Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 4 / 12H / Solution

A prior distribution assigns uncertainty to a parameter before observing data. Multiplying its density by the likelihood and normalizing gives the posterior distribution. A Bayes estimator minimizes the posterior expected loss function for each observed sample, and therefore minimizes integrated Bayes risk when the expectations exist.

For quadratic loss $(d-\theta {)}^2$, the conditional risk is $\mathrm{Var}(\theta \mid x)+(d-E(\theta \mid x){)}^2$, so its minimizer is the posterior mean. For absolute loss $|d-\theta |$, the one-sided derivatives or subgradients show that a minimizer is any posterior median, characterized by $P(\theta <d\mid x)\le 1/2$ and $P(\theta >d\mid x)\le 1/2$.

For the stated uniform distribution sampling model, let $x_{\min }={\min }_i{x}_i$, $x_{\max }={\max }_i{x}_i$. The likelihood is $2^{-n}$ on $x_{\max }-1<\theta <x_{\min }+1$ and zero elsewhere. Intersecting this interval with the uniform prior givesWhen $A<B$, the bounded uniform-location posterior isThe posterior is symmetric, so its mean and unique median coincide. If $A\ge B$, the data have zero marginal likelihood under the stipulated prior/sampling model and this posterior density cannot be normalized. An empty or singleton intersection is not an ordinary uniform posterior of positive width.