Combining Stratum Scores And Information

Why Combine Within-Stratum Evidence?

In a stratified logrank analysis, participants are grouped by a baseline factor such as disease stage or clinical site. The comparison of treatment groups is made within each stratum, using its own risk sets. At an interim look, the analysis needs one score and one variance that summarize the evidence across all strata. The key operation is to add the stratum-specific scores and add their variances, then standardize the total. It is not to average the separate standardized statistics.

This construction preserves the meaning of the logrank score: observed treatment-group events minus the number expected under the null, conditional on the risk sets. It also accounts for how much information each stratum contributes. A stratum with many informative risk sets generally has more influence than a stratum with little information.

Definition: Suppose there are $H$ mutually exclusive strata. At a specified analysis look, let $U_h$ be the logrank score in stratum $h$, and let $V_h$ be its null variance (its contribution to information), for $h=1,\ldots,H$. The combined stratified score and variance are $$ U_{\mathrm{str}}=\sum_{h=1}^{H}U_h,\qquad V_{\mathrm{str}}=\sum_{h=1}^{H}V_h. $$ The combined standardized statistic is $$ Z_{\mathrm{str}}=\frac{U_{\mathrm{str}}}{\sqrt{V_{\mathrm{str}}}}, $$ provided $V_{\mathrm{str}}>0$.

Use one sign convention throughout. Here, $U_h$ is observed treatment-group events minus expected treatment-group events under the null. Thus a positive score indicates an excess of events in the treatment group; if the treatment is intended to reduce the event hazard, a negative score points in the anticipated direction. Reversing the score convention reverses the sign of every $U_h$ and of $Z_{\mathrm{str}}$, but does not change a two-sided p-value.

What Is Being Added?

At an event time in stratum $h$, let $Y_{1h}$ and $Y_{0h}$ be the numbers at risk in the treatment and control groups, and let $Y_h=Y_{1h}+Y_{0h}$. If there is one event, the null expected treatment-group event contribution is $Y_{1h}/Y_h$. The observed-minus-expected contribution is therefore the treatment event indicator minus $Y_{1h}/Y_h$. Summing these contributions over the event times included at the look gives $U_h$. The corresponding null variance contributions, calculated from the risk-set allocation under the selected logrank tie convention, sum to $V_h$.

For example, with $d_h$ tied events at one time and the usual conditional hypergeometric calculation, the expected treatment event count is $d_hY_{1h}/Y_h$, and its variance contribution is $$ d_h\frac{Y_{1h}}{Y_h}\frac{Y_{0h}}{Y_h} \frac{Y_h-d_h}{Y_h-1}, $$ when $Y_h>1$. The tied-event score contribution is the observed treatment event count minus that expectation. The chosen tie method must be applied consistently in the score and variance within each stratum. The combination rule does not depend on using one particular tie convention.

The stratified logrank scores at interim analyses are cumulative within their respective strata up to the current data cutoff. Once those stratum-specific quantities have been calculated, the combination step is simply addition. The definition of $V_{\mathrm{str}}$ matters: it is the variance of the total score, not the sum of stratum standard errors.

Why Scores and Variances Add

The usual justification assumes that strata are a fixed, mutually exclusive partition of participants; participants in different strata are independent; and the within-stratum logrank scores and variances are calculated under the same null hypothesis and at the same analysis look. Under the global null of no treatment-group difference in event experience within any stratum, each $U_h$ has null expectation zero and null variance $V_h$. Independent participants in disjoint strata give independent stratum scores, so their covariance is zero.

Add the scores.
The score for all strata is $U_{\mathrm{str}}=\sum_h U_h$, because the total observed-minus-expected event contrast is the sum of its within-stratum contrasts.

Calculate the variance of the sum.
Since $\operatorname{Var}(U_h)=V_h$ and $\operatorname{Cov}(U_h,U_k)=0$ for distinct independent strata, $\operatorname{Var}(\sum_h U_h)=\sum_h V_h$.

Standardize the total.
When the null distribution is adequately approximated by a normal distribution, divide the total score by its standard deviation, $\sqrt{\sum_h V_h}$.

