 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/28l/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Given

X = x

d

E [θ^{- 1} (θ - d)^{2} ∣ X = x] = E (θ ∣ X = x) - 2 d + d^{2} E (θ^{- 1} ∣ X = x) .

This is a strictly convex quadratic in

d

whenever the conditional inverse moment is finite and positive. Differentiating with respect to

d

gives

- 2 + 2 d E (θ^{- 1} ∣ X = x) = 0,

δ_{π} (x) = (E (θ^{- 1} ∣ X = x))^{- 1} .

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