Why a Single Cumulative Score Can Miss a Time-Varying Effect
The ordinary logrank test is designed to compare event experience across the full follow-up period. At each event time, it compares the number of treatment-group events with the number expected under equal hazards, given who remains at risk. It then adds these contributions into one score. If the treatment reduces the hazard early but increases it later, the early and late contributions can have opposite signs. A substantial change in survival over time can therefore coexist with a small final score.
This is a property of the test’s cumulative summary, not evidence that the survival curves are identical. To understand the loss of power, it is useful to track the score’s contributions over time and ask whether they reinforce one another or cancel.
These formulas are written for event times without tied events; standard tie adjustments are used when events are tied. The null calibration of the ordinary logrank test does not require proportional hazards: under equal hazards, the treatment-group allocation among events is governed by the risk-set allocation. But its usual power interpretation is simplest when the treatment effect has a stable direction. Under time-varying hazards, the direction and size of the expected score contribution can change.
How an Event-Time Contribution Gets Its Sign
Suppose, for illustration, the treatment and control hazards at a particular time are \(\lambda_T\) and \(\lambda_C\). Let \(\pi_T=Y_T/Y\) and \(\pi_C=Y_C/Y\) denote the treatment and control fractions at risk. Conditional on an event occurring at that time, the probability that it is in the treatment group is approximately $$ q_T=\frac{\pi_T\lambda_T}{\pi_T\lambda_T+\pi_C\lambda_C}. $$ The logrank contribution for one event is the treatment-event indicator minus \(\pi_T\). Thus its expected value is \(q_T-\pi_T\), or equivalently $$ q_T-\pi_T = \frac{\pi_T\pi_C(\lambda_T-\lambda_C)} {\pi_T\lambda_T+\pi_C\lambda_C}. $$ The denominator is positive, so the expected contribution is negative when the treatment hazard is lower and positive when it is higher. If hazards cross, the expected contribution can switch sign. Adding the contributions across follow-up can then hide the changing pattern.
Worked Example: Opposing Expected Contributions
Consider a hypothetical study with equal numbers at risk in both groups within each of two broad follow-up periods. For a transparent approximation, suppose there are 60 events in an early period with hazard ratio \(\lambda_T/\lambda_C=0.5\), followed by 100 events in a late period with hazard ratio \(1.5\). Use the event-allocation calculation above to find the expected score contributions.
With equal risk-set fractions, \(\pi_T=\pi_C=0.5\). In the early period, $$ q_T=\frac{0.5(0.5)}{0.5(0.5)+0.5(1)}=\frac{1}{3}. $$ The expected score contribution per event is \(q_T-\pi_T=1/3-1/2=-1/6\), so the early expected contribution is \(60(-1/6)=-10\).
In the late period, the treatment hazard is \(1.5\) times the control hazard: $$ q_T=\frac{0.5(1.5)}{0.5(1.5)+0.5(1)}=0.6. $$ The expected contribution per event is \(0.6-0.5=0.1\), so the late expected contribution is \(100(0.1)=10\). The total expected score is \(-10+10=0\). These are simplified expected contributions, not a claim that every realized study will have exactly these counts or risk-set fractions. They show how substantial effects in opposite directions can yield little expected cumulative score.
For a survival-curve illustration, suppose the control hazard is 0.02 per time unit in both periods, while the treatment hazard is 0.01 in the first 10 time units and 0.03 in the next 10. At time 10, the hypothetical survival probabilities are \(S_C(10)=e^{-0.2}=0.8187\) and \(S_T(10)=e^{-0.1}=0.9048\). At time 20, they are \(S_C(20)=e^{-0.4}=0.6703\) and \(S_T(20)=e^{-0.1-0.3}=0.6703\). The curves separate early and then meet at time 20, even though the simplified expected logrank score contributions cancel.
Cancellation and the Power of the Test
A useful way to connect this mechanism to power is through the score’s approximate mean and variance. Under a specified alternative, write the mean score as \(E(U)\). When the score is approximately normal with variance close to \(V\), the standardized statistic has approximate mean $$ \delta=\frac{E(U)}{\sqrt{V}}. $$ For a two-sided test at significance level \(\alpha\), the approximate power is $$ P(|Z|>z_{1-\alpha/2}) \approx \Phi\!\left(-z_{1-\alpha/2}-\delta\right) + 1-\Phi\!\left(z_{1-\alpha/2}-\delta\right), $$ where \(\Phi\) is the standard normal cumulative distribution function. Power depends on the size of the standardized cumulative signal, not simply on whether the survival curves differ somewhere during follow-up. Opposing contributions can reduce \(E(U)\) while the variability remains substantial, pulling \(|\delta|\) toward zero and reducing power.
Worked Example: A Delayed Benefit with a Modest Cumulative Signal
Suppose a hypothetical equal-allocation study has 100 early events while the hazards are equal, followed by 60 late events when the treatment hazard is half the control hazard. Use the same event-allocation approximation, and use \(V\approx D/4\) for \(D=160\) events under equal allocation. This variance approximation is for illustrating power, not an exact alternative variance calculation.
