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Conditional and Joint Association · Tutorial 599 of 989

Omitted-Variant Bias in Association

Learn to calculate and interpret the shift in a variant’s marginal association when a correlated variant is left out of the regression model.

Advanced 10 min read

What You'll Learn

  • State a two-variant linear association model and its assumptions
  • Derive the marginal coefficient when one variant is omitted
  • Express omitted-variant bias using genotype covariance or correlation
  • Calculate the bias under positive and negative linkage disequilibrium
  • Distinguish a marginal association from a conditional variant effect
  • Identify when the omitted variant does not shift the marginal coefficient

Why Leaving Out a Correlated Variant Matters

At a locus, two variants can carry overlapping information because their genotype dosages are correlated. A regression of phenotype on just one variant therefore describes its marginal association, not necessarily its effect after accounting for the other variant. If the omitted variant is associated with the phenotype and correlated with the included variant, its association can contribute to the marginal coefficient.

The key result is simple: in a two-variant linear model, the shift in the coefficient for the included variant is the omitted variant’s conditional coefficient multiplied by the relationship between the two genotypes. The relationship can be written as a covariance for unstandardized genotypes or a correlation for standardized genotypes. This is a statement about regression coefficients; by itself, it does not establish a causal effect.

Two-Variant Model and Omitted-Variant Bias

Definition: Let \(Y\) be a quantitative phenotype, and let \(G_1\) and \(G_2\) be the genotype dosages for two variants, each centered at its sample or population mean. In the two-variant linear model, \(Y=\beta_1G_1+\beta_2G_2+\varepsilon\), \(\beta_1\) and \(\beta_2\) are the conditional regression coefficients: each describes the change in the mean phenotype per one-unit increase in that genotype dosage, holding the other dosage fixed. The marginal coefficient for \(G_1\) is the slope from a regression of \(Y\) on \(G_1\) alone. The difference between that marginal coefficient and \(\beta_1\) is the omitted-variant bias.

For the formula below, assume the conditional mean is correctly specified by the two-variant model, both genotype variances are nonzero, and the residual has mean zero conditional on both genotypes: \(E(\varepsilon\mid G_1,G_2)=0\). There is no intercept in the displayed equation because the genotypes are centered; equivalently, an intercept can be included when genotypes are not centered. The coefficient is a linear association coefficient under this model. Interpreting it as a causal effect requires additional assumptions not supplied by this regression setup.

Write \(\operatorname{Var}(G_1)\) for the variance of \(G_1\), and \(\operatorname{Cov}(G_1,G_2)\) for the covariance of the two genotype dosages. The population marginal slope is the covariance of \(G_1\) with \(Y\), divided by the variance of \(G_1\). Substituting the two-variant model gives the central identity.

$$ \begin{aligned} \beta_{1,\mathrm{marg}} &= \frac{\operatorname{Cov}(G_1,Y)}{\operatorname{Var}(G_1)}\\ &= \frac{\operatorname{Cov}(G_1,\beta_1G_1+\beta_2G_2+\varepsilon)} {\operatorname{Var}(G_1)}\\ &= \beta_1+ \beta_2\frac{\operatorname{Cov}(G_1,G_2)}{\operatorname{Var}(G_1)}. \end{aligned} $$

The last equality uses \(E(\varepsilon\mid G_1,G_2)=0\), which implies \(\operatorname{Cov}(G_1,\varepsilon)=0\). Thus the omitted-variant bias, defined as \(\beta_{1,\mathrm{marg}}-\beta_1\), is \(\beta_2\operatorname{Cov}(G_1,G_2)/\operatorname{Var}(G_1)\). It is zero if the omitted variant has no conditional association with the phenotype (\(\beta_2=0\)) or if the genotypes are uncorrelated. Otherwise, its direction and size depend on both the conditional coefficient and genotype covariance.

