Under $M_1$, <Bayes theorem> at the nested value $\psi=0$ gives
$$
p(\psi=0\mid D,M_1)
=\frac{p(D\mid\psi=0,M_1)p(\psi=0\mid M_1)}{p(D\mid M_1)}.
$$
Separability of the prior and equality of the $\phi$ priors imply
$$
p(D\mid\psi=0,M_1)
=\int p(D\mid\phi,\psi=0,M_1)p(\phi\mid M_1)\,d\phi
=p(D\mid M_0).
$$
Rearranging proves the <Savage-Dickey density ratio>
$$
B_{01}
=\frac{p(D\mid M_0)}{p(D\mid M_1)}
=\left.
\frac{p(\psi\mid D,M_1)}{p(\psi\mid M_1)}
\right|_{\psi=0},
$$
which is the <Bayes factor> in favor of the nested model.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 219 / 4 / b / Solution

{c}
{wiki=Savage–Dickey_density_ratio}

For suitable nested models whose common nuisance parameters have matching priors, the Savage-Dickey density ratio expresses a <Bayes factor> as posterior density divided by prior density at the nested parameter value.

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