Why Event Counts Are Not Always Information
In group sequential monitoring, an interim analysis is often described by the fraction of planned events observed. That is a useful approximation when each event contributes about the same amount of information. For a weighted logrank statistic, however, an event's contribution depends on its weight and on the treatment and control risk sets at that time. A large event count can therefore correspond to a small fraction of the total information, or the reverse.
Information time puts interim looks on a scale tied to the variance of the statistic being monitored, rather than to the calendar or the raw event count. This matters because the covariance structure of sequential weighted logrank statistics is governed by their accumulated variance under the null. The earlier material on independent increments for weighted logrank monitoring establishes the key connection: under its assumptions, the covariance of the score at an earlier look with the score at a later look equals the earlier look's variance.
Score and Variance Contributions
Suppose a two-group time-to-event study compares treatment with control. Immediately before distinct event time \(j\), let \(Y_{1j}\) and \(Y_{0j}\) be the numbers at risk in the two groups, and let \(Y_j=Y_{1j}+Y_{0j}\). Let \(d_j\) be the total number of events at that time and \(d_{1j}\) the number in the treatment group. Let \(w_j\) be the prespecified weight applied at that time. The cumulative weighted logrank score through look \(k\) is
The term in parentheses compares the observed treatment events with the number expected if, conditional on the risk set and total events at that time, the events were allocated between groups according to the risk-set proportions. Under the null hypothesis of equal event hazards, the usual conditional variance contribution for tied events is
This expression assumes \(Y_j>1\), and uses the conditional hypergeometric variance for the allocation of \(d_j\) events. If there are no ties, \(d_j=1\), it simplifies to \(v_j=w_j^2Y_{1j}Y_{0j}/Y_j^2\). Thus, even with one event at each time, the contribution depends both on the squared weight and on the balance of the risk set. An event when one group has very few people at risk generally contributes less than an event in a balanced risk set, all else equal.
Why the Variance Formula Has This Form
Conditional on the risk set and on \(d_j\), under the null the treatment event count has a hypergeometric distribution: \(d_j\) events are selected from \(Y_j\) people, of whom \(Y_{1j}\) are in treatment. Its expected value is \(d_jY_{1j}/Y_j\), and its variance is \(d_j(Y_{1j}/Y_j)(Y_{0j}/Y_j)(Y_j-d_j)/(Y_j-1)\). The weighted score contribution multiplies the centered treatment event count by \(w_j\), so its variance is multiplied by \(w_j^2\). Substituting the risk-set proportions gives
Summing these conditional variances gives the cumulative variance used for information time when increments are uncorrelated under the null. The formula also shows directly why weights cannot be ignored: doubling a weight quadruples that time point's variance contribution.
Worked Example: A Half of the Events, but Only a Fifth of the Information
Consider a hypothetical design with 100 planned events and no tied event times. Suppose treatment and control risk sets are balanced at every event, so \(Y_{1j}/Y_j=1/2\), and the fixed weight is \(0.25\) for the first 40 events and \(1\) for the remaining 60. Assume these weights and risk-set proportions are the same in the planned information calculation and at the interim look.
For one event with a balanced risk set, the unweighted variance contribution is \((1/2)(1/2)=0.25\). The weight makes the contribution for each of the first 40 events \(0.25^2(0.25)=0.015625\). The first 40 events therefore contribute \(40(0.015625)=0.625\). Each of the remaining 60 contributes \(1^2(0.25)=0.25\), for a total of \(60(0.25)=15\).
The planned total information is \(V_K=0.625+15=15.625\). At the interim look after 50 events, the information is \(V_k=0.625+10(0.25)=3.125\). Consequently,
The event fraction is \(50/100=0.5000\), but the information fraction is \(0.2000\). Most of the observed events occurred while the weight was small, so they contributed relatively little variance. The calculation assumes balanced risk sets and the stated fixed weights; if the risk-set proportions change, the contribution per event changes as well.
Estimating Information Time From Interim Data
At an interim look, an estimate of cumulative information is obtained by summing the variance contributions calculated from the observed risk sets, event counts, and weights through that look. In a design calculation, the same formula can instead be applied to planned event patterns or to a model-based expectation. The information time used for monitoring should be the ratio of comparable cumulative quantities: the numerator is information through the current look, and the denominator is the planned or estimated information at the final look.
