Tutorials › Group Sequential Methods and Sample Size Re-Estimation › MaxCombo Tests for Uncertain Effect Timing

Non-Proportional Hazards and Delayed Effects · Tutorial 532 of 699

MaxCombo Tests for Uncertain Effect Timing

Construct and calibrate a MaxCombo test using its joint covariance, then interpret its error rate, power, assumptions, and limitations.

Advanced 12 min read

What You'll Learn

  • Define a one-sided MaxCombo statistic from multiple weighted logrank scores.
  • Calculate correlations among the standardized scores from their covariance matrix.
  • Calibrate a maximum statistic under the joint null distribution rather than treating its component tests as independent by default.
  • Explain how a prespecified set of time weights can protect power when the timing of benefit is uncertain.
  • Identify assumptions, diagnostics, and limits relevant to interpretation.

Why Combine Several Weighted Logrank Tests?

A weighted logrank test can emphasize events at different parts of follow-up. An early-weighted test may be sensitive to an early treatment effect, while a late-weighted test may be more sensitive to a delayed effect. If the timing is uncertain, choosing one weight after looking at the observed treatment pattern risks overstating evidence. A MaxCombo test instead specifies several weights in advance and combines their test statistics by taking their maximum. Its calibration accounts for the fact that the statistics usually use the same patients and events, so they are correlated.

The procedure is aimed at a prespecified, one-sided question: is there evidence of a treatment benefit in the direction encoded by the score? It is not a license to try many weights and report whichever looks most favorable without adjusting for that search. Here, “maximum” means the largest of the signed, benefit-oriented standardized statistics—not the largest absolute statistic.

Definition: Suppose \(K\) weighted logrank scores \(U_1,\ldots,U_K\) are specified before examining the treatment results. Let \(V_j\) be the variance estimate for \(U_j\), and define \(Z_j=U_j/\sqrt{V_j}\). The one-sided MaxCombo statistic is \(M=\max_{1\leq j\leq K} Z_j\). The component scores are oriented so that a positive \(Z_j\) indicates fewer observed events in the treatment group than expected under the null.

For clarity, consider two randomized groups and an event time \(t_i\). Let \(Y_1(t_i)\) and \(Y_0(t_i)\) be the numbers at risk in the treatment and control groups just before that time, \(Y(t_i)=Y_1(t_i)+Y_0(t_i)\), and \(d(t_i)\) the total number of events. If \(O_1(t_i)\) is the treatment-group event count, its null expected count is \(Y_1(t_i)d(t_i)/Y(t_i)\). A weighted score can be written as

$$ U_j=\sum_i w_j(t_i)\left\{\frac{Y_1(t_i)}{Y(t_i)}d(t_i)-O_1(t_i)\right\}. $$

Thus, fewer treatment events than expected contribute positively. The variance \(V_j\) is the usual variance for that weighted logrank score, accounting for the risk sets, weights, and ties as appropriate. In practice, the scores and their covariance are computed together; one should not assume the component statistics are independent merely because their weights differ.

Null Calibration: The Joint Distribution Matters

The usual large-sample justification is that, under the null hypothesis of no treatment difference in the event-time distributions, the vector of prespecified weighted logrank scores is approximately multivariate normal with mean zero. Let \(C_{jk}=\operatorname{Cov}(U_j,U_k)\), estimated under the null, and let \(R\) be the correlation matrix of the standardized statistics. Its entries are

$$ R_{jk}=\frac{C_{jk}}{\sqrt{V_jV_k}}. $$

Under this approximation, \((Z_1,\ldots,Z_K)\) has a multivariate normal null distribution with mean vector zero and correlation matrix \(R\). If the observed maximum is \(m\), the one-sided p-value is

$$ p_{\mathrm{MaxCombo}}=P\left(\max_j Z_j\geq m\mid H_0\right) =1-P(Z_1<m,\ldots,Z_K<m\mid H_0). $$

The probability on the right is a joint multivariate normal probability. It can be evaluated with a multivariate normal calculation or approximated by simulation from the specified joint null distribution. The same principle gives a level-\(\alpha\) critical value \(c\): choose \(c\) such that \(P(\max_j Z_j\leq c\mid H_0)=1-\alpha\). This calibration controls the probability of rejecting \(H_0\) at the prespecified analysis under the null, subject to the approximation and assumptions.

Worked Example: Constructing the Standardized Statistics and Their Correlation

Suppose a hypothetical trial prespecifies two benefit-oriented weights. At the planned analysis, their weighted score summaries are \(U_1=6\), \(U_2=8\), with variance estimates \(V_1=9\), \(V_2=16\), and estimated covariance \(C_{12}=6\). These summaries are hypothetical and are supplied to illustrate construction.

