Tutorials › Group Sequential Methods and Sample Size Re-Estimation › Reconstructing Cumulative Scores Across Looks

Group Sequential Logrank Tests · Tutorial 505 of 699

Reconstructing Cumulative Scores Across Looks

You will reconstruct the cumulative logrank score and variance at each look from interval increments, calculate the corresponding standardized statistics, and diagnose inconsistencies.

Advanced 10 min read

What You'll Learn

  • Define interval score and variance increments under a consistent logrank score convention.
  • Recover cumulative scores and variances by summing increments through each analysis look.
  • Standardize each cumulative score using its cumulative variance.
  • Check whether interval records and reported look statistics are arithmetically consistent.
  • Recognize assumptions and warning signs that can undermine reconstruction.

Why Reconstruct the Cumulative Quantities?

A group sequential logrank analysis reports a statistic at each planned look, but the underlying evidence accrues between looks. Keeping the interval-specific score and variance increments separate from their cumulative totals makes it possible to verify what was available at each analysis, reproduce the standardized statistic, and locate inconsistencies in an analysis record. A standardized value at one look cannot generally be obtained by adding standardized values from earlier looks: the score and its variance must be accumulated first.

The reconstruction assumes nested looks: each later look includes all event information available at earlier looks, plus information from the intervening interval. Each event contribution must be assigned once, and the score convention and variance calculation must be consistent across intervals. The earlier material on independent logrank increments supplies the conditions for treating interval increments as independent or, in the usual large-sample formulation, approximately so. Here the emphasis is on reconstructing and checking the quantities at the looks.

Definition: Let \(\Delta U_j\) be the logrank score increment and \(\Delta V_j\) its estimated null variance increment for interval \(j\), the period between look \(j-1\) and look \(j\). Define the cumulative score and variance at look \(k\) by \(U_k=\sum_{j=1}^{k}\Delta U_j\) and \(V_k=\sum_{j=1}^{k}\Delta V_j\). If \(V_k>0\), the standardized logrank statistic at that look is \(Z_k=U_k/\sqrt{V_k}\).

For a clear sign convention, take each logrank score contribution to be the observed number of treatment-arm events minus the number expected under the null, conditional on the risk sets and event totals. Thus a positive score indicates more treatment-arm events than expected. When events are undesirable, that direction generally represents harm rather than benefit; conclusions must follow the convention and endpoint in the analysis plan. Score increments are dimensionless contrasts in event counts, and variance increments are in squared score units. If the analysis is stratified, each interval’s score and variance increments are the corresponding sums across strata.

Reconstructing and Standardizing at Each Look

The reconstruction is a cumulative sum, not an average. At look \(k\), add every interval increment through \(k\), then divide the cumulative score by the square root of the cumulative variance. Because a variance increment cannot be negative, the cumulative variance should not decrease. A zero variance increment is possible when an interval contributes no score information; a standardized statistic is undefined if the cumulative variance is zero.

$$ U_k=U_{k-1}+\Delta U_k,\qquad V_k=V_{k-1}+\Delta V_k,\qquad Z_k=\frac{U_k}{\sqrt{V_k}}, \qquad U_0=V_0=0. $$

The sums also explain why the method is appropriate under the stated increment assumptions. If the interval scores have mean zero under \(H_0\), are independent (or satisfy the relevant asymptotic independent-increment conditions), and have variances \(\Delta V_j\), then the variance of their sum is the sum of their variances:

$$ \operatorname{Var}(U_k\mid H_0) =\operatorname{Var}\left(\sum_{j=1}^{k}\Delta U_j\right) =\sum_{j=1}^{k}\operatorname{Var}(\Delta U_j) =\sum_{j=1}^{k}\Delta V_j =V_k. $$

The middle equality follows because independent increments have zero covariance, so all cross-covariance terms in the variance of the sum are zero. Under the usual large-sample normal approximation, \(U_k/\sqrt{V_k}\) is therefore the standardized score at look \(k\). The approximation is not a claim that every finite-sample logrank statistic is exactly normal. It relies on adequate information and the applicable event-process assumptions.

