 Solution

The order prior and probabilities of choosing upward or downward moves are not specified, so include them explicitly. Let ${\rho }_k$ be the model-order prior probability, $b_k$ the birth-selection probability at order $k$, and $d_{k+1}$ the reverse death-selection probability. Denote the correctly normalized coefficient prior by $p_k(a)$ and the probability density function of $z\sim N(0,{\sigma }_a^2)$ by $q(z)$. Common variance-prior factors cancel since $v={\sigma }^2$ is unchanged.

Put $m=n^{-1}{\sum }_i{x}_i^{k+1}$, $r=y-X_{k}a$, and $w_i=x_i^{k+1}-m$. The centered polynomial birth move in reversible-jump sampling gives $X_{k+1}{a}^{\prime }=X_{k}a+zw$, so the residual-sum-of-squares change isThe map $(a_0,a_1,\dots ,a_k,z)\mapsto (a_0-mz,a_1,\dots ,a_k,z)$ is triangular with determinant one. Thus the reversible-jump Markov chain Monte Carlo ratio isFor an entirely explicit Gaussian expression, define $Q_k=(a-{\mu }_k{)}^T{\Sigma }_k^{-1}(a-{\mu }_k)$ and $Q_{k+1}=(a^{\prime }-{\mu }_{k+1}{)}^T{\Sigma }_{k+1}^{-1}(a^{\prime }-{\mu }_{k+1})$. Dividing the two normalized Gaussian priors and the proposal probability density function givesEqual model-order priors and equal birth/death selection probabilities remove the first ratio, if that is the intended convention. Dropping the order-dependent normal-prior constants instead would give the wrong acceptance rule.