Why Unequal Stage Information Changes the Weights
In a multi-stage trial, each stage contributes evidence to a final test. If the stages have different planned information, giving each the same weight is a deliberate choice, not a neutral default. A stage with more information generally provides a more precise estimate and, under a common-effect model, tends to contribute more evidence against the null hypothesis.
The inverse normal combination test makes this choice explicit. Each stage supplies a one-sided p-value, which is transformed to a standard normal score; the scores are then combined using prespecified weights. The key design question here is how to choose those weights when the planned information differs by stage.
For a one-sided test in a prespecified beneficial direction, write the stagewise score as \(Z_j=\Phi^{-1}(1-p_j)\), where \(p_j\) is the stagewise p-value and \(\Phi\) is the standard normal cumulative distribution function. The combination statistic is \(Z_C=\sum_{j=1}^K w_jZ_j\). With information-proportional weights, \(\sum_j w_j^2=\sum_jq_j=1\).
Under the global null, the standard inverse normal calibration assumes that each stagewise p-value is continuous and uniform and that the stagewise p-values are independent. Then each \(Z_j\) has a standard normal distribution, and independence gives \(Z_C\sim N(0,1)\) when the weights are fixed and their squared sum is one. An upper-tail test at level \(\alpha\) rejects when \(Z_C>\Phi^{-1}(1-\alpha)\). The assumptions concern the validity and joint behavior of the stagewise p-values, not just the arithmetic of the weights.
Why the Square Root of Information Appears
A useful justification comes from a common-effect model. Suppose each stage estimates the same treatment effect \(\theta\), and its estimator has variance approximately \(1/I_j\). If the estimator is approximately normal, its standardized score has approximate mean \(\theta\sqrt{I_j}\) and variance 1. Thus, under a common positive effect, a more informative stage has a larger expected stagewise Z-score.
Among nonnegative weights satisfying \(\sum_j w_j^2=1\), the common-effect model gives a direct criterion: choose weights that maximize the expected combined score. The argument is a consequence of the Cauchy–Schwarz inequality.
For \(\theta>0\), equality holds when the weight vector is proportional to \((\sqrt{I_1},\ldots,\sqrt{I_K})\). Normalizing it to have squared weights summing to one gives \(w_j=\sqrt{I_j/I_{\mathrm{tot}}}=\sqrt{q_j}\). This derivation explains the weights as a reasonable power-oriented choice under a common-effect model. It does not show that they are best for every possible pattern of stage effects or every scientific objective.
Worked Example: Three Stages With Unequal Information
Hypothetical design. Suppose a three-stage trial plans information \(I_1=100\), \(I_2=300\), and \(I_3=600\) units. The total planned information is \(I_{\mathrm{tot}}=100+300+600=1000\) units. The information fractions are \(q_1=0.10\), \(q_2=0.30\), and \(q_3=0.60\).
The information-proportional weights are the square roots of these fractions:
The squared weights sum to one: \(0.10+0.30+0.60=1\). Suppose the stagewise one-sided p-values correspond to \(Z_1=0.40\), \(Z_2=1.10\), and \(Z_3=1.60\). Using unrounded weights, the combination statistic is
The upper-tail p-value is \(1-\Phi(1.968341)\approx0.02452\). At \(\alpha=0.05\), the critical value is approximately 1.6449, so the test rejects \(H_0\). In this result, stage 3 contributes \(1.2394\) to the statistic, compared with \(0.1265\) from stage 1. The larger contribution reflects both its larger weight and its observed Z-score; a weight alone does not determine a stage’s realized contribution.
What the Weights Do to Each Stage’s Influence
A stage’s direct contribution to the realized statistic is \(w_jZ_j\). Increasing \(w_j\) makes a given positive stage score contribute more, but it also makes a given negative score contribute more negatively. Weights do not change the stagewise p-values or the evidence within a stage; they change how the prespecified test combines that evidence.
Information fractions are not themselves the inverse normal weights. For example, a stage with 60% of total information receives weight \(\sqrt{0.60}\approx0.7746\), not 0.60. The square root is required by the variance normalization: under independent standard normal stage scores, the variance of the weighted sum is \(\sum_jw_j^2\).
Worked Example: Comparing Information Weights With Equal Weights
Hypothetical design. Consider two stages with planned information \(I_1=100\) and \(I_2=400\). Then \(q_1=0.20\) and \(q_2=0.80\), so the information-proportional weights are \(w_1=\sqrt{0.20}=0.4472\) and \(w_2=\sqrt{0.80}=0.8944\). Suppose the observed stage scores are \(Z_1=0.80\) and \(Z_2=1.40\).
With information-proportional weights,
The one-sided p-value is \(1-\Phi(1.609969)\approx0.0537\), so this test does not reject at the 0.05 level. For comparison, equal weights are \(1/\sqrt{2}\) for each stage, so their squared sum is also one. The equal-weight statistic is
with upper-tail p-value approximately \(0.0599\). Both tests give the same decision here, but their statistics differ because they allocate different influence to the stages. Equal weights give each stage equal influence per unit of Z-score, whereas the information weights give stage 2 twice the weight of stage 1.
