 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2/6h/a/solution

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

ℓ (θ) = - \frac{1}{2 σ ^{2}} \sum_{i = 1}^{n} (X_{i} - θ)^{2} .

Differentiating gives

ℓ^{'} (θ) = n (\overline{X} - θ) / σ^{2}

, so

θ_{MLE} = \overline{X} .

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