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Group Sequential Logrank Tests · Tutorial 491 of 699

Event-Level Information In A Logrank Trial

Learn to calculate event-level logrank information from the risk set just before each failure, and to interpret its limits and assumptions.

Advanced 11 min read

What You'll Learn

  • Define logrank information as the accumulated conditional variance of score increments under the null.
  • Calculate an untied event’s information from the two group-specific risk-set sizes.
  • Distinguish an event’s direct contribution from the way censoring changes later risk sets.
  • Apply a specified hypergeometric variance for tied events and compare it with an independence approximation.
  • Check event-level calculations for boundary cases, assumptions, and common errors.

Why an Event Does Not Always Add the Same Information

In a time-to-event trial, an event count alone does not tell us how much information the logrank score has accumulated. Each failure contributes according to the treatment-group composition of the risk set just before it occurs. An event with many participants at risk in both groups can be more informative about a treatment difference than an event occurring when one group is barely represented.

This tutorial defines information at the level of individual event times. The focus is the conditional variance of the logrank score increment under the null hypothesis, not a calculation of required sample size or a comparison of planned information across analysis looks. The earlier material on the two-sample logrank statistic and risk-set contributions supplies the score framework; here we identify and interpret the variance attached to each event.

Definition: Immediately before event time \(t_j\), let \(Y_{1j}\) and \(Y_{0j}\) be the numbers at risk in the treatment and control groups, respectively, and let \(Y_j=Y_{1j}+Y_{0j}\). For a single untied event, define \(D_{1j}=1\) if the event is in the treatment group and \(D_{1j}=0\) otherwise. The logrank score increment is \(D_{1j}-Y_{1j}/Y_j\). Its null conditional variance, \(I_j=Y_{1j}Y_{0j}/Y_j^2\), is the event-level information. For a set of untied events, accumulated logrank information is the sum of their event-level information contributions.

Assumptions and the Event-Level Formula

For the basic two-group formula, consider a randomized comparison with a treatment and a control group. Participants are followed until failure or censoring, and the risk set at each event is defined immediately before that event. Under the null hypothesis of equal event hazards in the two groups, and conditional on one event occurring at a time when \(Y_j\) participants are at risk, each of those participants has the same instantaneous chance of being the one who fails. Thus the conditional probability that the event is in the treatment group is \(Y_{1j}/Y_j\).

This calculation assumes the usual conditions for the logrank comparison: event and censoring times are observed appropriately, and censoring is non-informative for the event process within the groups being compared. In practical terms, censoring must not selectively remove participants in a way that carries unaccounted information about their future event risk. Randomization alone does not repair severely informative loss to follow-up. The formula also presumes that the risk sets and event classifications have been constructed correctly.

Conditional on the risk set, \(D_{1j}\) is a Bernoulli variable under the null with success probability \(p_j=Y_{1j}/Y_j\). The score increment subtracts its conditional expectation, so its variance is the Bernoulli variance. Substitution gives the event-level formula:

$$ \begin{aligned} I_j &=\operatorname{Var}\left(D_{1j}-\frac{Y_{1j}}{Y_j}\,\middle|\,\text{risk set at }t_j,H_0\right)\\ &=p_j(1-p_j)\\ &=\frac{Y_{1j}}{Y_j}\frac{Y_{0j}}{Y_j}\\ &=\frac{Y_{1j}Y_{0j}}{(Y_{1j}+Y_{0j})^2}. \end{aligned} $$

The observed group in which the event occurs determines the score increment, but not this conditional variance: once the risk set is fixed, the formula for \(I_j\) is the same whichever group experiences the event. For distinct event times \(t_1,\ldots,t_m\), the information through the \(m\)th event is \(\sum_{j=1}^{m} I_j\). This accumulated variance is the logrank information used to scale the cumulative score; the corresponding standardized statistic has the familiar form \(U/\sqrt{I}\), where \(U\) is the cumulative score and \(I\) is its null variance.

Worked Example: Three Untied Events in Changing Risk Sets

Suppose a hypothetical trial has six treatment and four control participants at risk before the first of three distinct event times. The first event is in the control group, the second in the treatment group, and the third in the treatment group. Assume no censoring between these events. Calculate the event-level and accumulated information.

Before the first event, \(Y_{1,1}=6\) and \(Y_{0,1}=4\). Therefore,

$$ I_1=\frac{6(4)}{(6+4)^2}=\frac{24}{100}=0.2400. $$

The first event is in the control group, so the risk set before the second event is five treatment and three control participants. Hence

$$ I_2=\frac{5(3)}{(5+3)^2}=\frac{15}{64}=0.2344. $$

The second event is in the treatment group, leaving four treatment and three control participants before the third event. Thus

$$ I_3=\frac{4(3)}{(4+3)^2}=\frac{12}{49}=0.2449. $$

The information accumulated over these three event times is \(0.2400+0.2344+0.2449=0.7193\). The event-group labels were needed to update the risk sets, but they did not enter the variance calculation directly. These values are in logrank score-variance units, not numbers of participants or events.

Risk-Set Balance, Censoring, and Information

For one untied event, \(I_j=p_j(1-p_j)\), where \(p_j=Y_{1j}/Y_j\). This quantity is largest when the groups are equally represented in the risk set and becomes zero if either group has no one at risk. It is bounded by \(0.25\), since \((p_j-0.5)^2\geq 0\) implies \(p_j(1-p_j)\leq 0.25\). So, under this model, a single untied event cannot contribute more than \(0.25\) units of information.

Censoring is not itself a failure and therefore does not create a logrank score increment or a direct event-level information contribution. But censoring changes who remains at risk. If it changes the balance between groups before a later event, it can change that event’s information. This distinction is important: “no direct contribution” does not mean “no effect on subsequent information.”

