Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 28L (Principles of Statistics)

Since $X/n$ is unbiased and has variance $\theta (1-\theta )/n$,for $0<\theta <1$.

The $\pi$-Bayes risk isConditioning on the observation and applying Tonelli givesThus choosing, for every observed $x$, an action minimizing the posterior expected loss minimizes each integrand and hence minimizes the Bayes risk.

Under the uniform prior, the posterior density after $X=x$ is proportional to ${\theta }^x(1-\theta {)}^{n-x}$. For $0<x<n$, differentiating posterior risk gives the unique stationary pointStrict convexity makes it the unique minimum. For $x=0$, every $a\ne 0$ gives infinite posterior risk at zero and $a=0$ is optimal; symmetrically $a=1$ is optimal for $x=n$. Hence the unique Bayes rule is $\delta (x)=x/n$.

The constant risk from part (a) givesEvery estimator has maximum risk at least its uniform-prior Bayes risk, and the minimum possible Bayes risk is $1/n$ by part (c). Hence no estimator has smaller maximum risk, so the MLE is minimax.

An estimator is admissible if no other estimator has risk no larger at every parameter and strictly smaller somewhere. If an estimator dominated the MLE, continuity of its binomial risk would make the inequality strict on a set of positive prior measure, lowering its uniform-prior Bayes risk below $1/n$. This contradicts Bayes optimality. Equivalently, uniqueness of the Bayes action at every $x$ rules out equality for a distinct estimator. Thus the MLE is admissible.