With constant hazard $\theta$, observation $i$ contributes
$$
\theta^{v_i}e^{-\theta x_i}.
$$
Writing $D=\sum_i v_i$ and $T=\sum_i x_i$, the <log-likelihood> is $D\log\theta-\theta T$, whose derivative is $D/\theta-T$. Hence
$$
\widehat\theta=\frac{\sum_i v_i}{\sum_i x_i},
$$
the number of events divided by total person-time.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 207 / 6 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-207/6/a/solution