 Solution

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: