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Fine-Mapping Across Ancestries · Tutorial 647 of 989

Interpreting Heterogeneity Diagnostics in Fine-Mapping

Use ancestry-specific estimates and heterogeneity statistics to judge whether observed effect differences are compatible with sampling variation or challenge a shared-effect model.

Advanced 12 min read

What You'll Learn

  • Define the shared-effect model and the assumptions needed for a heterogeneity test
  • Calculate inverse-variance pooled estimates and Cochran’s Q
  • Interpret confidence intervals and heterogeneity p-values in context
  • Distinguish evidence of heterogeneity from evidence about its cause
  • Recognize how low precision and few ancestry groups limit conclusions

Why Examine Effect Heterogeneity?

Trans-ancestry fine-mapping uses information from multiple ancestry groups, but a combined analysis can rely on different assumptions about genetic effects. A shared-effect model assumes that a variant has the same effect on the chosen outcome scale in each group. An ancestry-stratified analysis makes it possible to inspect that assumption rather than treating it as automatic.

Ancestry-specific estimates will rarely be identical. Even if the underlying effects are equal, sampling variation produces different estimates, especially when standard errors differ. Heterogeneity diagnostics help assess whether the observed differences are larger than expected under a shared-effect model. They do not, by themselves, establish why effects differ or identify which variant is causal.

Definition: Let \(\hat{\beta}_k\) be the estimated effect for ancestry group \(k\), with standard error \(SE_k\), for the same variant, harmonized effect allele, phenotype, and effect scale. A shared-effect model assumes that all groups estimate one common effect \(\beta\). Effect heterogeneity means that the underlying effects are not all equal on that scale. Differences among the estimates alone are not enough to establish heterogeneity, because estimates vary through sampling error.

Check Comparability Before Calculating

A heterogeneity statistic is meaningful only if its inputs can be compared. Each estimate should refer to the same variant and the same effect allele, with alleles aligned so that a positive estimate has the same direction in every group. The outcome and effect scale must also match. For example, log-odds coefficients should not be compared directly with risk differences as though they were measurements in the same units.

For the usual inverse-variance calculation, assume the ancestry-specific estimates are approximately unbiased and normally distributed around their true effects, and that their standard errors are valid. The basic Cochran \(Q\) calculation also assumes the estimates are independent across groups. If participants overlap or estimates have other dependence, their covariance must be included; treating dependent estimates as independent can miscalibrate the test. These diagnostics assess effect estimates, not differences in linkage disequilibrium (LD) patterns, which can change association patterns even when the causal effect is shared.

Model: Under the independent shared-effect model, \(\hat{\beta}_k \sim N(\beta, SE_k^2)\) for \(k=1,\ldots,K\). The \(SE_k\) values are treated as known or sufficiently well estimated for this approximation. The alternative for a heterogeneity test is that at least one group’s true effect differs from the common value.

Inverse-Variance Summary and Cochran’s Q

Under the shared-effect model, each estimate receives weight \(w_k=1/SE_k^2\). An estimate with a smaller standard error receives more weight because it provides more precise information. The inverse-variance pooled estimate and its standard error are

$$ \hat{\beta}_{\mathrm{FE}} =\frac{\sum_{k=1}^{K} w_k\hat{\beta}_k}{\sum_{k=1}^{K}w_k}, \qquad SE(\hat{\beta}_{\mathrm{FE}}) =\sqrt{\frac{1}{\sum_{k=1}^{K}w_k}}. $$

The subscript \(\mathrm{FE}\) denotes a fixed-effect calculation: it summarizes estimates under the assumption of one common effect. Cochran’s heterogeneity statistic measures the weighted dispersion of the ancestry-specific estimates around that pooled value:

$$ Q=\sum_{k=1}^{K}w_k(\hat{\beta}_k-\hat{\beta}_{\mathrm{FE}})^2. $$

When the shared-effect model and the stated sampling assumptions hold, \(Q\) is approximately distributed as chi-square with \(df=K-1\). A large \(Q\), relative to that reference distribution, is evidence against a common effect on the analyzed scale. The p-value is the probability, assuming the shared-effect model and its assumptions, of obtaining a \(Q\) at least as large as the observed one. It is not the probability that the shared-effect model is true.

