Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 207 / 6 / a / iv

Table of contents
- Solution iv

Solution

With cause-specific hazards

h_{A}

and

h_{B}

, survival free of either event through time

u

is

S (u) = exp [- \int_{0}^{u} {h_{A} (s) + h_{B} (s)} d s] .

The probability of remaining event-free to

u

and then experiencing

A

in

d u

is

S (u) h_{A} (u) d u

. Therefore

F_{A} (t) = \int_{0}^{t} exp [- \int_{0}^{u} {h_{A} (s) + h_{B} (s)} d s] h_{A} (u) d u .

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