Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 40 / 4 / a / iii / Solution

Use the Gaussian likelihood Gibbs annealing target ${\pi }_T(\mu ,v)\propto L(\mu ,v{)}^{1/T}$ relative to $d\mu dv$, with $S>0$. Completing the normal square and recognizing an inverse-gamma kernel giveThe shape is $n/(2T)-1$, because the probability density function in $v$ is $v^{-n/(2T)}{e}^{-[S+n(\mu -\bar{x}{)}^2]/(2Tv)}$. The base measure matters: inserting a prior factor $1/v$ would change this shape.

Integrating out $\mu$ contributes a factor proportional to $v^{1/2}$, so the marginal is $v\sim \mathrm{IG}(n/(2T)-3/2,S/(2T))$. Thus the joint target is proper for $T<n/3$. Choose a decreasing sequence $0<T_m\le n/10$ tending to zero and a finite starting value $v_0>0$. At iteration $m$, sample ${\mu }_m$ from the first conditional using $v_{m-1}$, then sample $v_m$ from the second using ${\mu }_m$. This is an explicit annealing algorithm entirely using Gibbs sampling, with all its inverse-gamma moments finite for the convergence proof.