= 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 $m=n^{-1}\sum_i x_i^{k+1}$, $r=y-X_ka$, and $w_i=x_i^{k+1}-m$. The <centered polynomial birth move in reversible-jump sampling> gives $X_{k+1}a'=X_ka+zw$, so the residual-sum-of-squares change is
$$
\|r-zw\|^2-\|r\|^2=z^2w^Tw-2zw^Tr.
$$
The map $(a_0,a_1,\ldots,a_k,z)\mapsto(a_0-mz,a_1,\ldots,a_k,z)$ is triangular with <determinant> one. Thus the <reversible-jump Markov chain Monte Carlo> ratio is
$$
R=\frac{\rho_{k+1}d_{k+1}}{\rho_k b_k}\frac{p_{k+1}(a')}{p_k(a)q(z)}\exp\left(\frac{2zw^Tr-z^2w^Tw}{2v}\right),\qquad\alpha_\uparrow=\min(1,R).
$$
For an entirely explicit Gaussian expression, define $Q_k=(a-\mu_k)^T\Sigma_k^{-1}(a-\mu_k)$ and $Q_{k+1}=(a'-\mu_{k+1})^T\Sigma_{k+1}^{-1}(a'-\mu_{k+1})$. Dividing the two normalized Gaussian priors and the proposal <probability density function> gives
$$
\boxed{\alpha_\uparrow=\min\left\{1,\frac{\rho_{k+1}d_{k+1}}{\rho_kb_k}\sigma_a\sqrt{\frac{|\Sigma_k|}{|\Sigma_{k+1}|}}\exp\left[\frac{2zw^Tr-z^2w^Tw}{2v}-\frac{Q_{k+1}-Q_k}{2}+\frac{z^2}{2\sigma_a^2}\right]\right\}.}
$$
Equal 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.
Back to article page
