Why discreteness matters when combining evidence
Combination tests often use stagewise p-values as inputs to a final statistic. A familiar calibration argument treats each stagewise p-value as uniform under the null hypothesis. That argument is exact for continuous test statistics under suitable conditions, but many clinical trial outcomes are discrete: event counts, binary responses, and rank statistics with ties can yield only a finite set of possible p-values. Exact tail probabilities then tend to be conservative: small p-values occur no more often than their nominal probabilities permit.
This distinction matters because a combination statistic may be calibrated against the distribution it would have if its inputs were independent uniform random variables. If actual stagewise p-values are independent but discrete and conservative, the nominal rejection threshold will generally control the type I error rate at or below its target, rather than attain it. This is a size issue, not a reason to replace valid p-values with smaller ones. Calibration can account for the discrete null distribution when attaining the nominal size is important.
For example, a nonrandomized one-sided exact p-value may be defined as the null probability of an outcome at least as extreme as the observed outcome, with ties included in the tail. Because outcomes arrive in jumps, a tail probability may move from above a chosen cutoff to below it without ever equalling it. The resulting test rejects at a rate no greater than its nominal level. For a composite null, validity requires the super-uniform inequality under every null parameter value, or under the relevant supremum over those values.
Effect on the inverse normal combination test
For two stages, let \(P_1\) and \(P_2\) be stagewise p-values, and use the upper-tail normal score \(Z_i=\Phi^{-1}(1-P_i)\), where \(\Phi\) is the standard normal cumulative distribution function. With positive weights \(w_1,w_2\) satisfying \(w_1^2+w_2^2=1\), the inverse normal statistic is \(C=w_1Z_1+w_2Z_2\). The nominal continuous-reference test rejects for \(C>z_{1-\alpha}\), where \(z_{1-\alpha}=\Phi^{-1}(1-\alpha)\).
If the p-values are independent and uniform under \(H_0\), then \(Z_1,Z_2\) are independent standard normal variables and \(C\) is standard normal. If instead each p-value is independent and super-uniform, the transformed scores are stochastically no larger than standard normal scores. Since the weights are positive, the weighted sum is likewise stochastically no larger than the standard normal reference. Thus the continuous-reference cutoff controls size, but may be conservative. The argument depends on independence under the null; it does not establish validity for arbitrary dependence.
The key step can be justified by a quantile coupling. Let \(U_i\) be independent uniform random variables on \([0,1]\). Super-uniformity means that a draw from the distribution of \(P_i\) is stochastically at least as large as a uniform draw. Using the same uniform variable to construct both draws through their quantile functions gives \(P_i\geq U_i\) in this coupling. Since \(p\mapsto\Phi^{-1}(1-p)\) is decreasing, the corresponding normal score from \(P_i\) is no larger than the score from \(U_i\). The latter scores are independent standard normals. Multiplying by positive weights and adding preserves the inequality, which proves the stated size bound.
Worked Example: A Discrete Inverse Normal Test
Suppose, under a hypothetical null model, each stagewise p-value has this distribution: \(P=.1\) with probability .1, \(P=.5\) with probability .4, and \(P=1\) with probability .5. The stages are independent. This is super-uniform: its cumulative probabilities at .1, .5, and 1 are .1, .5, and 1, respectively, and between these values its cumulative probability does not exceed the argument.
Use equal weights \(w_1=w_2=1/\sqrt{2}\) and nominal \(\alpha=.20\). The nominal critical value is \(z_{.80}=\Phi^{-1}(.80)\approx .8416\). The transformed scores are \(1.2816\) for \(p=.1\), \(0\) for \(p=.5\), and \(-\infty\) for \(p=1\). A stagewise value of 1 therefore makes the combined statistic non-rejecting under this upper-tail rule.
If both p-values are .1, then \(C=(1.2816+1.2816)/\sqrt{2}\approx1.812\), so the test rejects. If exactly one is .1 and the other is .5, then \(C=1.2816/\sqrt{2}\approx.9061\), so it also rejects. If both are .5, \(C=0\), and no other pair can reject. Therefore the actual null rejection probability is \(P(C>.8416)=.1(.1)+2(.1)(.4)=.09\).
The nominal continuous-reference test has size at most .20, but in this hypothetical discrete model its actual size is .09. The discrepancy arises from the limited set of possible stagewise p-values, not from an arithmetic error in the reference critical value.
Calibration with a discrete null distribution
When the trial design and null distribution are sufficiently specified, the combination statistic can instead be calibrated directly. Enumerate or otherwise compute its null distribution, then choose a rejection region whose probability under \(H_0\) is no greater than \(\alpha\). Because the distribution is discrete, the attainable rejection probabilities may skip over \(\alpha\). A nonrandomized test should use the largest available rejection region that stays within the target; it may remain conservative. Randomization at a boundary can attain the target exactly, but must be specified in advance and is not always desirable in clinical practice.
Worked Example: Exact Calibration by Enumeration
Continue the hypothetical inverse normal example. The possible finite values of \(C\) are approximately 1.812 when both p-values are .1 (probability .01), .9061 when exactly one is .1 and the other is .5 (probability .08), and 0 when both are .5 (probability .16). Any pair containing \(p=1\) gives \(C=-\infty\). These probabilities sum to .25 for the finite outcomes; the remaining .75 is at \(-\infty\).
At the nominal cutoff .8416, the rejection probability is .01+.08=.09. Lowering the cutoff to 0 and rejecting only when \(C>0\) does not add the \(C=0\) outcome, so the size remains .09. A cutoff below 0 would add the probability .16 at \(C=0\), giving size .25, which exceeds the target .20. Thus no nonrandomized cutoff in this ordering attains .20 exactly.
