I think it’s not going to work out. The expected posterior is equal to the prior, but the expected log Bayes factor will have the form p log(K1) + (1-p) log(K2), which for general p is just a mess. Only when p=1/2 does it simplify to log(K1 K2), and when p=1/2, K2=1/K1, so the whole thing is zero.
I think it’s not going to work out. The expected posterior is equal to the prior, but the expected log Bayes factor will have the form p log(K1) + (1-p) log(K2), which for general p is just a mess. Only when p=1/2 does it simplify to log(K1 K2), and when p=1/2, K2=1/K1, so the whole thing is zero.