Deriving the Pooled Association Under the Null

Why a Null Within-Group Association May Not Be a Pooled Null

Suppose genotype dosage and a quantitative phenotype are unrelated within each ancestry group. If the groups differ both in their average genotype dosage and in their average phenotype, the combined data can nevertheless show a genotype–phenotype association. The key is that a pooled comparison mixes two sources of covariance: variation among individuals within groups and variation among the group means.

We use \(A\) for ancestry group, \(G\) for genotype dosage, and \(Y\) for phenotype. For a biallelic variant, \(G\) is typically coded as 0, 1, or 2 copies of the allele of interest. The phenotype \(Y\) may be a measurement such as a concentration in milligrams per deciliter. The calculations below describe a target population or an empirical data set in which each individual has one observed group, genotype, and phenotype. We assume the relevant means and covariances exist.

This definition concerns statistical association, not causation. In particular, a zero within-group covariance does not by itself establish that genotype has no causal effect. The focus here is how a pooled covariance can arise even when the within-group covariances are zero.

The Covariance Decomposition

Let \(\pi_a=P(A=a)\) be the proportion of the population in group \(a\), \(g_a=E[G\mid A=a]\) its mean genotype dosage, and \(\mu_a=E[Y\mid A=a]\) its mean phenotype. The pooled means are \(\bar g=E[G]=\sum_a\pi_a g_a\) and \(\bar\mu=E[Y]=\sum_a\pi_a\mu_a\). The law of total covariance gives:

The first term is the weighted average of within-group covariances. The second is the covariance of the group-specific means, weighted by group proportions. Under the within-group null just defined, the first term is zero, so the pooled covariance is entirely the between-group component:

Thus, the pooled covariance is nonzero when the weighted deviations of genotype means and phenotype means have a nonzero net product. In two groups, this condition is especially transparent. If group 1 has proportion \(\pi\), and the two groups have means \(g_1,g_0\) and \(\mu_1,\mu_0\), then the between-group term is \(\pi(1-\pi)(g_1-g_0)(\mu_1-\mu_0)\). It is nonzero if both mean differences are nonzero; its sign is the sign of their product.

Worked Examples

Why the Decomposition Holds

The identity follows by splitting each variable into its group mean and its deviation from that mean. For an individual in group \(a\), write \(G=g_a+(G-g_a)\) and \(Y=\mu_a+(Y-\mu_a)\). Subtracting the pooled means and multiplying gives four terms. On averaging, the two cross-terms vanish because the within-group deviations have conditional mean zero.

For the second term, conditioning on \(A\) gives \(E[Y-\mu_A\mid A]=0\); for the third, \(E[G-g_A\mid A]=0\). Both expectations are therefore zero. The first term is the between-group covariance of the means. The final term is the average within-group covariance, since:

This proves the decomposition and shows precisely where the two components come from. For two groups, the between-group term simplifies to \(\pi(1-\pi)(g_1-g_0)(\mu_1-\mu_0)\), making its direction and magnitude explicit. With more groups, individual contributions may have different signs, as in the third example.

Connection to a Pooled Regression Slope

For a simple linear regression of \(Y\) on \(G\) with an intercept, the population least-squares slope is \(\operatorname{Cov}(G,Y)/\operatorname{Var}(G)\), provided \(\operatorname{Var}(G)>0\). In a sample, the corresponding slope is the sample covariance divided by the sample variance, using the same covariance convention in numerator and denominator. A nonzero pooled covariance therefore gives a nonzero pooled slope; its units are phenotype units per dosage unit.

This slope summarizes the pooled linear association. It does not say that each ancestry group has that slope, and it does not identify a causal effect. If the within-group covariances are zero, the pooled slope’s numerator comes from group mean differences; the denominator also includes both within- and between-group genotype variation. Consequently, the magnitude of the slope is not simply the between-group covariance.

When to Expect a Nonzero Pooled Association

Under zero within-group covariance, a nonzero pooled covariance requires a nonzero weighted covariance between group genotype means and group phenotype means. In the two-group case, both group-mean differences must be nonzero. If the genotype means match, the between-group term is zero even if phenotype means differ; if phenotype means match, it is also zero even if genotype means differ. With three or more groups, nonzero differences can still sum to zero.

The decomposition also clarifies what to inspect in a population-genetics analysis: the ancestry-specific genotype means, phenotype means, and proportions, as well as the within-group relationships. A pooled statistic alone cannot reveal which component generated the observed association. These calculations describe the covariance structure; they do not prescribe a particular adjustment method.

Common Mistake

• Wrong reasoning: “The genotype and phenotype have zero covariance within every ancestry group, so their pooled covariance must also be zero.”
• Correction: The pooled covariance includes a between-group term. If ancestry groups differ in mean genotype and mean phenotype, the products of those mean deviations can add to a nonzero value even when every within-group covariance is zero.
• A second caution: “The pooled slope is nonzero, so the variant must affect the phenotype.” A pooled association is not a causal conclusion. In the null examples, differences between group means alone generated the pooled slope.