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Crossing hazards with a zero observed logrank score

Statement

Use follow-up horizon $τ = 2.5$ . Let the event times be mutually independent, with group hazards $λ_{1} (t) = {\begin{cases} 2, & 0 \leq t \leq 1, \\ 1, & t > 1, \end{cases} λ_{0} (t) = {\begin{cases} 1, & 0 \leq t \leq 1, \\ 2, & t > 1. \end{cases}$ Let all censoring times be mutually independent and independent of the entire event-time family. In group 1, let each censoring time equal $1.5$ with probability $2 / 3$ and $2.5$ with probability $1 / 3$ ; in group 0 let it equal $2.5$ . Observe each subject through $U_{i} = \min (T_{i}, C_{i})$ and $δ_{i} = 1 {T_{i} \leq C_{i}}$ , using the event-censor tie convention in Two-group logrank and weighted logrank scores. Since every $C_{i} \leq τ$ , these records lie in $[0, τ] \times {0, 1}$ . A possible sample has $n_{1} = 4$ : one event at $t_{1} \in (0, 1)$ , two subjects censored at $1.5$ , and one subject censored at $2.5$ . It has $n_{0} = 2$ : one event at $t_{2} \in (2, 2.5)$ and one subject censored at $2.5$ . Compute the unweighted logrank score, its plug-in variance, and the empirical product-limit curves. Verify that this sample pattern has positive probability and that the population survival curves cross even though the observed score is zero.

Facts & Assumptions

Given: The group hazards, mutually independent event and censoring times with the stated censoring laws, the sample pattern, and the definitions in Two-group logrank and weighted logrank scores and Nelson–Aalen, Kaplan–Meier, and Greenwood estimators.

[F1]

Two-group logrank and weighted logrank scores: at event time $t_{j}$ , the score contribution is $O_{j} - d_{j} Y_{1 j} / Y_{j}$ and the plug-in variance contribution is the displayed risk-set formula; with one failure it is $Y_{1 j} Y_{0 j} / Y_{j}^{2}$ .

[F2]

Nelson–Aalen, Kaplan–Meier, and Greenwood estimators: in each group, the empirical product-limit estimate is ${\hat{S}}_{g} (t) = \prod_{0 \leq u \leq t : d_{g} (u) > 0} (1 - d_{g} (u) / Y_{g} (u))$ , where $Y_{g} (u)$ is the number at risk and $d_{g} (u)$ the number of failures at $u$ .

Proof

1.1givenalgebra

Integrating the hazards gives, for $t \geq 1$ , $S_{1} (t) = e^{- (t + 1)}$ and $S_{0} (t) = e^{- (2 t - 1)}$ ; for $0 \leq t \leq 1$ , $S_{1} (t) = e^{- 2 t}$ and $S_{0} (t) = e^{- t}$ . On $[0, 1]$ , $S_{1} (t) < S_{0} (t)$ for $t > 0$ . For $t \geq 1$ their ratio is $S_{1} (t) / S_{0} (t) = e^{t - 2}$ , so they meet at $t = 2$ and reverse order after it. Each event-time interval $(0, 1)$ and $(2, 2.5)$ has positive probability, and the specified censor-time values have positive mass. For every subject required to be censored at $c \in {1.5, 2.5}$ , $P (T > c) = S_{g} (c) > 0$ under its group hazard. Mutual independence of all event and censoring times therefore makes the finite intersection defining the sample pattern have positive probability.

2.1F1F2step 1.1givenalgebra∎

At $t_{1}$ , $(Y_{1}, Y_{0}, d, O) = (4, 2, 1, 1)$ , so the score contribution is $1 - 4 / 6 = 1 / 3$ and the plug-in variance contribution is $4 \cdot 2 / 6^{2} = 2 / 9$ . The two censorings at $1.5$ leave one group-1 subject at risk. At $t_{2}$ , $(Y_{1}, Y_{0}, d, O) = (1, 2, 1, 0)$ , so the score contribution is $0 - 1 / 3 = - 1 / 3$ and the plug-in variance contribution is $1 \cdot 2 / 3^{2} = 2 / 9$ . Hence $U = 0$ and $\hat{V} = 4 / 9 > 0$ , giving standardized score zero. By [F2], ${\hat{S}}_{1} (t) = 1 - 1 / 4 = 3 / 4$ for $t_{1} \leq t \leq 2.5$ because the only group-1 failure is at $t_{1}$ ; and ${\hat{S}}_{0} (t) = 1$ for $t < t_{2}$ , then $1 - 1 / 2 = 1 / 2$ for $t_{2} \leq t \leq 2.5$ . The censorings contribute no product factors. Thus the empirical curves reverse order at $t_{2}$ . The cancellation of the score and the crossing population hazards show why this one sample statistic does not establish equality of the survival laws.

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