= Solution
With equal model <prior probabilities>, <Bayes factor> updating gives $w_0=B_{01}/(1+B_{01})$ and $w_1=1/(1+B_{01})$. The <Bayesian model averaging> <posterior density> is therefore
$$
\boxed{p(\beta\mid y)=\frac{B_{01}}{1+B_{01}}p_0(\beta\mid y)+\frac1{1+B_{01}}p_1(\beta\mid y).}
$$
This is a two-component <mixture model> of the <normal distributions> already derived. For the numerical observation above, \b[$w_0\simeq0.10829$ and $w_1\simeq0.89171$]. Consequently it is predominantly the wide-model posterior, not predominantly the narrow one.
Back to article page