The standardized statistics are

$$ Z_1=\frac{6}{\sqrt{9}}=2.00,\qquad Z_2=\frac{8}{\sqrt{16}}=2.00. $$

Their estimated correlation is

$$ R_{12}=\frac{6}{\sqrt{9\cdot16}}=\frac{6}{12}=0.500. $$

Therefore \(M=\max(2.00,2.00)=2.00\), and the null calibration must use a bivariate normal distribution with correlation \(0.500\). The two marginal statistics are not independent. In particular, multiplying their marginal probabilities as though the correlation were zero would use the wrong null distribution.

What the Maximum Buys—and What It Costs

Taking the maximum allows the data to obtain evidence through whichever prespecified weight is most responsive to the actual effect pattern. If treatment benefit is delayed, for example, a late-emphasizing statistic might be larger than an ordinary logrank statistic. But maximizing over several statistics creates more opportunities for a large value under the null. The joint calibration raises the rejection threshold, or equivalently adjusts the p-value, to account for those opportunities.

The ordinary logrank test is a single component with a particular time weighting and can be efficient when its assumptions about the effect pattern are appropriate. A single prespecified weighted logrank test can be more focused when the relevant timing is well justified in advance. MaxCombo trades some of that focused power for protection against uncertainty across its chosen weight patterns. It does not guarantee greater power than every single test for every alternative.

Worked Example: Independent Components and the Multiplicity Adjustment

For a simple hypothetical calculation, suppose two standardized component statistics are independent under \(H_0\), and both observed values are \(2.10\). Independence is a special case for illustration, not a default assumption for weighted logrank statistics. Then \(M=2.10\), and

$$ p_{\mathrm{MaxCombo}} =1-P(Z_1<2.10,Z_2<2.10) =1-\Phi(2.10)^2 \approx 0.03541. $$

Thus, the one-sided, multiplicity-adjusted p-value is about \(0.0354\). It is larger than the single-component one-sided tail probability, \(1-\Phi(2.10)\), because the test considers either component capable of producing a large value. For a level-\(0.05\) test in this independent two-statistic case, the critical value solves \(\Phi(c)^2=0.95\), so

$$ c=\Phi^{-1}(\sqrt{0.95})\approx1.9545. $$

Since \(2.10>1.9545\), the maximum exceeds this critical value. The calculation shows why comparing \(M\) with the ordinary one-sided normal cutoff \(1.645\) would be incorrect for this two-component test.

For two standard normal components with correlation \(\rho\), the null probability is instead \(P(Z_1<m,Z_2<m)\) under that correlation. With more components, the full matrix \(R\) matters. A common practical approach is to simulate many vectors from the estimated joint null distribution, calculate the maximum for each vector, and compare the observed maximum with this simulated reference distribution. Simulation error should be small relative to the desired precision of the p-value or critical value.

Worked Example: How Correlation Changes the Null Tail

Consider two hypothetical standard normal null statistics with correlation \(\rho=0.500\). Suppose the observed maximum is \(m=0\), so both observed component statistics are nonpositive. For a centered bivariate normal vector,

$$ P(Z_1\leq0,Z_2\leq0)=\frac14+\frac{\arcsin(\rho)}{2\pi}. $$

Here \(\arcsin(0.500)=\pi/6\), giving

$$ P(Z_1\leq0,Z_2\leq0)=\frac14+\frac{\pi/6}{2\pi} =\frac14+\frac1{12}=\frac13. $$

Consequently, \(P(M\geq0)=1-1/3=2/3\). If the statistics were incorrectly treated as independent, the corresponding probability would be \(1-\Phi(0)^2=1-0.25=0.75\). Even this simple boundary calculation shows that dependence changes the calibration. For an observed value above zero, the appropriate probability is obtained from the bivariate normal distribution at that value and correlation, not from the independence formula.

Power When Effect Timing Is Uncertain

Under an alternative, the standardized vector is often approximated by a multivariate normal distribution with a nonzero mean vector \(\boldsymbol{\delta}\) and a covariance matrix that may be approximated by its null value for planning. The power of a test rejecting when \(M>c\) is then \(P(\max_j Z_j>c)\) under that alternative distribution. If the effect timing matches one of the prespecified weights, the corresponding component may have a relatively large mean. If it falls between the patterns represented by the weights, no component is guaranteed to be highly responsive.

