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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-32/5/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 32 / 5 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-07

At day 165 there are two A and two B individuals at risk. The event is in A, with expected A count

2/4 = 1/2

, so its score contribution is

1 - 1/2 = 1/2

and its variance contribution is

1/4

. The remaining A individual is censored at 173 and leaves the risk set before either later event. The risk-set calculations are

Event day	$n_{A}$	$n_{B}$	$d_{A}$	$e_{A}$	$d_{A} - e_{A}$	Variance
165	2	2	1	$1/2$	$1/2$	$1/4$
180	0	2	0	0	0	0
191	0	1	0	0	0	0

Thus

U_{A, > 160} = \frac{1}{2}, V_{> 160} = \frac{1}{4} .

(1)

The positive score indicates higher A event hazard in this late contribution. Although B has more observed events, its two events occur with no A individual at risk, so those times do not compare the groups. This is why raw event totals alone are insufficient for the log-rank test.

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