= Solution
Divide the two <normal distribution> predictive densities to obtain the <Bayes factor>
$$
\boxed{B_{01}(y)=\sqrt{\frac{V_1}{V_0}}\exp\left[-\frac{y^2}{2}\left(\frac1{V_0}-\frac1{V_1}\right)\right].}
$$
At $y=0$ this becomes
$$
\boxed{B_{01}(0)=\sqrt{\frac{n_0/n+c^2}{n_0/n+1}}.}
$$
The wider <prior distribution> spreads its predictive mass over more possible means, giving the narrower model more <Bayesian model evidence> for observations very near zero. Away from zero, the exponential term opposes that factor.
Back to article page
