 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-2/29k/b/solution

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

Write

Z_{i} = (X_{i}, Y_{i})^{T}

for the independent observations and put

Ω = Σ^{- 1}

ℓ (Ω) = \frac{n}{2} lo g ∣Ω∣ - \frac{1}{2} \sum_{i = 1}^{n} Z_{i}^{T} Ω Z_{i} .

\frac{\partial ℓ}{\partial Ω} = \frac{n}{2} Ω^{- 1} - \frac{1}{2} \sum_{i = 1}^{n} Z_{i} Z_{i}^{T} .

At a root of the score,

Ω^{- 1} = n^{- 1} \sum_{i} Z_{i} Z_{i}^{T}

Σ = \frac{1}{n} i = 1 \sum n Z_{i} Z_{i}^{T} = (n^{- 1} \sum_{i} X_{i}^{2} n^{- 1} \sum_{i} X_{i} Y_{i} n^{- 1} \sum_{i} X_{i} Y_{i} n^{- 1} \sum_{i} Y_{i}^{2}) .

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