Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 28K (Principles of Statistics)

The risk function of an estimator $\hat{\theta }$ under the stated quadratic loss isThe multivariate normal density is proportional, as a function of $\theta$, toso maximum likelihood estimation gives ${\hat{\theta }}_{\mathrm{MLE}}=X$. Writing $X=\theta +Z$ with $Z\sim N_p(0,I_p)$,for every $\theta \in {\mathbb{R}}^p$.

An admissible estimator is one for which there is no estimator $\delta$ satisfyingfor every $\theta$, with strict inequality for at least one $\theta$. For $p\ge 3$, the James–Stein estimatorhas riskIt strictly dominates $X$, and hence

Combining the Gaussian likelihood with the prior $\theta \sim N_p(0,c^2{I}_p)$ gives the posterior precision and meanThus the posterior in the Gaussian normal-mean shrinkage Bayes estimator isThe Bayes estimator under squared error loss is the posterior mean, soIts posterior expected loss is the trace of the posterior covariance and does not depend on $X$. Therefore its Bayes risk is

Since the MLE has constant risk $p$, its worst-case risk is $p$, so the minimax risk is at most $p$. On the other hand, the worst-case risk of every estimator is at least its average risk under any prior, and therefore at least the optimal Bayes risk for that prior. Using ${\pi }_c=N_p(0,c^2{I}_p)$ givesLetting $c\to \infty$ yields a lower bound $p$. The bounds agree, proving the minimaxity of the usual multivariate normal mean estimator: