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The Combination Test Principle · Tutorial 197 of 799

Deriving Weights From Stage Information

Learn how planned stage information determines inverse normal weights, why the square-root rule is optimal under stated assumptions, and when that weighting choice may fail.

Advanced 11 min read

What You'll Learn

  • Define stage information and relate it to the precision of a stage-specific treatment estimate.
  • Derive information-proportional inverse normal weights by maximizing the combined noncentrality.
  • Distinguish exact null calibration from approximate power calculations under alternatives.
  • Calculate weights from information values, standard errors, and per-participant information.
  • Diagnose when equal weights or sample-size-proportional weights are more or less appropriate.
  • Identify conditions under which the information-based weighting argument does not apply.

Why Derive Weights From Information?

In a planned multi-stage trial, the inverse normal method combines evidence from each stage using weights. The weights determine how strongly each stage contributes to the final statistic. If the stages are expected to provide different amounts of information about the same treatment effect, equal weights need not be an efficient choice. The information-proportional rule gives greater weight to stages expected to estimate that effect more precisely.

The derivation depends on more than knowing the sample sizes. It assumes a common treatment-effect scale, independent stage contributions, and a suitable normal approximation for the stagewise evidence under alternatives. Separately, exact null calibration of the inverse normal statistic requires independent uniform stagewise p-values. Validity alone does not imply uniformity, so these conditions must not be conflated.

Definition: The stage information \(I_i\) is a measure of the precision that stage \(i\) provides for estimating the treatment effect on a specified common scale. In a regular model, it is often the inverse of the variance of an efficient stage-specific estimate: if that estimate has variance \(1/I_i\), then larger \(I_i\) means greater precision. Information is additive across independent increments when the model and estimand are comparable across stages.

Assumptions Behind the Information Rule

Consider \(K\) planned stages. Let \(Z_i\) denote the oriented, one-sided stagewise normal score, with larger values supporting the prespecified beneficial direction. The information-based derivation uses the following working model under a local alternative:

$$ Z_i \ \mathrel{\dot\sim}\ N(\delta\sqrt{I_i},1), \qquad i=1,\ldots,K, $$

where \(\delta\) is the common treatment effect on the model's standardized scale and the stage contributions are independent. The dot indicates an approximation when the stage statistics are only approximately normal. For a normally distributed stage estimate with variance \(1/I_i\), standardizing its estimate gives this mean structure exactly under the corresponding normal model. In other settings it is a large-sample or local-alternative approximation.

For an inverse normal test, a one-sided p-value \(p_i\) is transformed as \(Z_i=\Phi^{-1}(1-p_i)\), where \(\Phi\) is the standard normal cumulative distribution function. If the p-values are independent and each is exactly uniform under a continuous exact null, the transformed scores are independent standard normal variables under that null. For discrete or conservative p-values, or at interior points of a composite null, this exact standard normal statement need not hold. A design must use appropriate null calibration for its actual stagewise tests.

The information rule is appropriate when stagewise evidence concerns the same estimand and direction, the planned \(I_i\) values represent the stages' relative precision, and independence is justified by independent increments or another suitable design argument. It is not a general rule for choosing weights when stages measure different effects, have dependent evidence, or have substantially different alternative mean structures.

Deriving the Information-Proportional Weights

Write the weighted combination statistic as \(C=\sum_{i=1}^K w_i Z_i\). Under independent standard normal stage scores at the exact null, its variance is \(\sum_i w_i^2\). The usual normalization \(\sum_i w_i^2=1\) therefore makes \(C\) standard normal under that null. Under the working alternative model, its expected value is \(\delta\sum_i w_i\sqrt{I_i}\). For a beneficial effect \(\delta>0\), maximizing this mean maximizes the normal-approximation noncentrality for a fixed one-sided critical value.

Result: Among nonnegative weights satisfying \(\sum_i w_i^2=1\), the weights that maximize \(\sum_i w_i\sqrt{I_i}\) are
$$ w_i=\sqrt{\frac{I_i}{\sum_{j=1}^K I_j}}, \qquad i=1,\ldots,K. $$

The restriction to nonnegative weights aligns each stage's contribution with the same beneficial direction. The normalization fixes the null variance; without it, multiplying all weights by a positive constant would change the statistic's scale without changing its ordering.

