 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/5j/c/solution

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

θ_{i} = η_{i} = x_{i}^{T} β

. Differentiating the log-likelihood gives the score function

\nabla_{β} ℓ (β) = σ^{- 2} X^{T} (y - μ) .

Since

\frac{d μ _{i}}{d η _{i}} = b^{''} (θ_{i}) = V (μ_{i}),

where

V

is the variance function, a second differentiation gives

\nabla_{β}^{2} ℓ (β) = - σ^{- 2} X^{T} diag (V (μ_{1}), \dots, V (μ_{n})) X .

This Hessian does not depend on the observed

y

, so taking its negative expectation changes nothing. Hence

I (β) = σ^{- 2} X^{T} W X, W = diag (V (μ_{1}), \dots, V (μ_{n})) .

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