Past exam of the statistics course of the University of Oxford / 2017 / 3 / d / Solution

Let $y_{(n)}={\max }_i{y}_i$ and $c=\max (\beta ,y_{(n)})$. For the observed positive data, the uniform distribution likelihood function isMultiplying by the Pareto distribution prior density gives the posterior density kernel ${\theta }^{-(\alpha +n+1)}{1}_{\{\theta >c\}}$. Sincethe normalized posterior distribution isThis is uniform-Pareto conjugacy. The posterior density is strictly decreasing over its support, so the highest posterior density interval starts at $c$, where the admissible density is largest in the limiting sense. Its upper endpoint $u$ satisfiesHence a 95% highest posterior density interval isInclusion of endpoints does not change its posterior probability. Every point inside has higher posterior density than every admissible point to its right, and points below $c$ have density zero. This is the Pareto highest density interval; equal endpoint densities are unnecessary because one endpoint is the support boundary.