Why the Square-Root Rule Is Optimal

The justification is the Cauchy–Schwarz inequality. For any weights with \(\sum_i w_i^2=1\),

$$ \sum_{i=1}^K w_i\sqrt{I_i} \leq \sqrt{\sum_{i=1}^K w_i^2}\sqrt{\sum_{i=1}^K I_i} = \sqrt{\sum_{i=1}^K I_i}. $$

Equality holds precisely when the weight vector is proportional to the vector with entries \(\sqrt{I_i}\). Normalizing that vector to have squared length one gives the stated weights. Thus, under the working model, the maximum alternative mean of \(C\) is \(\delta\sqrt{\sum_i I_i}\). This result is a justification for a planned weighting choice, not a claim that this choice is optimal for every possible alternative or every combination function.

Worked Example: Three Planned Stages

Suppose a hypothetical trial plans stage information values \(I_1=40\), \(I_2=90\), and \(I_3=160\), in common information units. Assume the stage contributions are independent, address the same treatment effect, and satisfy the normal-score working model above. The total information is \(40+90+160=290\).

The weights are the square roots of each stage's fraction of total information:

$$ w_1=\sqrt{\frac{40}{290}}=0.3714,\qquad w_2=\sqrt{\frac{90}{290}}=0.5571,\qquad w_3=\sqrt{\frac{160}{290}}=0.7428. $$

Their squared values sum to \(40/290+90/290+160/290=1\), as required. Under a hypothetical effect \(\delta=0.05\) per square-root information unit, the stage means are \(0.05\sqrt{40}=0.3162\), \(0.05\sqrt{90}=0.4743\), and \(0.05\sqrt{160}=0.6325\). The expected combined statistic is

$$ \sum_{i=1}^3 w_i\delta\sqrt{I_i} = 0.05\frac{40+90+160}{\sqrt{290}} = 0.05\sqrt{290} = 0.8515. $$

For comparison, equal weights would be \(1/\sqrt{3}=0.5774\), giving expected combined statistic \(0.05(\sqrt{40}+\sqrt{90}+\sqrt{160})/\sqrt{3}=0.8216\). This comparison concerns the stated common-effect model and planned information values; it does not establish that equal weights are generally inefficient or invalid.

Obtaining Information From Standard Errors

In many designs, information is available through planned standard errors rather than a direct information table. If a stage-specific treatment-effect estimate has variance \(SE_i^2\), then \(I_i=1/SE_i^2\) under the information convention used here. These quantities carry the reciprocal squared units of the effect estimate, while the weights themselves are dimensionless.

Worked Example: Weights From Planned Standard Errors

Imagine two independent hypothetical stages estimating the same treatment difference in millimetres of mercury. Their planned standard errors are \(SE_1=0.10\) mmHg and \(SE_2=0.08\) mmHg. Assume these are standard errors for comparable, unbiased stage estimates and that the normal approximation is suitable.

The information values are \(I_1=1/(0.10)^2=100\) mmHg\(^{-2}\) and \(I_2=1/(0.08)^2=156.25\) mmHg\(^{-2}\). Their total is \(256.25\) mmHg\(^{-2}\), so

$$ w_1=\sqrt{\frac{100}{256.25}}=0.6247,\qquad w_2=\sqrt{\frac{156.25}{256.25}}=0.7809. $$

The second stage receives the larger weight because its planned estimate is more precise. The weights are not the fractions of information themselves: those fractions are approximately \(0.3902\) and \(0.6098\), while the weights are their square roots. Confusing these two quantities would fail the required normalization \(\sum_i w_i^2=1\).

Information Is Not Always Sample Size

For equal per-participant information and otherwise comparable observations, information is proportional to sample size. In that special case, square-root information weights are also proportional to the square roots of planned sample sizes. If per-participant information differs across stages, using sample size alone can misrepresent the stages' relative precision.

