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Stagewise Boundaries and Familywise Error · Tutorial 170 of 499

Inverse-Normal Boundaries Across Arms and Stages

Learn to combine stagewise evidence within each arm and calibrate boundaries across arms and stages while controlling familywise error.

Advanced 12 min read

What You'll Learn

  • Construct an inverse-normal combination statistic from valid stagewise p-values.
  • State the independence and weighting conditions that give the statistic a standard normal null distribution.
  • Calibrate a joint armwise boundary using the dependence among treatment-arm statistics.
  • Extend calibration to repeated looks by accounting for covariance across arms and stages.
  • Use closed testing or Bonferroni calibration to obtain strong familywise error control.
  • Recognize when arm selection or misspecified dependence makes a nominal boundary invalid.

Why combine evidence across arms and stages?

In a multi-arm, multi-stage trial, each treatment arm has its own efficacy hypothesis, but evidence for that hypothesis may arrive in several stages. A stagewise p-value summarizes the evidence from one stage; an inverse-normal combination statistic puts those p-values onto a common scale and combines them within an arm. A second calibration is then needed: testing several arms, and possibly looking more than once, increases the chance of at least one false rejection.

The goal here is to separate these two tasks. First construct a valid arm-specific combination statistic. Then choose rejection boundaries that control the probability of falsely rejecting one or more true arm hypotheses. The introductory material on stagewise familywise error boundaries provides useful context; the calculations below focus on inverse-normal combination and its joint calibration.

Definition: For arm \(j\) and stage \(l\), let \(p_{jl}\) be a valid one-sided stagewise p-value for the efficacy hypothesis \(H_j\), with smaller values supporting efficacy. Define \(Z_{jl}=\Phi^{-1}(1-p_{jl})\), where \(\Phi\) is the standard normal cumulative distribution function. Given prespecified weights \(w_l\) satisfying \(\sum_{l=1}^L w_l^2=1\), the inverse-normal combination statistic is \(Z_j=\sum_{l=1}^L w_l Z_{jl}\). The arm-specific combined p-value is \(1-\Phi(Z_j)\).

The standard exact calibration assumes that, under \(H_j\), each stagewise p-value is uniformly distributed on \([0,1]\), and that the stagewise p-values are independent across stages. Then each \(Z_{jl}\) has a standard normal distribution, and the weighted sum \(Z_j\) is standard normal because its mean is zero and its variance is \(\sum_l w_l^2=1\). If stagewise p-values are only super-uniform, or the stages are dependent, this exact normal calibration need not hold; a valid null distribution or a conservative procedure is needed instead.

The arm statistics need not be independent. Shared controls, for example, can induce positive correlation between treatment-versus-control statistics. That dependence affects the probability of at least one rejection across arms, even when each individual \(Z_j\) has the correct standard normal null distribution.

Combining stagewise evidence within an arm

The weights describe how stagewise evidence contributes to the final statistic. Equal weights across two stages are \(w_1=w_2=1/\sqrt{2}\). Unequal weights are also possible, but must be chosen before examining the stagewise p-values for the analysis at hand. A larger weight gives that stage more influence, while the sum-of-squares condition preserves unit variance under the stated independence assumptions.

Worked Example: Combining Two Stages for One Arm

Suppose a hypothetical two-stage trial has valid, independent stagewise p-values \(p_{11}=0.040\) and \(p_{12}=0.200\) for arm 1. Use prespecified weights \(w_1=0.6\) and \(w_2=0.8\). These satisfy \(0.6^2+0.8^2=1\).

Convert each p-value to its standard normal score: \(Z_{11}=\Phi^{-1}(0.960)\approx1.7507\) and \(Z_{12}=\Phi^{-1}(0.800)\approx0.8416\).

$$ Z_1=0.6(1.7507)+0.8(0.8416) =1.0504+0.6733 =1.7237. $$

The combined one-sided p-value is \(1-\Phi(1.7237)\approx0.0424\). Under the stated null assumptions, this is the p-value for the prespecified inverse-normal combination test of arm 1. It is not, by itself, an adjustment for having tested multiple arms.

Stage independence is a substantive assumption, not a consequence of applying the inverse-normal transformation. It is most plausible when stages use nonoverlapping data and stagewise statistics are constructed without reusing earlier observations. If the same patients contribute to more than one stagewise test, or if a later analysis uses information from an earlier one in a way that compromises independent increments, the variance and null distribution of the weighted sum may differ from the standard normal distribution.