Worked Example: Censoring Changes a Later Risk Set

In a hypothetical trial, four treatment and six control participants are initially at risk. First, a control participant has an event. Before the next event, one treatment participant is censored. Next, a treatment participant has an event, followed by a control event, with no further censoring. Calculate the information at the three event times and explain the censoring step.

At the first event, the risk set is \(4\) treatment and \(6\) control, so \(I_1=4(6)/(10^2)=0.2400\). The control event leaves four treatment and five control participants. Censoring one treatment participant then leaves three treatment and five control participants at risk. The censoring contributes no event information, but this changed risk set is used at the next event.

At the second event, \(I_2=3(5)/(8^2)=15/64=0.2344\). Because this event is in the treatment group, the next risk set has two treatment and five control participants. At the third event, \(I_3=2(5)/(7^2)=10/49=0.2041\). The total over the three events is \(0.2400+0.2344+0.2041=0.6785\). The censoring affected the risk-set composition, and thus the later contributions, even though it had no contribution of its own.

Tied Event Times

The single-event formula treats one failure at a time. If \(d_j\) failures occur at the same recorded time, a tie-handling convention is needed. One standard conditional calculation treats the \(d_j\) failing participants as a sample drawn without replacement from the \(Y_j\) people at risk. If all such subsets are equally likely under the null, the number of treatment failures has a hypergeometric distribution. Its variance is the tied-time information:

$$ I_j = d_j\frac{Y_{1j}}{Y_j}\frac{Y_{0j}}{Y_j} \frac{Y_j-d_j}{Y_j-1}, \qquad 0<d_j<Y_j. $$

The final factor is the finite-population correction for sampling without replacement. This is a particular tied-event convention; calculations using other approximations for tied times must use the variance consistent with that convention. A tie should not be handled by automatically applying the single-event formula \(d_j\) times to an unchanged risk set.

Worked Example: A Tied Event Under the Hypergeometric Convention

Suppose a hypothetical trial has five treatment and five control participants at risk, and two failures occur at the same recorded time. Use the hypergeometric tied-event variance.

Here \(Y_j=10\), \(Y_{1j}=Y_{0j}=5\), and \(d_j=2\). Therefore,

$$ I_j = 2\left(\frac{5}{10}\right)\left(\frac{5}{10}\right) \left(\frac{10-2}{10-1}\right) =2(0.25)\frac{8}{9} =\frac{4}{9} =0.4444. $$

If the two failures were incorrectly treated as independent single events with the unchanged risk set, the calculation would be \(2(0.25)=0.5000\). That ignores the without-replacement structure of two failures from the same ten-person risk set and overstates the variance under the stated hypergeometric convention.

Checks, Alternatives, and Limitations

Before interpreting an event-level information calculation, check that each risk set refers to the instant just before the event, that event and censoring records have been distinguished, and that all participants who should be at risk are counted exactly once. For an untied event, useful arithmetic checks are \(I_j\geq 0\), \(I_j\leq 0.25\), and \(I_j=0\) if one group has no one at risk. If a purported single-event contribution exceeds \(0.25\), the risk-set denominator or arithmetic is wrong.

Logrank information is a variance measure, not an effect estimate. A large value does not show that treatment is beneficial, and it does not say how many events occurred without reference to the risk-set balance. Nor does the formula guarantee that the logrank comparison is appropriate: its standard interpretation depends on the event and censoring assumptions and on the chosen score procedure. The formula describes null information from the risk sets; it does not by itself establish that the treatment effects are proportional over time.

Event count is a simple alternative design or monitoring scale, but it assigns the same weight to events that may have very different risk-set compositions. Participant count is also not a substitute for score variance: a participant who has not yet had an event does not generate an event-level score increment, while participants entering or leaving the risk sets affect the composition at later failures. For analysis, the score and its variance must remain paired: the standardized cumulative score uses the logrank variance associated with the same risk sets and tie convention as the score.

Common pitfall: “There were five events, so the logrank information is five.” This assumes every event contributes one unit. Under the untied two-group formula each event contributes at most \(0.25\), and its actual contribution depends on the risk-set proportions. Calculate each event’s variance from the risk set rather than equating information with the event count.

Check Your Understanding

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

  1. At an untied event, there are eight treatment and two control participants at risk. What is the event-level information?
  2. Why does a censoring time have no direct event-level logrank information contribution, yet still affect a later event’s contribution?
  3. At an untied event, there are no control participants at risk. What is the information, and why?
  4. Under the hypergeometric convention, calculate the information for three tied failures among six treatment and six control participants at risk.
  5. A reviewer says that two tied failures among ten participants at risk contribute \(0.50\) because each failure has a \(0.25\) contribution. What qualification or correction is needed under the hypergeometric convention?

Solutions. (1) \(I=8(2)/(10^2)=16/100=0.1600\). (2) Censoring is not a failure, so it produces no failure score increment; it can remove someone from a group’s risk set and change the group proportions at a later event. (3) \(I=0\), because \(Y_{1}Y_{0}=0\); with no control participant at risk, the null event-group allocation has no variation. (4) Here \(Y=12\), \(Y_1=Y_0=6\), and \(d=3\). Thus \(I=3(6/12)(6/12)(12-3)/(12-1)=3(0.25)(9/11)=27/44=0.6136\). (5) The value \(0.50\) treats the failures as independent single events with an unchanged risk set. Under the stated hypergeometric convention, \(I=2(0.5)(0.5)(8/9)=4/9=0.4444\); the finite-population correction accounts for sampling without replacement.

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