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. )
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?
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."
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