaystyle=$	$\\displaystyle-\\frac{n^{2}K^{2}}{M^{2}}+\\frac{nK(n-1)(K-1)}{M(M-1)}+\\frac{nK}{M}$
	$\\displaystyle=$	$\\displaystyle\\frac{-n^{2}K^{2}(M-1)+Mn(n-1)K(K-1)+KnM(M-1)}{M^{2}(M-1)}$
	$\\displaystyle=$	$\\displaystyle\\frac{nK(M^{2}+(-K-n)M+nK)}{M^{2}(M-1)}$
	$\\displaystyle=$	$\\displaystyle\\frac{nK(M-K)(M-n)}{M^{2}(M-1)}$
	$\\displaystyle=$	$\\displaystyle n\\frac{K}{M}\\left(1-\\frac{K}{M}\\right)\\frac{M-n}{M-1}.$

This is the one we wanted to prove.