 Solution

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

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

For

δ (X) = X

and

X \sim Pois (θ)

R (θ, δ) = \frac{1}{θ} E_{θ} [(X - θ)^{2}] = \frac{Var _{θ} ( X )}{θ} = 1.

Thus this rule has constant risk one.

To show that no rule has smaller worst-case risk, fix any

α > 1

and let

π_{j} = Gamma (α, λ_{j}), λ_{j} ↓ 0.

Part (a) gives

θ ∣ X = x \sim Gamma (α + x, λ_{j} + 1) .

Since the reciprocal moment of a gamma variable with shape

k > 1

and rate

s

s / (k - 1)

, part (b)(iii) gives the Bayes rule

δ_{j} (x) = \frac{α + x - 1}{λ _{j} + 1} .

E (θ ∣ X = x) - (E (θ^{- 1} ∣ X = x))^{- 1} = \frac{α + x}{λ _{j} + 1} - \frac{α + x - 1}{λ _{j} + 1} = \frac{1}{λ _{j} + 1} .

Hence the Bayes risk is

r_{j} = \frac{1}{λ _{j} + 1} ⟶ 1.

The criterion from part (b)(ii) now proves

δ (X) = X is minimax .

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