Worked Example: Estimates Consistent With a Shared Effect

Suppose three hypothetical ancestry groups have estimated log-odds effects of \(0.12\), \(0.10\), and \(-0.02\) per effect allele, with standard errors \(0.08\), \(0.08\), and \(0.20\), respectively. Assume the alleles and outcome are harmonized, the groups do not overlap, and the estimates are approximately unbiased and normal with the stated standard errors.

The inverse-variance weights are \(1/0.08^2=156.25\), \(156.25\), and \(1/0.20^2=25\). Therefore,

$$ \hat{\beta}_{\mathrm{FE}} =\frac{156.25(0.12)+156.25(0.10)+25(-0.02)} {156.25+156.25+25} =\frac{18.75+15.625-0.5}{337.5} =\frac{33.875}{337.5} \approx 0.1004. $$

The pooled standard error is \(\sqrt{1/337.5}\approx0.05443\). A 95% normal-approximation confidence interval is \(0.1004\pm1.96(0.05443)\), or approximately \((-0.0063,\,0.2071)\) log-odds units per effect allele.

For heterogeneity, substitute the pooled estimate into \(Q\):

$$ \begin{aligned} Q &=156.25(0.12-0.1003704)^2 +156.25(0.10-0.1003704)^2 +25(-0.02-0.1003704)^2\\ &\approx 0.0602066+0.00002143+0.3622257 =0.4224537\approx 0.4225. \end{aligned} $$

There are \(df=3-1=2\) degrees of freedom. For two degrees of freedom, the upper-tail probability is \(e^{-Q/2}\), so \(p=e^{-0.4225/2}\approx0.8096\). These data do not provide evidence against a shared effect. They do not prove that the effects are identical; in particular, the third group’s estimate is imprecise.

Confidence Intervals and the Heterogeneity Test

A confidence interval for one ancestry-specific effect describes uncertainty about that group’s effect, under the model used to construct the interval. It is useful to inspect the intervals, but visual overlap is not a formal heterogeneity test. Conversely, the absence of overlap is not itself the definition of heterogeneity. The test of a difference should account for the standard errors of the estimates and, where relevant, their covariance.

For two independent estimates, the estimated difference is \(\hat{\beta}_1-\hat{\beta}_2\), and its standard error is \(\sqrt{SE_1^2+SE_2^2}\). This makes the role of precision explicit: a numerical gap can be unremarkable when its uncertainty is large, while a smaller gap can be compelling when estimated precisely.

Worked Example: A Precise Difference Challenges a Shared Effect

Consider two hypothetical, independent ancestry groups with harmonized log-odds estimates \(0.20\) and \(0.02\), each with standard error \(0.04\). Assume the same variant, outcome, and effect scale are used. The estimated difference is \(0.20-0.02=0.18\), with standard error

$$ SE(\hat{\beta}_1-\hat{\beta}_2) =\sqrt{0.04^2+0.04^2} =\sqrt{0.0032} \approx0.05657. $$

A 95% interval for the difference is \(0.18\pm1.96(0.05657)\), approximately \((0.0691,\,0.2909)\) log-odds units per effect allele. It excludes zero. The individual 95% intervals are approximately \((0.1216,\,0.2784)\) and \((-0.0584,\,0.0984)\), which do not overlap.

The fixed-effect estimate is \((625(0.20)+625(0.02))/(625+625)=0.11\), because each weight is \(1/0.04^2=625\). Thus

$$ Q=625(0.20-0.11)^2+625(0.02-0.11)^2 =625(0.0081)+625(0.0081) =10.125. $$

With \(df=1\), the chi-square upper-tail p-value is approximately \(0.0015\). Under the stated assumptions, the estimates provide evidence against a shared effect on the log-odds scale. This result does not reveal the cause of the difference: it could reflect genuine effect modification or a problem with comparability or estimation.