More explicitly, the variance identity follows from $$ \operatorname{Var}\left(\sum_{h=1}^{H}U_h\right) =\sum_{h=1}^{H}\operatorname{Var}(U_h) +2\sum_{h<k}\operatorname{Cov}(U_h,U_k) =\sum_{h=1}^{H}V_h. $$ Under the null and a suitable large-sample approximation, $Z_{\mathrm{str}}$ is approximately standard normal. A two-sided p-value is then $2P(N(0,1)\geq |Z_{\mathrm{str}}|)$. A one-sided p-value is appropriate only when the direction of the alternative was specified in advance. This approximation can be unreliable when the total information is very small or the score distribution is highly discrete.

Independence is an important part of the argument, not a bookkeeping convenience. If scores from different groups are correlated, the variance of their sum includes covariance terms. In that setting, merely adding stratum variances is not justified; the analysis must account for the dependence. For the ordinary stratified logrank test, each participant belongs to one stratum, and independent participants with appropriately handled censoring support the usual calculation.

Worked Example: Combining Two Strata at One Look

Consider a hypothetical randomized trial with two prespecified baseline strata and an interim cutoff applied to both. Assume participants are independent, censoring is noninformative within treatment group and stratum, the tie convention has been applied consistently, and both strata have enough information for a normal approximation to be reasonable. Suppose the calculated stratum results, in score units, are:

Stratum 1: $U_1=-2.4$, $V_1=5.76$. Stratum 2: $U_2=-4.8$, $V_2=12.25$.

First add the scores and variances: $$ U_{\mathrm{str}}=-2.4+(-4.8)=-7.2,\qquad V_{\mathrm{str}}=5.76+12.25=18.01. $$ Then standardize: $$ Z_{\mathrm{str}}=\frac{-7.2}{\sqrt{18.01}} =\frac{-7.2}{4.244}= -1.697. $$

The negative sign points toward fewer treatment-group events under the stated convention. The approximate two-sided p-value is $2P(N(0,1)\geq 1.697)\approx 0.0898$. For a prespecified lower-tailed alternative, the approximate p-value is $P(N(0,1)\leq -1.697)\approx 0.0449$. These are different tests: the one-sided conclusion is warranted only if that direction was specified before examining the interim results.

Why Averaging Stratum Statistics Is Wrong

A stratum-specific standardized statistic is $Z_h=U_h/\sqrt{V_h}$. Averaging the $Z_h$ values gives each stratum equal weight, regardless of how much information it contains. It also does not produce the variance-standardized sum of the scores. The correct relationship is $$ Z_{\mathrm{str}}= \frac{\sum_{h=1}^{H}\sqrt{V_h}\,Z_h} {\sqrt{\sum_{h=1}^{H}V_h}}. $$ The weights arise because the scores, not their standardized versions, are added. The formula also shows that the combined test is not generally the arithmetic mean of the stratum $Z$ statistics.

Worked Example: A Simple Average Can Mislead

Suppose a hypothetical trial has two independent strata with $U_1=-3$, $V_1=4$, $U_2=2$, and $V_2=25$. Assume the quantities are calculated at the same look under the same null and that the large-sample approximation is adequate.

The individual standardized values are $$ Z_1=\frac{-3}{\sqrt{4}}=-1.5,\qquad Z_2=\frac{2}{\sqrt{25}}=0.4. $$ Their arithmetic average is $(-1.5+0.4)/2=-0.55$, but this is not the stratified statistic. The correct calculation is $$ U_{\mathrm{str}}=-3+2=-1,\qquad V_{\mathrm{str}}=4+25=29,\qquad Z_{\mathrm{str}}=\frac{-1}{\sqrt{29}}=-0.1857. $$

The second stratum has more information, as shown by $V_2=25$, and its positive score offsets some of the first stratum’s negative score. The unweighted average of the two $Z_h$ values treats their precision as equal and does not preserve the variance of the total score. Averaging raw scores would also be unjustified: it would change the scale without giving the corresponding variance of that average.

Checking the Calculation and Its Interpretation

Before reporting a combined result, verify that the stratum calculations refer to the same analysis cutoff, treatment coding, null hypothesis, and tie-handling convention. Check that the stated strata form a non-overlapping partition and that no participant contributes to more than one stratum. For each stratum, confirm that the score and variance use the same risk sets and event data. Then check the two totals separately: the total score should equal the sum of the stratum scores, and the total variance should equal the sum of their variances under the independence assumption.