Joint calibration across arms

For a set \(I\) of arms whose hypotheses form an intersection null, a joint max test rejects that intersection when at least one corresponding statistic exceeds a critical value. If the joint null distribution is known, the critical value \(c_I\) can be calibrated to satisfy \(P(\max_{j\in I} Z_j>c_I\mid H_I)=\alpha\), where \(H_I\) states that every hypothesis in \(I\) is true. This controls the probability of falsely rejecting that intersection at level \(\alpha\).

A single common critical value is convenient when the statistics have the same marginal null distribution. Its value depends on their joint dependence. With independent arms, the probability that all \(|I|\) statistics are at most \(c_I\) is \(\Phi(c_I)^{|I|}\). With correlated arms, calibration instead uses their joint multivariate normal distribution, provided that model is justified. A numerical calibration should use the correct correlation matrix rather than treating shared-control statistics as independent.

Worked Example: Exact Calibration for Two Independent Arms

Consider two hypothetical arm statistics \(Z_1,Z_2\) that are independent standard normal variables under the global null. We want a one-sided global familywise error rate of \(\alpha=0.10\), using the rule that rejects if either statistic exceeds a common boundary \(c\).

The probability of no rejection is \(P(Z_1\leq c,Z_2\leq c)=\Phi(c)^2\). Therefore the familywise error is \(1-\Phi(c)^2\), and the boundary must solve \(1-\Phi(c)^2=0.10\).

$$ \Phi(c)=\sqrt{0.90}=0.948683, \qquad c=\Phi^{-1}(0.948683)\approx1.6322. $$

Indeed, \(\Phi(1.6322)\approx0.9487\), so the joint rejection probability is approximately \(1-(0.9487)^2=0.1000\). The corresponding marginal tail probability is about \(1-0.9487=0.05132\). Thus each arm has a nominal boundary that is less stringent than \(0.05\), but the probability of rejecting at least one of the two independent true hypotheses is \(0.10\), as targeted. This calculation is for the global intersection; strong familywise control for all configurations of true and false hypotheses requires valid tests for the other intersections as well.

Closed testing supplies that broader guarantee. For every nonempty subset \(I\) of arm hypotheses, test the intersection \(H_I\) at level \(\alpha\). Reject an individual \(H_j\) only if every intersection \(H_I\) containing \(H_j\) is rejected. If each local intersection test has type I error no greater than \(\alpha\) whenever its intersection null is true, the closed procedure strongly controls familywise error: whenever any true hypothesis is rejected, at least one true intersection containing that hypothesis must also have been rejected incorrectly.

For the local test, a max statistic over \(j\in I\) is natural. Its critical value may differ across subsets because the subset's size and correlation structure can differ. Under subset pivotality, the joint null distribution of statistics for a true subset is unaffected by which hypotheses outside that subset are false, so its global-null subset distribution can be used for local calibration. If that assumption is not justified, the local test must instead be calibrated for the relevant intersection null.

Boundaries across repeated stages

A trial may also permit efficacy rejection at an interim look. Let \(q_t=\sum_{l=1}^t w_l^2\), the cumulative weight variance through stage \(t\), and define the standardized partial statistic \(U_{jt}=\sum_{l=1}^t w_l Z_{jl}/\sqrt{q_t}\). Under independent standard normal stagewise scores, each \(U_{jt}\) is standard normal under \(H_j\). A sequential rule can reject when \(U_{jt}\) crosses a boundary at any stage.

The repeated looks are correlated: \(U_{jt}\) and \(U_{js}\) share early-stage evidence. If the stagewise cross-arm correlation at stage \(l\) is \(R_{l,ij}\), then under independent stages the covariance is \(\operatorname{Cov}(U_{it},U_{js})=\sum_{l\leq\min(t,s)}w_l^2R_{l,ij}/\sqrt{q_tq_s}\). Thus a boundary for a maximum over arms and stages must account for both sources of dependence, not just the number of arms.

A simple simultaneous-boundary construction uses a common boundary \(b_I\) for the statistics in an intersection: \(P(\max_{j\in I,\,t\leq L} U_{jt}\geq b_I\mid H_I)=\alpha\). This probability can be evaluated from the joint multivariate normal distribution when its covariance assumptions are appropriate. More flexible stage-specific boundaries \(b_{It}\) can be calibrated to a specified error-spending plan, but their joint crossing probability must still be controlled. In either case, local tests for all intersections can be combined by closed testing for strong familywise control.

Justification of the sequential calibration: Under the intersection null, collect the standardized partial statistics \(U_{jt}\) for all \(j\in I\) and all looks \(t\) into one vector. Under independent normal stagewise increments, this vector has a multivariate normal distribution with the covariance given above. Choosing \(b_I\) so the probability that any component exceeds \(b_I\) is \(\alpha\) makes the local intersection test level \(\alpha\). Closed testing then limits the probability of any false individual rejection to \(\alpha\).

