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Group Sequential Logrank Tests · Tutorial 484 of 699

Martingale Derivation Of Sequential Logrank Increments

Use counting-process martingales to connect risk-set logrank contributions with mean-zero sequential increments and accumulated variance.

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What You'll Learn

  • State the null hazard and counting-process assumptions behind the martingale representation.
  • Derive the logrank score increment as an integral with respect to martingales.
  • Relate the predictable variance to the treatment and control sizes of each risk set.
  • Calculate sequential scores and variance estimates from hypothetical event data.
  • Explain why increments are uncorrelated across looks under the null, but need not be independent.
  • Recognize diagnostics and assumptions that can undermine the martingale argument.

Why derive sequential increments?

At an interim analysis, the logrank score is cumulative: it collects observed-minus-expected event contributions over the risk sets encountered so far. For sequential use, it matters not only how that total is calculated, but also why the contribution accumulated after one look has mean zero under the null, conditional on the history available at that look. A counting-process martingale makes this property explicit and gives a principled account of how variance builds over time.

The derivation below concerns a two-arm trial with a single event type and no tied event times. It explains the score and its variance process; it does not provide a method for handling ties. Risk-set allocation and the ordinary logrank variance can also be described directly through event-level comparisons. The martingale representation shows how those comparisons fit into a process evolving over calendar or follow-up time.

Model, notation, and assumptions

Index participants by \(i=1,\ldots,n\). Let \(Z_i=1\) for a participant assigned to treatment and \(Z_i=0\) for a participant assigned to control. Let \(N_i(t)\) count that participant’s events by time \(t\), and let \(Y_i(t)\) indicate whether the participant is at risk just before \(t\). Write \(Y_i(t-)\) when the left limit is needed to emphasize that the risk status is known immediately before the event time. Define the treatment and control risk-set sizes by \(Y_1(t-)=\sum_iY_i(t-)Z_i\) and \(Y_0(t-)=\sum_iY_i(t-)(1-Z_i)\), with \(Y(t-)=Y_1(t-)+Y_0(t-)\).

Let \(\mathcal{F}_t\) denote the observed history through time \(t\), including who remains at risk and which events have occurred. Under the null hypothesis \(H_0\), assume the two arms have the same event hazard \(\lambda_0(t)\) among participants still at risk. More precisely, conditional on the observed history, participant \(i\)’s event-counting process has intensity \(Y_i(t-)\lambda_0(t)\). Assume the intensity is well defined and integrable over the follow-up interval, and that participants’ event processes have no simultaneous jumps. Censoring must be non-informative for this event-intensity model, so that the stated intensity remains appropriate given the observed history. These assumptions are used to construct the martingales and their variance; violations require a different model or analysis.

Definition: Under \(H_0\), the compensated counting process for participant \(i\) is $$ M_i(t)=N_i(t)-\int_0^t Y_i(u-)\lambda_0(u)\,du. $$ It is a martingale with respect to \(\mathcal{F}_t\): for \(s<t\), \(E\{M_i(t)-M_i(s)\mid\mathcal{F}_s\}=0\). In words, after accounting for the event intensity and risk time, the future compensated event contribution has conditional mean zero.

From the logrank score to a martingale integral

At each event time, the logrank contribution is the observed treatment event indicator minus the treatment fraction in that event’s risk set. In continuous time, define the predictable treatment fraction just before \(t\) as \(\bar Z(t-)=Y_1(t-)/Y(t-)\), whenever \(Y(t-)>0\). The cumulative score through time \(t\) is

$$ U(t)=\sum_{i=1}^n\int_0^t\{Z_i-\bar Z(u-)\}\,dN_i(u). $$

This formula is the cumulative sum of the risk-set contributions: an event in treatment contributes \(1-\bar Z\), and an event in control contributes \(-\bar Z\). To see why it is a martingale under \(H_0\), replace each \(dN_i\) by its martingale part \(dM_i\) plus its compensator \(Y_i(u-)\lambda_0(u)\,du\). The compensator contribution to the score is zero at every time:

