 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-2/6h/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Put

k = m + 1

and

L = λ + X

. The posterior risk is

\frac{1}{2} {e^{- r a} M (r) + e^{r a} M (- r)},

where

M (s) = (L / (L - s))^{k}

. Differentiating in

a

gives

e^{2 r a} = \frac{M ( r )}{M ( - r )} = (\frac{L + r}{L - r})^{k} .

Thus

\hat{θ} = \frac{m + 1}{2 r} lo g \frac{λ + X + r}{λ + X - r} .

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