Worked Stratified Sequential Logrank Analysis

Why Stratification Changes the Calculation

A survival trial may be stratified at randomization by factors such as disease stage or study center. The strata matter to a logrank analysis because each stratum has its own event history and risk sets. The stratified test combines the evidence from those strata; it does not first average their standardized statistics. At sequential looks, it also combines the new evidence with the information accumulated at earlier looks.

This tutorial uses a score oriented so that positive values favor treatment: at an event time, the score contribution is the number of treatment events expected under the null minus the number observed. Thus, fewer treatment events than expected produce a positive contribution. A negative statistic points in the harm direction under this convention. The convention must be fixed before applying efficacy boundaries.

Definition: For stratum $h$, let $U_h(t)$ be the cumulative benefit-oriented logrank score through look $t$, and let $V_h(t)$ be its estimated variance, called the stratum information. The stratified cumulative score and information are $$ U(t)=\sum_{h=1}^{H} U_h(t), \qquad V(t)=\sum_{h=1}^{H} V_h(t). $$ Provided $V(t)>0$, the standardized stratified statistic is $$ Z(t)=\frac{U(t)}{\sqrt{V(t)}}. $$

For a simple event-time calculation with no tied failures, suppose that immediately before an event there are $Y_{1hj}$ treatment participants and $Y_{0hj}$ control participants at risk in stratum $h$. If the event is a treatment event, then $O_{1hj}=1$; if it is a control event, $O_{1hj}=0$. Under the null of equal event hazards between groups within the stratum, the expected number of treatment events at this event time is $E_{1hj}=Y_{1hj}/(Y_{1hj}+Y_{0hj})$. The score contribution is $E_{1hj}-O_{1hj}$, and the corresponding variance contribution is $Y_{1hj}Y_{0hj}/(Y_{1hj}+Y_{0hj})^2$. Summing these contributions over event times gives $U_h$ and $V_h$. With tied failures, a standard tie-adjusted logrank calculation is needed instead.

Assumptions for Sequential Stratified Analysis

The trial’s strata should be defined in advance, and randomization should be carried out within them. Participants contribute to the risk set for their assigned stratum, not to a pooled risk set across strata. The usual logrank null calculation assumes that, under $H_0$, the treatment and control event hazards are equal within each stratum. Censoring must be non-informative for the event process conditional on the variables used in the analysis; in particular, censoring should not selectively remove participants in a way that invalidates the risk-set comparisons.

For a sequential analysis, each look must use a prespecified data cutoff and the score and information must be cumulative through that cutoff. Under the null and regularity conditions for the logrank process, increments across nonoverlapping event intervals are approximately independent with mean zero and variance equal to their information increments. The earlier material on independent logrank increments and canonical joint distributions gives the basis for treating the standardized statistics at different looks as a correlated sequence. That approximation can be poor with very sparse events or unusual monitoring patterns, so the design’s calibration and realized information should be checked.

Proportional hazards are not required merely to define the ordinary logrank statistic or to control its null rejection probability under its null assumptions. They are often used when interpreting its power and when translating a treatment effect into a single hazard ratio. Therefore, a significant result is not by itself proof that the hazards have a constant ratio over time. The analysis here concerns the stratified sequential statistic and its boundary decisions, not an assessment of the shape of the treatment effect over time.

Combining Strata and Looks

Stratum scores are added because each is a signed contribution to the same treatment comparison. Their variances are added when the stratum contributions are independent under the design, as they ordinarily are for disjoint participants in distinct strata. The resulting standardization uses the square root of the total information. This gives more influence to strata contributing more information, as a variance-based combination should.

Z(t)=\frac{\sum_h U_h(t)}{\sqrt{\sum_h V_h(t)}}.

At a later look, calculate the score and variance contributions from the newly accrued event interval in each stratum, then add those increments to the stratum’s cumulative values. Equivalently, first sum increments across strata and then add them to the previous combined cumulative score and information. Do not add standardized statistics across looks: the denominator changes as information accumulates.

