 Solution

Under the reference convention ${\alpha }_1=0$, the hypothesis becomes absence of the free group contrast ${\alpha }_2$. Introduce an inclusion indicator $J$: set ${\alpha }_2=0$ when $J=0$, and use its stated proper normal prior when $J=1$. Assign positive prior probabilities $q_0,q_1$ to those models, and retain $\mu ,{\beta }_2$ in both.

A birth draws $u$ from a proposal probability density function $g(u)$ and appends it as ${\alpha }_2^{\prime }=u$; the reverse death deletes it. The Jacobian is one. At fixed $\theta ,{\beta }_2$, the likelihood ratio isThus the birth acceptance ratio before capping iswhere $p_{\alpha }$ is the normalized $N(0,{\sigma }_1^2)$ prior. Again, taking $g=p_{\alpha }$ cancels that factor. Use the reciprocal death rule and ordinary within-model parameter updates.

After warm-up, the Markov chain Monte Carlo occupation fraction estimates the desired probability:This requires the chain to explore both models and satisfy its usual ergodic convergence conditions. Under the original continuous normal prior alone, the equality event has posterior probability zero; the explicit model indicator supplies the necessary prior atom. If intercept selection from part (v) is also performed, use two indicators and count all sampled states with $J=0$, regardless of the intercept indicator.