$$ \begin{aligned} \sum_i\{Z_i-\bar Z(u-)\}Y_i(u-)\lambda_0(u) &=\lambda_0(u)\left\{Y_1(u-)-\bar Z(u-)Y(u-)\right\}\\ &=0, \end{aligned} $$

because \(\bar Z(u-)=Y_1(u-)/Y(u-)\). Thus

$$ U(t)=\sum_{i=1}^n\int_0^t\{Z_i-\bar Z(u-)\}\,dM_i(u). $$

The integrands depend only on the risk set immediately before time \(u\), so they are predictable. A sum of these predictable integrals against martingales is itself a martingale, provided the integrals are well defined and have finite variance over the time interval. Consequently, for an interim time \(t_k\) after a previous look at \(t_{k-1}\),

$$ \Delta U_k=U(t_k)-U(t_{k-1}),\qquad E(\Delta U_k\mid\mathcal{F}_{t_{k-1}})=0. $$

This is the key sequential statement: under \(H_0\), the next score increment has conditional mean zero given the information at the previous look. It does not say that a particular observed increment must be near zero.

How the variance accumulates

For counting-process martingales with no simultaneous events, the different participants’ event martingales have zero cross-variation. The predictable quadratic variation of \(U(t)\), denoted \(\langle U\rangle(t)\), is therefore the sum of the integrated squared weights times the event intensities:

$$ \langle U\rangle(t) =\int_0^t\lambda_0(u)\sum_iY_i(u-)\{Z_i-\bar Z(u-)\}^2\,du. $$

The sum inside the integral simplifies to a quantity determined by the two risk-set sizes. Since \(Z_i\) is binary, the treatment participants contribute \(Y_1(1-\bar Z)^2\) and the control participants contribute \(Y_0\bar Z^2\). Substituting \(\bar Z=Y_1/Y\) gives

$$ \begin{aligned} Y_1(1-\bar Z)^2+Y_0\bar Z^2 &=Y_1\left(\frac{Y_0}{Y}\right)^2+ Y_0\left(\frac{Y_1}{Y}\right)^2\\ &=\frac{Y_1Y_0}{Y}. \end{aligned} $$

Hence the predictable variance process is

$$ \langle U\rangle(t) =\int_0^t\lambda_0(u)\frac{Y_1(u-)Y_0(u-)}{Y(u-)}\,du. $$

This process increases as event intensity accumulates in risk sets containing participants in both arms. It contributes no variance when one arm has no one at risk, because then \(Y_1Y_0/Y=0\). The variance is predictable: at each instant, its rate depends on the risk set just before that instant and the null event intensity.

The usual event-based logrank variance is the corresponding observable risk-set calculation. In the no-ties setting, estimating the pooled cumulative hazard increment in a risk set by \(dN(u)/Y(u-)\), where \(dN(u)=\sum_i dN_i(u)\), yields the accumulated estimate

$$ V(t)=\sum_{u\le t:\,\text{event}} \frac{Y_1(u-)Y_0(u-)}{Y(u-)^2}. $$

Thus each event adds the treatment risk-set fraction times the control risk-set fraction. This event-based \(V(t)\) estimates the predictable variance; it is not the same object as the continuously accumulated bracket \(\langle U\rangle(t)\). Under the martingale assumptions, \(E\{U(t)^2\}=E\{\langle U\rangle(t)\}\) when the score starts at zero and has finite second moment.

Worked Example: One Event’s Mean and Variance

Suppose a hypothetical risk set just before an event contains 6 treatment participants and 9 control participants. Assume the common-hazard null model and no tied event at this time. Conditional on an event occurring in this risk set, the probability that its participant is from treatment is \(6/15=0.4\), and the probability of a control event is \(9/15=0.6\).

