Why Combine Discovery and Replication Evidence?
A discovery cohort may identify a genetic association, while an independent replication cohort tests whether the finding is supported in new data. Combining the two p-values can summarize evidence across cohorts, but it answers a different question from asking whether the replication result itself meets a prespecified criterion. A combined result can be small even when the replication p-value is not.
The distinction matters because a p-value combination method usually tests a global null: the hypothesis that none of the component studies contains an association. Rejecting that null does not show that both studies support the association, that their effects have the same direction, or that the replication effect meets a specified size threshold. The earlier material on replication criteria and replication failure distinguishes these questions; p-value combination does not replace those criteria.
Assumptions Before Calculation
For the standard Fisher and Stouffer reference distributions, each component p-value must be valid under its own null: under that null, it is uniformly distributed on \([0,1]\) for a continuous test, or at least super-uniform for a conservative or discrete test. The component tests must also be independent under the joint null for the usual independent-study formulae to have their stated calibration. Independent cohorts make this assumption more plausible, but independence can still fail if participants overlap, data are shared, or study results are otherwise dependent.
The studies should address compatible association questions. For example, a p-value for one variant and outcome should not be combined as though it tested the same question as a p-value for a different variant or phenotype. Alleles, effect direction, and outcome definitions need to be aligned when direction is used. Fisher’s method uses p-value magnitudes and does not require aligned directions; that feature is also a limitation. Stouffer’s signed version requires a consistent definition of positive direction.
A further concern is selection. If a discovery result was chosen because its p-value was unusually small among many tested variants, that selected p-value is not distributed like an unselected uniform p-value, conditional on selection. Applying an ordinary combination reference distribution to that selected discovery p-value can therefore overstate combined evidence. An independent, prespecified replication p-value remains directly informative for its replication test; the discovery selection process does not make the replication result significant or nonsignificant by itself.
Fisher’s Method
Fisher’s method combines p-value magnitudes. For \(k\) independent, continuous, valid p-values under the joint null, define the statistic \(X_F\) as follows.
Under the joint null, \(X_F\) has a chi-square distribution with \(2k\) degrees of freedom. Thus, the combined p-value is the upper-tail probability \(P(\chi^2_{2k}\geq X_F)\). Small component p-values contribute larger positive values to \(X_F\), making the combined result more significant.
The reference distribution follows directly from the uniform null distribution. If \(U\sim\text{Uniform}(0,1)\), then for \(t\geq0\),
This is the cumulative distribution function of a chi-square random variable with 2 degrees of freedom. Each independent term \(-2\ln(p_i)\) therefore has a \(\chi^2_2\) distribution; their sum has a \(\chi^2_{2k}\) distribution. Independence is essential to this sum result.
Worked Example: Fisher’s Method With Unequal Evidence
Suppose independent discovery and replication tests give two-sided p-values \(p_D=0.02\) and \(p_R=0.30\). Assume both p-values are valid for the same association question and that the cohorts are independent. Calculate the Fisher result.
There are two p-values, so the reference distribution has \(2k=4\) degrees of freedom. The test statistic is
For a chi-square distribution with 4 degrees of freedom, the upper-tail probability at \(x\) is \(e^{-x/2}(1+x/2)\). Here this gives
The combined p-value is approximately 0.0367, so this calculation rejects the joint null at a 0.05 significance level, given the assumptions. But the replication p-value is 0.30. The combined result is not evidence that the replication test itself was significant; the discovery p-value contributes much of the combined evidence. Nor does the replication p-value of 0.30 prove that the association is absent in the replication population.
Fisher’s method is sensitive to very small p-values and treats all p-values as evidence in the same direction. It can be useful when the question is whether there is evidence against the global null in at least one study. It is not a directional consistency test. Even a very small Fisher p-value cannot establish that discovery and replication effects agree.
