Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 47 / 5 / a / ii / Solution

Use the shape-scale convention for the inverse-gamma distribution: $\mathrm{IG}(a,b)$ has density proportional to $s^{-a-1}{e}^{-b/s}$ for $s>0$. Write $X=X_k$, $\beta ={\beta }_k$, $\mu ={\mu }_k$, $\Sigma ={\Sigma }_k$ and $s={\sigma }^2$. The independent normal and inverse-gamma regression priors make the conditional log-likelihood plus log prior in $\beta$ equal, up to a constant, toDefine $V=(X^{T}X/s+{\Sigma }^{-1}{)}^{-1}$ and $m=V(X^{T}y/s+{\Sigma }^{-1}\mu )$. Completing the square turns this expression into $-(\beta -m{)}^T{V}^{-1}(\beta -m)/2$ plus a constant. Thus the conditional posterior distribution is the multivariate normal distributionPositive definiteness of the prior covariance matrix makes this well defined even if the design matrix is rank deficient.

Holding $\beta$ fixed, the factors depending on $s$ areConsequentlyThere is no additional $(k+1)/2$ in the shape: the normal distribution prior for ${\beta }_k$ is independent of ${\sigma }^2$. A prior scaled by ${\sigma }^2$ would be a different model and would change that calculation.