Ignoring factors that do not depend on $\theta$, each observed side-effect contributes $\theta e^{-\theta x_i}$ and each censored observation contributes $e^{-\theta x_i}$. Thus
$$
L(\theta)\propto\theta^{v_+}e^{-\theta x_+},
\qquad
\ell(\theta)=v_+\log\theta-\theta x_++\text{constant}.
$$
The <score equation> $v_+/\theta-x_+=0$ gives
$$
\widehat\theta=\frac{v_+}{x_+}.
$$
This event-count divided by person-time estimator is the sample analogue of the expectation ratio in part iv.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 207 / 6 / b / v / Solution

 Solution

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