 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/13k/c/solution

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

Write

Y_{i} = X_{i}^{T} β + ε_{i}

, where conditionally

E (ε_{i} ∣ X_{i}) = 0

and

Var (ε_{i} ∣ X_{i}) = σ^{2} v (X_{i})

. Then

β (Σ) - β = (\frac{1}{n} \sum_{i} \frac{X _{i} X _{i}^{T}}{v ( X _{i} )})^{- 1} \frac{1}{n} \sum_{i} \frac{X _{i} ε _{i}}{v ( X _{i} )} .

A = E [\frac{X X ^{T}}{v ( X )}],

while the vector tends to zero, proving consistency. The central limit theorem and Slutsky's theorem give

n (β (Σ) - β) \Rightarrow N (0, σ^{2} A^{- 1}) .

Similarly, with

B = E (X X^{T})

and

C = E (v (X) X X^{T})

n (β (I) - β) \Rightarrow N (0, σ^{2} B^{- 1} C B^{- 1}) .

These conclusions require the displayed matrices to be finite and nonsingular.

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