Hypergeometric distribution - expected value derivation

**GTaiko** · 05-22-2012 06:40 PM

Hello,

I just had a question regarding deriving the expected value of the Hypergeometric(N,M,n) distribution. I know that the expected value is nM/N. So, I used the standard definition for determining expected value for discrete distribution (ie. $\sum_{x} x P(X=x)$ )

So far, I have been able to extract the nM/N from the main term as shown below, but this would mean that the remaining sum would need to equal 1. How would I be able to prove this?

Thanks.

$E(X) = \frac{nM}{N}\sum_{x=max(1,n-N+M)}^{min(n,m)} \frac{\frac{(M-1)!}{(x-1)!(M-x)!} \binom{N-M}{n-x}}{\frac{(N-1)!}{(n-1)!(N-n)!} }$

**Dragan** · 05-22-2012 08:40 PM

You're approaching this in the wrong way. In short, change variables where k = x - 1. Further, let M = N1 where N1 + N2 = N as follows:

$\frac{\sum_{k=0}^{n-1}\binom{N_{1}-1}{k}\binom{N_{2}}{n-1-k}}{\binom{N-1}{n-1}}=1$

**amd** · 06-15-2012 02:46 PM

Right. The straightforward way to derive this value is to set up the weighted sum as you've done, fiddle with the binomial coefficients a bit to absorb the summation variable into one of them and then realize that you're staring at an instance of Vandermonde's identity.

A slicker way that doesn't involve this nasty-looking sum is to use indicator random variables and take advantage of symmetry. Hint: consider the r.v.s I_k which indicate that the kth draw was one of the M possible "successes."

**Dragan** · 06-16-2012 01:30 AM

Originally Posted by amd

Right. The straightforward way to derive this value is to set up the weighted sum as you've done,...

Well, no, it is not easier. Rather, it is easier to derive the expectation by using the following fact:

$\binom{N}{n}=\sum_{x=0}^{n}\binom{N_{1}}{x}\binom{N_{2}}{n-x}$

where

$N = N_{1}+N_{2}$

and it is understood that $\binom{k}{j}=0$ if $j>k$ .

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