Justification: If the stratum scores have mean zero under $H_0$, variances $V_h$, and zero covariance across strata, then $$ \operatorname{Var}\left(\sum_h U_h\right)=\sum_h\operatorname{Var}(U_h)=\sum_h V_h. $$ Standardizing the combined score therefore gives $\sum_h U_h/\sqrt{\sum_h V_h}$. In particular, writing $Z_h=U_h/\sqrt{V_h}$ gives $$ Z=\frac{\sum_h \sqrt{V_h}Z_h}{\sqrt{\sum_h V_h}}, $$ not the unweighted average of the $Z_h$'s. Across looks, the same score-and-variance addition applies to increments; cumulative statistics are then compared with the boundaries specified for their respective looks.

Worked Sequential Analysis

Worked Example: Two Strata and Two Event-Driven Looks

Hypothetical trial. A randomized survival trial has two prespecified strata, A and B, and monitors a one-sided efficacy statistic at two event-driven looks. Assume the risk sets were updated within each stratum, there were no tied failures (or the displayed summaries used the appropriate tie adjustment), and censoring was non-informative conditional on treatment and stratum. The table gives the score and information increments for events since the previous look. Positive score means fewer treatment events than expected under $H_0$.

Look	Stratum	Score increment	Information increment
1	A	8	40
1	B	4	20
2	A	7	35
2	B	3	25

Look 1. The combined score and information are $$ U(1)=8+4=12,\qquad V(1)=40+20=60. $$ Therefore, $$ Z(1)=\frac{12}{\sqrt{60}}=\frac{12}{7.746}=1.549. $$ The stratum-specific calculations are $8/\sqrt{40}=1.265$ for A and $4/\sqrt{20}=0.8944$ for B. Their average is not the stratified statistic; the combined statistic is obtained from the total score and total information above.

Look 2. The new increments sum to $7+3=10$ score units and $35+25=60$ information units. Adding these to the first-look totals gives $$ U(2)=12+10=22,\qquad V(2)=60+60=120, \qquad Z(2)=\frac{22}{\sqrt{120}}=\frac{22}{10.954}=2.008. $$ For a check, the cumulative stratum values are $U_A=15,V_A=75$ and $U_B=7,V_B=45$. They give the same combined totals: $15+7=22$ and $75+45=120$.

Suppose, for this illustration, the design’s prespecified positive efficacy boundaries are 2.50 at look 1 and 2.00 at look 2. At look 1, $1.549<2.50$, so the efficacy boundary is not crossed. At look 2, $2.008>2.00$, so the positive boundary is crossed and the trial stops for efficacy under this rule. These illustrative cutoffs are assumed to have been chosen for the design; the arithmetic alone does not establish their type I error calibration.

Why an Unweighted Average Is Not the Stratified Test

Worked Example: Unequal Stratum Information

Consider a hypothetical look with two independent stratum contributions. Stratum A has $U_A=8.55$ and $V_A=50$; stratum B has $U_B=2.6955$ and $V_B=10$. The corresponding displayed stratum statistics are $$ Z_A=\frac{8.55}{\sqrt{50}}=1.209,\qquad Z_B=\frac{2.6955}{\sqrt{10}}\approx 0.8521. $$ The unweighted average of these displayed values is $$ \frac{1.209+0.8521}{2}=1.03055\approx 1.031. $$ This is not the stratified statistic, and the arithmetic average itself rounds to 1.031, not 1.030.

The correct calculation combines the scores and information: $$ U=8.55+2.6955=11.2455,\qquad V=50+10=60, \qquad Z=\frac{11.2455}{\sqrt{60}}=1.452. $$ Stratum A contributes substantially more information than B, so its evidence receives more weight in the variance-based combination. Averaging the two $Z$ values equally would discard that difference and would not have the variance of the combined score.

Checking an Increment from Risk Sets

Worked Example: Updating the Combined Statistic

Suppose that, just before a new event interval in a hypothetical trial, the cumulative stratified values are $U=3.0$ and $V=15.0$. During the interval, one event occurs in each of two strata, with no ties within either event time.

