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

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

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

Keeping the event-free A individual under observation through day 193 leaves them in the risk set at both later B events. Now the 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	1	2	0	$1/3$	$- 1/3$	$2/9$
191	1	1	0	$1/2$	$- 1/2$	$1/4$

Therefore

U_{A, > 160} = \frac{1}{2} - \frac{1}{3} - \frac{1}{2} = - \frac{1}{3}, V_{> 160} = \frac{1}{4} + \frac{2}{9} + \frac{1}{4} = \frac{13}{18} .

(1)

The negative score now indicates lower A event hazard, relative to B, in the contribution after day 160. The observed event counts have not changed. The change is in what events A was expected to contribute: its additional event-free exposure makes the two later B events informative comparisons, producing negative A score terms. No conclusion about the complete study's significance follows from this late contribution without the earlier scores and variances.

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