Correlation Form and the Role of Linkage Disequilibrium

Let \(\sigma_1\) and \(\sigma_2\) be the standard deviations of the two genotype dosages, and let \(r_{12}=\operatorname{Corr}(G_1,G_2)\). Since \(\operatorname{Cov}(G_1,G_2)=r_{12}\sigma_1\sigma_2\), the same identity can be written \(\beta_{1,\mathrm{marg}}=\beta_1+\beta_2r_{12}\sigma_2/\sigma_1\). If both genotypes are standardized to have variance one, the ratio of standard deviations is one, so the bias term simplifies to \(r_{12}\beta_2\).

In association analysis, linkage disequilibrium (LD) is the non-independence of alleles at different variants. Genotype-dosage correlation is a convenient way to quantify how the observed dosages co-vary in this regression. Positive correlation means individuals with above-average dosage at one variant tend also to have above-average dosage at the other; negative correlation means that above-average dosage at one tends to accompany below-average dosage at the other. Neither sign alone determines the bias: the sign of \(\beta_2\) matters as well.

Theorem: Under the stated two-variant model and conditional-mean assumption, the omitted-variant bias in the marginal coefficient for \(G_1\) is \(\beta_2\operatorname{Cov}(G_1,G_2)/\operatorname{Var}(G_1)\). For standardized genotypes it is \(r_{12}\beta_2\). Positive genotype correlation shifts the marginal coefficient in the direction of \(\beta_2\); negative genotype correlation shifts it in the opposite direction.

Worked Calculations

Worked Example: Positive Linkage Disequilibrium

Consider hypothetical data on \(n=5{,}000\) individuals. Suppose the genotype dosages have been standardized, their correlation is \(r_{12}=0.60\), and a two-variant model gives \(\beta_1=0.20\) and \(\beta_2=0.50\) phenotype units per genotype standard deviation. What marginal coefficient is expected when \(G_2\) is omitted?

With standardized genotypes, the marginal coefficient is \(\beta_{1,\mathrm{marg}}=\beta_1+r_{12}\beta_2\). Substituting the hypothetical values, \(\beta_{1,\mathrm{marg}}=0.20+(0.60)(0.50)=0.20+0.30=0.50\).

The omitted-variant bias is \(0.50-0.20=0.30\) phenotype units per genotype standard deviation. Because the correlation is positive and \(\beta_2\) is positive, the shift is positive. The marginal slope of \(0.50\) includes the conditional association represented by \(\beta_1\) and the contribution associated with \(G_2\); it is not the conditional coefficient for \(G_1\).

Worked Example: Negative Linkage Disequilibrium

In a second hypothetical data set, both genotype dosages are standardized, \(r_{12}=-0.50\), and the two-variant model gives \(\beta_1=0.35\) and \(\beta_2=0.40\) phenotype units per genotype standard deviation. Calculate the marginal coefficient for \(G_1\).

The formula is \(\beta_{1,\mathrm{marg}}=\beta_1+r_{12}\beta_2\). Therefore, \(\beta_{1,\mathrm{marg}}=0.35+(-0.50)(0.40)=0.35-0.20=0.15\).

The marginal coefficient is \(0.15\), compared with the conditional coefficient \(0.35\). The bias is \(-0.20\) phenotype units per genotype standard deviation. Here the negative genotype correlation shifts the marginal coefficient opposite to the positive conditional coefficient for \(G_2\). The calculation illustrates attenuation in this example; its numerical result depends on the specified coefficients and correlation.

Worked Example: Unequal Genotype Variances

Suppose a hypothetical dosage-based analysis gives \(\operatorname{Var}(G_1)=0.36\) dosage-squared, \(\operatorname{Var}(G_2)=0.49\) dosage-squared, and \(\operatorname{Cov}(G_1,G_2)=0.126\) dosage-squared. In the two-variant model, let \(\beta_1=0.08\) and \(\beta_2=0.20\), both in phenotype units per dosage. Find the marginal slope for \(G_1\).

Use the covariance form, which applies directly to dosages on their original scales: \(\beta_{1,\mathrm{marg}}=\beta_1+\beta_2\operatorname{Cov}(G_1,G_2)/\operatorname{Var}(G_1)\). Substitution gives \(\beta_{1,\mathrm{marg}}=0.08+(0.20)(0.126/0.36)\). The ratio is \(0.126/0.36=0.35\), so \(\beta_{1,\mathrm{marg}}=0.08+(0.20)(0.35)=0.08+0.07=0.15\) phenotype units per dosage.