In stratum A, $Y_{1A}=30$ and $Y_{0A}=20$, and the event is in treatment. Thus $E_{1A}=30/50=0.6$, $O_{1A}=1$, and the score increment is $0.6-1=-0.4$. Its variance increment is $$ \frac{30(20)}{(30+20)^2}=\frac{600}{2500}=0.24. $$ In stratum B, $Y_{1B}=10$ and $Y_{0B}=40$, and the event is in control. Then $E_{1B}=10/50=0.2$, $O_{1B}=0$, so the score increment is $0.2-0=0.2$; the variance increment is $$ \frac{10(40)}{(10+40)^2}=\frac{400}{2500}=0.16. $$

Across the strata, the new score increment is $-0.4+0.2=-0.2$, and the new information is $0.24+0.16=0.40$. The updated totals are $U=3.0-0.2=2.8$ and $V=15.0+0.40=15.40$, so $$ Z=\frac{2.8}{\sqrt{15.40}}\approx 0.7135. $$ The negative increment reduces the positive evidence but does not make the cumulative statistic negative. One event interval need not point in the same direction as the overall accumulated evidence.

Diagnostics, Alternatives, and Limitations

Before acting on a boundary comparison, verify that the score direction, data cutoff, strata, and information calculation match the prespecified analysis. Useful checks include confirming that the cumulative score at a later look equals the previous score plus all new stratum increments, and that cumulative information is nondecreasing. A nonpositive information total, a missing stratum contribution, or a sudden unexplained change in the score’s sign convention indicates a calculation or data-handling problem.

Pooling risk sets across strata: this replaces within-stratum expected event counts with pooled expectations and can change the test. Use each stratum’s own risk sets before summing scores.
Averaging stratum $Z$ statistics: this ignores differences in information. Sum scores and variances, then standardize once.
Using event count as information: the number of events is not generally the logrank variance. Information depends on the treatment-control composition of risk sets at event times.
Mixing cumulative and incremental values: adding a cumulative stratum score to a combined cumulative score counts earlier events twice. Keep a clear record of whether every quantity is an increment or a cumulative total.
Ignoring sparse strata or ties: very small risk sets can make asymptotic approximations unstable; tied-event data require a consistent tie-handling method. Check the analysis specification rather than using the no-tie increment formula indiscriminately.

An unstratified logrank analysis is simpler, but it does not make the same within-stratum comparisons and may be less appropriate when the stratification factors are important to the design. Separate stratum tests preserve the within-stratum comparisons but do not, by themselves, supply one prespecified combined sequential statistic. The stratified logrank test provides that combined statistic by adding scores and information. Its limitations remain: it does not estimate a separate treatment effect for every stratum, does not establish that effects are identical across strata, and can have limited precision when event counts are sparse. Sequential validity also depends on using the planned monitoring rule and its calibrated boundaries, not merely on computing the right $Z$ values.

Key takeaway: At every look, calculate each stratum’s score and information using its own risk sets, add the stratum contributions, and standardize the total score by the square root of total information. Compare that single combined statistic with the boundary for the current look.

Check Your Understanding

Try the calculations first, then compare them with the solutions below.

At one look, strata A and B have scores 6 and 2 and information 16 and 9. Calculate the combined $Z$ statistic.
Why is the arithmetic mean of two stratum-specific $Z$ statistics generally not the stratified logrank statistic?
A stratum has 20 treatment and 30 control participants at risk immediately before one untied event, which occurs in control. Calculate its benefit-oriented score and variance increment.
At a later look, the combined cumulative score is 14 and information is 49. The positive boundary is 2.10. Is the boundary crossed?

Solutions. (1) $U=6+2=8$, $V=16+9=25$, so $Z=8/\sqrt{25}=1.600$. (2) An unweighted mean ignores the different stratum variances; the correct statistic is $\sum_h U_h/\sqrt{\sum_h V_h}$, or equivalently a variance-weighted combination of the stratum statistics. (3) $E_1=20/50=0.4$ and $O_1=0$, so the score is $0.4-0=0.4$. The variance increment is $20(30)/50^2=600/2500=0.240$. (4) $Z=14/\sqrt{49}=14/7=2.000$, which is below 2.10, so the boundary is not crossed.