Worked Example: Calculating the Statistic at Three Looks

Suppose a hypothetical trial has the following interval increments. The values are illustrative, not data from a real study. The score uses observed minus expected treatment-arm events.

Interval ending at look\(\Delta U_j\)\(\Delta V_j\)
1-4.225
2-5.130
31.820

At look 1, \(U_1=-4.2\) and \(V_1=25\), so \(Z_1=-4.2/\sqrt{25}=-4.2/5=-0.840\).

At look 2, add the second interval’s increments: \(U_2=-4.2+(-5.1)=-9.3\), and \(V_2=25+30=55\). Therefore, \(Z_2=-9.3/\sqrt{55}=-9.3/7.416=-1.254\).

At look 3, \(U_3=-9.3+1.8=-7.5\) and \(V_3=55+20=75\). Thus \(Z_3=-7.5/\sqrt{75}=-7.5/8.660=-0.866\). The score moves toward zero during the third interval, while the variance continues to accumulate. Under the stated sign convention, the negative scores indicate fewer treatment-arm events than expected; the statistic alone does not establish a boundary decision or a causal interpretation.

Recovering Increments from Cumulative Records

Sometimes a record gives cumulative values at each look rather than the interval increments. In that case, recover an interval increment by subtracting the preceding cumulative value. The initial cumulative values are \(U_0=0\) and \(V_0=0\). This subtraction works only when the reported cumulative values refer to the same score definition, strata, event inclusion rules, and analysis population at successive looks.

$$ \Delta U_k=U_k-U_{k-1},\qquad \Delta V_k=V_k-V_{k-1}. $$

Worked Example: Recovering the Missing Interval Contributions

A hypothetical analysis archive records cumulative scores and variances at three looks:

LookCumulative score \(U_k\)Cumulative variance \(V_k\)
12.09
2-1.025
3-2.050

For interval 1, the increments are \(\Delta U_1=2.0-0=2.0\) and \(\Delta V_1=9-0=9\). For interval 2, they are \(\Delta U_2=-1.0-2.0=-3.0\) and \(\Delta V_2=25-9=16\). For interval 3, they are \(\Delta U_3=-2.0-(-1.0)=-1.0\) and \(\Delta V_3=50-25=25\). Summing these recovered increments reproduces the recorded cumulative values: the scores are \(2.0\), \(-1.0\), and \(-2.0\); the variances are \(9\), \(25\), and \(50\).

The standardized statistics are \(Z_1=2.0/\sqrt{9}=0.667\), \(Z_2=-1.0/\sqrt{25}=-0.200\), and \(Z_3=-2.0/\sqrt{50}=-0.283\). The change from positive to negative score is not an arithmetic contradiction: the second interval contributes a negative score large enough to reverse the cumulative direction. The cumulative statistic must be recalculated at each look; a change in direction is not found by simply adding the earlier \(Z\) values to a new interval statistic.

Checks, Assumptions, and Failure Modes

A reconstruction is meaningful only if each interval’s quantities belong to the same sequential analysis. In particular, the looks must be nested, interval boundaries must not omit or double-count events, and the score’s treatment-arm convention must remain unchanged. The variance increments must be computed consistently with the logrank score, including the trial’s handling of strata and tied event times. The risk sets and event contributions used to produce the increments must also be valid for the specified analysis cutoffs.

Under the independent-increment conditions, the cumulative variance is the sum of the interval variances. More generally, the exact identity for the variance of a sum includes covariance terms; simply adding interval variances is justified only when those cross-covariances are zero or negligible under the adopted approximation. If increments are correlated in a way the analysis has not accounted for, the sum of their marginal variance estimates need not be the variance of the cumulative score. Reconstructing the arithmetic cannot repair a failure of the underlying assumptions.