Worked Example: Why Planned Sample Size Can Mislead

Consider a hypothetical two-stage design with 80 participants in stage 1 and 100 in stage 2. Suppose the design calculations imply information of \(0.5\) units per participant in stage 1 and \(1.0\) unit per participant in stage 2. The planned information values are therefore \(I_1=80(0.5)=40\) and \(I_2=100(1.0)=100\). Assume the information units refer to the same treatment effect, and the stages are independent.

The information-based weights are

$$ w_1=\sqrt{\frac{40}{140}}=0.5345,\qquad w_2=\sqrt{\frac{100}{140}}=0.8452. $$

Weights based only on sample size would instead be \(\sqrt{80/180}=0.6667\) and \(\sqrt{100/180}=0.7454\). Those would not reflect the stated difference in per-participant information. The sample-size weights would be appropriate only if the design assumptions made information proportional to sample size across both stages.

Diagnostics, Alternatives, and Limitations

Before adopting information-proportional weights, check whether the design assumptions match the planned trial. In particular, verify that the stagewise statistics use the same effect scale and direction; that information estimates represent the intended precision; and that stagewise evidence is independent under the planned design. If stages share observations or otherwise have correlated statistics, the variance of the weighted sum includes covariance terms, so the unit-squared-weight normalization and derivation above no longer apply unchanged.

Also distinguish planned weights from weights chosen after seeing stage results. The derivation uses information values fixed by the design and does not by itself establish validity for outcome-dependent weight selection. If information changes, its source matters: a change known from design features is different from a change selected using observed treatment effects. Any adaptation requires a method whose calibration accounts for that adaptation.

Weighting choiceWhen it may be appropriateMain caution
Information-proportional weightsStages target the same effect, have independent evidence, and differ in planned precision.Depends on the common-effect and information assumptions used in the derivation.
Equal weightsStages are intended to contribute equally, or a design objective supports equal influence.May sacrifice power under the common-effect model when planned information differs.
Sample-size-proportional square-root weightsPer-participant information is comparable across stages.Can be misleading when outcome variability, allocation, or measurement precision differs.

The information rule is a power-oriented choice under a specified alternative structure, not a universal guarantee of greatest power. If treatment effects vary by stage, endpoints differ, or information estimates are poorly specified, an alternative set of prespecified weights may better reflect the scientific objective. The earlier material on selecting weights distinguishes such design goals from the algebraic normalization of a combination statistic.

Common Mistake

A frequent error is to reason, “Stage 2 has twice as much information, so it should get twice the weight.” That confuses information fractions with inverse normal weights. The information fractions are proportional to \(w_i^2\), not \(w_i\), because the statistic's variance is determined by squared weights. The correction is to calculate \(w_i=\sqrt{I_i/\sum_j I_j}\) and then check that \(\sum_i w_i^2=1\).

Key takeaway: Under independent, comparable stage contributions with a common effect scale and the stated normal approximation, weights proportional to \(\sqrt{I_i}\) maximize the combined mean subject to unit null variance. Exact inverse normal null calibration separately requires independent uniform stagewise p-values.

Check Your Understanding

Answer each question using the assumptions stated in the tutorial.

  1. Two independent planned stages have information \(I_1=25\) and \(I_2=75\). What are their information-proportional weights?
  2. Why does a stage with four times the information not receive four times the inverse normal weight?
  3. When can sample-size-proportional square-root weights equal information-proportional weights?
  4. Does validity of each stagewise p-value by itself guarantee that the inverse normal combination statistic is standard normal under the null? Explain.

Solutions: 1. Total information is \(100\), so \(w_1=\sqrt{25/100}=0.5000\) and \(w_2=\sqrt{75/100}=0.8660\); their squared weights sum to one. 2. Weights are proportional to square roots of information, since the variance of a weighted sum depends on squared weights. Four times the information therefore gives twice the weight, when comparing those stages under this rule. 3. They coincide when per-participant information is the same across stages, so total stage information is proportional to planned sample size. 4. No. Independent uniform p-values yield independent standard normal scores; valid but conservative p-values need not be uniform, and the exact standard normal conclusion does not follow from validity alone.

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