Low Precision Can Hide Heterogeneity

A non-small heterogeneity p-value does not demonstrate that effects are equal. When one ancestry group has a large standard error, its estimate can differ substantially from the others without making \(Q\) large. This is especially important with few groups: the test may have limited power to detect moderate differences, and the chi-square reference distribution does not turn a weak diagnostic into evidence of equivalence.

Worked Example: A Wide Interval Limits the Conclusion

Suppose two hypothetical independent groups have log-odds estimates \(0.12\) and \(0.08\), with standard errors \(0.04\) and \(0.16\). The estimated difference is \(0.04\), but its standard error is

$$ \sqrt{0.04^2+0.16^2} =\sqrt{0.0272} \approx0.1649. $$

The 95% interval for the difference is \(0.04\pm1.96(0.1649)\), approximately \((-0.2832,\,0.3632)\) log-odds units per effect allele. It is wide and includes zero; the data are compatible with both little difference and appreciable differences in either direction.

The weights are \(625\) and \(39.0625\). The pooled estimate is

$$ \hat{\beta}_{\mathrm{FE}} =\frac{625(0.12)+39.0625(0.08)}{625+39.0625} =\frac{78.125}{664.0625} \approx0.1176. $$

Then \(Q=625(0.12-0.117647)^2+39.0625(0.08-0.117647)^2\approx0.05882\). With \(df=1\), \(p\approx0.808\). This is not evidence against a shared effect, but it also does not establish that the effects are equal: the less precise group contributes little information about heterogeneity.

What a Heterogeneity Signal Does—and Does Not—Mean

When \(Q\) is large and its p-value is small, first verify that the estimates are genuinely comparable. Check allele alignment, variant identity, phenotype definition, effect scale, imputation and measurement quality, and the standard-error calculations. Also check whether the independence assumption is reasonable. An apparent difference caused by a mismatched allele or scale is not biological effect heterogeneity.

If those checks are satisfactory, the result is evidence that the common-effect model may be inadequate for these estimates. It is not a test of whether the same causal variant is present across ancestries, nor does it identify a biological mechanism. A shared causal variant can have different estimated effects, and a heterogeneity test alone cannot distinguish genuine effect modification from residual bias or other model misspecification. Likewise, a non-significant test does not establish effect portability.

The fixed-effect pooled estimate is most defensible when a common effect is a reasonable working assumption. If that assumption is doubtful, do not present the pooled value as though it necessarily describes every group. Report the ancestry-stratified estimates and intervals, the heterogeneity statistic and its degrees of freedom, and the assumptions behind the comparison. A p-value should be read alongside the size and precision of the differences, not as a stand-alone measure of their importance.

Common mistake: “The ancestry-specific confidence intervals overlap, so the effects are the same,” or “the heterogeneity p-value is not significant, so the shared-effect model is proven.” Neither conclusion follows. Interval overlap is not an equivalence test, and a non-significant heterogeneity test may simply have low power. Describe the observed estimates and uncertainty, then state whether the test found evidence against the shared-effect model under its assumptions.
1
Confirm comparability.
Verify the same variant, aligned effect allele, phenotype, effect scale, and valid standard errors.
2
Inspect estimates and intervals.
Describe directions, magnitudes, and precision in each ancestry group; do not treat visual overlap as a formal test.
3
Calculate and interpret heterogeneity.
Use \(Q\) and \(df=K-1\) under the independent shared-effect model, and interpret its p-value conditionally on those assumptions.
4
Qualify the conclusion.
Evidence against a common effect is not an explanation of why effects differ; absence of such evidence is not proof of equality.

Check Your Understanding

Answer each question by referring to the assumptions and interpretation of the diagnostics.

  1. Why must effect alleles be aligned before comparing ancestry-specific estimates?
  2. For four independent ancestry groups, what degrees of freedom are used for Cochran’s \(Q\) under the shared-effect model?
  3. What does a small heterogeneity p-value indicate, and what does it not identify?
  4. Why can a non-significant heterogeneity test be uninformative when one group has a large standard error?
  5. Two ancestry-specific confidence intervals overlap. Does this establish equal effects? Explain.
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