Choosing Inverse-Normal Weights for Unequal Stages

Why Stage Weights Matter

In a seamless phase 2/3 trial, the stages may contribute different amounts of information to the final evidence. Phase 2 might enroll fewer participants, use a different allocation, or measure an endpoint less precisely than phase 3. An inverse-normal combination test converts each stage’s valid p-value into a normal-score scale and combines those scores using weights. The weights determine how strongly each stage contributes to the final statistic.

Unequal stage sizes make a simple default such as equal weights potentially inefficient for a common treatment effect: a more informative stage may reasonably contribute more. But sample size is only a proxy for information. The relationship depends on the endpoint, variability, allocation, event rate, and analysis. A defensible choice therefore starts with the planned information in each stage and is fixed before the stage results are observed.

Definition: For prespecified nonnegative weights $w_1,\ldots,w_K$ satisfying $\sum_{k=1}^{K} w_k^2=1$, and stagewise one-sided p-values $p_k$ oriented so smaller values favor the alternative, the inverse-normal combination statistic is $$ Z_{\mathrm{comb}}=\sum_{k=1}^{K} w_k\Phi^{-1}(1-p_k), $$ where $\Phi$ is the standard normal cumulative distribution function. For an upper-tail test, the combined p-value under standard-normal calibration is $1-\Phi(Z_{\mathrm{comb}})$.

The squared-weight normalization is essential to the usual standard-normal calibration. The weights are not probabilities and need not sum to one; it is their squares that sum to one. This tutorial focuses on two stages, with subscripts 1 and 2 denoting phase 2 and phase 3, respectively.

Assumptions Behind the Calibration

The combination test uses stagewise p-values that are valid for the same null hypothesis and the same direction of benefit. Each p-value must be constructed using an analysis appropriate to its stage, including any adjustment required by treatment selection or other design features. The earlier material on valid stagewise p-values and independent increments explains why those properties matter at a phase boundary.

For the exact standard-normal reference distribution, assume under the boundary null that $p_1$ and $p_2$ are independent and each is exactly uniform on $(0,1)$. Then $Z_k=\Phi^{-1}(1-p_k)$ are independent standard normal random variables. If the p-values are only valid or super-uniform, they need not be exactly uniform; in that case, exact normality of the transformed scores and of their weighted sum does not follow.

Calibration result: If the stagewise p-values are independent and exactly uniform under the boundary null, and $w_1^2+w_2^2=1$, then $Z_{\mathrm{comb}}\sim N(0,1)$. If they are independent and super-uniform instead, with nonnegative weights, the upper-tail combination test is generally conservative: its null distribution need not be exactly standard normal.

The conservative case follows from stochastic ordering. A super-uniform p-value satisfies $P(p_k\le u)\le u$ for every $u$ between zero and one. Thus $Z_k=\Phi^{-1}(1-p_k)$ has no larger upper-tail probabilities than a standard normal variable. With independent stagewise p-values, their transformed scores can be compared jointly to independent standard normals; a weighted sum with nonnegative weights is an increasing function of each score. Its upper-tail probability is therefore no larger than the corresponding standard-normal upper-tail probability. This argument does not make the sum exactly normal.

Deriving Information-Based Weights

A common design-based choice gives each stage a weight proportional to the square root of its information. To see why, suppose the two stages provide independent score increments $U_1$ and $U_2$. Under a common effect model, let their planned information be $I_1$ and $I_2$, with each score increment having variance $I_k$ under the null. The standardized stage scores are $Z_k=U_k/\sqrt{I_k}$.

\frac{U_1+U_2}{\sqrt{I_1+I_2}} = \sqrt{\frac{I_1}{I_1+I_2}}\,Z_1 + \sqrt{\frac{I_2}{I_1+I_2}}\,Z_2.

The identity follows by substituting $U_k=\sqrt{I_k}Z_k$ into the left-hand side. It shows that the standardized combined score uses weights $w_k=\sqrt{I_k/(I_1+I_2)}$. These weights satisfy the required normalization because $w_1^2+w_2^2=(I_1+I_2)/(I_1+I_2)=1$. This derivation motivates the choice when the aim is to reflect the relative information in independent stage scores for a common effect. It does not establish that these are the best weights for every estimand or every possible pattern of effects.

If information per participant is the same in both stages and other design features are comparable, information is proportional to sample size. In that special case, the weights reduce to the square roots of the sample-size fractions. If per-participant information differs, use the planned information rather than the raw sample counts. Information may be expressed as inverse variance for an estimator, or as the appropriate model-based or design-based information for the endpoint and analysis.

