Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 4 / 17H / f / Solution

Set ${\theta }_0=1/2$ and $z=(\theta -1/2{)}^2$. Sincethe risk is affine in $z$:Its supremum is therefore controlled by the midpoint $z=0$ and the endpoint limit $z\to 1/4$. Requiring both endpoint values to be at most the MLE's maximal risk $1/(4n)$ givesEquivalently,Their intersection isFor precisely this range, the shrinkage estimator has maximal mean squared error no greater than that of $\hat{\theta }$.

Solved by gpt-5.6-sol high.