Why the Weights Matter
An inverse-normal combination test converts each stage’s one-sided p-value to a normal score and combines the scores with prespecified weights. The weights determine how much each stage contributes to the final statistic. If the stagewise evidence is expected to differ under a particular treatment alternative, weights tailored to that alternative can produce greater power than a default choice. That gain is specific to the alternative used to choose the weights; it does not guarantee greater power for other effects.
This tutorial derives the maximizing weights for a fixed, specified alternative and fixed stagewise information. The derivation assumes a one-sided test in which larger normal scores favor the alternative. It concerns the power of the combination test, not a rule for changing weights after observing stagewise results.
Model and Optimization Target
Under the global null, if the stagewise p-values are independent and continuous uniform random variables, each \(Z_i\) is standard normal. The unit-length condition on the weights then makes \(Z_w\) standard normal under the null. Thus a one-sided level-\(\alpha\) test rejects when \(Z_w>c_\alpha\), where \(c_\alpha=\Phi^{-1}(1-\alpha)\). For the power calculation, independence and unit variances give
Since \(\Phi\) is increasing and \(c_\alpha\) does not depend on the weights, maximizing power is equivalent to maximizing \(w_1\lambda_1+w_2\lambda_2\) over nonnegative unit-length weights. More generally, for \(K\) stages, replace the weighted sum by \(\sum_{i=1}^K w_i\lambda_i\), with \(w_i\geq0\) and \(\sum_{i=1}^K w_i^2=1\).
Deriving the Power-Maximizing Weights
Write \(\lambda_+=\max(\lambda,0)\) for a single value, and let \(\lambda_i^+=\max(\lambda_i,0)\). If at least one stage has a positive alternative mean, negative or zero means cannot improve the objective: assigning weight to them uses part of the unit-length constraint without adding a positive contribution. The Cauchy–Schwarz inequality then gives
When the positive part is nonzero, equality holds by taking weights proportional to the positive parts. This yields the maximum mean \(\left(\sum_i(\lambda_i^+)^2\right)^{1/2}\). If all means are nonpositive, that proportionality argument does not apply. The maximum is instead attained by assigning all weight to a stage with the largest mean, except that when the largest mean is zero, any unit-length nonnegative weights supported only on zero-mean stages attain the maximum.
For the all-nonpositive cases, the equality conditions matter. Let \(M=\max_i\lambda_i\leq0\). For any admissible weights, \(\sum_i w_i\lambda_i\leq M\sum_iw_i\). If \(M<0\), then \(\sum_iw_i\geq1\), so \(M\sum_iw_i\leq M\). The bound \(M\) is attained by putting weight one on a stage with mean \(M\). Equality requires \(\sum_iw_i=1\); together with \(\sum_iw_i^2=1\), this permits only a single nonzero weight. If \(M=0\), the objective is at most zero and equals zero exactly when positive weights are placed only on zero-mean stages.
Worked Example: Different Stagewise Effects
Suppose a hypothetical two-stage trial has specified alternative means \(\lambda_1=1.2\) and \(\lambda_2=0.8\). Use a one-sided test with \(\alpha=0.025\). Compare the maximizing weights with square-root information weights for stage information in the ratio \(1:4\).
Both means are positive, so the maximizing weights are the normalized alternative means:
The alternative mean of the combination statistic is \(\sqrt{2.08}\approx1.442\). With \(c_{0.025}\approx1.960\), the power is \(\Phi(1.442-1.960)=\Phi(-0.518)\approx0.302\).
The square-root information weights for information ratio \(1:4\) are \((1/\sqrt{5},2/\sqrt{5})\approx(0.4472,0.8944)\). Their alternative mean is \(0.4472(1.2)+0.8944(0.8)\approx1.252\), giving power \(\Phi(1.252-1.960)\approx0.240\). In this hypothetical setting, the alternative-specific weights have higher power because they put more weight on the stage with the larger specified mean. This is a comparison at the stated alternative, not a claim that those weights are preferable for every possible treatment effect.
When the Optimal Weights Match Information Weights
In a common canonical model, a common treatment effect \(\delta>0\) gives stagewise means \(\lambda_i=\delta\sqrt{I_i}\), where \(I_i\) is the stage’s information and the effect and information are defined on compatible scales. The maximizing weights are then \(w_i=\sqrt{I_i}/\sqrt{\sum_j I_j}\). These are square-root information weights, also called weights proportional to the square root of the information fraction. They are not weights proportional to \(I_i\) itself.
Worked Example: A Common Effect and Square-Root Information
For a hypothetical trial, suppose the common specified effect is \(\delta=0.10\) on the model’s effect scale and the two stagewise information values are \(I_1=100\) and \(I_2=300\). Assume the canonical model and independent stagewise scores. Compare square-root information weights with weights proportional directly to information.
The stagewise means are \(\lambda_1=0.10\sqrt{100}=1\) and \(\lambda_2=0.10\sqrt{300}\approx1.732\). Since both are positive, the optimal weights are
Equivalently, these are \((\sqrt{100},\sqrt{300})/\sqrt{400}\). The resulting alternative mean is \(0.5000(1)+0.8660(1.732)\approx2.000\). At one-sided \(\alpha=0.025\), the power is \(\Phi(2.000-1.960)\approx0.516\).
Directly normalizing the information values gives \((100,300)/\sqrt{100^2+300^2}\approx(0.3162,0.9487)\). This is a valid unit-length vector, but it is not the canonical square-root information choice. Its alternative mean is approximately \(0.3162(1)+0.9487(1.732)=1.960\), and its power is approximately \(\Phi(1.960-1.960)=0.500\). The distinction between information and square-root information is therefore consequential even in this simple example.
