Why Tied Events Need a Convention
The logrank calculation compares the observed number of events in a treatment group with the number expected from its share of the risk set. When event times are tied, several events occur at the same recorded time, so the calculation must specify whether those events are treated as one block or approximated as separate events. This choice can affect the variance, and therefore the standardized statistic, at an interim look.
The focus here is the usual two-group logrank calculation with tied events. Let group 1 be the group whose event count is being scored, and let group 0 be the comparison group. A positive score means that group 1 had more events than expected from its representation in the risk set. For an outcome where events are undesirable, that direction is evidence of more events in group 1, not a benefit.
The expected count \(dY_1/Y\) is the number of events group 1 would contribute, on average, if the \(d\) events were allocated among the \(Y\) people at risk in proportion to group size. The key question is how to calculate the variation around that expectation when the events are tied.
Assumptions and the Hypergeometric Convention
For the standard tied-block calculation, condition on the risk set immediately before the event time and on the total number \(d\) of events in the block. Under the null hypothesis of equal event hazards in the two groups, each person in that risk set has the same chance of being among those \(d\) events. Thus the \(d\) event cases can be viewed as a sample without replacement from the \(Y\) people at risk. The usual logrank setting also requires risk sets and event counts to be defined consistently; event and censoring records at the same recorded time need a prespecified ordering convention.
Under this null model, the number \(X\) of group 1 events in the block has a hypergeometric distribution. Its mean gives the expected count used in the score, and its variance gives the tied-block variance contribution.
The factor \((Y-d)/(Y-1)\) is the finite-population correction. Once one person is selected as an event, that person cannot be selected again within the same tied block. Treating event indicators as independent would miss this dependence and produce a different variance.
Imagine drawing \(d\) people without replacement from the \(Y\) people at risk; \(Y_1\) of them belong to group 1.
Each draw has group 1 proportion \(p=Y_1/Y\), so the expected number in group 1 is \(dp=dY_1/Y\).
Let each draw indicator be 1 for a group 1 person and 0 otherwise. Each has variance \(p(1-p)\), and indicators for distinct draws have covariance \(-p(1-p)/(Y-1)\). Summing the \(d\) variances and \(d(d-1)\) covariances gives \(dp(1-p)(Y-d)/(Y-1)\).
Add the score contributions and the variance contributions for all event-time blocks observed by the interim look.
The covariance in Step 3 follows because sampling is without replacement: after one person is drawn, the group composition of the remaining risk set changes. Substituting \(p=Y_1/Y\) and \(1-p=Y_0/Y\) yields the stated variance. When \(d=1\), the correction is 1, and the formula reduces to the familiar single-event contribution \(Y_1Y_0/Y^2\).
Worked Calculations at an Interim Look
Worked Example: One Tied Block
Suppose a hypothetical trial has \(Y_1=12\) and \(Y_0=8\) people at risk at an event time. There are \(d_1=3\) events in group 1 and \(d_0=1\) in group 0. Calculate the score and variance contribution using the hypergeometric convention.
The total risk set is \(Y=12+8=20\), and the tied block has \(d=3+1=4\) events. Group 1's expected event count is \(4(12/20)=2.4\). Therefore,
The variance contribution is
If this were the only block observed at the look, the standardized score would be \(Z=U/\sqrt{V}=0.6000/\sqrt{0.8084}\approx0.6673\). The positive value reflects more group 1 events than expected, but this calculation alone does not determine a trial-level conclusion; the design's testing procedure and error control also matter.
Worked Example: Accumulating Two Tied Blocks
Consider two hypothetical event-time blocks observed by an interim look. In the first, \(Y_1=10\), \(Y_0=10\), \(d_1=2\), and \(d_0=1\). In the second, \(Y_1=8\), \(Y_0=7\), \(d_1=2\), and \(d_0=2\). Calculate the cumulative score, cumulative variance, and standardized score.
For the first block, \(Y=20\) and \(d=3\). Its score and variance are
For the second block, \(Y=15\) and \(d=4\). Its expected group 1 count is \(4(8/15)=32/15\), so
Adding the block contributions gives \(U=0.5000-0.1333=0.3667\) and \(V=0.6711+0.7822=1.4533\). Thus the standardized score is \(Z=0.3667/\sqrt{1.4533}\approx0.3042\). The cumulative score remains positive, but the second block partly offsets the first.
Worked Example: A Look at a Tied Event Time
Suppose a hypothetical interim cutoff coincides with a recorded event time at which \(Y_1=6\), \(Y_0=4\), \(d_1=2\), and \(d_0=1\). Under a prespecified rule that includes the whole tied block at that cutoff, calculate its contribution.
