Why Information Comes From Risk Sets
A logrank analysis compares the observed treatment-arm event count with the count expected under equal event hazards. Its information is not simply the number of events: at each event time, the comparison is more informative when both arms are represented in the risk set, and less informative when one arm is sparse. The calculation follows directly from the variance of treatment allocation among those who experience an event.
Here, \(n_{1j}\) and \(n_{0j}\) denote the numbers at risk in the treatment and control arms immediately before event time \(j\); \(N_j=n_{1j}+n_{0j}\). Let \(d_j\) be the number of events at that time, and \(x_{1j}\) the observed number of those events in the treatment arm. The event-time logrank score contribution is observed minus expected treatment events.
This definition treats the risk sets as known immediately before each event time. It assumes that, under \(H_0\), the event allocation between arms is exchangeable within a risk set after conditioning on its size and the total number of events. The usual logrank interpretation also relies on appropriate censoring: within the analysis groups, censoring should not selectively remove people according to their unobserved event risk in a way that invalidates the risk-set comparison. Risk sets must correctly account for entry, censoring, and prior events.
One Event: The Basic Risk-Set Variance
First suppose there is one event at time \(j\), so \(d_j=1\). Under the null, the event is in the treatment arm with conditional probability \(p_j=n_{1j}/N_j\), and in the control arm with probability \(1-p_j=n_{0j}/N_j\). The treatment event indicator is therefore Bernoulli with variance \(p_j(1-p_j)\). Hence
This contribution is largest when the two arms are balanced in the risk set and is zero if either arm has no one at risk. An event in a risk set containing only one arm cannot distinguish the arms: its treatment allocation is already determined.
Worked Example: One Untied Event
Suppose a hypothetical risk set immediately before an event contains 3 treatment participants and 5 control participants. Thus \(N_j=8\) and \(d_j=1\). Assume the event occurs in the treatment arm, so \(x_{1j}=1\).
Under \(H_0\), the expected treatment event count is \(d_jn_{1j}/N_j=1(3/8)=0.375\). The score contribution is \(1-0.375=0.625\) treatment events.
The information contribution is
The observed event contributes a positive score, but its information is determined by the risk-set allocation, not by which arm happened to experience the event. The calculation presumes that the risk set is correct and that the null makes the event allocation exchangeable between its members.
Tied Events and the Finite-Population Variance
When \(d_j>1\) events occur at the same recorded time, the treatment event count is not a single Bernoulli outcome. Conditional on \(d_j\), the standard logrank calculation treats the event participants as a sample of size \(d_j\) from the \(N_j\) people at risk, with \(n_{1j}\) treatment participants among them. Thus \(X_{1j}\), the treatment event count, has a hypergeometric distribution under the null.
To derive its variance, write \(X_{1j}\) as the sum of \(d_j\) indicators for whether each selected event participant is in the treatment arm. Each indicator has mean \(p_j=n_{1j}/N_j\) and variance \(p_j(1-p_j)\). For two distinct selections, sampling without replacement gives covariance \(-p_j(1-p_j)/(N_j-1)\). Summing the \(d_j\) individual variances and the \(d_j(d_j-1)\) ordered covariance terms yields
Consequently, for \(N_j>1\), the tied-event contribution is
The last factor is the finite-population correction: selecting several event participants without replacement makes their treatment indicators negatively associated. When \(d_j=1\), the factor is 1 and the formula reduces to the untied-event result. When \(d_j=N_j\), everyone at risk has an event, the treatment event count is fixed at \(n_{1j}\), and the variance is zero.
There is one boundary case the displayed formula cannot cover. If \(N_j=1\), the only possible event total is \(d_j=1\); the treatment event count is fixed, so define \(V_j=0\). The formula would have a \(0/0\) finite-population factor and must not be used for that case. If \(d_j=0\), there is no event-time score contribution and \(V_j=0\).
Worked Example: A Tied Event Time
Consider a hypothetical risk set with 3 treatment and 4 control participants, so \(N_j=7\). Two events are recorded at this time (\(d_j=2\)). Under the null, the expected number of treatment events is \(2(3/7)=6/7\).
Since \(N_j>1\), the tied-event variance formula applies:
For comparison, omitting the finite-population factor would give \(24/49\approx0.4898\), which is too large for this conditional, without-replacement allocation model. If instead a risk set had \(N_j=1,d_j=1\), its contribution would be 0 by the boundary rule, not a value obtained by substituting into a \(0/0\) expression.
Adding Contributions Across Event Times
At each event time, the logrank score contribution is \(x_{1j}-d_jn_{1j}/N_j\). Its conditional variance under the null is \(V_j\). Across event times, the usual logrank variance is obtained by summing the risk-set contributions. This is the risk-set form of the total information used to standardize the cumulative score:
This variance addition is justified by the logrank score's sequential, mean-zero event-time increments under the null: distinct increments are orthogonal in the martingale formulation, so their variance contributions add. The risk-set variance is calculated from the allocation just before each event time, not from the arm counts after the event. Tied-event handling must be consistent with the time resolution and the chosen logrank calculation.
