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

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 ProofOfExpectedValueOfTheHypergeometricDistribution 2013-03-22 13:27:44 2013-03-22 13:27:44 mathwizard (128) mathwizard (128) 8 mathwizard (128) Proof msc 62E15