Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 27I / Solution

Write $R(\theta ,\delta )={\mathbb{E}}_{\theta }L(\theta ,\delta (X))$ and $r(\pi ,\delta )=\int R(\theta ,\delta )d\pi (\theta )$. An equaliser rule has constant risk $R(\theta ,\delta )=c$. An Extended Bayes rule has, for every $\epsilon >0$, a proper prior $\pi$ such that $r(\pi ,\delta )\le r^{\ast }(\pi )+\epsilon$, where $r^{\ast }(\pi )={\inf }_{d}r(\pi ,d)$. If a rule has both properties, every competitor $d$ satisfiesLetting $\epsilon \to 0$ proves that its maximum risk $c$ is the minimax risk.

For the given normal distribution model, completing the square in prior times likelihood gives the posterior $N(m,H^{-1})$, with $H=h_0+nh$ and $m=(h_0{m}_0+h{\sum }_i{x}_i)/H$. Another completion of the square givesThis is maximized uniquely at $a=m$. Consequently the Bayes act and posterior posterior expected loss areThe posterior loss is independent of $x$, so averaging over the predictive distribution gives the same value for the Bayes risk of this Bayes decision rule.

The sample mean obeys $\bar{X}-\theta \sim N(0,(nh{)}^{-1})$, so its risk is the constant $c=1-\sqrt{nh/(nh+k)}$. Take any fixed $m_0$ and let the proper prior precisions $h_0\downarrow 0$. The optimal Bayes risks just calculated tend to $c$, while the integrated risk of $\bar{X}$ is exactly $c$ for every prior. Thus its excess Bayes risk tends to zero: it is extended Bayes as well as equaliser. The preceding argument proves