 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/13j/b/solution

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

Let

Λ (t) = \int_{0}^{t} λ (s) d s

h_{i} (t) = λ (t) e^{β^{T} x_{i}}, S_{i} (t) = exp [- e^{β^{T} x_{i}} Λ (t)] .

Thus

f_{i} (Y_{i}) = h_{i} (Y_{i}) S_{i} (Y_{i}) = λ (Y_{i}) e^{β^{T} x_{i}} exp [- e^{β^{T} x_{i}} Λ (Y_{i})] .

Independence makes the log likelihood

ℓ (β) = i = 1 \sum n [lo g λ (Y_{i}) + β^{T} x_{i} - e^{β^{T} x_{i}} Λ (Y_{i})] .

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