I don't follow the math. WLOG, for N total datapoints, let y_i = y_N. Then the leave-one-out posterior predictive is \int p(y_N|θ)p(θ|{y_1...y_{N-1}}) dθ = p

I don't follow the math. WLOG, for N total datapoints, let y_i = y_N. Then the leave-one-out posterior predictive is \int p(y_N|θ)p(θ|{y_1...y_{N-1}}) dθ = p(y_N|{y_1...y_{N-1}) by the law of total probability. Expanding the leave-one-out posterior (via Bayes' rule), we have p(θ|{y_1...y_{N-1}}) = p({y_1...y_{N-1}}|θ)p(θ)/\int p({y_1...y_{N-1}}|θ')p(θ') dθ' which when plugged back into the first equation is \int p(y_N|θ) p({y_1...y_{N-1}}|θ)p(θ) dθ/(\int p({y_1...y_{N-1}}|θ')p(θ') dθ') I don't see how this simplifies to the harmonic mean expression in the post. Regardless, the author is asserting that p(y_N|{y_1...y_{N-1}}) ≤ p(y_N|{y_1...y_N}) which seems intuitively plausible for any trained model — given a model trained on data {y_1...y_N}, performing inference on any datapoint y_1...y_N in the training set will generally be more accurate than performing inference on a datapoint y_{N+1} not in the training set.