LD, Precision, and Conditional Power

Why LD Affects Conditional Precision

At a locus, two variants can be correlated because of linkage disequilibrium (LD). If the variants are included together in a regression model, the coefficient for one variant describes its association with the outcome conditional on the other included variants. When the tested genotype is strongly predictable from the other genotypes, little of its variation remains to distinguish its conditional effect. The consequence is less information, a larger standard error, a wider confidence interval, and often lower power to detect an additional signal.

This is a precision issue, not a rule that the conditional effect must be smaller than the marginal effect. The conditional coefficient is the model’s estimate of a different quantity. Its value depends on the joint association structure, while its standard error also depends on how much variation in the tested genotype remains after adjustment.

Model and Precision Formula

Consider a quantitative outcome measured on \(n\) individuals. Let \(G_2\) be the tested genotype, \(G_1\) another genotype included in the model, and \(C\) the other covariates. A linear regression model is

Here, \(\beta_2\) is the conditional mean difference in outcome per unit increase in \(G_2\), holding \(G_1\) and \(C\) fixed. For the usual standard-error formula, assume the mean model is correctly specified, observations are independent, and errors have mean zero, common variance \(\sigma^2\), and are uncorrelated with the predictors. For the exact variance conditional on the observed design matrix, also assume this homoskedastic linear model. In practice, robust standard errors or other models may be appropriate when these assumptions do not hold.

To see the role of LD, regress \(G_2\) on the intercept, \(G_1\), and \(C\), and let \(\widetilde G_{2i}\) be the resulting residual for individual \(i\). The Frisch–Waugh–Lovell result expresses the coefficient for \(G_2\) as the regression of the outcome residualized on the other predictors on \(\widetilde G_2\). Its conditional variance, given the predictors, is

Thus, a genotype with little residual variation after adjustment has a larger standard error. In a random-design approximation, if \(G_2\) has variance \(v_2\) and its multiple \(R^2\) from regression on the other predictors is \(R^2_{G_2\mid\text{others}}\), then

For just two genotypes, each standardized to variance 1, with correlation \(r\) and no additional covariates, the regression \(R^2\) is \(r^2\). The conditional standard error is therefore inflated relative to the unadjusted standard error by the factor \(1/\sqrt{1-r^2}\). The variance inflation factor is \(1/(1-r^2)\). This factor depends on the magnitude of \(r\), not its sign.

A Short Derivation of the Two-Variant Result

Assume the two genotype columns are centered and standardized, so their sums of squares are \(n\) and their cross-product is \(nr\). Regressing \(G_2\) on \(G_1\) gives slope \(r\), so the residualized genotype is \(\widetilde G_2=G_2-rG_1\). Its sum of squares is

Substituting this residual sum of squares into the conditional variance formula gives \(\sigma^2/[n(1-r^2)]\). Without \(G_1\) in the model, the corresponding variance is \(\sigma^2/n\). The ratio is therefore \(1/(1-r^2)\), and taking square roots gives the standard-error inflation factor \(1/\sqrt{1-r^2}\). As \(|r|\) approaches 1, the residualized genotype variation approaches zero and the conditional coefficient becomes very imprecise; at perfect collinearity, the two coefficients cannot be separately estimated.

Conditional Power for an Additional Signal

Power is the probability that a specified test rejects \(H_0\) when a particular alternative is true. For a two-sided Wald test of \(H_0:\beta_2=0\) at significance level \(\alpha\), an approximate calculation uses the standard normal distribution. If the standard error is approximately constant under the alternative, the test statistic has approximate distribution \(N(\delta,1)\), where \(\delta=\beta_2/SE(\hat\beta_2)\) is the noncentrality parameter. For two standardized genotypes and one other genotype with correlation \(r\), this gives

The first probability is the chance that the statistic exceeds the positive critical value; the second is the chance it falls below the negative critical value. This is an approximation based on the normal model, the stated effect size, and the assumed residual variance and genotype distribution. It is not a probability that the null hypothesis is true.

Solving the approximation for sample size at target power \(1-\beta\) gives

For two standardized genotypes, \(v_2=1\) and \(R^2_{G_2\mid\text{others}}=r^2\). Holding effect size, residual variance, and target power fixed, stronger LD increases the approximate sample size by \(1/(1-r^2)\) compared with no genotype correlation.

Interpreting Precision and Power Carefully

The relationship between LD and conditional precision matters when asking whether a variant contributes an additional association signal beyond variants already in the model. A large conditional standard error can make a confidence interval include zero and a test have low power even if the conditional effect is not zero. Conversely, strong evidence from a conditional test depends on both the estimated effect and its uncertainty. The size of \(r\) alone does not determine whether an additional signal exists.

In real locus models, the relevant quantity is generally the multiple \(R^2\) for the tested genotype after regression on all other modeled predictors, not just its pairwise correlation with one variant. Genotype variance also matters: a less variable genotype provides less information even without LD. For example, allele frequency affects genotype variance under common genetic coding, so equal effect sizes do not necessarily imply equal standard errors across variants. The simple two-variant factor is a useful interpretation, not a substitute for calculating precision under the fitted design.

Common Mistake

Wrong reasoning: “The genotypes have high LD, so the conditional effect must be small, and a nonsignificant test shows there is no additional signal.” This confuses the estimated effect with its precision. LD can inflate the standard error without forcing the coefficient toward zero; a nonsignificant result may reflect low power.

Correction: Interpret the conditional coefficient and its uncertainty together. Report the effect in its stated units, the standard error or confidence interval, and the conditional test result. Assess whether the genotype retains enough variation after adjustment to estimate its effect precisely. A wide interval that includes zero is compatible with a range of conditional effects and is not proof of no additional association.