When the same outcome, estimand, and analysis scale are used in each stage, information is often approximately proportional to sample size. In that special case, information weights are square roots of sample-size fractions. But sample-size fractions and information fractions need not agree: allocation ratios, event rates, outcome variability, follow-up, and analysis methods can all change information per participant. The design should therefore use a justified measure of planned information, rather than substitute sample size automatically.
Assumptions, Diagnostics, and Alternatives
The information-weight rationale has two distinct parts. First, the null calibration requires valid stagewise p-values, the required independence (or another correctly specified joint calibration), and fixed weights with \(\sum_jw_j^2=1\). Second, the power motivation for information-proportional weights assumes a common treatment effect on a common analysis scale and stage scores whose means are approximately \(\theta\sqrt{I_j}\). If stages target different populations or use materially different endpoints, this common-effect interpretation may not apply.
Before applying the weights, check that the planned information values refer to the same estimand and are calculated on a comparable scale. Verify that all information fractions are nonnegative and sum to one, and that the weights are their square roots. Then check \(\sum_jw_j^2=1\). For a design using other prespecified weights, the normalization and null reference distribution must be matched to those weights; the usual standard normal reference does not follow from unequal weights unless their squared sum is one and the stage-score assumptions hold.
Also distinguish planned from realized information. Planned information determines the design weights described here. Replacing those weights after observing stage results is not a harmless recalculation: it can change the test’s operating characteristics. If realized information departs from plan, the analysis must use a prespecified method that maintains valid calibration; this tutorial’s simple calculation does not by itself establish that calibration.
Equal weights are a transparent alternative when equal stage influence is the goal, or when no defensible information-based distinction is available. Other fixed weights can target a particular design objective, such as emphasizing a stage with a distinct scientific role. The choice should be made before examining stagewise results and stated with its rationale. Fisher’s combination test is another way to combine stagewise p-values, but it uses a different statistic and does not allocate influence through inverse normal weights in the same way.
Worked Example: When Information Weights Are Not Automatically Preferable
Hypothetical design. Return to the two-stage information pattern \(I_1=100\), \(I_2=400\), with weights \(0.4472\) and \(0.8944\). Suppose instead that stage scores are \(Z_1=2.00\) and \(Z_2=0.20\). The information-weighted statistic is
Its one-sided p-value is approximately \(1-\Phi(1.073312)=0.1416\), so it does not reject at the 0.05 level. With equal weights, the statistic would be \((2.00+0.20)/\sqrt{2}=1.5556\), with p-value approximately \(0.0599\), also not significant at 0.05. The difference shows that the information-weighted design gives the lower-information first stage less influence, even though its observed evidence is stronger.
This is not a defect in the arithmetic. It illustrates the scope of the design rationale: information-proportional weights are motivated by a common effect whose expected standardized signal grows with the square root of information. If effects vary substantially by stage, or if a particular stage has special scientific importance, a different prespecified weighting objective may be more appropriate. The observed scores in this example cannot be used to select weights retrospectively without accounting for that selection.
Common Mistakes and Failure Modes
- Using information fractions as weights. The incorrect reasoning is “stage 2 has 80% of the information, so its weight is 0.80.” The normalized inverse normal weight is \(\sqrt{0.80}\), because the squared weights—not the weights—sum to one.
- Confusing a large weight with a guaranteed large contribution. The contribution is \(w_jZ_j\). A high-weight stage with a small or negative score may contribute less than a lower-weight stage with a large positive score.
- Assuming sample-size proportions always equal information proportions. This holds only under suitable comparability of per-participant information. Check the outcome model and design assumptions.
- Updating weights after seeing stage results. The incorrect reasoning is that weights should be increased for stages with stronger evidence. That makes the final procedure depend on the observed results; use weights specified by the design or a separately calibrated method.
- Applying the standard normal reference without checking dependence or p-value validity. The weighted sum is standard normal under the stated null assumptions, not merely because the inputs have been transformed to Z-scores.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- Two stages have planned information 150 and 450 units. Find the information fractions and inverse normal weights.
- Why is the weight for a stage with 75% of total information \(\sqrt{0.75}\), rather than 0.75?
- With weights \(0.60\) and \(0.80\), and stage scores \(Z_1=1.00\) and \(Z_2=0.50\), calculate the combination statistic. Does it exceed the one-sided 0.05 critical value of 1.6449?
- State two assumptions needed for the standard normal null calibration of the weighted combination statistic.
Solutions. For question 1, total information is \(150+450=600\), so the fractions are \(0.25\) and \(0.75\), and the weights are \(\sqrt{0.25}=0.5000\) and \(\sqrt{0.75}=0.8660\). For question 2, normalization requires squared weights to sum to one; the weight is the square root of the information fraction, so the squared weight equals 0.75. For question 3, the weights are already normalized because \(0.60^2+0.80^2=1\). Thus \(Z_C=(0.60)(1.00)+(0.80)(0.50)=0.60+0.40=1.000\), which is below 1.6449, so the test does not reject at the one-sided 0.05 level. For question 4, the stagewise p-values must be valid under the null and independent across stages, and the fixed weights must have squared sum one. Any two of these stated requirements address the central calibration assumptions.