Why Stratify an Interim Logrank Score?
A survival trial may prespecify strata, such as disease stage or recruiting region, because patients in different strata can have different baseline event risks. A stratified logrank calculation respects these differences: at each event time, it compares the observed treatment-arm events with the number expected from that arm’s share of the risk set within the same stratum. The stratum-level observed-minus-expected contributions are then summed to obtain the interim score.
This calculation is not the same as pooling everyone into one risk set. Pooling can compare treatment groups whose risk-set composition differs across strata, allowing baseline-risk differences to affect the comparison. Nor does the raw score alone provide a standardized test statistic or a stopping decision: it is the cumulative score component, not a score divided by its standard error.
Assumptions and Risk-Set Expectations
The calculation uses the event and censoring records available at the interim analysis, together with the prespecified stratum membership. The comparison is meaningful under the null hypothesis of equal event hazards between A and B within each stratum. In a randomized trial, treatment assignment should also be implemented as planned. As in the ordinary logrank test, censoring must be non-informative for the event process within the relevant treatment and stratum groups: conditional on the information represented in the analysis, censoring should not select patients according to their unobserved event times. See the earlier material on censoring assumptions behind the logrank test for the implications of this condition.
At an event time, the risk set consists of patients who have not had the event and have not been censored before that time, according to the trial’s prespecified event-time and censoring convention. Under the null, the event allocation between A and B is proportional to their representation in this risk set. Thus the expected number of A events at that time is the total number of events multiplied by A’s fraction of the risk set. For tied events, this expectation remains \(d_{rk}Y_{Ark}/(Y_{Ark}+Y_{Brk})\); the tied events are treated as a group for this expected-count calculation.
The analysis is cumulative: include event times observed by the interim cutoff and do not include events that occur later. At each included time, calculate the expectation from that time’s risk set, not from the original randomized group sizes or from the risk set at another time. Apply the same risk-set convention for tied event and censoring times throughout.
Why the Within-Stratum Contributions Have Mean Zero Under the Null
The score’s observed-minus-expected form follows directly from the risk-set allocation under the null. For a single event time in stratum \(r\), write \(Y_{rk}=Y_{Ark}+Y_{Brk}\). Conditional on the risk set and the total \(d_{rk}\) events, the expected number of A events is \(d_{rk}Y_{Ark}/Y_{rk}\). Therefore:
Adding these contributions over event times gives a stratum score whose conditional expected increment is zero under the null; summing across prespecified strata preserves that observed-minus-expected construction. This justification explains why each expectation must be calculated within its own risk set and stratum. It does not establish the variance or distribution of the sum, which are separate ingredients in standardized inference.
Calculating the Score at an Interim Look
A reliable calculation keeps a separate running total for every stratum. At each event time, record both treatment-arm risk-set sizes and both event counts, calculate the expected A events, subtract that expectation from the observed A events, and then add the contribution to the appropriate stratum total. Only after this step are the stratum scores summed.
Include only events observed by the interim cutoff, and use the stratum categories specified for the trial.
For each included event time, determine \(Y_{Ark}\) and \(Y_{Brk}\) immediately before the event time under the agreed censoring convention.
Use \(d_{Ark}-d_{rk}Y_{Ark}/(Y_{Ark}+Y_{Brk})\), including all tied events at that time in \(d_{rk}\).
Keep the stratum totals visible so that the final sum can be checked and interpreted with the stated sign convention.
Worked Example: Summing Two Strata at an Interim Analysis
Consider a hypothetical randomized trial comparing A with B. At the interim cutoff, the following are the risk sets immediately before each listed event time. Counts at different times are separate snapshots; each risk set already reflects prior events and censoring.
| Stratum | Event time | \(Y_{Ark}\) | \(Y_{Brk}\) | \(d_{Ark}\) | \(d_{rk}\) |
|---|---|---|---|---|---|
| 1 | 1 | 8 | 12 | 1 | 2 |
| 1 | 2 | 6 | 9 | 2 | 3 |
| 2 | 1 | 10 | 10 | 1 | 2 |
| 2 | 2 | 7 | 8 | 0 | 1 |
For stratum 1 at time 1, A’s expected event count is \(2(8/20)=0.8000\), so the contribution is \(1-0.8000=0.2000\). At time 2, the expected count is \(3(6/15)=1.200\), giving contribution \(2-1.200=0.8000\). Thus \(U_1=0.2000+0.8000=1.000\).
For stratum 2 at time 1, the expected count is \(2(10/20)=1.000\), giving contribution \(1-1.000=0\). At time 2, it is \(1(7/15)=0.4667\), giving contribution \(0-0.4667=-0.4667\). Thus \(U_2=-0.4667\).
