Stratum-Specific Information Imbalance

Why Stratum-Specific Information Matters

A stratified logrank test combines score contributions from several strata, such as age groups or disease stages. Its precision depends on how much information each stratum contributes, not simply on how many events occur there. Two strata with the same event count can contribute quite different information if their treatment and control risk sets differ. Conversely, a stratum with relatively few events can dominate the combined information if its event risk sets are especially informative.

This distinction matters in an event-driven trial: the overall event target can be reached while information is concentrated in only a subset of the intended strata. The combined test may still be valid under its assumptions, but the information distribution affects how the result represents the enrolled population and how precisely stratum-specific effects can be assessed.

Definition and Assumptions

Consider $K$ disjoint strata. At event time $j$ in stratum $k$, let $Y_{1kj}$ and $Y_{0kj}$ be the numbers at risk in the treatment and control groups, respectively, and let $Y_{kj}=Y_{1kj}+Y_{0kj}$. For a standard, unweighted logrank score with a binary treatment indicator, the event-level information under the null is the treatment-indicator variance in that risk set.

Definition: The stratum-specific logrank information $I_k$ is the sum of its event-level information contributions. With distinct event times and the usual unweighted logrank score, $$ I_k=\sum_j \frac{Y_{1kj}Y_{0kj}}{Y_{kj}^2}. $$ The combined information is $I=\sum_{k=1}^{K} I_k$, and the information share of stratum $k$ is $q_k=I_k/I$. Thus $0\le q_k\le 1$ and $\sum_k q_k=1$.

This formula assumes that each event is analyzed using its risk set, that the treatment indicator is the score covariate, and that the null-model variance is being calculated. The risk sets must be defined consistently with the trial’s follow-up and censoring rules. If there are tied event times, the analysis must use the chosen standard tie method and its corresponding information calculation; the distinct-event formula above should not be applied as though tied events occurred one at a time.

A useful simplifying representation is $Y_{1kj}/Y_{kj}=r_{kj}$, the treatment fraction at risk. Then an event contributes $r_{kj}(1-r_{kj})$, which is largest at $r_{kj}=0.5$ and becomes small when one treatment group is scarce in the risk set. This is why event counts alone cannot determine information. The calculations below assume the stated average risk-set fractions summarize event contributions; in an actual analysis, information is calculated from the observed risk sets using the specified method.

Why Information Adds Across Strata

Let $U_k$ be the logrank score from stratum $k$, using the convention that positive scores correspond to more treatment-group events than expected under the null. The stratified score is $U=\sum_k U_k$. Under the null, each stratum’s score has mean zero, and its variance is $I_k$. When subjects belong to only one stratum and the strata are independent, the score contributions have zero covariance across strata. Therefore, the variance of the sum is the sum of the variances.

\operatorname{Var}_0(U) =\operatorname{Var}_0\left(\sum_{k=1}^{K} U_k\right) =\sum_{k=1}^{K}\operatorname{Var}_0(U_k) =\sum_{k=1}^{K} I_k =I.

The standardized combined statistic is consequently $Z=U/\sqrt{I}$, with the usual large-sample standard normal reference distribution under the null. The stratum information shares $q_k$ describe how that total variance is apportioned. They do not, by themselves, say whether a treatment effect is present or whether the test’s assumptions hold.

A compact descriptive summary of concentration is $H=\sum_k q_k^2$. Its reciprocal, $1/H$, can be read as an effective number of equally informative strata: it equals $K$ when all $K$ shares are equal and approaches 1 when one stratum supplies nearly all the information. This is a diagnostic summary, not a replacement for the logrank statistic or a formal test.

Worked Information Calculations

Worked Example: Unequal Event Counts, Unequal Information

Suppose a hypothetical trial has two strata. Stratum A has 80 distinct events, and its average treatment fraction at risk is 0.40. Stratum B has 10 events, and its average treatment fraction at risk is 0.20. Assume these fractions summarize the event-level risk sets closely enough for the illustrative calculation.

For a risk-set treatment fraction $r$, the average event contribution is $r(1-r)$. Thus, for A, the contribution is $0.40(0.60)=0.24$, giving $I_A=80(0.24)=19.2$. For B, it is $0.20(0.80)=0.16$, giving $I_B=10(0.16)=1.60$.

The total information is $I=19.2+1.60=20.8$. The information shares are $q_A=19.2/20.8=0.9231$ and $q_B=1.60/20.8=0.07692$. A supplies 80 of 90 events, or 88.89% of the events, but 92.31% of the information. The concentration summary is $H=0.9231^2+0.07692^2=0.8580$, so the effective number of equally informative strata is $1/0.8580=1.166$. The combined information is strongly concentrated in A.

This calculation describes information allocation, not a comparison of observed treatment effects. It is consistent with the stated risk-set assumptions: both groups are represented at risk in each stratum, and the fractions used are explicitly treated as averages for this illustration.

Worked Example: Equal Event Counts, Unequal Information

Consider a second hypothetical trial with 40 distinct events in each of two strata. Suppose the average treatment fraction at risk is 0.50 in stratum A and 0.10 in stratum B. Both groups are represented in both strata, but the risk-set allocation differs.

