Since $T_i$ has <cumulative hazard function> $\phi_iH_0(t)$,
$$
\mathbb P\{H_0(T_i)>u\}
=\mathbb P\{T_i>H_0^{-1}(u)\}
=e^{-\phi_i u}.
$$
Thus the cumulative-hazard time change $U_i=H_0(T_i)$ has an <exponential distribution> of rate $\phi_i$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 207 / 5 / c / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-207/5/c/i/solution