Fixed Versus Random Effects With Ancestry Differences

Why the Choice of Effect Model Matters

A genetic meta-analysis may combine cohorts whose participants differ in ancestry. Even after allele harmonization and alignment of phenotype and effect scales, the estimated association can vary across cohorts. The choice between fixed-effect and random-effects meta-analysis determines what pooled quantity is being estimated; it does not explain why the cohort estimates differ.

Ancestry differences can matter because allele frequencies and linkage disequilibrium (LD) patterns vary across populations. For a typed or imputed marker that tags a causal variant, its marginal association can therefore differ even when the causal variant’s effect is the same. Effects may also differ for other reasons, including genuine effect modification or differences in study design. Heterogeneity statistics can indicate that estimates vary more than expected from sampling error, but they do not identify the cause.

Definition: In an inverse-variance fixed-effect model, each cohort estimates the same underlying effect \(\beta\), and differences between estimates arise from sampling error. In a random-effects model, cohort-specific effects \(\beta_i\) are treated as draws from a distribution with mean \(\mu\) and between-cohort variance \(\tau^2\). The fixed-effect pooled estimate targets the common effect under its model; the random-effects estimate targets the mean of the assumed distribution of cohort effects.

These models require estimates of the same association to be combined: the effect allele must be aligned, and the phenotype definition and effect scale must be comparable. For example, coefficients measured in different units cannot be pooled directly. The calculations below assume independent cohorts and approximately normal cohort estimates \(\hat{\beta}_i\) with known or sufficiently well-estimated standard errors \(SE_i\). If cohorts overlap, their estimate covariance must be accounted for rather than treating them as independent.

What the Two Models Estimate

Write \(w_i=1/SE_i^2\) for the inverse-variance weight of cohort \(i\). Under the fixed-effect model, the pooled estimate and its standard error are

\hat{\beta}_{\mathrm{FE}} = \frac{\sum_{i=1}^k w_i\hat{\beta}_i}{\sum_{i=1}^k w_i}, \qquad SE(\hat{\beta}_{\mathrm{FE}}) = \sqrt{\frac{1}{\sum_{i=1}^k w_i}}.

The estimate is a precision-weighted average. A larger cohort or a more precise estimate receives more weight. If cohort effects truly differ, this number is still a weighted average of the observed effects, but it is not automatically a common biological effect or the average effect in every possible target population. Its value depends on which cohorts are included and their precision.

In the random-effects model, the usual inverse-variance weights are instead \(w_i^*=1/(SE_i^2+\tau^2)\), giving

\hat{\mu}_{\mathrm{RE}} = \frac{\sum_{i=1}^k w_i^*\hat{\beta}_i}{\sum_{i=1}^k w_i^*}, \qquad SE(\hat{\mu}_{\mathrm{RE}}) = \sqrt{\frac{1}{\sum_{i=1}^k w_i^*}}.

Here \(\tau^2\) represents between-cohort variation in the effects. As it grows, weights become less dominated by the most precise cohort, so smaller cohorts can contribute relatively more than under fixed-effect weighting. The random-effects mean is meaningful for a specified distribution of exchangeable cohort effects; it is not a guarantee that the mean describes every ancestry group or a new population outside the studies.

Worked Example: A Fixed-Effect Estimate Under a Common-Effect Model

Suppose two hypothetical, independent cohorts estimate the effect of the same aligned allele on a quantitative trait measured in standard deviation units per allele. Cohort A reports \(\hat{\beta}_1=0.12\) with \(SE_1=0.04\); cohort B reports \(\hat{\beta}_2=0.10\) with \(SE_2=0.05\). Assume the trait definition and scale are comparable and, for this example, a common underlying effect is plausible.

The inverse-variance weights are \(w_1=1/0.04^2=625\) and \(w_2=1/0.05^2=400\). Thus,

\hat{\beta}_{\mathrm{FE}} = \frac{625(0.12)+400(0.10)}{625+400} = \frac{75+40}{1025} = 0.1122.

Its standard error is \(\sqrt{1/1025}=0.03123\). Under the common-effect model, the pooled estimate is an estimated increase of 0.1122 trait standard deviations per copy of the aligned allele. The more precise cohort contributes more, but cohort B still contributes to the estimate. This interpretation relies on the common-effect assumption; the arithmetic alone does not establish that assumption.

Heterogeneity: Evidence of Variation, Not Its Explanation

Cochran’s \(Q\) compares cohort estimates with the fixed-effect pooled estimate:

Q=\sum_{i=1}^k w_i(\hat{\beta}_i-\hat{\beta}_{\mathrm{FE}})^2.