Worked Example: Total Information From a Two-Arm Event History
The following is a fully hypothetical event history. The risk-set counts are just before each event time. For tied events, \(x_{1j}\) is the number of treatment events among the \(d_j\) events.
| Event time | \(n_{1j}\) | \(n_{0j}\) | \(d_j\) | \(x_{1j}\) |
|---|---|---|---|---|
| 1 | 4 | 5 | 1 | 1 |
| 2 | 3 | 4 | 2 | 1 |
| 3 | 2 | 3 | 1 | 0 |
| 4 | 1 | 0 | 1 | 1 |
The changing risk sets are possible, for example, when earlier events and censoring remove participants between event times. For time 1, the untied contribution is \(V_1=(4/9)(5/9)=20/81\), and the score contribution is \(1-4/9=5/9\).
At time 2, two events occur in a risk set of size 7. The tied contribution is \(V_2=2(3/7)(4/7)(5/6)=20/49\), and its score contribution is \(1-2(3/7)=1/7\).
At time 3, \(V_3=(2/5)(3/5)=6/25\), and the score contribution is \(0-2/5=-2/5\). At time 4, only one treatment participant remains at risk. The event allocation and expected treatment event count are both 1, so the score contribution is 0 and \(V_4=0\).
Adding the information contributions gives
The total score is \(U=5/9+1/7-2/5=94/315\approx0.2984\). Therefore the standardized score is \(Z=0.2984/\sqrt{0.8951}\approx0.3154\). This is a small positive standardized score: the observed treatment event count is modestly above its null expectation in aggregate. The arithmetic assumes each row gives the correct pre-event risk set and that the stated tied-event model is appropriate.
Checks, Alternatives, and Limitations
A practical audit is to calculate each row's expected treatment count and variance separately before summing. For every event time, check that \(N_j=n_{1j}+n_{0j}\), that \(0\le d_j\le N_j\), and that \(0\le x_{1j}\le d_j\). For the tied formula, also verify \(N_j>1\), unless using the \(N_j=1,d_j=1\) boundary rule. The contributions must be nonnegative; a negative variance signals an arithmetic or data-definition error.
The risk-set derivation makes clear why total events alone do not specify information. Events occurring when both arms are represented can contribute variance; events in a one-arm risk set contribute none. Similarly, using the untied formula once for each tied event ignores the dependence induced by sampling without replacement and can overstate the conditional variance. Conversely, treating tied events as one event discards the number of events and does not implement the stated tied-event calculation.
The hypergeometric calculation is the standard conditional tied-event convention. If recorded ties conceal distinct event ordering, another tie-handling convention may be appropriate, but it must be specified and used consistently; the variance formula here is not a universal correction for every way of resolving ties. The result also depends on valid risk sets and the logrank null allocation model. Informative censoring, incorrectly recorded event times, or an inappropriate comparison across groups can undermine the interpretation even when the arithmetic is flawless.
The risk-set method gives a transparent calculation that can be checked event by event. A model-based score calculation can provide an alternative route to a logrank variance, but it should agree when it uses the same null model, risk sets, and tie convention. A naive binomial variance may coincide for one untied event, yet it is not a substitute for the finite-population variance for tied events.
A frequent mistake is to say, “There were two tied events, so the information is twice the single-event variance.” That reasoning assumes independent allocation of the two events. Conditional on the total event count, the participants are sampled without replacement, so the treatment indicators are negatively associated. Use the finite-population factor \((N_j-d_j)/(N_j-1)\); and when a single person is the entire risk set, assign zero information directly.
Check Your Understanding
Answer each question using the risk-set formulas, then compare with the solutions below.
- A risk set contains 6 treatment and 4 control participants, and one event occurs. What is its information contribution?
- A risk set contains 2 treatment and 3 control participants, with 2 tied events. Find the null expected treatment event count and the variance contribution.
- All 8 people in a risk set have an event at the same recorded time. What is the information contribution, and why?
- Why is the tied-event formula not evaluated by direct substitution when \(N_j=1,d_j=1\)? What contribution should be used?
Solutions: (1) \(V_j=(6/10)(4/10)=0.2400\). (2) The expected treatment count is \(2(2/5)=0.8000\), and \(V_j=2(2/5)(3/5)(3/4)=0.3600\). (3) The contribution is 0: when \(d_j=N_j\), the treatment event count is fixed by the risk-set composition, so its variance is zero. (4) The finite-population factor has denominator \(N_j-1=0\), producing an undefined \(0/0\) expression. The event allocation is fixed, so set \(V_j=0\).