Why the Weights Matter After Re-Estimation
In a two-stage trial, an interim result may determine how many participants are enrolled in stage two. That change can alter the amount of information collected, but it does not automatically make a fixed-design critical value appropriate for the final analysis. An inverse-normal combination test instead transforms each stage’s p-value to a normal score and combines the scores using specified weights.
The key distinction in this example is between planned information used to define the weights and information actually collected after the interim decision. When the sample-size rule depends on interim data, using the realized information to change the weights can also make the weights depend on those data. The resulting statistic is not automatically standard normal. Here, the weights are fixed before outcomes are observed, while the second-stage sample size follows a prespecified interim rule.
Model, Assumptions, and Weight Choice
Suppose each stage estimates the same treatment effect \(\theta\), measured in millimeters of mercury (mmHg). For clarity, take the stagewise estimators to be normal with known standard errors: \(\hat{\theta}_j\sim N(\theta,1/I_j)\), where \(I_j\) is the information in stage \(j\). Thus \(Z_j=\hat{\theta}_j\sqrt{I_j}\) has variance 1 and mean \(\theta\sqrt{I_j}\). The one-sided stagewise p-value is \(p_j=1-\Phi(Z_j)\).
The validity argument requires independent stagewise data, a correctly specified normal test (or a stagewise procedure that gives valid p-values), and an interim rule that uses only information available before stage two. In particular, no stage-two participant or outcome may be reused in stage one. Conditional on the interim data and the chosen stage-two size, the stage-two test must still give a valid one-sided p-value under the null. At the boundary \(\theta=0\), this p-value is uniform conditional on the interim data in the continuous normal model.
A common information-based choice uses the original planned information values \(I_1^*\) and \(I_2^*\):
These weights are nonnegative and satisfy \(w_1^2+w_2^2=1\). For an adaptation that changes the second-stage size, the procedure below keeps these original weights fixed. It does not replace \(I_2^*\) with the outcome-dependent realized information when calculating the weights. A procedure using other weight rules requires its own calibration argument.
Worked Example: Information Weights and an Increased Second Stage
Consider a hypothetical normal trial with known outcome variance \(1\ \text{mmHg}^2\) per participant. Stage one has \(n_1=100\), so its information is \(I_1=100\ \text{mmHg}^{-2}\). The original plan called for \(n_2^*=300\), giving \(I_2^*=300\ \text{mmHg}^{-2}\). The prespecified rule is: if \(Z_1<2.5\), enroll 500 participants in stage two; otherwise enroll 300. Suppose the observed stage-one estimate is \(0.2400\) mmHg.
The stage-one standard error is \(1/\sqrt{100}=0.1000\) mmHg, so \(Z_1=0.2400/0.1000=2.400\). Its one-sided p-value is \(p_1=1-\Phi(2.400)=0.008198\). Since \(2.400<2.5\), the rule selects \(n_2=500\), and hence actual stage-two information \(I_2=500\ \text{mmHg}^{-2}\).
The weights are still calculated from the original plan:
Suppose the independent stage-two estimate is \(0.1162\) mmHg. Its standard error is \(1/\sqrt{500}=0.04472\) mmHg, giving \(Z_2=0.1162/0.04472\approx2.598\) and \(p_2=1-\Phi(2.598)\approx0.00469\). Combining the stagewise normal scores gives:
The combined p-value is \(1-\Phi(3.4499)\approx0.0002804\). At a one-sided significance level of \(\alpha=0.025\), this result is statistically significant because \(0.0002804<0.025\). The inference is evidence against \(H_0:\theta\leq0\) in favor of a positive treatment effect; it does not give the probability that \(H_0\) is true.
Why the Fixed-Weight Combination Is Calibrated
The justification can be seen by conditioning on the interim data. Under the boundary null \(\theta=0\), \(Z_1\sim N(0,1)\). Given the stage-one data and the selected stage-two size, the independent stage-two statistic has \(Z_2\sim N(0,1)\). Because this conditional distribution does not depend on the interim result, \(Z_2\) is independent of \(Z_1\) under the boundary null. With fixed weights, the sum is therefore normal with mean zero and variance one:
Thus the rule that selects the second-stage size does not, by itself, alter the null distribution of this combination statistic. The argument relies on the stage-two p-value remaining conditionally valid and on weights that do not respond to the interim result. For the composite null \(H_0:\theta\leq0\), the stagewise one-sided p-values are super-uniform under the null, and the conditional-validity argument gives Type I error control at the boundary and no larger rejection probability under the null.
This is a short but important derivation: the standard normal reference distribution follows from the independent, standard normal stagewise scores and the sum-of-squared-weights condition. It is not a consequence of merely having two normally distributed estimates. Selection, reuse of data, invalid stagewise p-values, or weights that vary with interim outcomes can break the argument.
