proof of expected value 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)

Now we can start with the definition of the expected value:

$E[X]=\sum_{x=0}^{n}\frac{x{K\choose x}{M-K\choose n-x}}{{M\choose n}}.$

Since for $x=0$ we add a $0$ in this we can say

$E[X]=\sum_{x=1}^{n}\frac{x{K\choose x}{M-K\choose n-x}}{{M\choose n}}.$

Applying equation (1) we get:

$E[X]=\frac{nK}{M}\sum_{x=1}^{n}\frac{{K-1\choose x-1}{M-1-(K-1)\choose n-1-(x-% 1)}}{{M-1\choose n-1}}.$

Setting $l:=x-1$ we get:

$E[X]=\frac{nK}{M}\sum_{l=0}^{n-1}\frac{{K-1\choose l}{M-1-(K-1)\choose n-1-l}}% {{M-1\choose n-1}}.$

The sum in this equation is $1$ as it is the sum over all probabilities of a hypergeometric distribution. Therefore we have

$E[X]=\frac{nK}{M}.$

Title	proof of expected value of the hypergeometric distribution
Canonical name	ProofOfExpectedValueOfTheHypergeometricDistribution
Date of creation	2013-03-22 13:27:44
Last modified on	2013-03-22 13:27:44
Owner	mathwizard (128)
Last modified by	mathwizard (128)
Numerical id	8
Author	mathwizard (128)
Entry type	Proof
Classification	msc 62E15