--

Hi Andy,

BF(n) is equal to P(n|M1) / P(n|M0).

For P(n|M1), we have:

P(n|M1) = int_{0.5}^{1} 2 Binom(n,N) p^n (1-p)^{N-n} dp,

and for the other, we have:

P(n|M0) = Binom(n,N) 1 / 2^N.

So we have

BF(n) = 2^{N+1} int_{0.5}^{1} p^n (1-p)^{N-n} dp

whose logarithm is the same what written in the text.