If exact equality is required, one could reject with probability \(r\) when \(C=0\). Solving \(.09+r(.16)=.20\) gives \(r=.11/.16=.6875\). The resulting size is \(.09+.6875(.16)=.20\). This calculation illustrates the role of randomization; it does not imply that such a procedure is preferable to a transparent, slightly conservative nonrandomized test.
Direct calibration is only as reliable as the null distribution used to obtain it. Enumeration must reflect the actual stagewise test definitions, sample sizes, allocation, nuisance parameters where relevant, and any adaptation that affects the second stage. If adaptation changes the stage-two design, its p-value must still be valid under the appropriate null conditional on the information used to adapt the design. Independence or conditional independence assumptions used by the calculation must also hold. A nominal cutoff cannot repair a misspecified null distribution.
Consequences for Fisher calibration
The same discreteness issue applies when a Fisher combination test is compared with its continuous-uniform reference distribution. For two independent uniform p-values, the statistic \(F=-2\{\ln(P_1)+\ln(P_2)\}\) has a chi-square reference distribution with four degrees of freedom. If the stagewise p-values are instead independent and super-uniform, the decreasing transformation \(-2\ln(p)\) gives contributions stochastically no larger than those from uniform p-values. Consequently, use of the continuous chi-square cutoff is conservative or exact, under those conditions. This statement concerns size calibration; it does not require treating the continuous reference as the exact discrete null distribution.
Worked Example: A Conservative Fisher Reference Cutoff
Use the same independent hypothetical p-value distribution, with masses .1, .4, and .5 at .1, .5, and 1. At nominal \(\alpha=.10\), the four-degree-of-freedom chi-square cutoff is approximately 7.779. If both p-values are .1, then \(F=-2\ln(.1\times.1)=-2\ln(.01)\approx9.210\), and the test rejects. This outcome has probability \(.1\times.1=.01\).
If one p-value is .1 and the other is .5, then \(F=-2\ln(.05)\approx5.991\), which is below 7.779. If both are .5, \(F=-2\ln(.25)\approx2.773\). A p-value of 1 contributes \(-2\ln(1)=0\), so those pairs do not create any additional rejection at this cutoff. The actual null rejection probability is therefore .01, below the nominal .10. As with the inverse normal example, the gap is due to the discrete input distribution.
Diagnostics, alternatives, and limitations
Before applying a continuous-reference calibration, check whether the stagewise p-values are valid under the relevant null and whether the assumptions about their joint distribution are justified. Useful diagnostics include enumerating the possible stagewise outcomes in a small design, comparing attainable p-values with their null probabilities, and simulating the full null design—including adaptation—when exact enumeration is impractical. Simulation estimates size with Monte Carlo error; it is not a proof of validity, so report the number of replicates and uncertainty when relying on it.
The main approaches differ in what they prioritize:
| Approach | Strength | Limitation |
|---|---|---|
| Continuous-reference cutoff with valid exact p-values | Simple and controls size under the required validity and independence conditions | May be conservative when the inputs are discrete |
| Direct exact or simulated calibration | Reflects the design-specific null distribution and can improve attainable size | Requires a correctly specified null model and accurate computation |
| Randomization at a boundary | Can attain a target size that a nonrandomized test cannot attain | Adds random decision-making and must be prespecified |
| Mid-p substitution | Can reduce conservatism in some discrete single-stage settings | May not be super-uniform and therefore does not automatically preserve combination-test type I error |
A common mistake is to reason, “Exact p-values are conservative, so replacing them by mid-p values will make the combination test less conservative without consequence.” That conclusion is not justified: a mid-p value can have \(P(P\leq u)>u\) for some \(u\), so it need not be a valid p-value. The stochastic-order argument for the inverse normal and Fisher calibrations then no longer applies. Any alternative input or calibration requires its own type I error justification.
A second limitation is that discreteness is not the only source of size error. Dependence between stages, invalid stagewise tests, unaccounted selection, or use of a reference distribution that does not match the design can all undermine calibration. Conversely, a discrete p-value does not by itself imply that a test is unusable: a conservative reference test can be valid, and a design-specific calibration can be used when justified. The choice should be made with the intended error-rate guarantee and the trial’s operational constraints in view.
Check Your Understanding
Answer each question, then compare with the solutions below.
- What inequality defines a super-uniform p-value, and why can a discrete exact test satisfy it strictly?
- For an inverse normal combination test with independent super-uniform p-values and positive normalized weights, is the continuous-reference test anti-conservative, exact, or conservative in general?
- In the hypothetical distribution with masses .1, .4, and .5 at p-values .1, .5, and 1, what is the probability that two independent stages both produce \(p=.1\)?
- Why does direct enumeration sometimes fail to produce a nonrandomized test with size exactly equal to \(\alpha\)?
- Why does substituting mid-p values require a new validity argument?
Solutions. (1) Super-uniformity means \(P(P\leq u\mid H_0)\leq u\) for every \(u\in[0,1]\). A discrete test has jumps in its tail probabilities, so its cumulative probability can remain below \(u\) between attainable values. (2) It is conservative or exact: the rejection probability is at most \(\alpha\), but equality is not guaranteed. (3) Independence gives \(.1\times.1=.01\). (4) A discrete null distribution has only particular attainable rejection probabilities, which can fall on either side of \(\alpha\) without equalling it. (5) Mid-p values need not be super-uniform, so the stochastic-order proof that supports the original size bound may not apply.