Nonpositive Means and Ties
A negative stagewise mean means that, under the specified alternative, that stage’s score tends on average in the direction opposite to rejection. Giving it positive weight can reduce the combination mean. When every stagewise mean is negative, the constrained optimum is to use just one stage: the stage with the least negative mean. Spreading weight across tied negative means is not generally optimal, because a unit-length combination of several negative means can have a more negative mean than a single stage.
Worked Example: A Tie at a Negative Mean
Suppose the specified means are \(\lambda_1=\lambda_2=-1\). The largest mean is \(-1<0\), so an optimal choice puts weight one on either stage: \((1,0)\) or \((0,1)\). The maximum alternative mean is \(-1\).
Equal weights instead give \((1/\sqrt{2},1/\sqrt{2})\), with alternative mean \(-1/\sqrt{2}-1/\sqrt{2}=-\sqrt{2}\approx-1.414\), which is smaller. At one-sided \(\alpha=0.025\), the optimal power is \(\Phi(-1-1.960)\approx0.00154\); the equal-weight power is \(\Phi(-1.414-1.960)\approx0.00037\). The weights are tied as individual stages, but an arbitrary unit-length mixture supported on tied stages is not maximizing when their tied mean is negative.
For contrast, if \((\lambda_1,\lambda_2,\lambda_3)=(-0.4,0,0)\), the maximum mean is zero. Any nonnegative unit-length weights supported only on stages 2 and 3 attain it, including \((0,1,0)\) and \((0,1/\sqrt{2},1/\sqrt{2})\). Assigning positive weight to stage 1 lowers the mean.
Assumptions, Diagnostics, and Alternatives
The result is exact for the stated normal-score model: independent stages, unit stagewise variances, and a specified alternative giving fixed means \(\lambda_i\). For the usual exact null calibration, the stagewise p-values must also be independent and continuous uniform under \(H_0\), or the test must otherwise be calibrated for its actual null distribution. A valid adaptive design must preserve the required null calibration; in particular, the formula does not license treating dependent or invalid stagewise p-values as independent standard normal scores.
In adaptive re-estimation, stage-two sample size can depend on stage-one data. The result applies directly when the stagewise alternative means and stagewise score distributions are specified for the design being evaluated. If stage-two information is random, or if its selection changes the conditional distribution of the stage-two score under the alternative, a single fixed vector \((\lambda_1,\lambda_2)\) may not describe unconditional power. In that case, power must be evaluated under the actual adaptation rule, and maximizing the fixed-information expression alone is not sufficient.
Useful checks include confirming that weights are nonnegative and sum to one in squared length, that the specified effect direction agrees with the one-sided p-values, and that the alternative means reflect the stated stagewise information. A major warning sign is using \(\lambda_i=\delta\sqrt{I_i}\) despite a model with unequal stage effects, different variances, or a different score scaling. Discrete or conservative p-values can also make the transformed scores non-normal, so the displayed power formula may then be an approximation rather than an exact calculation.
| Weight choice | Basis | Main consideration |
|---|---|---|
| Power-maximizing weights | Specified stagewise alternative means | Maximizes power at that alternative under the stated model; can be sensitive to misspecification. |
| Square-root information weights | Stagewise information | Matches the optimum under a common positive effect in the canonical model; does not require choosing different weights for each alternative effect size. |
| Equal weights | Equal contribution by stage | Simple, but not generally power-maximizing when stagewise means differ. |
The main practical trade-off is between targeting and robustness. Alternative-specific weights can improve power for a scientifically justified target, but if the actual effect pattern differs, the improvement may disappear or reverse. Square-root information weights offer a natural benchmark under a common-effect model. Any chosen weights should be specified before the stagewise evidence they combine is observed; selecting them after looking at that evidence changes the procedure and requires appropriate calibration.
Common Mistake
- Wrong reasoning: “The two stages have tied largest means, so any unit-length nonnegative weights supported on those stages are optimal.”
- Correction: This is true for a tied maximum of zero, but not for a tied negative maximum. If the tied maximum is negative, the maximizing choices put weight one on a single stage attaining it. For example, with \((-1,-1)\), equal weights produce mean \(-\sqrt{2}\), less than the attainable mean \(-1\).
- Another common error: Calling weights proportional to \(I_i\) the canonical information weights. Under the common-effect canonical model, the inverse-normal weights are proportional to \(\sqrt{I_i}\).
Check Your Understanding
Answer each question, then compare with the solutions below.
- For \((\lambda_1,\lambda_2)=(2,1)\), find the maximizing weights and the maximum alternative mean.
- For \((\lambda_1,\lambda_2)=(-2,-0.5)\), identify a maximizing weight vector and the maximum mean.
- For \((\lambda_1,\lambda_2)=(0,0)\), describe all maximizing weights.
- If \(I_1=9\), \(I_2=16\), and \(\lambda_i=0.2\sqrt{I_i}\), what are the maximizing weights?
Solutions. 1. Both means are positive, so \(w=(2,1)/\sqrt{5}\approx(0.8944,0.4472)\), and the maximum mean is \(\sqrt{5}\approx2.236\). 2. The largest mean is \(-0.5\), so put weight one on stage 2: \(w=(0,1)\), with maximum mean \(-0.5\). 3. Every nonnegative unit-length vector maximizes, because every such vector has alternative mean zero. 4. The means are \(0.6\) and \(0.8\), so the maximizing weights are \((3,4)/5=(0.6000,0.8000)\).