The bias is \(0.07\) phenotype units per dosage. The covariance and variance both have units of dosage-squared, so their ratio is dimensionless; multiplying by \(\beta_2\) leaves the correct coefficient units. The dosage correlation is \(0.126/(\sqrt{0.36}\sqrt{0.49})=0.30\), consistent with positive covariance.

Why the Formula Holds in a Sample

The population identity follows from covariance algebra, but the same idea can be seen directly in a finite sample. Let \(x_1\) and \(x_2\) be vectors of centered genotype dosages, and let \(y\) and \(e\) be the centered phenotype and residual vectors. The simple-regression slope of \(y\) on \(x_1\) is \(x_1^\mathsf{T}y/(x_1^\mathsf{T}x_1)\). If the observed phenotype satisfies \(y=\beta_1x_1+\beta_2x_2+e\), substitution gives the exact decomposition below.

$$ \begin{aligned} \hat{\beta}_{1,\mathrm{marg}} &=\frac{x_1^\mathsf{T}y}{x_1^\mathsf{T}x_1}\\ &=\beta_1+\beta_2\frac{x_1^\mathsf{T}x_2}{x_1^\mathsf{T}x_1} +\frac{x_1^\mathsf{T}e}{x_1^\mathsf{T}x_1}. \end{aligned} $$

The second term is the sample analogue of omitted-variant bias: the sample genotype cross-product divided by the included genotype sum of squares, multiplied by \(\beta_2\). The final term is the sample association between \(G_1\) and the residual. Under the conditional-mean assumption it has expectation zero across repeated samples, but it need not be exactly zero in any one sample. Consequently, the population formula describes the systematic shift; a realized sample coefficient can also differ because of residual variation.

Interpretation, Scope, and a Common Pitfall

Omitted-variant bias is important because a marginal association can reflect more than the conditional association assigned to the included variant in a joint model. The formula identifies the ingredients of that difference: the omitted variant’s conditional coefficient and the degree of genotype co-variation. A marginal coefficient is therefore not automatically an estimate of a variant-specific conditional coefficient. In particular, a strong marginal association does not by itself establish which correlated variant accounts for the association.

Common mistake: “The variants are correlated, so omitting one must increase the coefficient of the other.” This reasoning ignores both the sign of the genotype correlation and the sign of the omitted variant’s conditional coefficient. The correction is to calculate the product: \(\beta_2\operatorname{Cov}(G_1,G_2)/\operatorname{Var}(G_1)\). For standardized genotypes, calculate \(r_{12}\beta_2\). The product can be positive, negative, or zero.

The identity also depends on the model and assumptions stated above. If the conditional mean is not adequately represented by the two-variant linear model, or if residuals are associated with genotype, the displayed decomposition does not isolate the entire difference in the same way. Additional covariates and more variants change the regression setting; the two-variant formula should not be applied without checking that the intended model matches its setup.

Key takeaway: In a two-variant model, the marginal coefficient for \(G_1\) equals its conditional coefficient plus the omitted variant’s conditional coefficient multiplied by the genotype covariance-to-variance ratio. With standardized genotypes, that ratio is the genotype correlation.

Check Your Understanding

Use the two-variant model and assumptions stated above. Keep genotype and phenotype units in mind when interpreting each result.

  1. For standardized genotypes, suppose \(r_{12}=0.40\), \(\beta_1=0.10\), and \(\beta_2=0.25\). Calculate the marginal coefficient for \(G_1\) and the omitted-variant bias.
  2. If \(\operatorname{Cov}(G_1,G_2)=0\), what is the marginal coefficient for \(G_1\) under the model, and why?
  3. For standardized genotypes with \(r_{12}=-0.30\) and \(\beta_2=-0.50\), is the bias positive, negative, or zero? Calculate it.
  4. Why does a nonzero marginal coefficient for \(G_1\) not, on its own, show that \(G_1\) has the same conditional coefficient in the two-variant model?
  5. In the finite-sample decomposition, what does the residual cross-product term represent, and why can it differ from zero in a particular sample?
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