Worked Example: Unequal Sample Sizes With Equal Per-Participant Information

Hypothetical design. Phase 2 plans to enroll 120 participants and phase 3 plans to enroll 480. Assume equal per-participant information across the stages, so the planned information fractions are $120/600=0.20$ and $480/600=0.80$. The design-based weights are $$ w_1=\sqrt{0.20}=0.4472,\qquad w_2=\sqrt{0.80}=0.8944. $$ Their squared sum is $0.20+0.80=1.00$, as required.

Suppose the hypothetical stagewise one-sided p-values are $p_1=0.0800$ and $p_2=0.0150$. Under the independent, exactly uniform boundary-null calibration, their normal scores are $$ Z_1=\Phi^{-1}(0.9200)\approx1.405,\qquad Z_2=\Phi^{-1}(0.9850)\approx2.170. $$ Therefore, $$ Z_{\mathrm{comb}}=(0.4472)(1.405)+(0.8944)(2.170) \approx0.6283+1.9408\approx2.569. $$

Rechecking the arithmetic is important: $(0.8944)(2.170)\approx1.941$, and $(0.4472)(1.405)\approx0.6283$, giving $Z_{\mathrm{comb}}\approx2.569$. The upper-tail p-value is $1-\Phi(2.569)\approx0.0051$. Thus, under the stated calibration assumptions, this prespecified combination rejects at the 0.05 level. The result is driven more by phase 3 because its planned information is four times phase 2’s.

Comparing Weighting Choices

Several choices can be valid, but they answer different design priorities. Information-based weights reflect relative planned precision. Sample-size-based weights are a convenient approximation when information per participant is comparable. Equal weights give both stages equal squared contribution to the null variance, regardless of their sizes. Provided the weights are fixed independently of the observed stage results, nonnegative, and normalized, changing the choice changes the test’s power profile, not the exact standard-normal null calibration under independent exact-uniform stage p-values.

Choice	Formula for two stages	When it may be reasonable
Information-based	$w_k=\sqrt{I_k/(I_1+I_2)}$	Planned information differs or can be estimated from endpoint and design features
Sample-size-based	$w_k=\sqrt{n_k/(n_1+n_2)}$	Per-participant information is comparable across stages
Equal weights	$w_1=w_2=1/\sqrt{2}$	The design intentionally gives stages equal weight, or a simple prespecified choice is preferred

Worked Example: Sample Size Is Not Always Information

Hypothetical design. Phase 2 plans $n_1=100$ participants with information $0.5$ units per participant; phase 3 plans $n_2=300$ with information $1.0$ unit per participant. Thus $I_1=50$ and $I_2=300$, so the information-based weights are $$ w_1=\sqrt{\frac{50}{350}}=0.3780,\qquad w_2=\sqrt{\frac{300}{350}}=0.9258. $$ For comparison, sample-size-based weights are $0.5000$ and $0.8660$, and equal weights are $0.7071$ and $0.7071$. Each pair has squared sum one.

Suppose the hypothetical stagewise p-values are $p_1=0.0200$ and $p_2=0.2000$, giving $Z_1=\Phi^{-1}(0.9800)\approx2.054$ and $Z_2=\Phi^{-1}(0.8000)\approx0.8416$. Under independent exact-uniform null calibration, the information-based result is $$ Z_{\mathrm{comb}}=(0.3780)(2.054)+(0.9258)(0.8416) \approx0.7764+0.7792=1.556, \qquad 1-\Phi(1.556)\approx0.0599. $$

Using sample-size weights instead gives $$ Z_{\mathrm{comb}}=(0.5000)(2.054)+(0.8660)(0.8416) \approx1.027+0.7288=1.756, \qquad 1-\Phi(1.756)\approx0.0396. $$ With equal weights, the result is $$ Z_{\mathrm{comb}}=(0.7071)(2.054)+(0.7071)(0.8416) \approx2.047, \qquad 1-\Phi(2.047)\approx0.0203. $$ These are different tests applied to the same hypothetical stage results. The example shows why the weighting rule must be specified before seeing the p-values: choosing the weights that produce the smallest combined p-value after observing results is not the prespecified combination test.

Computing Information From Standard Errors

For an approximately normal stage-specific treatment-effect estimate with known or design-based standard error $SE_k$, information is proportional to $1/SE_k^2$. The inverse-variance relationship is useful when phase 2 and phase 3 have different endpoint variability or allocation. In applications, the assumptions behind the standard errors and their comparability must be checked; a standard error estimated from the observed data may not equal the planned information used to set weights.