Stouffer’s Method and Direction
Stouffer’s method transforms p-values to standard-normal scores and combines those scores. For two-sided p-values, the unsigned transformation is \(z_i=\Phi^{-1}(1-p_i/2)\), where \(\Phi\) is the standard normal cumulative distribution function. This score is nonnegative and measures the magnitude of evidence, but it discards effect direction.
When effects have a meaningful, harmonized direction, use a signed score. Define \(s_i=1\) if the estimated effect in study \(i\) is in the prespecified positive direction, and \(s_i=-1\) if it is in the opposite direction. The signed score is \(z_i=s_i\Phi^{-1}(1-p_i/2)\). For two independent studies with equal weights, the Stouffer statistic is
Under the joint null, each signed score is standard normal when the underlying tests have the appropriate continuous null calibration and the signs are defined consistently. The sum of two independent standard normal scores has variance 2, so division by \(\sqrt{2}\) gives a standard normal statistic. A two-sided combined p-value is \(2[1-\Phi(|Z_S|)]\); for a prespecified one-direction alternative, the corresponding one-sided tail is used instead.
More generally, fixed weights \(w_i\) give \(Z_S=(\sum_i w_i z_i)/\sqrt{\sum_i w_i^2}\). Under independent standard normal scores, the numerator has mean zero and variance \(\sum_i w_i^2\) under the joint null, so the normalized statistic is standard normal. Weights and the direction of the test should be chosen before examining the results. Sample-size-based weights are not automatically appropriate for every test; the scores must have a compatible interpretation for the weighting scheme to represent the intended evidence.
Worked Example: Opposing Directions Under Signed Stouffer Combination
Suppose the discovery p-value is 0.01 with an estimated effect in the prespecified positive direction. The independent replication p-value is 0.20, but its estimated effect is in the negative direction. Assume two-sided tests, aligned effect scales, and independent cohorts.
The discovery signed score is \(z_D=\Phi^{-1}(0.995)\approx2.5758\). The replication score has the opposite sign: \(z_R=-\Phi^{-1}(0.90)\approx-1.2816\). Therefore,
The two-sided combined p-value is \(2[1-\Phi(0.9152)]\approx0.360\). This is not significant at 0.05. The opposing directions reduce the signed combined evidence. That result does not establish that the effects differ; it says this particular signed combination does not provide strong evidence for a common prespecified direction. The replication result also does not meet a 0.05 threshold.
Worked Example: Same-Direction Stouffer Combination
Now suppose independent discovery and replication tests give two-sided p-values \(p_D=0.03\) and \(p_R=0.08\), and both estimated effects are in the prespecified positive direction. Assume the same variant and outcome are tested, with aligned effect direction.
The normal scores are \(z_D=\Phi^{-1}(0.985)\approx2.1701\) and \(z_R=\Phi^{-1}(0.96)\approx1.7507\). With equal weights,
The two-sided combined p-value is approximately \(2[1-\Phi(2.7724)]=0.0056\). This indicates strong combined evidence against the joint null under the stated assumptions. The replication p-value alone is 0.08, however, so it does not pass a 0.05 two-sided threshold. The small combined p-value is not a substitute for a prespecified replication criterion, nor does it demonstrate that the replication effect exceeds a meaningful effect-size threshold.
What a Combined Result Establishes
A correctly calibrated combined p-value is evidence against the joint null that all component tests are null. The evidence can arise from one study, several studies, or a mixture, depending on the method and the observed results. It does not report how much evidence came from each cohort. Always show the component p-values and, when direction matters, the cohort-specific effect estimates and directions alongside the combined result.
For a claim specifically about replication, the replication test and its prespecified success rule remain central. For example, if replication was defined as a one-sided test in the discovery direction at a specified \(\alpha\), the replication p-value must be evaluated against that criterion. If success instead requires an effect at least as large as a meaningful threshold, a combined p-value does not answer that effect-size question. A small combined p-value cannot retroactively change the replication rule.