In A, each event contributes $0.50(1-0.50)=0.25$, so $I_A=40(0.25)=10.0$. In B, each contributes $0.10(1-0.10)=0.09$, so $I_B=40(0.09)=3.60$. Hence $I=10.0+3.60=13.6$, with shares $q_A=10.0/13.6=0.7353$ and $q_B=3.60/13.6=0.2647$.

The event counts are equal, but A contributes about 2.78 times as much information as B, since $10.0/3.60=2.778$. The reason is not that A has more events; it is that its risk sets have a more balanced allocation, yielding more treatment-indicator variance per event. These numbers are internally consistent with the stated distinct-event and average-risk-set assumptions.

Worked Example: Information Weights Under Different Effects

Information imbalance also matters when the treatment effects differ across strata. Suppose, hypothetically, that two strata contribute $I_1=18$ and $I_2=2$, and their log hazard ratios are $\beta_1=\log(0.80)=-0.2231$ and $\beta_2=\log(1.10)=0.09531$. Under local alternatives, the expected stratum score is approximately $I_k\beta_k$. A first-order information-weighted summary of the combined effect is therefore $$ \frac{I_1\beta_1+I_2\beta_2}{I_1+I_2} =\frac{18(-0.2231)+2(0.09531)}{20} =\frac{-4.016+0.1906}{20} =-0.1913. $$

Exponentiating gives $\exp(-0.1913)=0.826$, an approximate combined hazard ratio of 0.826 under this local-effect calculation. The information shares are 0.90 and 0.10, so the summary is much more influenced by stratum 1’s effect. It is not the simple average of the two hazard ratios, $(0.80+1.10)/2=0.95$, nor should it be described as an equal-stratum average. This approximation illustrates the role of information weights; with effects of this size, it is not an exact identity for every fitted stratified Cox model.

Reading Imbalance and Diagnosing Problems

Information concentration is not automatically a defect. If the treatment effect is the same in every stratum and the stratified logrank model is appropriate, a dominant stratum contributes a large share of precision rather than invalidating the combined test. The concern is whether the observed information distribution matches the design’s purpose and whether the interpretation remains appropriate.

For an interim or final analysis, inspect at least the following:

Information shares: calculate $q_k=I_k/I$, rather than inferring contributions from event counts.
Event shares: compare each stratum’s proportion of events with its information share. A large difference can reveal risk-set imbalance or different allocation patterns.
Very small contributions: a stratum with little information may have a highly variable stratum-specific estimate, even if its event count appears nontrivial.
Dominant contributions: check whether one stratum supplies most information because of the planned allocation, accumulated follow-up, censoring, or unexpectedly sparse risk sets elsewhere.
Consistency over time: at a sequence of analyses, calculate cumulative stratum-specific information at each cutoff. The distribution can change as events and censoring accumulate.

These checks should be based on the same event definition, risk-set conventions, and tie handling as the primary analysis. A discrepancy between planned and observed shares can be a sign to investigate trial conduct or accrual and follow-up patterns; it is not, on its own, evidence that the treatment effect has changed.

Comparison, Limitations, and Common Mistake

The stratified logrank test combines stratum-specific scores while retaining the stratification in the risk-set calculations. A pooled, unstratified logrank test instead uses pooled risk sets and may fail to account adequately for differences in baseline event patterns across strata. The choice between them is a design and modeling decision; information shares do not determine that choice. Within a stratified analysis, reporting only the total event count is also inadequate for diagnosing information imbalance. Event counts describe how many failures occurred, while information incorporates the risk-set treatment allocation at those failures.

The simple event-contribution expression assumes distinct events and a binary treatment indicator. Weighted logrank tests, tied-event methods, covariate adjustment, or other score definitions require their corresponding information calculations. Also, $q_k$ describes the allocation of statistical information; it does not account for every consideration in transporting a result to a target population. In particular, it cannot resolve effect heterogeneity, selection into strata, or whether the chosen strata capture clinically important differences.

Common mistake: “Each stratum has the same number of events, so each contributes the same information.” This is wrong because information per event depends on the risk-set composition. Calculate the stratum-specific information from its risk sets, then compare $I_k$ or $q_k$; do not substitute event counts for information.

Check Your Understanding

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

At an event time, 30 treatment and 70 control participants are at risk. What is the event’s logrank information contribution under the stated formula?
Two strata contribute $I_1=12$ and $I_2=8$. Find the total information and each stratum’s information share.
Why can equal event counts produce unequal stratum-specific information?
If one stratum has $q_k=0.95$, what does that say about the combined information, and what does it not establish?
In the local-alternative approximation, two strata have $I_1=9$, $I_2=1$, $\beta_1=-0.10$, and $\beta_2=0.20$. Calculate the information-weighted summary.

Solutions. (1) The treatment fraction is $30/100=0.30$, so the contribution is $0.30(0.70)=0.2100$. (2) $I=12+8=20$; the shares are $12/20=0.60$ and $8/20=0.40$. (3) Information per event depends on the treatment and control risk-set proportions, through $r(1-r)$, not only on the number of events. (4) The stratum supplies 95% of the total information, so the combined result is heavily informed by it. This alone does not establish that the test is invalid, that the treatment effect is homogeneous, or that a causal effect exists. (5) The summary is $\{9(-0.10)+1(0.20)\}/10=(-0.90+0.20)/10=-0.0700$. It is an information-weighted local approximation, not necessarily an exact model-based effect estimate.