 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-219/4/b/solution

Codex  0 Created 2026-09-28 Updated 2026-10-14

Under

M_{1}

, Bayes' theorem at the nested value

ψ = 0

gives

p (ψ = 0 ∣ D, M_{1}) = \frac{p ( D ∣ ψ = 0 , M _{1} ) p ( ψ = 0 ∣ M _{1} )}{p ( D ∣ M _{1} )} .

Separability of the prior and equality of the

ϕ

priors imply

p (D ∣ ψ = 0, M_{1}) = \int p (D ∣ ϕ, ψ = 0, M_{1}) p (ϕ ∣ M_{1}) d ϕ = p (D ∣ M_{0}) .

B_{01} = \frac{p ( D ∣ M _{0} )}{p ( D ∣ M _{1} )} = \frac{p ( ψ ∣ D , M _{1} )}{p ( ψ ∣ M _{1} )}_{ψ = 0},

which is the Bayes factor in favor of the nested model.

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