I'm on my phone so I haven't tried to work through the math to see where the error is, but the author's conclusion is wrong and the counterexample is simple. >

I'm on my phone so I haven't tried to work through the math to see where the error is, but the author's conclusion is wrong and the counterexample is simple. > Unless in degenerating cases (the posterior density is point mass), then the harmonic mean inequality guarantees a strict inequality p ( y i | y − i ) < p ( y i | y ) , for any point i and any model. Let y_1, ... y_n be iid from a Uniform(0,theta) distribution, with some nice prior on theta (e.g. Exponential(1)). Then the posterior for theta, and hence the predictive density for a new y_i, depends only on max(y_1, ..., y_n). So for all but one of the n observations the author's strict inequality does not hold.