Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 4 / 28J (Principles of Statistics)

The posterior is $\mathrm{Gamma}(\alpha +X,\beta +1)$, so under quadratic loss the Bayes estimator is $\delta (X)=(\alpha +X)/(\beta +1)$. At fixed $\theta$ its risk is

Put $c=(\beta +1)/(\beta +2)$. The posterior mean of $e^{-\theta }$ is $\delta (X)=c^{\alpha +X}$. Using the Poisson probability-generating function, its risk is

An admissible Bayes estimator whose Bayes risk equals its maximum risk is minimax. Indeed every estimator has maximum risk at least its prior-average risk, which is at least the Bayes risk; equality for the stated estimator attains this lower bound. Admissibility excludes a distinct estimator improving it everywhere.

Neither proper gamma prior yields the condition in part (c). The risk in part (a) is unbounded as $\theta \to \infty$ because $\beta >0$. The risk in part (b) is analytic, tends to zero at infinity, and is not identically zero, so its prior average is strictly below its supremum for a gamma prior with positive density everywhere. Thus this sufficient condition proves neither estimator minimax for any $\alpha ,\beta >0$.