Here \(Y=10\) and \(d=3\). The expected group 1 count is \(3(6/10)=1.8\), so \(U=2-1.8=0.2000\). The variance is
The block's standardized contribution, if considered alone, is \(0.2000/\sqrt{0.5600}\approx0.2673\). The events must be handled as one block under this convention. Splitting the three events into an arbitrary sequence would require inventing intermediate risk sets and could make the result depend on the chosen order.
What Changes Under an Approximation?
A common approximation uses the same score expectation but omits the finite-population correction, giving \(V_{\mathrm{approx}}=d(Y_1/Y)(Y_0/Y)\). It can be reasonable when the number of tied events is small relative to the risk set, because then \((Y-d)/(Y-1)\) is close to 1. It is not the same as the hypergeometric variance, and the difference can matter when a large fraction of the risk set has tied events.
| Convention | Score contribution | Variance contribution | Practical implication |
|---|---|---|---|
| Hypergeometric tied block | \(d_1-dY_1/Y\) | \(d(Y_1/Y)(Y_0/Y)(Y-d)/(Y-1)\) | Accounts for sampling without replacement within the tied block. |
| Approximation without correction | \(d_1-dY_1/Y\) | \(d(Y_1/Y)(Y_0/Y)\) | Convenient when ties are sparse relative to the risk set; generally gives a larger variance for \(d>1\). |
| Arbitrary sequential ordering | Depends on the invented intermediate risk sets | Depends on the ordering and update rule | Can make tied-block results order-dependent; it is not the same as treating the tied block jointly. |
For the first worked example, omitting the correction would give \(V_{\mathrm{approx}}=4(0.6)(0.4)=0.9600\), rather than \(0.8084\). The standardized score would then be \(0.6000/\sqrt{0.9600}\approx0.6124\), rather than \(0.6673\). The score itself is unchanged in this example, but its standardization is not.
Tie handling in a Cox partial-likelihood calculation may use methods such as Breslow or Efron. Those likelihood approximations are related to logrank calculations, but their definitions should not be substituted automatically for a specified tied-block logrank variance. For an interim analysis, the score and variance must use a coherent, documented convention at every look.
Diagnostics and Common Pitfalls
A useful check is to compare the tied-block variance with and without its correction. Since \((Y-d)/(Y-1)\leq1\) for \(1\leq d\leq Y\), the hypergeometric variance cannot exceed the uncorrected approximation. The two agree when \(d=1\); for a large tied block, the correction can be substantial. A computed correction greater than 1, a negative variance, or a total event count exceeding the risk set signals an error in the inputs or formula.
- Using \(d_1\) instead of \(d\) in the expected count. The allocation expectation uses all events in the block: \(dY_1/Y\).
- Forgetting the finite-population correction. This changes the variance and therefore the standardized score, even when the score contribution is unchanged.
- Ordering tied events arbitrarily. The tied-block convention conditions on the total \(d\); it does not assign an artificial order to those events.
- Splitting a tied block at a look without a rule. If the cutoff coincides with the tied time, specify in advance whether the whole block is included. Applying inconsistent rules across looks can undermine comparability.
- Interpreting \(Z\) without its sign or design context. A positive score means more group 1 events than expected, not automatically a statistically significant result or a beneficial treatment effect.
At an interim look, the cumulative score is the sum of the tied-block score contributions through the cutoff, and its corresponding variance is the sum of their variance contributions under the sequential logrank construction. The standardized cumulative statistic is \(Z=U/\sqrt{V}\), provided \(V>0\). Its inferential interpretation depends on the prespecified monitoring procedure; the ordinary normal approximation should not be treated as a complete group-sequential decision rule.
Check Your Understanding
Work each problem using the hypergeometric tied-block convention. Solutions are included below.
- A block has \(Y_1=9\), \(Y_0=6\), \(d_1=2\), and \(d_0=1\). Find its score and variance.
- Why does the standard tied-block variance include a finite-population correction?
- For \(Y_1=5\), \(Y_0=5\), and \(d=1\), explain how the tied-block variance relates to the single-event logrank variance.
- A colleague says that a tied block can be split into events in any order because the total event count is unchanged. What is wrong with this reasoning?
Solutions. (1) \(Y=15\), \(d=3\), so \(U=2-3(9/15)=0.2000\). The variance is \(3(9/15)(6/15)(12/14)=0.6171\). (2) The \(d\) events are selected without replacement from the risk set, so their indicators are negatively dependent; the correction accounts for that dependence. (3) With \(d=1\), the correction is \((Y-1)/(Y-1)=1\), giving \(V=(5/10)(5/10)=0.2500=Y_1Y_0/Y^2\). (4) An arbitrary ordering invents intermediate risk sets and can change the calculated contributions; a specified tied-block convention treats the events jointly.