Question

A10. If X and Y are independent binomial random variables with identical parameters n

and p, show analytically that the conditional probability of X, given that X + Y = m

is the hypergeometric distribution.

Answer 1

$X,Y\sim Bin\left( n,p \right) \Rightarrow X+Y\sim Bin\left( 2n,p \right) \\P\left( X=k|X+Y=m \right) =\frac{P\left( X=k,Y=m-k \right)}{P\left( X+Y=m \right)}=\\=\frac{C_{n}^{k}p^k\left( 1-p \right) ^{n-k}C_{n}^{m-k}p^{m-k}\left( 1-p \right) ^{n-m+k}}{C_{2n}^{m}p^m\left( 1-p \right) ^{2n-m}}=\\=\frac{C_{n}^{k}C_{n}^{m-k}}{C_{2n}^{m}}-hypergeometric\,\,with\,\,parameters\,\,2n,n,m$