Worked Example: Weights From Stage Standard Errors

Hypothetical model. Suppose independent stage estimates use the same treatment-effect scale, and their planned standard errors are $SE_1=0.24$ and $SE_2=0.12$. Assume the estimates are approximately normal, the standard errors represent planned precision, and both stages target the same effect. Information is proportional to inverse variance: $$ I_1=\frac{1}{0.24^2}=17.36,\qquad I_2=\frac{1}{0.12^2}=69.44. $$ The information ratio is $1:4$, giving $$ w_1=\sqrt{\frac{17.36}{17.36+69.44}}=0.4472,\qquad w_2=\sqrt{\frac{69.44}{17.36+69.44}}=0.8944. $$

For a numerical combination, suppose the hypothetical one-sided stage statistics are $Z_1=1.250$ and $Z_2=1.500$, corresponding to $p_1\approx0.1056$ and $p_2\approx0.0668$. The combined statistic and p-value are $$ Z_{\mathrm{comb}}=(0.4472)(1.250)+(0.8944)(1.500) \approx0.5590+1.342=1.901, \qquad 1-\Phi(1.901)\approx0.0287. $$ The combination is significant at 0.05 under independent standard-normal stage statistics. The more precise phase 3 estimate receives the larger weight. This conclusion depends on the assumed common effect scale and independence; if those assumptions do not hold, the information calculation may not represent the intended design.

Prespecification, Diagnostics, and Limitations

A practical choice begins with the endpoint and estimand carried across the stage boundary, then identifies each stage’s planned information under the intended analysis. If information is proportional to enrollment, sample-size weights can be justified; otherwise, the information calculation should account for relevant differences in precision. The selected weights, their rationale, and the p-value calibration should be documented before unblinded stage results are available.

Diagnostics include checking that weights are nonnegative and their squared sum is one; verifying that the stagewise p-values are on the same one-sided scale and test the same null; and confirming that their dependence assumptions match the design. If information estimates change during planning, recalculate the proposed weights and document the final prespecified rule. Do not substitute observed stage effects or p-values into the weight calculation.

Information-based weights are not automatically optimal if stage effects differ, if the target estimand changes, or if one stage has a distinct role such as screening. Nor does a valid combination rule fix invalid stagewise p-values or inappropriate endpoint definitions. The inverse-normal test’s calibration and power both depend on the design features represented in its inputs and its prespecified weights.

Common pitfall: “A valid p-value is uniform, so the transformed score is standard normal.” This is wrong: validity generally means super-uniformity, not exact uniformity. Exact standard-normal calibration of the inverse-normal statistic requires independent, exactly uniform stagewise p-values under the boundary null. With independent super-uniform p-values and nonnegative normalized weights, the upper-tail test can be conservative, but the combined statistic need not have an exact standard-normal distribution.

Another common error is to use raw sample-size fractions as the weights. The inverse-normal weights are square roots of fractions, not the fractions themselves. For example, stage information fractions $0.20$ and $0.80$ imply weights $0.4472$ and $0.8944$, not $0.20$ and $0.80$. Using the fractions directly would produce a squared sum of $0.68$, violating the stated normalization.

Check Your Understanding

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

Two stages have planned information $I_1=36$ and $I_2=64$. What are their information-based weights?
When do sample-size-based weights coincide with information-based weights?
What calibration conclusion is justified if independent stagewise p-values are super-uniform but not known to be exactly uniform?
Why is it not acceptable to compare several weight choices on the observed stage p-values and report whichever gives the smallest combined p-value?

Solutions. (1) The total information is 100, so $w_1=\sqrt{36/100}=0.6000$ and $w_2=\sqrt{64/100}=0.8000$; their squared sum is one. (2) They coincide when information per participant is comparable across stages and the other relevant design features make information proportional to sample size. (3) With independence and nonnegative normalized weights, the upper-tail combination test is generally conservative; exact standard-normal calibration cannot be claimed solely from super-uniformity. (4) Choosing weights in response to observed results makes the rule data-dependent and can invalidate the calibration for the prespecified test; the choice must be fixed independently of those results.

Choice	Formula for two stages	When it may be reasonable
Information-based	\(w_k=\sqrt{I_k/(I_1+I_2)}\)	Planned information differs or can be estimated from endpoint and design features
Sample-size-based	\(w_k=\sqrt{n_k/(n_1+n_2)}\)	Per-participant information is comparable across stages
Equal weights	\(w_1=w_2=1/\sqrt{2}\)	The design intentionally gives stages equal weight, or a simple prespecified choice is preferred