Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 29L (Principles of Statistics)

The MLE is $\hat{\theta }=X_{(n)}$. Since $Y=X_{(n)}/\theta \sim \mathrm{Beta}(n,1)$, its risk is $E(Y-1{)}^2=2/[(n+1)(n+2)]$.

It is inadmissible. For $c{X}_{(n)}$, the risk is $c^{2}n/(n+2)-2cn/(n+1)+1$, minimized by $c=(n+2)/(n+1)$ with risk $1/(n+1{)}^2$, strictly smaller for every $\theta$.

The Bayes risk is $r(\pi ,\delta )=\int R(\delta ,\theta )\pi (d\theta )$.

A Bayes estimator minimizes Bayes risk, equivalently minimizes posterior expected loss almost surely.

The posterior is Pareto with shape $\alpha +n$ and lower endpoint $m=\max \{\ell ,X_{(n)}\}$. Minimizing $E[(\delta -\theta {)}^2/{\theta }^2\mid X]$ gives $\delta =E({\theta }^{-1}\mid X)/E({\theta }^{-2}\mid X)$, hence