 Solution

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

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

S_{i} (t) = 1 - F_{i} (t)

h_{i} (t) = \frac{f _{i} ( t )}{S _{i} ( t )} = - \frac{S _{i}^{'} ( t )}{S _{i} ( t )} = - \frac{d}{d t} lo g S_{i} (t) .

Since

S_{i} (0) = 1

, integration yields

S_{i} (t) = exp [- \int_{0}^{t} h_{i} (s) d s] .

Therefore

F_{i} (t) = 1 - exp [- \int_{0}^{t} h_{i} (s) d s] .

The exponent is the negative cumulative hazard.

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