If the event is in treatment, its logrank contribution is \(1-0.4=0.6\). If it is in control, its contribution is \(-0.4\). The conditional expected contribution is

$$ 0.4(0.6)+0.6(-0.4)=0.24-0.24=0. $$

The conditional variance is the expected squared contribution, since the mean is zero:

$$ 0.4(0.6^2)+0.6\{(-0.4)^2\} =0.4(0.36)+0.6(0.16)=0.24. $$

The risk-set variance contribution is also \(Y_1Y_0/Y^2=(6)(9)/(15^2)=54/225=0.24\). This agreement is the event-level counterpart of the martingale variance calculation. It relies on equal hazards under \(H_0\); unequal hazards would change the event-arm probabilities.

Worked sequential score and variance

For successive looks, each interval contributes the score and variance accumulated from the event risk sets in that interval. The score increments are martingale differences under \(H_0\), while the variance increments are nonnegative and add across intervals. The following event sequence illustrates the calculations. Assume no censoring between the listed events, no ties, and a common hazard under the null.

Worked Example: Two Looks With Four Events

In a hypothetical trial, the risk sets immediately before four successive events are as follows. An event in treatment contributes \(Y_0/Y\); an event in control contributes \(-Y_1/Y\). Each event adds \(Y_1Y_0/Y^2\) to the estimated variance.

Event\(Y_1\)\(Y_0\)Event armScore contributionVariance contribution
146Control\(-4/10=-0.4000\)\(24/100=0.2400\)
245Treatment\(5/9=0.5556\)\(20/81=0.2469\)
335Control\(-3/8=-0.3750\)\(15/64=0.2344\)
434Treatment\(4/7=0.5714\)\(12/49=0.2449\)

Let the first look occur after event 2 and the second after event 4. The first cumulative score and variance estimate are

$$ U(t_1)=-0.4000+0.5556=0.1556,\qquad V(t_1)=0.2400+0.2469=0.4869. $$

The next interval’s increments are

$$ \Delta U_2=-0.3750+0.5714=0.1964,\qquad \Delta V_2=0.2344+0.2449=0.4793. $$

By the second look, the accumulated values are \(U(t_2)=0.1556+0.1964=0.3520\) and \(V(t_2)=0.4869+0.4793=0.9662\), subject to rounding. The positive observed score means the treatment arm had more events than expected under the pooled risk-set allocation over these events. It is a small, hypothetical sequence and does not by itself establish a treatment effect. The martingale result concerns the conditional distribution under \(H_0\), not the interpretation of this particular realized score as a causal effect.

What the martingale property implies across looks

Martingale increments from non-overlapping intervals are uncorrelated under the stated assumptions. For two successive increments, with \(t_0<t_1<t_2\), the justification is

$$ \begin{aligned} E\{\Delta U_1\Delta U_2\} &=E\left[\Delta U_1\,E(\Delta U_2\mid\mathcal{F}_{t_1})\right]\\ &=E(\Delta U_1\cdot 0)=0. \end{aligned} $$

Here \(\Delta U_1\) is known by \(t_1\), and the conditional mean of the later increment is zero. This does not imply that the increments are independent: their distributions can depend on the risk sets and event intensities created by the trial’s evolving history. Nor does it imply that standardized statistics at different looks are independent; the cumulative score at a later look contains the earlier score.

A useful additional consequence is that the cumulative score has mean zero under \(H_0\). Its variance is determined by the expected predictable quadratic variation, and its variance process accumulates over time rather than resetting at each look. These properties motivate sequential analyses that retain the increments or their cumulative score and account for their shared history. A normal approximation may be useful when the information is sufficiently spread across many events, but the martingale mean-zero statement alone does not guarantee that an increment has a normal distribution.

