Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 4 / 17H (Statistics)

Here $X$ has the binomial distribution $\mathrm{Bin}(n,\theta )$, so, up to a factor independent of $\theta$, the likelihood function isFor $0<X<n$, differentiating the log-likelihood givesand the boundary cases give the same formula. Thus the binomial proportion maximum-likelihood estimator is

The estimator is unbiased because ${\mathbb{E}}_{\theta }X=n\theta$, andThe bias-variance decomposition of mean squared error therefore givesThe quadratic $\theta (1-\theta )$ is maximized at $\theta =1/2$, so

The uniform prior is the Beta distribution $\mathrm{Beta}(1,1)$. By Beta-binomial conjugacy, after observing $X$ heads the posterior isThe Bayes estimator under squared error loss is the posterior mean, hence

For the weighted loss, the Bayes estimator under parameter-weighted squared error is the mean of the posterior after multiplication bySince the original posterior density is proportional tothe reweighted density is $\mathrm{Beta}(X+\alpha ,n-X+\beta )$. Its mean givesWriting this in the required form,

Forthe bias and variance areThe bias-variance decomposition of mean squared error yieldsThis is the risk formula for an affine shrinkage estimator for a binomial proportion.

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 $\hat{\theta }$.