Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 27J (Principles of Statistics)

A loss function $L(\theta ,a)$ quantifies the cost of taking action $a$ when the parameter is $\theta$. A decision rule $d$ is a measurable map from the observation space to the action space (or, for randomized rules, a conditional action distribution). Its risk function is $R(\theta ,d)={\mathbb{E}}_{\theta }L(\theta ,d(X))$. For a prior probability distribution $\pi$, its Bayes risk isA minimax rule minimizes ${\sup }_{\theta }R(\theta ,d)$.

For any rule $d$, the average risk is at most its supremum, and the specified Bayes-optimality property then givesTherefore $d^{\ast }$ is minimax.

Interpret the printed expression as $d^{\ast }(x)=x/({\sigma }^2\lambda )$. If $X\sim N(\theta ,{\sigma }^2)$, the Gaussian exponential-moment formula givesIts supremum is one, attained at zero. For every rule $d$, $L(0,d(X))=1$, so $R(0,d)=1$ and ${\sup }_{\theta }R(\theta ,d)\ge 1$. HenceIn the language of part (b), the point-mass prior ${\delta }_0$ makes every rule Bayes with Bayes risk one, and $d^{\ast }$ has exactly that worst-case risk.