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

Table of contents
- Solution iv

Solution

After correction, the records from age

60

are

(0, 2), (0, 7), (0, 1 4^{+}), (6, 35), (8, 3 9^{+}), (13, 4 1^{+}),

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

L (q_{2}, q_{7}, q_{35}) = q_{2} (1 - q_{2})^{2} q_{7} (1 - q_{7})^{2} q_{35} (1 - q_{35})^{2} .

Each estimated event hazard is

1/3

.

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