Worked Example: A Dependence-Robust Boundary for Three Arms and Two Looks

Suppose a hypothetical trial has three treatment arms and two planned efficacy looks. At each arm-look pair, assume the one-sided stagewise p-value is valid under that arm's null. No particular dependence among arms or looks is assumed. Consider the global intersection null and target \(\alpha=0.05\).

There are \(3\times2=6\) opportunities for a false crossing. By the union bound, if each arm-look test uses tail probability \(\alpha/6=0.008333\), then the probability of at least one false crossing is no greater than the sum of the six probabilities:

$$ P(\text{at least one false crossing}) \leq 6(0.008333) \approx0.0500. $$

Reject the global intersection if any of the six valid arm-look p-values is at most \(0.05/6\). By validity, each true-null crossing probability is at most \(0.05/6\), so the union bound gives the stated guarantee. The union-bound guarantee does not require independence, but it is generally conservative because crossings may be dependent. To obtain strong familywise control through closure, construct a level-\(0.05\) local test for every intersection, using the relevant number of arms and looks in the Bonferroni allocation or a justified joint calibration. This example uses repeated-look tests directly; it is an alternative to a calibrated inverse-normal crossing rule, not a change to the combination statistic's formula.

Diagnostics, alternatives, and limitations

Before using an inverse-normal boundary, check the ingredients that determine its null distribution. Confirm that stagewise p-values are valid for the stated hypotheses, that the weighting rule was specified without using the current stage's p-values, and that the assumed independence across stages is defensible. For joint calibration, inspect whether the arm correlation matrix reflects the design, including shared controls and any planned adaptations. A boundary calibrated under independence can be too liberal if the true dependence differs in a way that increases the relevant maximum-tail probability.

Adaptation requires particular care. If arms are selected or dropped using interim evidence, simply treating the surviving stagewise p-values as if the original fixed set had been tested independently can invalidate the combination or its joint calibration. Stagewise p-values must remain valid under the adaptation, and the joint rejection rule must account for the selection procedure. Conditional-error methods provide one established way to preserve error after adaptations; the inverse-normal weights alone do not repair invalid selection.

ApproachWhat it usesMain trade-off
Inverse-normal combinationValid stagewise p-values and prespecified weightsFlexible across stages, but exact normal calibration requires suitable stage independence and p-value validity
Joint max calibrationJoint distribution of arm and stage statisticsCan be efficient, but depends on a correctly specified covariance model
Bonferroni calibrationValid marginal tests and an error allocationValid under arbitrary dependence, often conservative
Closed testingLevel-controlled local tests for every intersectionProvides strong familywise control, but may require many local calibrations

A common mistake is to calibrate each arm at level \(\alpha\) and conclude that the familywise error is also \(\alpha\). That only controls each arm's marginal error; it does not generally control the chance of at least one false rejection across arms or repeated looks. Another mistake is to combine p-values and then ignore their dependence when calibrating the maximum. The correction is to define the joint rejection event explicitly and calibrate its probability under each relevant intersection, or use a dependence-robust allocation such as Bonferroni. These procedures can lose power, and exact joint calibration can be sensitive to covariance misspecification; neither limitation is removed merely by using inverse-normal weights.

Check Your Understanding

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

  1. For two independent stagewise p-values, why must the inverse-normal weights satisfy a sum-of-squares condition for a standard normal combined statistic?
  2. For two independent standard normal arm statistics, what equation defines a common boundary giving global familywise error \(0.10\)?
  3. What dependence information is needed to calibrate a max boundary over several arms and looks?
  4. Why does testing every intersection matter when the aim is strong familywise error control?

Solutions. (1) Each transformed stage score is standard normal under its null, so the weighted sum has variance \(\sum_l w_l^2\); setting that sum to 1 gives unit variance and hence a standard normal distribution under the independent-normal assumptions. (2) The no-rejection probability is \(\Phi(c)^2\), so solve \(1-\Phi(c)^2=0.10\), giving \(c=\Phi^{-1}(\sqrt{0.90})\approx1.6322\). (3) One needs the joint covariance structure across arms and across looks, as well as valid marginal null distributions; shared controls and overlapping stage data can affect it. (4) A false rejection of an individual true hypothesis must also entail a false rejection of a true intersection containing it. Level-\(\alpha\) tests for all intersections therefore limit the probability of any false individual rejection to \(\alpha\).

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