Worked Example: Illustrating Power Across Two Timing Patterns

Suppose a hypothetical planning calculation uses two independent components and a level-\(0.05\) critical value \(c=1.9545\). Under one specified delayed-effect alternative, assume the approximate means are \(\delta_1=1.50\) and \(\delta_2=0.20\), with unit variances. Under this model, \(Z_1\) and \(Z_2\) are independent normal variables with those means. Then

$$ \begin{aligned} \operatorname{Power} &=P(Z_1>c\text{ or }Z_2>c)\\ &=1-P(Z_1\leq c)P(Z_2\leq c)\\ &=1-\Phi(1.9545-1.50)\Phi(1.9545-0.20)\\ &=1-\Phi(0.4545)\Phi(1.7545)\approx0.352. \end{aligned} $$

In this illustration, the first component carries most of the alternative signal, while the second provides some additional opportunity to reject. This is a model-based power calculation, not a universal power guarantee: different dependence, effect sizes, censoring patterns, or timing can change the result. In a real design, the means and covariance should be justified under the assumed event-time model or evaluated by suitable simulation.

Assumptions, Diagnostics, and Limitations

The large-sample null calibration relies on standard weighted logrank conditions. Treatment assignment should be randomized or otherwise support the intended group comparison; subjects should be independent; and censoring should be noninformative for the event process within the analysis groups or strata. The weights and the component family must be specified before examining the treatment contrasts. The score variances and cross-covariances must be calculated consistently with the chosen weights and design. A small number of events, sparse risk sets, or extreme weights can make a normal approximation less reliable.

Proportional hazards are not required for the test to be defined or for its null calibration. Non-proportional effects are part of the motivation for using multiple time weights. However, a single MaxCombo p-value does not summarize the shape, timing, or clinical size of an effect; effect estimates and survival summaries still need context. Nor does a significant result establish which component represents the “true” timing. The largest observed component is selected by construction and may be an unstable description of the effect pattern.

Useful checks include confirming that the intended weights were fixed in advance, verifying the signs and scales of all scores, inspecting the estimated correlation matrix for valid entries and plausible dependence, and checking that the computation uses the full joint distribution. If the correlation matrix is nearly singular, some components may be redundant; this is not automatically an error, but it can signal that the nominal collection of weights supplies little distinct information. If estimated correlations are invalid or numerical calculations are unstable, investigate the covariance calculation, ties, risk sets, and simulation procedure rather than silently substituting independence.

MaxCombo is also distinct from repeated interim testing. Its adjustment handles the maximum across the specified weighted statistics at the analysis being calibrated; it does not by itself account for repeatedly examining accumulating data at multiple analysis times. Any sequential monitoring plan requires an appropriate sequential design and calibration in addition to choosing the weighted statistics.

1
Specify the candidates.
Choose the benefit direction and the finite set of weights before examining treatment-effect results, based on plausible timing patterns.
2
Compute the joint score information.
Calculate each weighted score, its variance, and the covariance matrix for all component scores using the same analysis data and consistent methods.
3
Form and calibrate the maximum.
Standardize the scores, take their maximum, and obtain its p-value or critical value from the joint null distribution, not an unadjusted single-test reference.

Common mistake: “The largest component is \(2.10\), so its ordinary one-sided normal p-value is the MaxCombo p-value.” That ignores the chance that any one of several statistics could be large under \(H_0\). The correction is to calculate \(P(\max_j Z_j\geq2.10\mid H_0)\) using the joint null distribution and the component correlations. A second error is to use the maximum absolute statistic for a benefit-only test; that changes the rejection region and must not be presented as the same one-sided procedure.

Key takeaway: MaxCombo addresses uncertainty in effect timing by taking the maximum over a prespecified collection of weighted logrank statistics. Its validity depends on calibrating that maximum under the joint null distribution, with the dependence among component statistics properly represented.

Check Your Understanding

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

  1. Two prespecified standardized statistics have values \(1.2\) and \(1.7\). What is the MaxCombo statistic, and what further information is needed for its p-value?
  2. Why is it generally inappropriate to calculate a MaxCombo p-value by multiplying marginal probabilities as if the component statistics were independent?
  3. For two independent standard normal components, write the equation defining the level-\(0.05\) critical value.
  4. Does MaxCombo require proportional hazards? Name two assumptions that still matter for its usual null calibration.
  5. Does the maximum component identify the true timing of treatment benefit?

Solutions. (1) The maximum is \(M=1.7\). Its p-value requires the joint null distribution, including the correlation matrix of the two statistics. (2) The statistics usually share subjects and events, so their dependence affects the probability that at least one crosses a threshold; ignoring it miscalibrates the test. (3) The equation is \(\Phi(c)^2=0.95\), giving \(c=\Phi^{-1}(\sqrt{0.95})\approx1.9545\). (4) No. Independence of subjects and noninformative censoring are examples of assumptions that matter; prespecification of the weights and valid score covariance estimation also matter. (5) No. The largest component is selected by the maximum and may be unstable; it is evidence from a responsive weight, not proof of the true effect-timing pattern.

Clinical Biostats