Worked Example: Checking the Cumulative Variance Arithmetic
In a hypothetical no-ties study, suppose four event times have the risk sets and weights shown below. The risk-set sizes can change over time as participants enter or leave follow-up. Each event time has \(d_j=1\), so the contribution is \(w_j^2Y_{1j}Y_{0j}/Y_j^2\).
| Event time | \(Y_{1j}\) | \(Y_{0j}\) | \(w_j\) | Variance contribution |
|---|---|---|---|---|
| 1 | 1 | 1 | 1 | \(1^2(1)(1)/2^2=20/80=0.2500\) |
| 2 | 19 | 20 | 1 | \(1^2(19)(20)/39^2=380/1521\approx0.2498\) |
| 3 | 18 | 20 | 3 | \(3^2(18)(20)/38^2=3240/1444\approx2.2438\) |
| 4 | 18 | 19 | 3 | \(3^2(18)(19)/37^2=3078/1369\approx2.2484\) |
For the first event, the displayed fraction \(20/80\) is simply \(1/4\). At look 2, adding the first two unrounded contributions gives \(V_2=0.25+380/1521\approx0.4998\). At look 4, use the unrounded fractions:
The information time at look 2 relative to look 4 is therefore \(t_2=V_2/V_4\approx0.4998356/4.9919594\approx0.1001\). Although two of the four events have occurred, only about 10.01% of the final information has accumulated. The larger weights at the third and fourth event times explain the difference.
Worked Example: A Tied Event Time
At one hypothetical interim look, suppose a tied event time has \(Y_{1j}=30\), \(Y_{0j}=20\), \(d_j=4\), and \(w_j=2\). The total risk set is \(Y_j=50\). The tied-event variance contribution is
The event count is four, but this block contributes about 3.6049 units of null variance. For comparison, omitting the finite tied-event correction would give \(w_j^2d_j(Y_{1j}/Y_j)(Y_{0j}/Y_j)=4(4)(0.6)(0.4)=3.8400\). The exact conditional tied-event formula reduces that value by the factor \((Y_j-d_j)/(Y_j-1)=46/49\). The tied-event formula is appropriate here because four events occur at the same recorded time and the risk set has 50 members.
Interpreting, Diagnosing, and Using Information Time
Under the null and the usual independent-increment approximation, \(V_k\) is the variance of \(U_w(k)\), and for looks \(i\leq j\), \(\operatorname{Cov}_0\{U_w(i),U_w(j)\}=V_i\). Thus, if \(Z_k=U_w(k)/\sqrt{V_k}\), the null correlation between looks \(i\) and \(j\) is approximately \(\sqrt{V_i/V_j}\). Information time summarizes this variance-based progress in a way that is relevant to sequential monitoring. It does not, by itself, specify a stopping rule or a boundary.
Useful checks include confirming that the weights used in the score and in the variance calculation agree, that each risk set is defined immediately before its event time, and that all planned or observed variance contributions are nonnegative. With nonzero weights and nondegenerate risk sets, cumulative information should not decrease as events accumulate. A decrease usually points to a coding or bookkeeping error, such as subtracting rather than adding a variance increment. If some weights are zero, events at those times contribute no information to this weighted score.
Information time also has alternatives, but they answer different questions. Calendar-time scheduling is simple to communicate but does not reflect how many informative events have accrued. Event-count fraction is convenient and often adequate when weights and risk-set allocation make per-event contributions similar. A direct variance calculation is more appropriate when the weights vary substantially or risk-set balance changes. The unweighted logrank test uses the same general risk-set framework with weights equal to one; its information pattern can differ from that of a weighted test. These summaries should not be treated as interchangeable when planning or describing interim looks.
Assumptions and Limitations
The variance calculation above is conditional on the observed risk sets and event totals at each time and uses the null allocation of events between randomized groups. Its sequential interpretation relies on the weighted score having the appropriate independent-increment structure under the null, and on weights being specified in a manner compatible with that structure. A planned deterministic weight function is a straightforward case. Data-adaptive weights or changes to weights based on accumulating results can alter the variance and dependence structure; simply inserting such weights into the formula does not automatically preserve the stated monitoring properties.
An estimated information fraction can be noisy, especially early in follow-up when event counts are small or risk sets are changing quickly. The denominator also matters: a ratio based on observed interim variance divided by a model-based planned final variance is only as reliable as the assumptions behind that final projection. In addition, the weighted logrank variance measures precision for the chosen weighted score; it is not a universal measure of information about every treatment effect or estimand. When hazards vary over time, the score's sensitivity depends on where its weights place emphasis, so a high information fraction does not guarantee high power for every possible pattern of treatment effect.
Check Your Understanding
Try each question, then compare your reasoning with the solutions.
- With no ties, what happens to an event's variance contribution if its weight is doubled while the risk set is unchanged?
- A look has information \(V_k=6\), and the planned final information is \(V_K=15\). What is its information time?
- Why can an event fraction of 50% correspond to an information fraction below 50%?
- For one untied event with \(Y_{1j}=12\), \(Y_{0j}=8\), and \(w_j=2\), calculate the variance contribution.
Solutions. (1) It is multiplied by four because the variance contribution contains \(w_j^2\). (2) \(t_k=6/15=0.4000\), so 40.00% of planned information has accumulated. (3) Events can occur when weights are small or risk sets are imbalanced, making their variance contributions smaller than those of later events; information time sums those contributions rather than counting events equally. (4) \(Y_j=20\), so \(v_j=2^2(12)(8)/20^2=768/400=1.920\) units of variance.