 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-35/3/d/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

Divide the two normal distribution predictive densities to obtain the Bayes factor

B_{01} (y) = \frac{V _{1}}{V _{0}} exp [- \frac{y ^{2}}{2} (\frac{1}{V _{0}} - \frac{1}{V _{1}})] .

y = 0

this becomes

B_{01} (0) = \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.

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