If the cohorts are independent, their standard errors are appropriate, and the common-effect model holds, \(Q\) is approximately chi-squared with \(k-1\) degrees of freedom. A small p-value indicates that the observed dispersion is difficult to reconcile with sampling variation alone under that model. It does not say which cohort differs, quantify the cause, or show that ancestry caused the differences. \(I^2\), commonly calculated as \(\max(0,(Q-(k-1))/Q)\times100\%\), summarizes the proportion of observed dispersion attributed to between-cohort heterogeneity under this framework; it is not a measure of the size or cause of effect differences.

Worked Example: Interpreting Heterogeneity Across Three Cohorts

Consider three hypothetical independent cohorts with aligned effect estimates on the same trait scale: A has \(\hat{\beta}_1=0.20\), \(SE_1=0.05\); B has \(\hat{\beta}_2=0.22\), \(SE_2=0.05\); and C has \(\hat{\beta}_3=0.50\), \(SE_3=0.10\). Suppose A and B have one broad ancestry profile and C another. This grouping is descriptive; it does not establish a cause.

The fixed-effect weights are \(400,400,100\), respectively. The pooled estimate is

\hat{\beta}_{\mathrm{FE}} = \frac{400(0.20)+400(0.22)+100(0.50)}{900} = \frac{218}{900} = 0.2422.

Using this estimate, the heterogeneity statistic is

\begin{aligned} Q &=400(0.20-0.2422)^2+400(0.22-0.2422)^2+100(0.50-0.2422)^2\\ &\approx 0.7131+0.1975+6.645=7.556. \end{aligned}

There are \(k-1=2\) degrees of freedom. For a chi-squared distribution with 2 degrees of freedom, the upper-tail probability is approximately \(P(\chi^2_2\geq7.556)=0.0229\). Thus, under the stated independence and common-effect assumptions, the estimates show evidence of heterogeneity at the 0.05 level. The corresponding \(I^2\) is

I^2=\frac{7.556-2}{7.556}\times100\%\approx73.5\%.

This indicates substantial relative heterogeneity by this summary, not that 73.5% of the effect is caused by ancestry. The high estimate in cohort C contributes strongly to \(Q\), but neither \(Q\) nor \(I^2\) identifies whether the difference reflects LD, effect modification, phenotype or design differences, or another source.

Why Allele Frequency and LD Can Change a Marker Effect

A cohort’s effect estimate may be for a marker that tags, rather than is, the causal variant. A simple quantitative-trait model shows how LD can alter the marker’s marginal association even when the causal effect is held constant. Let \(G_c\) and \(G_t\) be allele dosages at a causal variant and a tag marker, respectively, and suppose \(Y=\beta_cG_c+\varepsilon\), with residual \(\varepsilon\) uncorrelated with both genotypes. The marginal regression coefficient of \(Y\) on \(G_t\) is \(\operatorname{Cov}(G_t,Y)/\operatorname{Var}(G_t)\). Substituting the model gives

\beta_t = \frac{\operatorname{Cov}(G_t,Y)}{\operatorname{Var}(G_t)} = \beta_c\frac{\operatorname{Cov}(G_t,G_c)}{\operatorname{Var}(G_t)}.

The covariance between the two dosages depends on their allele frequencies and LD. Therefore the marginal tag-marker coefficient can differ across ancestry groups even if \(\beta_c\) is identical. Allele frequency also affects genotype variance and thus the standard error and power. A frequency difference by itself does not necessarily imply a different causal per-allele effect.

Worked Example: Same Causal Effect, Different Tag-Marker Effects

For illustration, suppose two hypothetical ancestry groups share a causal per-allele effect \(\beta_c=0.40\) trait standard deviations. Assume Hardy–Weinberg genotype variance \(2p(1-p)\) for each variant, the same trait scale, and an additive linear model with no confounding. The tag-marker effect can be written as

\beta_t = \beta_c r \sqrt{\frac{p_c(1-p_c)}{p_t(1-p_t)}},

where \(p_c\) and \(p_t\) are the causal and tag allele frequencies, and \(r\) is their signed LD correlation. This follows from \(\operatorname{Cov}(G_t,G_c)=r\sqrt{\operatorname{Var}(G_t)\operatorname{Var}(G_c)}\) and the marginal-coefficient formula above.

In group A, let \(p_c=0.20\), \(p_t=0.30\), and \(r=0.80\). Then

\beta_t = 0.40(0.80)\sqrt{\frac{0.20(0.80)}{0.30(0.70)}} = 0.32\sqrt{\frac{0.16}{0.21}} \approx 0.2793.

