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.

Title	proof of variance of the hypergeometric distribution
Canonical name	ProofOfVarianceOfTheHypergeometricDistribution
Date of creation	2013-03-22 13:27:41
Last modified on	2013-03-22 13:27:41
Owner	mathwizard (128)
Last modified by	mathwizard (128)
Numerical id	13
Author	mathwizard (128)
Entry type	Proof
Classification	msc 62E15