Why Model Genetic Effects Across Ancestries?
When a candidate variant is studied in multiple ancestry groups, a central modeling decision is whether its genetic effect should be treated as the same in every group. A common effect can summarize evidence efficiently, but it may hide genuine differences. Separate ancestry-specific estimates allow those differences, but can be imprecise. A random-effects model instead represents effects as varying around a mean, with a parameter describing the extent of variation.
These models concern the effect of the candidate variant, not whether the variant is causal. A high association statistic or a good fit under one effect model does not establish causality. The interpretation also depends on what effect is being modeled: here, assume an additive per-allele effect on a specified outcome scale, estimated within each ancestry group using the same covariate adjustment.
The effect allele must have the same meaning in each group, and the effect scale and outcome definition must be comparable. If the outcome is measured in millimeters of blood pressure, for example, each \(\beta_a\) is in millimeters per additional effect allele. For a binary outcome, the effect could instead be a log odds ratio per allele; effects on different scales should not be compared as if they were interchangeable.
In practice, ancestry-specific association estimates may also differ because of sampling error, allele-frequency differences, imputation quality, or model misspecification. In addition, marginal effects at a tag variant can differ across groups because linkage disequilibrium differs, even when the causal variant's direct effect is shared. The models below describe the effect parameter for the candidate being modeled; their interpretation depends on whether that candidate is itself the causal variant or a proxy.
Three Ways to Represent Effects
Suppose \(\hat{\beta}_a\) is the estimated effect in group \(a\), with standard error \(s_a\). A basic sampling model is \(\hat{\beta}_a\mid\beta_a\sim N(\beta_a,s_a^2)\). This approximation assumes the estimates are approximately normal and, for the formulas below, independent across ancestry groups. If samples overlap or estimates have other dependence, their covariance must be included instead of assuming independence.
| Model | Effect parameters | What it assumes | Useful when |
|---|---|---|---|
| Common effect | One effect, \(\beta\) | Every \(\beta_a=\beta\) on the chosen scale | A shared effect is scientifically plausible and a pooled estimate is desired |
| Ancestry-specific fixed effects | \(\beta_1,\ldots,\beta_A\) | No equality constraint or distribution across group effects | The group effects themselves are the target or differences are important |
| Random effects | Mean \(\mu\), variance \(\tau^2\), and group effects | Group effects vary around \(\mu\) according to a specified distribution | Effects may vary, and a mean and extent of variation are useful summaries |
Under the common-effect model and independent estimates with known standard errors, the inverse-variance estimate is a weighted average. The weights \(w_a=1/s_a^2\) give more influence to more precise estimates. Its standard error follows from the variance of a weighted sum:
A confidence interval based on a normal approximation is \(\hat{\beta}_{\mathrm{common}}\pm1.96\,SE(\hat{\beta}_{\mathrm{common}})\) for a 95% interval. It describes uncertainty in the common effect under the common-effect model and its sampling assumptions; it is not an interval for every group effect if effects truly vary.
Worked Example: Pooling Under a Common-Effect Model
Suppose two hypothetical ancestry groups have per-allele estimates of \(0.20\) and \(0.35\) outcome standard deviations, with standard errors \(0.20\) and \(0.1414\), respectively. Assume the effect allele and outcome scale are harmonized, the estimates are independent, and the normal approximation is adequate. Under the common-effect model, calculate the pooled estimate and its 95% confidence interval.
The precisions are \(w_1=1/(0.20^2)=25\) and \(w_2=1/(0.1414^2)\), approximately \(50\) (using the unrounded value \(SE_2=\sqrt{0.02}\)). Thus,
The standard error is \(SE=\sqrt{1/75}=0.11547\). Therefore the 95% interval is
On the stated scale, the estimated common effect is \(0.30\) outcome standard deviations per additional effect allele. Under the common-effect model, the interval is compatible with a positive effect. It does not show that both group-specific effects equal exactly \(0.30\), nor does it establish causality.
The ancestry-specific fixed-effects model does not pool the effects into one shared parameter. Its parameters are the \(A\) group effects themselves. For two groups, a useful heterogeneity parameter is the contrast \(\Delta_{12}=\beta_1-\beta_2\); for more groups, all pairwise contrasts or a set of \(A-1\) independent contrasts can describe departures from equality. The no-heterogeneity null is \(H_0:\beta_1=\cdots=\beta_A\). A fixed-effects model makes no claim that these effects are sampled from a wider population distribution.
Worked Example: Interpreting Ancestry-Specific Effects
Consider hypothetical estimates for three ancestry groups, all in outcome standard deviations per effect allele: \(\hat{\beta}_1=0.10\), \(\hat{\beta}_2=0.30\), and \(\hat{\beta}_3=0.50\). Each has standard error \(0.10\). Assume harmonized alleles, the same outcome and adjustment set, approximate normal sampling, and independent estimates. What do the fixed-effects estimates say, and how much do they differ from a common-effect description?
The fixed-effects estimates are the three values as observed; the estimated contrast between groups 3 and 1 is \(0.50-0.10=0.40\) outcome standard deviations per allele. For a descriptive check of equality, the inverse-variance pooled estimate is \(0.30\), since all three estimates have equal weight \(1/0.10^2=100\). The weighted discrepancy from that estimate is
If the common-effect model is true and the standard errors are known, this statistic has approximately a chi-square distribution with \(A-1=2\) degrees of freedom. Here \(P(\chi^2_2\geq8)=0.0183\), so these estimates show evidence against exact equality under those assumptions. This does not identify the cause of the differences: chance, measurement issues, model differences, or real effect modification could contribute. The fixed-effects model preserves the group estimates rather than forcing a single effect.
