With cause-specific hazards $h_A$ and $h_B$, survival free of either event through time $u$ is
$$
S(u)=\exp\!\left[-\int_0^u\{h_A(s)+h_B(s)\}\,ds\right].
$$
The probability of remaining event-free to $u$ and then experiencing $A$ in $du$ is $S(u)h_A(u)du$. Therefore
$$
F_A(t)=\int_0^t
\exp\!\left[-\int_0^u\{h_A(s)+h_B(s)\}\,ds\right]
h_A(u)\,du.
$$


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-207/6/a/iv/solution