Worked Example: The Other Branch of the Interim Rule
Use the same hypothetical trial, original information plan, and fixed weights \(w_1=0.5000\) and \(w_2=0.8660\). Suppose instead that \(Z_1=2.800\). Since \(2.800\) is not less than \(2.5\), the rule selects the original second-stage size, with \(I_2=300\). Suppose the stage-two result is \(Z_2=0.500\).
The stagewise p-values are \(p_1=1-\Phi(2.800)=0.002555\) and \(p_2=1-\Phi(0.500)=0.3085\). The combined statistic is:
The combined one-sided p-value is \(1-\Phi(1.8330)\approx0.0334\). At \(\alpha=0.025\), the result is not statistically significant. A small first-stage p-value alone does not determine the final decision: the prespecified combination uses evidence from both stages.
Realized-Information Weights: A Calibration Warning
Information-based weights are appealing because, in a fixed design, they reflect how much information each stage contributes. But in an adaptive design, the realized second-stage information may be a function of \(Z_1\). Recomputing both weights from that realized information makes the weights functions of the interim data. The fixed-weight derivation above then no longer applies.
Worked Example: What Changes if the Weights Follow the Adaptation?
Return to the first branch of the hypothetical rule: \(I_1=100\), \(I_2=500\), \(Z_1=2.400\), and \(Z_2=2.598\). If one substitutes the realized information into the information-weight formulas, the resulting values are:
Using these weights gives the numerical statistic \( (0.4082)(2.400)+(0.9129)(2.598)\approx3.351\). That arithmetic is valid, but it does not establish that \(1-\Phi(3.351)\) is a calibrated p-value. Here the rule selects \(I_2\), and thus the weights, using \(Z_1\); consequently the combined statistic need not have a standard normal null distribution. The observed statistic alone cannot resolve that calibration problem.
Checks, Alternatives, and Limitations
Before reporting an inverse-normal result, verify that the implemented analysis matches the prespecified design. In particular, check that the rule’s branch and stage-two size are correct, that both stagewise p-values correspond to the intended one-sided hypothesis, and that the weight values were fixed using the stated reference information. Also check that the second-stage data are independent of stage one and that any changes to the endpoint, analysis population, or test preserve conditional validity.
A naive pooled test is not a substitute: its usual fixed-design reference distribution may fail when sample size is selected using interim outcomes. Likewise, using a fixed-design critical value after re-estimation does not automatically control the overall Type I error. The inverse-normal method combines stagewise evidence under its stated conditions; it does not make every adaptive analysis valid. A conditional-error recalibration is a different approach when the planned final testing rule must be modified, and other combination functions use different ways of combining p-values.
Fixed weights also have a cost. They need not reflect the realized information split, so they may not be the most powerful choice for every adaptation rule or alternative effect. The method’s validity does not guarantee optimal power, and the weights and sample-size rule should be specified and justified before the relevant outcomes are observed. If the analysis uses estimated variance, discrete outcomes, or a test with ties, stagewise p-value validity and calibration require particular care.
Common Mistakes
- Recalculating weights from the realized second-stage size without justification. If that size depends on interim data, the resulting weights are data-dependent; the fixed-weight standard normal argument no longer applies automatically.
- Combining the p-values directly as if they were normal scores. The inverse-normal rule first uses \(Z_j=\Phi^{-1}(1-p_j)\), then combines those scores.
- Using a two-sided p-value in a one-sided formula without a prespecified direction convention. The stagewise p-values here are upper-tail p-values for evidence that \(\theta>0\).
- Assuming that independent stagewise data alone guarantee validity. The weights, stagewise tests, and interim rule must also satisfy the conditions used in the calibration argument.
Check Your Understanding
Answer each question, then compare your reasoning with the brief solutions.
- If \(I_1^*=150\) and \(I_2^*=450\), what are the information-based weights?
- With \(Z_1=1.6\), \(Z_2=2.0\), and weights \(0.5000\) and \(0.8660\), calculate the combined statistic and state the one-sided decision at \(\alpha=0.025\).
- Why can a stage-two p-value remain valid when the sample size is chosen using stage-one data?
- What is the main calibration concern if the inverse-normal weights are recalculated from an outcome-dependent realized \(I_2\)?
Solutions. (1) \(w_1=\sqrt{150/600}=0.5000\) and \(w_2=\sqrt{450/600}=0.8660\). (2) \(Z_{\mathrm{comb}}=(0.5000)(1.6)+(0.8660)(2.0)=0.8000+1.732=2.532\). Its upper-tail p-value is about \(0.00567\), so reject \(H_0\) at the one-sided \(0.025\) level. (3) Given the interim data and selected size, independent stage-two observations still yield a valid stage-two test under the null. (4) The weights then depend on interim outcomes, so the fixed-weight standard normal derivation does not establish the reference distribution; separate calibration is needed.