Random Effects and Heterogeneity
A random-effects model treats the group effects as varying around a mean. A standard specification is \(\beta_a\sim N(\mu,\tau^2)\), together with the sampling model \(\hat{\beta}_a\mid\beta_a\sim N(\beta_a,s_a^2)\). The parameter \(\mu\) is the mean of the modeled distribution of group effects, and \(\tau^2\) is their between-group variance. Thus \(\tau\), not the standard error of an individual estimate, represents the scale of effect heterogeneity. When \(\tau^2=0\), the model reduces to a common effect.
This distributional assumption is a substantive choice. It treats the observed group effects as exchangeable draws from a specified distribution, rather than as unrelated fixed parameters. The mean \(\mu\) is not automatically the effect in every ancestry group, or the effect in a new group unless the distributional interpretation supports that prediction. With few groups, estimation of \(\tau^2\) is especially uncertain.
A short derivation clarifies the role of \(\tau^2\). Since the sampling error and the deviation of \(\beta_a\) from \(\mu\) are independent in this model, write \(\hat{\beta}_a-\mu=(\hat{\beta}_a-\beta_a)+(\beta_a-\mu)\). The two terms have variances \(s_a^2\) and \(\tau^2\), so their sum has variance \(s_a^2+\tau^2\). Integrating out \(\beta_a\) therefore gives:
For a given value of \(\tau^2\), this marginal model weights estimates by \(1/(s_a^2+\tau^2)\), rather than just by \(1/s_a^2\). As estimated heterogeneity increases, the most precise group does not dominate as strongly. Estimation procedures differ in how they estimate \(\tau^2\); the following calculation uses a simple method-of-moments estimate only to illustrate the model.
Worked Example: Estimating Random-Effects Variation
Suppose two hypothetical ancestry groups have estimates \(0.10\) and \(0.50\) outcome standard deviations per effect allele, each with standard error \(0.10\). Assume the same harmonization and outcome conditions as above, independent approximately normal estimates, and a normal random-effects distribution. Use the DerSimonian–Laird method-of-moments estimate of \(\tau^2\), then calculate the random-effects mean and a plug-in interval.
First calculate the common-effect weighted mean, which is \(0.30\), and the discrepancy statistic:
For the DerSimonian–Laird estimate, \(C=\sum w_a-\sum w_a^2/\sum w_a=200-20000/200=100\). With \(A=2\), the estimate is
This is in squared outcome-standard-deviation units per allele; the corresponding estimated heterogeneity standard deviation is \(\hat{\tau}=\sqrt{0.07}=0.2646\). The random-effects weights are \(1/(0.10^2+0.07)=12.5\) for each group. Their weighted mean remains \(0.30\), and its plug-in standard error is \(\sqrt{1/(12.5+12.5)}=0.20\). A normal-approximation 95% interval is \(0.30\pm1.96(0.20)=(-0.092,\ 0.692)\) outcome standard deviations per allele.
This interval summarizes uncertainty in the mean under the specified random-effects model, conditional on the plug-in heterogeneity estimate. It is not a prediction interval for an individual ancestry effect. With only two groups, \(\hat{\tau}^2\) and the interval are highly uncertain; methods that account more fully for heterogeneity-estimation uncertainty may give different intervals.
Choosing and Interpreting the Model
Model choice should follow the estimand—the quantity the analysis is intended to describe—not whichever model produces the smallest standard error. A common-effect model is appropriate when a shared per-allele effect is a defensible assumption on the chosen scale. An ancestry-specific model is useful when the group effects or their contrasts are the targets. A random-effects model is useful when variation across groups is part of the scientific description and a mean and between-group variance are meaningful summaries.
None of the models repairs incompatible measurements or unmodeled differences in analysis. Before interpreting heterogeneity as biological effect modification, check that the effect allele is aligned, the phenotype and scale are comparable, and covariate adjustment is sufficiently consistent. Ancestry labels describe groups used in the analysis; they are not themselves a complete explanation for any observed difference. Effects can vary with environment, ascertainment, phenotype definition, or other factors correlated with group membership.
Specify the candidate variant, harmonized effect allele, outcome scale, covariate adjustment, and whether the target is a shared effect, group effects, or a distribution of effects.
Confirm that estimates refer to compatible quantities. Account for covariance if ancestry-specific estimates are not independent.
Impose equality only when a common effect is scientifically justified; otherwise retain group effects or specify a random-effects distribution with its assumptions.
State whether an interval concerns a shared effect, a group-specific effect, or a random-effects mean, and report heterogeneity in the appropriate units.
Common Mistake: Confusing Shared Causality With a Shared Effect
A common error is to reason that if the same variant is causal in every ancestry, its effect must be identical in every ancestry. Shared causal status and effect portability are distinct claims. A variant can be causal in more than one group while its effect differs because of effect modification, measurement or modeling differences, or the scale on which effects are expressed. Conversely, different marginal associations at a tag variant can result from ancestry-specific linkage disequilibrium even when a causal variant has a common direct effect.
The correction is to state precisely which parameter is being compared. A common-effect model asserts equality of the modeled \(\beta_a\) values; it does not assert merely that one causal variant is shared. An ancestry-specific or random-effects model allows effect heterogeneity, but evidence of heterogeneity alone does not explain its mechanism or establish that the candidate variant is causal.
Check Your Understanding
Try these questions without looking back.
- In a common-effect model, what equality constraint is imposed on the ancestry-specific effects?
- For a random-effects model, distinguish the meaning and units of \(\mu\) from those of \(\tau^2\).
- Why can a tag variant have different marginal associations across ancestries even if a causal variant has the same direct effect?
- What assumptions justify using the inverse-variance common-effect estimate shown in this tutorial?
- Does evidence against equal ancestry-specific effects establish that the candidate variant is causal? Explain.