Worked Example: Checking Increment Covariance

Suppose a hypothetical two-look design has score increments \(\Delta U_1\) and \(\Delta U_2\), with the second increment observed after the first look. Under the null model and the assumptions above, the defining result is \(E(\Delta U_2\mid\mathcal{F}_{t_1})=0\). Therefore, even if the first increment happens to equal \(0.30\), the conditional expected product with the second increment is

$$ E\{\Delta U_1\Delta U_2\mid\mathcal{F}_{t_1}\} =\Delta U_1E(\Delta U_2\mid\mathcal{F}_{t_1}) =0.30(0)=0. $$

Averaging over possible first-look histories gives \(E(\Delta U_1\Delta U_2)=0\). This establishes zero covariance when the increments have finite second moments and mean zero. It does not assert that every later increment is zero, or that the two increments are independent.

Diagnostics, alternatives, and limitations

The derivation depends on the specified event intensity and risk-set history, so diagnostics should focus on whether those are credible and whether the calculations reflect the observed trial structure. Check that each event is assigned to the correct arm, that each risk set uses participants at risk immediately before the event, and that the score and variance contributions use the same risk set. Risk sets with only one arm present have zero variance contribution and cannot provide a two-arm comparison at that event.

Warning signs include unexpected simultaneous events when a no-ties derivation is being applied, missing or misclassified events, risk-set counts that do not reconcile with enrollment and censoring, or evidence that the common-hazard null intensity is an inadequate working model. Informative censoring can also invalidate the assumed intensity conditional on the observed history. These issues should be investigated rather than addressed by mechanically applying the event-level variance formula. The treatment of tied events requires a tie-aware calculation and is not derived here.

The counting-process route is especially useful for understanding conditional mean zero, variance accumulation, and why later increments are orthogonal to earlier ones. The risk-set allocation route provides a direct event-by-event calculation of observed-minus-expected contributions and their variance; it is often the more convenient computational description. Both connect to the same logrank score under the stated setting, but neither removes the need to check the risk sets and assumptions. The martingale derivation also has limits: it does not itself establish a normal approximation, calibrate a stopping boundary, or justify a particular censoring model.

  • Mistake: “The null makes every observed increment equal zero.” Correction: The null makes the increment’s conditional expectation zero; a realized increment can be positive or negative.
  • Mistake: “Zero covariance proves that increments are independent.” Correction: The martingale argument gives zero covariance for non-overlapping increments with finite second moments, not independence.
  • Mistake: “The predictable variance and the event-based variance estimate are identical.” Correction: The predictable quadratic variation integrates the null intensity; the usual event-based risk-set sum estimates it.
Key takeaway: Under a common event hazard and the stated counting-process assumptions, the logrank score is a martingale. Its successive increments have conditional mean zero, and its predictable variance accumulates at a rate proportional to \(Y_1Y_0/Y\), the risk-set allocation term.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

  1. State the null intensity for one participant and define the corresponding compensated counting process.
  2. A risk set contains 8 treatment and 12 control participants. What are the score contributions for a treatment event and a control event? Show that the conditional mean is zero under the common-hazard null.
  3. For the same risk set, calculate the event-based variance contribution.
  4. Why does a martingale increment after an interim look have conditional mean zero under \(H_0\)?
  5. Does zero covariance of successive score increments imply independence? Explain.

Solutions. (1) The intensity is \(Y_i(t-)\lambda_0(t)\), and \(M_i(t)=N_i(t)-\int_0^tY_i(u-)\lambda_0(u)\,du\) is the compensated martingale. (2) Here \(\bar Z=8/20=0.4\). A treatment event contributes \(1-0.4=0.6\), and a control event contributes \(-0.4\). Their null probabilities are 0.4 and 0.6, so the mean is \(0.4(0.6)+0.6(-0.4)=0\). (3) The variance contribution is \(Y_1Y_0/Y^2=(8)(12)/(20^2)=96/400=0.2400\). (4) The martingale property gives \(E\{U(t_k)-U(t_{k-1})\mid\mathcal{F}_{t_{k-1}}\}=0\) under the assumed null intensity. (5) No. The conditional-mean argument establishes zero covariance for finite-variance increments; it does not establish independence.

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