A practical audit should check the following before interpreting the resulting \(Z_k\) values:

  • Re-add the score increments. Each cumulative score should equal the sum through that look, with the same sign convention throughout.
  • Check variance monotonicity. Each \(\Delta V_j\) should be nonnegative, and \(V_k\) should not decrease. A decrease often signals a transcription error or a change in what is being reported.
  • Recalculate each standardization. Compare the reported statistic with \(U_k/\sqrt{V_k}\), using the cumulative values, not the latest interval alone.
  • Check the analysis record. Confirm that the intervals partition the events used in the looks and that the reported score and variance use a consistent method.
  • Distinguish rounding from a substantive mismatch. Small differences may arise from rounding displayed increments. Reconcile with unrounded values when available; do not silently alter an increment to force agreement.

Worked Example: Finding a Variance-Record Discrepancy

Suppose a hypothetical interval ledger gives score increments \(-1.5\), \(-2.5\), and \(1.0\), with variance increments \(9\), \(16\), and \(16\). A separate report lists cumulative values through look 2 as \(U_2=-4.0\), \(V_2=25\), and through look 3 as \(U_3=-3.0\), \(V_3=40\).

The scores reconcile: \(-1.5-2.5=-4.0\), then \(-4.0+1.0=-3.0\). The variance through look 2 also reconciles: \(9+16=25\). But through look 3, the ledger gives \(9+16+16=41\), not 40. Equivalently, subtracting the reported cumulative variances gives a third increment of \(40-25=15\), which conflicts with the ledger’s increment of 16.

Using the ledger, the look 3 statistic is \(Z_3=-3.0/\sqrt{41}=-3.0/6.403=-0.4685\). Using the separately reported cumulative variance gives \(-3.0/\sqrt{40}=-0.4743\). These are not the same calculation. The discrepancy should be traced to the source records and resolved before reporting a verified statistic; neither value should be selected merely because it seems more plausible.

What This Reconstruction Does—and Does Not—Establish

Reconstructing the cumulative quantities is a bookkeeping and verification step, not a test of efficacy by itself. The value \(Z_k\) summarizes the cumulative score relative to its estimated standard deviation at look \(k\). Whether it crosses a prespecified efficacy or futility boundary, and how repeated looks affect error rates, require the monitoring plan and its relevant decision rules. A statistic’s magnitude alone does not supply those rules.

This reconstruction is also not a substitute for recomputing event contributions from the underlying risk sets when those contributions are in doubt. Summing supplied increments verifies their arithmetic consistency, but cannot establish that the increments were calculated from the correct participants, events, strata, or cutoff-specific risk sets. The alternative of computing each look directly from all event-level records can help investigate a mismatch, but it is a distinct calculation that must use the same analysis definitions. When both approaches are valid, cumulative summation and a direct calculation should agree apart from rounding.

Key takeaway: At every look, add the interval score increments to obtain \(U_k\), add the interval variance increments to obtain \(V_k\), and then calculate \(Z_k=U_k/\sqrt{V_k}\). Verify the partition, sign convention, variance accumulation, and standardization before interpreting a sequential statistic.

Check Your Understanding

Answer each question, then check the concise solutions below.

  1. Two hypothetical intervals have \(\Delta U_1=-3\), \(\Delta V_1=16\), \(\Delta U_2=1\), and \(\Delta V_2=9\). Find \(U_2\), \(V_2\), and \(Z_2\).
  2. Cumulative variances at two successive looks are 18 and 17. What should be checked before calculating a standardized statistic at the second look?
  3. Why is \(Z_2\) generally not the sum of \(Z_1\) and a standardized statistic calculated from interval 2 alone?
  4. If the interval score increments are independent under \(H_0\) and have variances equal to their listed \(\Delta V_j\), why does the cumulative variance equal their sum?

Solutions. (1) \(U_2=-3+1=-2\), \(V_2=16+9=25\), and \(Z_2=-2/\sqrt{25}=-0.400\). (2) Since variance should not decrease under this cumulative reconstruction, check for a transcription error, inconsistent definitions, or a change in reported analysis quantities; do not proceed as though 17 were verified. (3) Standardization divides by the square root of variance, so separately standardized increments have different denominators from the cumulative statistic. Add scores and variances first, then standardize. (4) Independence makes the cross-covariances zero, so the variance of the sum is the sum of the interval variances.

Clinical Biostats