 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/5j/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

At a current value

β

, put

η_{i} = x_{i}^{T} β = g (μ_{i}), W_{ii} = \frac{1}{a _{i} σ ^{2} V ( μ _{i} ) { g ^{'} ( μ _{i} ) } ^{2}},

and define the working response

z_{i} = η_{i} + (Y_{i} - μ_{i}) g^{'} (μ_{i}) .

Then the displayed formulas in the question can be written compactly as

I (β) = X^{T} W X,

and

U (β) = X^{T} W (z - Xβ) .

Indeed, the

j

th component of the latter is

\sum_{i = 1}^{n} \frac{X _{ij} ( Y _{i} - μ _{i} )}{a _{i} σ ^{2} V ( μ _{i} ) g ^{'} ( μ _{i} )},

which is precisely the stated score.

Substitution into the Fisher-scoring update gives

β^{new} = β + (X^{T} W X)^{- 1} X^{T} W (z - Xβ) = (X^{T} W X)^{- 1} X^{T} W z .

By part (b), this is the weighted least squares fit of the working response

z

X

. Both

z

and

W

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