= Solution
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
$$
\boxed{U_{A,>160}=\frac12,\qquad V_{>160}=\frac14.}
$$
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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