 Solution

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.