= Solution
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
$$
\boxed{U_{A,>160}=\frac12-\frac13-\frac12=-\frac13,
\qquad V_{>160}=\frac14+\frac29+\frac14=\frac{13}{18}.}
$$
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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