 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/5j/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

Y

f_{θ} (y) = h (y) exp {θ y - K (θ)},

where

θ

K

μ (θ) = K^{'} (θ) .

θ_{1}

θ_{2}

D (θ_{1}, θ_{2}) = 2 E_{θ_{1}} [lo g \frac{f _{θ_{1}} ( Y )}{f _{θ_{2}} ( Y )}] .

The carrier

h (Y)

cancels from the likelihood ratio, so, with

μ_{1} = E_{θ_{1}} Y

D (θ_{1}, θ_{2}) = 2 {(θ_{1} - θ_{2}) μ_{1} - K (θ_{1}) + K (θ_{2})} .

μ

θ = lo g μ, K (θ) = e^{θ} = μ .

Therefore

D (μ_{1}, μ_{2}) = 2 {μ_{1} lo g \frac{μ _{1}}{μ _{2}} - μ_{1} + μ_{2}} .

μ_{2} = μ_{1} + δ

and applying the Taylor series of

lo g (1 + δ / μ_{1})

gives

D (μ_{1}, μ_{1} + δ) = \frac{δ ^{2}}{μ _{1}} + O (δ^{3}),

D (μ_{1}, μ_{2}) ≃ \frac{( μ _{2} - μ _{1} ) ^{2}}{μ _{1}} .

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