Zero or tiny stratum variance: A stratum with $V_h=0$ contributes no standardized evidence; if it has no informative event comparisons, it should not be assigned an artificial $Z_h$. A very small $V_h$ can make its individual $Z_h$ unstable, even though its contribution to the total score and variance remains the appropriate combination.
Sparse data: A very small total $V_{\mathrm{str}}$, few events, or highly discrete scores can make the normal approximation poor. The formula for combining scores remains the same, but the approximate p-value may not be reliable.
Opposing stratum scores: Scores with different signs can cancel in the total. This is not an arithmetic error; it may reflect genuinely different observed treatment contrasts. Report the stratum results when scientifically relevant, but do not replace the prespecified combined test with a post hoc choice of favorable strata.
Dependence across strata: If the score contributions are correlated, the total variance must include their covariances. Adding only $V_h$ would understate or overstate uncertainty.

The stratified test and the unstratified logrank test are not interchangeable calculations. The former compares treatment groups within strata and combines those contrasts; the latter uses pooled risk sets and may therefore reflect a different comparison when baseline event patterns differ across strata. Nor does the combined statistic assert that every stratum has the same estimated treatment effect. It is a single test of the specified stratified null, and its power and interpretation can be affected by heterogeneity in treatment effects. These limits are reasons to prespecify the stratification and interpret the overall result alongside relevant clinical context, not reasons to average stratum statistics.

Worked Example: Three Strata and a Direction Check

Suppose a hypothetical trial has three prespecified strata. At a common interim cutoff, their score and variance calculations are $U_1=-1.2, V_1=2.25$; $U_2=-2.0, V_2=4.00$; and $U_3=0.5, V_3=1.00$. Assume independent participants across strata, noninformative censoring within strata and treatment groups, consistent tie handling, and enough total information for an approximate normal calculation.

The combined score is $$ U_{\mathrm{str}}=-1.2-2.0+0.5=-2.7, $$ and the combined variance is $$ V_{\mathrm{str}}=2.25+4.00+1.00=7.25. $$ Thus $$ Z_{\mathrm{str}}=\frac{-2.7}{\sqrt{7.25}} =\frac{-2.7}{2.693}=-1.003. $$

The statistic is negative, so the total observed score points toward fewer treatment-group events; the positive score in stratum 3 partially offsets the negative scores in strata 1 and 2. The approximate two-sided p-value is about $0.316$, so this calculation does not provide strong evidence against the two-sided null at conventional significance levels. The sign alone should not be reported as proof of a treatment benefit.

Common Mistake

Wrong reasoning: “Each stratum has its own Z statistic, so I will average the Z values to obtain the overall Z.” This assigns equal weight to strata with potentially very different information and ignores the variance of the combined score.

Correction: Add the stratum scores, add their variances when the strata are independent, and divide the total score by the square root of the total variance. If the individual $Z_h$ values are all that is available, recover each score as $U_h=\sqrt{V_h}Z_h$, then use the combined formula. The stratum variances are essential; the $Z_h$ values alone do not determine the combined statistic.

Key takeaway: A stratified logrank statistic is the standardized sum of within-stratum logrank scores: $Z_{\mathrm{str}}=(\sum_h U_h)/\sqrt{\sum_h V_h}$, under the usual independence and large-sample conditions. It is not an average of stratum-specific standardized statistics.

Check Your Understanding

For each question, calculate or explain the requested result. Solutions follow the questions.

Two independent strata have $U_1=1.5, V_1=4$ and $U_2=-2.5, V_2=9$. Find the combined score, variance, and standardized statistic.
Why is $(Z_1+Z_2)/2$ generally not the combined stratified statistic?
If a stratum has $V_h=0$, what can be said about its contribution to the combined score and variance under the usual logrank calculation?
What additional variance terms are needed if scores from different strata are correlated?

Solutions. 1. $U_{\mathrm{str}}=1.5-2.5=-1.0$, $V_{\mathrm{str}}=4+9=13$, so $Z_{\mathrm{str}}=-1/\sqrt{13}\approx -0.2774$. 2. An arithmetic average gives equal weight to the standardized values rather than combining the scores with their variances; the correct statistic is $(\sum_h U_h)/\sqrt{\sum_h V_h}$. 3. With no information in that stratum, its score and variance contribution are zero; it supplies no evidence to the combined statistic and does not have a meaningful $Z_h=U_h/\sqrt{V_h}$. 4. The variance of the total must include $2\sum_{h<k}\operatorname{Cov}(U_h,U_k)$, as well as the individual variances.