The transformation is $T_k=a_kU_k^b$. Because $U_k$ has a unit-rate <exponential distribution>,
$$
S_k(t)=\mathbb P\!\left(U_k>(t/a_k)^{1/b}\right)
=\exp\!\left[-(t/a_k)^{1/b}\right].
$$
Thus $T_k$ has a <Weibull distribution>, with
$$
H_k(t)=(t/a_k)^{1/b},
\qquad
h_k(t)=\frac1b a_k^{-1/b}t^{1/b-1}.
$$
Consequently $h_2(t)/h_1(t)=(a_1/a_2)^{1/b}$ is constant, proving <proportional hazards>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 207 / 5 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-207/5/a/ii/solution