After correction, the records from age $60$ are
$$
(0,2),\ (0,7),\ (0,14^+),\ (6,35),\ (8,39^+),\ (13,41^+),
$$
where each pair is delayed-entry time and event or censoring time. The risk sets at event times $2,7,35$ all have size three. Thus the left-truncation-adjusted empirical likelihood is
$$
\boxed{L(q_2,q_7,q_{35})
=q_2(1-q_2)^2q_7(1-q_7)^2q_{35}(1-q_{35})^2.}
$$
Each estimated event hazard is $1/3$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 207 / 4 / b / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-207/4/b/iv/solution