The group hazards are $\lambda_0=\theta$ and $\lambda_1=\theta e^\beta$. Separate constant-hazard likelihoods give
$$
\widehat\lambda_0=\frac{D_0}{T_0},
\qquad
\widehat\lambda_1=\frac{D_1}{T_1}.
$$
Part (c) then verifies exactly that $\widehat\lambda_0=\widehat\theta$ and $\widehat\lambda_1=\widehat\theta e^{\widehat\beta}$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 207 / 5 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-207/5/d/solution