Fisher and Stouffer also differ in what they retain. Fisher uses p-value magnitudes and ignores signs. Stouffer can combine magnitudes without signs, or can use signed scores to evaluate evidence in a common direction. The signed version can lose evidence when estimates point in opposite directions, but that is appropriate if the target claim requires a shared direction. Neither method estimates a common effect size or quantifies between-study heterogeneity.
| Method | What it combines | Important limitation |
|---|---|---|
| Fisher | Products of independent p-values through \(-2\sum\ln(p_i)\) | Ignores effect direction; a strong result may be driven by one study |
| Unsigned Stouffer | Normal scores derived from p-value magnitudes | Does not distinguish concordant from opposing effects |
| Signed Stouffer | Normal scores with signs aligned to a defined direction | Requires harmonized directions; a combined result still does not prove replication success |
| Replication-only test | The prespecified replication cohort’s test statistic or p-value | Does not summarize overall evidence across cohorts |
These methods are not replacements for methods that combine effect estimates and their sampling uncertainties. Such approaches address estimation questions that a combined p-value does not. Likewise, if cohorts are dependent, the independent-study calibration cannot simply be assumed; the covariance among contributions must be accounted for by an appropriate method.
Diagnostics and Common Pitfalls
Before reporting a combined result, check that the tests concern the same prespecified association, that p-values are valid for their stated tests, and that the independence assumption is plausible. For signed Stouffer combination, verify allele coding and effect direction, and state whether the combined test is one-sided or two-sided. Inspect the component p-values and estimates rather than presenting only the combined number. A dramatic combined result paired with a weak or opposite-direction replication result is a signal to be precise about what has and has not been established.
- Wrong reasoning: “The Fisher p-value is below 0.05, so the association replicated.” Correction: Fisher rejects the joint null under its assumptions; it does not require the replication p-value to pass a replication threshold.
- Wrong reasoning: “A non-significant replication p-value proves no replication effect exists.” Correction: Failure to reject is not proof of a zero effect; precision and power affect the result.
- Wrong reasoning: “The two studies are independent, so their p-values can always be combined with the ordinary formula.” Correction: Independence is only one requirement. P-values must also be valid and the tests must address compatible hypotheses; selection and dependence among tests can invalidate calibration.
- Wrong reasoning: “Fisher’s method confirms that both effects have the same sign.” Correction: Fisher’s statistic uses p-value magnitudes only. Use harmonized signed evidence if direction is part of the scientific claim.
Check Your Understanding
Answer each question, then compare your reasoning with the brief solutions.
- Two independent, valid p-values are 0.05 and 0.10. Calculate Fisher’s statistic and approximate combined p-value using the \(\chi^2_4\) reference distribution.
- Two studies have equal-magnitude two-sided p-values of 0.04, but their effects point in opposite directions. What is the signed equal-weight Stouffer statistic, and what does it imply?
- A Fisher combined p-value is 0.01, while the replication p-value is 0.25. What conclusion is justified, and what conclusion is not?
- Why can ordinary Fisher calibration be inappropriate when the discovery variant was selected as the smallest p-value from thousands of tests?
Solutions. (1) \(X_F=-2\ln(0.05\times0.10)=-2\ln(0.005)\approx10.597\). The upper-tail probability for \(\chi^2_4\) is \(e^{-10.597/2}(1+10.597/2)\approx0.0315\). (2) The scores are approximately \(+2.0537\) and \(-2.0537\), so \(Z_S=0/\sqrt{2}=0\), with two-sided p-value 1. The opposing scores cancel; this is not evidence for a common direction. (3) Under its assumptions, the combined result is evidence against the joint null, but it does not show that replication succeeded; a replication p-value of 0.25 does not meet a conventional 0.05 threshold and does not prove the effect is absent. (4) Selection makes the chosen discovery p-value unusually small conditional on being selected, so it is not distributed as an unselected uniform p-value. The ordinary joint-null calibration can overstate evidence unless selection is accounted for.