In the early period, equal hazards imply \(q_T=0.5\), so the expected score contribution is \(100(0.5-0.5)=0\). In the late period, a hazard ratio of 0.5 gives \(q_T=1/3\), so the expected contribution is \(60(1/3-1/2)=-10\). Therefore \(E(U)\approx-10\), \(V\approx160/4=40\), and $$ \delta\approx\frac{-10}{\sqrt{40}}=-1.581. $$ For a two-sided test with \(\alpha=0.05\), \(z_{0.975}\approx1.960\). The approximate power is $$ \Phi(-1.960-(-1.581))+1-\Phi(1.960-(-1.581)) = \Phi(-0.379)+1-\Phi(3.541) \approx 0.352. $$ Thus the approximation gives power of about 35.2%. The delayed benefit contributes a signal, but the early period contributes many events with no expected treatment-control difference. The result is not a statement that the treatment effect is unimportant; it is a statement about the information and score accumulated by this test under these assumptions.
The approximation \(V\approx D/4\) follows when risk-set allocation is close to one half in each group: each event then contributes about \((0.5)(0.5)=0.25\) to the null variance. With unequal allocation, changing risk-set fractions, or substantial departures from the assumed alternative, this shortcut need not be accurate. Power calculations for a real design should reflect its allocation, censoring, accrual, event pattern, and anticipated hazard functions.
Worked Example: Reading Opposite Increments in a Score Trajectory
Suppose a hypothetical monitoring summary reports an early score increment of \(-8\) with variance increment \(16\), followed by a late score increment of \(+7\) with variance increment \(16\). Assume these increments can be added for this illustration, and use a large-sample normal approximation for the final test.
The cumulative score is \(U=-8+7=-1\), and the cumulative variance is \(V=16+16=32\). The final standardized statistic is $$ Z=\frac{-1}{\sqrt{32}}=-0.177. $$ The approximate two-sided p-value is \(2\Phi(-|{-0.177}|)\approx0.860\). This weak final statistic does not mean that neither period contained a pattern: the early score favored fewer treatment events, while the late score moved in the opposite direction. It does mean that, after combining the increments, the observed net score is small relative to its standard error. In a sequential design, the test’s repeated-look calibration still governs any boundary decision; a small final statistic alone does not reconstruct whether an earlier boundary was crossed.
Diagnostics, Alternatives, and Limits
When non-proportional hazards are plausible, inspect the survival curves and the cumulative logrank score trajectory together. A score that first moves strongly in one direction and later moves back toward zero is a direct warning of possible cancellation. Also examine the timing of events and the risk sets: a late effect may contribute little if few participants remain at risk or few events occur in that period. These patterns are descriptive diagnostics, not proof of a particular hazard model.
The ordinary logrank test is most efficient among common rank-based comparisons under proportional hazards, where treatment effects point consistently in one direction. When the effect changes over time, its unweighted accumulation may not match the scientific question. Alternatives depend on the intended target: a prespecified time-weighted test can emphasize selected periods, while a direct survival-summary estimand can describe survival differences over a chosen horizon. Such choices require careful prespecification and their own assumptions and calibration; choosing a method after seeing which curve pattern looks most favorable can distort type I error.
| Approach | What it summarizes | Key consideration |
|---|---|---|
| Ordinary logrank test | Unweighted cumulative comparison of event experience | Opposing time-specific contributions can cancel |
| Prespecified time-weighted test | Event contributions emphasized according to chosen time weights | Weight choice should reflect a prior scientific rationale |
| Survival difference over a fixed horizon | Difference in a survival summary defined through that horizon | Answers a different question from a single logrank comparison |
In a group sequential design, repeated looks do not remove this limitation. Boundaries control the overall type I error under the design’s assumptions, but they cannot create a larger cumulative treatment signal when increments offset. An early favorable trend might cross a boundary in a particular design, but whether that happens depends on the information at each look, the boundary, and the trajectory; monitoring itself is not a general remedy for non-proportional hazards.
- Mistake: “The final logrank p-value is large, so the treatment had no effect at any time.” A large p-value is not proof of no effect, and the final score can conceal time-specific differences.
- Correction: State what the test evaluates, then inspect the timing and direction of the score contributions and the survival curves. Do not treat a small net score as evidence that every period had a small effect.
- Mistake: “The curves separate, so the ordinary logrank test must have high power.” Curve separation alone does not determine the test’s power; the direction, timing, event counts, and risk sets all matter.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- With equal risk-set fractions and a treatment hazard below the control hazard, what is the expected sign of a treatment-minus-expected logrank contribution?
- A study has an early expected score contribution of \(-6\) and a late contribution of \(+5\). What is the cumulative expected score, and what does its small size fail to establish?
- Why can a study with visibly different survival curves still have low power for the ordinary logrank test?
- Does group sequential monitoring, by itself, eliminate cancellation of score contributions? Explain.
- Name one diagnostic and one alternative analysis strategy that may be relevant when treatment effects vary over time.
Solutions. (1) It is negative: fewer treatment events are expected than under equal hazards. (2) The cumulative expected score is \(-6+5=-1\). Its small size does not establish that effects were small in both periods or that the survival curves were identical. (3) The curve differences may occur in periods with few events, or effects in different directions may produce opposing score increments that nearly cancel. (4) No. Sequential boundaries address repeated testing and type I error; they do not guarantee a larger cumulative signal or remove opposing increments. (5) Inspect survival curves alongside the cumulative score trajectory; one possible alternative is a prespecified time-weighted test, chosen to match the scientific question.