Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 36 / 2 / e / Solution

Keep $y$ fixed. Integral bounds for the decreasing function $1/t$ give $H_M=\log M+O(1)$, so $H_M-H_{y-1}=\log M+O_y(1)$. ConsequentlyThe posterior mean thus grows like $M/\log M$ instead of approaching a data-determined limit. Even the posterior distribution mass below a fixed endpoint $n$ is $(H_n-H_{y-1})/(H_M-H_{y-1})\to 0$. Hence increasing the supposedly innocuous prior cutoff moves the inference to larger and larger endpoints. In an integral approximation its median is of order $\sqrt{yM}$, also depending strongly on the cutoff. This explains the sensitivity of the entire inference, not just of its posterior mean.