Reducing EMI

I'm trying to reduce the third term in the EMI equation for the AMI (see AMI on wikipedia, EMI is listed under "Adjustment for chance"). Sorry for the quasi-latex notation.

• Shorthand is used: a_i -> a, b_j -> b, n_ij -> n, N is the same. I'm only interested in the third term, which is the one with factorials.

• Remember that \frac{n!}{(n-k)!} to simplify the notation below, we use \frac{t!}{(t - u)!}

• Assume that a >= b (if b > a, simply reverse the roles of a and b in the following).

• As a >= b, then (N-a-b+n) <= (N-b), because n <= min(a, b) and therefore n <= b <= a.

• Letting t <- (N-b) and u <- (a - n), we can remove the (N-b)! and (N-a-b+n)! terms in the equation, and replace with (N-b) choose (a-n)

• Letting t <- N and u <- a, we remove the N! and (N-a)! and replace with {N choose a} ^ -1

• Letting t <- {a, b} and u <- n, we remove the {a, b}! and ({a, b} - n)! with (a choose n) and (b choose n) respectively.

The only term this leaves is (n!)^-1.

This replaces the third term of the equation with:

{(N-b) choose (a - n)} x {N choose a}^-1 x {a choose n} x {b choose n} x {n!}^-1