 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-37/2/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Interpret

h_{A}, h_{B}

as cause-specific hazards: their conditioning population consists of individuals still free of both events. Write

H_{A} (t) = \int_{0}^{t} h_{A} (u) d u, H_{B} (t) = \int_{0}^{t} h_{B} (u) d u .

In a short interval, either event removes an individual from that population. Hence the survivor function satisfies

S^{'} (t) = - S (t) [h_{A} (t) + h_{B} (t)]

, with

S (0) = 1

. Integrating gives

Pr (T > t) = S (t) = exp {- H_{A} (t) - H_{B} (t)} .

No independence assumption on hypothetical latent event times is needed: this follows directly from the specified cause-specific hazards.

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