In group B, let \(p_c=0.40\), \(p_t=0.30\), and \(r=0.40\). Then

\beta_t = 0.40(0.40)\sqrt{\frac{0.40(0.60)}{0.30(0.70)}} = 0.16\sqrt{\frac{0.24}{0.21}} \approx 0.1710.

The marginal tag effects differ, although the assumed causal effect is the same. This is a simplified illustration, not a method for identifying the causal variant from the pooled results. It shows why ancestry-linked variation in LD and frequency can contribute to heterogeneity at a marker without demonstrating that ancestry itself changes the causal effect.

Estimating a Random-Effects Mean

A random-effects analysis allows cohort effects to vary and estimates their mean. For example, the DerSimonian–Laird moment estimate of \(\tau^2\) is

\hat{\tau}^2 = \max\left(0,\frac{Q-(k-1)} {\sum_i w_i-\frac{\sum_i w_i^2}{\sum_i w_i}}\right).

This is one estimator of between-cohort variance, not a test of why the effects differ. With few cohorts, \(\hat{\tau}^2\) and the random-effects mean can be uncertain; a random-effects result should not be treated as automatically more generalizable simply because it permits heterogeneity.

Worked Example: Comparing Fixed and Random-Effects Pooling

Return to the three cohorts in the heterogeneity example. Their fixed-effect \(Q=7.556\), \(k=3\), and weights are \(400,400,100\). The denominator in the moment estimate is

900-\frac{400^2+400^2+100^2}{900} = 900-\frac{330000}{900} = 533.3.

Therefore,

\hat{\tau}^2 = \frac{7.556-2}{533.3} \approx 0.01042.

The random-effects weights are approximately \(1/(0.0025+0.01042)=77.42\) for A and B, and \(1/(0.0100+0.01042)=48.98\) for C. The resulting pooled mean is

\hat{\mu}_{\mathrm{RE}} = \frac{77.42(0.20)+77.42(0.22)+48.98(0.50)} {77.42+77.42+48.98} \approx 0.2797.

This random-effects mean is higher than the fixed-effect estimate \(0.2422\), because the less precise cohort C receives more relative weight when between-cohort variance is included. It estimates a mean under the assumed distribution of cohort effects; it does not resolve why C differs or establish that this mean is appropriate for a population not represented by these cohorts.

Choosing and Reporting an Interpretation

The fixed-effect model is appropriate when a common effect on a harmonized scale is the scientific target and the shared-effect assumption is defensible. The random-effects model is appropriate when cohort-specific effects are viewed as varying around a meaningful mean and the distributional assumptions are reasonable. Neither choice should be made solely by asking whether a heterogeneity test is statistically significant. With limited precision, a nonsignificant test does not prove homogeneity; with many precise cohorts, small differences can be detected.

When ancestry differs across cohorts, report the pooled estimand and the evidence of heterogeneity separately. Consider whether allele frequencies or LD could alter a marker’s tagging relationship, and whether cohort composition is entangled with study design or other differences. Ancestry categories are broad descriptors of population history, not mechanisms that by themselves explain an association. Statistical heterogeneity associated with ancestry groups is not sufficient evidence of an ancestry-specific causal effect.

Common mistake: “The heterogeneity test is significant, and the cohorts differ in ancestry, so ancestry caused the effect difference.”
Correction: \(Q\) provides evidence that estimates vary beyond what the common-effect sampling model predicts. It neither identifies the source nor separates ancestry-linked LD or frequency differences from other cohort differences.
Common mistake: “The random-effects estimate is the effect that applies to every ancestry group.”
Correction: It estimates a mean for the modeled distribution of cohort effects. Cohort-specific effects may lie above or below that mean, and the mean may not transport to an unrepresented population.

Key takeaway: Fixed-effect pooling targets a common effect; random-effects pooling targets a mean across varying cohort effects. Heterogeneity statistics describe dispersion under specified assumptions, while ancestry differences may suggest hypotheses about allele frequency or LD but do not, alone, explain the dispersion.

Check Your Understanding

Answer each question using the model and estimand in view.

Under what assumption does an inverse-variance fixed-effect estimate represent one common genetic effect?
If Cochran’s \(Q\) is significant, what conclusion is supported, and what conclusion is not supported?
Why can the marginal effect of a tag marker differ across ancestry groups when the causal-variant effect is the same?
What does the random-effects mean estimate, and why need it not apply to every population?
Why can a difference in allele frequency affect precision without necessarily implying a different causal per-allele effect?