The stratified score is \(U_{\mathrm{strat}}=U_1+U_2=1.000-0.4667=0.5333\) events. It is positive, so A had 0.5333 more events than expected in the summed within-stratum comparisons. This raw score alone does not say whether the difference is statistically significant.
Worked Example: A Tied Event Time
In a hypothetical stratum, immediately before a tied event time, 4 patients in A and 6 in B are at risk. Three events occur at that time: 2 in A and 1 in B. The expected A count is
The tied-time score contribution is \(2-1.200=0.8000\). The calculation uses all 3 events in the total and the risk-set proportion \(4/10\); it does not treat the tied events as three successive event times with risk sets that shrink after each event. The tied-time convention must be applied consistently, especially when censoring also occurs at that recorded time.
Worked Example: Respecting the Interim Cutoff
Suppose a hypothetical trial has two prespecified strata. By the first interim cutoff, stratum X has one event time with \(Y_A=12\), \(Y_B=8\), \(d_A=1\), and \(d=2\). Stratum Y has one event time with \(Y_A=5\), \(Y_B=5\), \(d_A=0\), and \(d=1\). A further event in stratum X is recorded after the cutoff and is not part of this look.
For X, the expected A count is \(2(12/20)=1.200\), so \(U_X=1-1.200=-0.2000\). For Y, the expected A count is \(1(5/10)=0.5000\), so \(U_Y=0-0.5000=-0.5000\). Therefore the interim stratified score is \(U_{\mathrm{strat}}=-0.2000-0.5000=-0.7000\) events.
The later event must not be added to this interim score, even if it is known when the dataset is eventually reviewed. It belongs to a later accumulation of follow-up, not the data available by this cutoff. The negative score indicates fewer events in A than expected in these within-stratum comparisons, which is favorable to A when events are undesirable.
Diagnostics, Alternatives, and Limitations
Several checks help detect calculation errors before the score is used in an interim analysis:
- Check each risk set. Both arm counts must be nonnegative, and \(Y_{Ark}+Y_{Brk}\) must be positive at an event time. Risk-set counts should agree with the trial’s event and censoring records.
- Check event counts. The total \(d_{rk}\) must equal \(d_{Ark}+d_{Brk}\), and each arm’s events cannot exceed its risk-set size.
- Check expectations. The expected count \(d_{rk}Y_{Ark}/Y_{rk}\) lies between 0 and \(d_{rk}\). If all patients at risk are in A, the expected A count equals all events; if none are in A, it is zero.
- Check the cutoff and stratum labels. Events after the cutoff should not appear, and each patient’s risk set and events must be assigned to the same prespecified stratum.
- Check the sign. A contribution is positive when observed A events exceed expected A events and negative when they fall short.
The main alternative is an unstratified logrank calculation, which uses the combined risk set rather than separate risk sets. That may be appropriate when stratification is not part of the design, but it does not make the same within-stratum comparison. Another possible analysis is to calculate separate stratum-specific tests; that produces multiple results rather than the single summed score defined here. Weighted logrank procedures change the event-time contributions by applying weights and target different patterns of time-varying effects; they are not obtained merely by stratifying this unweighted score.
Stratification is not a guarantee against every problem. Sparse strata can have few events or little treatment representation in risk sets, making their contributions unstable and their later variance calculations less reliable. A poorly chosen or post hoc stratification can also make interpretation and calibration difficult. The ordinary logrank score is most naturally interpreted under proportional hazards, although the score calculation itself can still be performed when hazards are non-proportional; its behavior may then reflect the timing and direction of effects. A positive or negative raw score is not by itself a p-value, a confidence interval, or evidence that the treatment caused an outcome difference.
Check Your Understanding
Use the definition and sign convention above. Each problem is followed by its solution.
- In one stratum at an event time, \(Y_A=9\), \(Y_B=6\), \(d=2\), and \(d_A=1\). Calculate the contribution.
- There are 5 patients at risk in A and 5 in B. Two tied events occur, both in B. Calculate the expected A events and the score contribution.
- Explain why a score computed from the original randomized sample sizes is generally not the required event-time contribution.
- A positive stratified score is reported for an endpoint where events are undesirable. State its direction in context, and name one inferential quantity not supplied by the raw score alone.
Solutions. (1) The expected A count is \(2(9/15)=1.200\), so the contribution is \(1-1.200=-0.2000\) events. (2) The expected A count is \(2(5/10)=1.000\); the contribution is \(0-1.000=-1.000\) event. (3) The expectation uses the arm proportions among patients still at risk immediately before that event time, not the original allocation proportions; events and censoring may have changed the risk sets. (4) A positive score means more events in A than expected from its within-stratum risk-set shares, an unfavorable direction for A when events are undesirable. The raw score alone does not provide a variance, standardized test statistic, p-value, or stopping decision.