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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.

For the reverse centered polynomial birth move in reversible-jump sampling, recover $z={\beta }_{k+1}^{\prime }$ from the current coefficient vector. Remove that highest-order coefficient and undo the intercept shift:No auxiliary random number is drawn for this death move: the removed coefficient supplies the auxiliary value needed to reconstruct the birth move. The reverse Jacobian determinant is also one. The inverse prediction change is $-zv$, hence the residual in the lower-order model is $r=r^{\prime }+zv$.

A reversible-jump Markov chain Monte Carlo calculation compares normalized model-specific densities. The fixed-order posterior written only up to proportionality in part (a) omits normalizing factors that depend on $k$ and therefore cannot simply be canceled between dimensions. Moreover, the printed question does not specify the model-order prior distribution or the probabilities of selecting birth and death moves. Denote those by $p_k$, $b_k$ and $d_k$, respectively, retaining them in the answer.

Put $s={\sigma }^2$, $c=n^{-1}{\sum }_i{x}_i^{k+1}$, $v_i=x_i^{k+1}-c$, and $r=y-X_k\beta$. The centered polynomial birth move in reversible-jump sampling changes the fitted vector by $zv$, so its new residual is $r^{\prime }=r-zv$. ThereforeThe map from $({\beta }_0,\dots ,{\beta }_k,z)$ to $({\beta }_0^{\prime },\dots ,{\beta }_{k+1}^{\prime })$ has ones down the diagonal and a single additional entry $-c$ in the intercept row. Its Jacobian determinant is one.

Write $Q_k=(\beta -{\mu }_k{)}^T{\Sigma }_k^{-1}(\beta -{\mu }_k)$ and $Q_{k+1}=({\beta }^{\prime }-{\mu }_{k+1}{)}^T{\Sigma }_{k+1}^{-1}({\beta }^{\prime }-{\mu }_{k+1})$. The normalized multivariate normal distribution prior ratio isDivide also by the forward proposal density $q(z)=(2\pi {\sigma }_{\beta }^2{)}^{-1/2}{e}^{-z^2/(2{\sigma }_{\beta }^2)}$. The common inverse-gamma distribution prior for $s$ and the common normal-likelihood normalization cancel, since $s$ stays fixed. The forward acceptance ratio is consequentlyHere ${\sigma }_{\beta }>0$ is the proposal standard deviation. If equal model priors and symmetric birth/death selection are explicitly adopted, their ratios become one; they are not determined by the PDF. The expression assumes the same hyperparameters for the variance prior in both models, as in part (a). A dimension-dependent variance prior would contribute its own ratio.