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Mixed-Model GWAS, Relatedness and Inflation Diagnostics · Tutorial 242 of 795

Why Relatedness Inflates Naive GWAS Evidence

Learn to quantify how relatedness changes the variance of a single-variant association estimate, and why naive GWAS evidence can be inflated—or sometimes deflated.

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What You'll Learn

  • Derive the OLS effect-estimate variance when related individuals have correlated residuals.
  • Compare the correct standard error with the ordinary independent-observation standard error.
  • Translate variance inflation into a change in a null test statistic and its p-value.
  • Recognize why relatedness does not guarantee inflation for every genotype pattern.
  • Separate covariance-driven miscalibration from a misspecified mean or population structure.
  • Identify diagnostic clues and understand the limits of simple corrections.

Why Relatedness Matters for Association Tests

A single-variant association test asks whether genotype is associated with a phenotype after accounting for the specified covariates. Ordinary regression calculations generally treat the participants’ residual phenotype values as independent. That assumption can fail when sampled individuals are related: relatives may share genetic or environmental influences that remain in their phenotypes after adjustment. Their residuals then tend to move together.

The consequence is not simply that an analysis has “less data.” The variance of the estimated genotype effect depends on the pattern of genotype values across the correlated individuals. If genotype values tend to be similar within related pairs, positive residual covariance can increase the estimator’s variance. The usual independent-observation standard error may then be too small, making the association statistic too large in magnitude and its p-value too small. Other genotype arrangements can yield less inflation or even reduced variance.

The calculations below use a quantitative trait and a linear association model to isolate this issue. They do not imply that every relatedness pattern produces inflation, nor do they establish a particular biological cause for residual covariance.

OLS Variance with Correlated Residuals

Consider the model \(y_i=\mu+\beta x_i+\varepsilon_i\), where \(y_i\) is the quantitative phenotype, \(x_i\) is the genotype dosage for the tested variant (0, 1, or 2 copies of an allele), and \(\beta\) is the per-allele association effect in phenotype units. Assume the mean model is correct, \(E(\varepsilon\mid X)=0\), and, conditional on the genotypes, the residual vector has covariance matrix \(\sigma^2R\). The diagonal entries of \(R\) are 1; its off-diagonal entries describe residual correlations. For the calculations, take \(\sigma^2\) as known so that the effect of covariance on the standard error is separated from the additional issue of estimating residual variance.

With an intercept and no other covariates, let \(x_i\) denote the centered genotype dosage, so \(\sum_i x_i=0\), and let \(S_{xx}=\sum_i x_i^2\). The OLS slope estimate is \(\hat{\beta}=\sum_i x_i y_i/S_{xx}\). Under the null hypothesis \(H_0:\beta=0\), its conditional variance is determined by the residual covariance, not just the residual variances.

$$ \operatorname{Var}(\hat{\beta}\mid X) =\frac{\operatorname{Var}\left(\sum_i x_i\varepsilon_i\mid X\right)}{S_{xx}^2} =\frac{\sigma^2 x^\mathsf{T}Rx}{S_{xx}^2}. $$

If residuals are independent, \(R=I\), so \(x^\mathsf{T}Rx=S_{xx}\) and the variance reduces to \(\sigma^2/S_{xx}\). That is the variance underlying the ordinary standard error when \(\sigma^2\) is known. When residuals are correlated, the off-diagonal terms contribute to \(x^\mathsf{T}Rx\). A convenient expression for the ratio of the covariance-aware variance to the independent-residual variance is

$$ \frac{\operatorname{Var}(\hat{\beta}\mid X)} {\sigma^2/S_{xx}} =\frac{x^\mathsf{T}Rx}{S_{xx}}. $$

This ratio depends on both the covariance matrix and the genotype pattern. In a model with additional fixed covariates, the corresponding calculation uses the genotype after removing its linear association with those covariates, along with the covariance of the residuals. The same principle applies: the independent-observation formula is justified only if the relevant covariance structure supports it.

Key result: Under a correctly specified linear mean model, the OLS coefficient can still target the intended effect when residuals are correlated, but its ordinary independent-observation standard error need not be valid. The variance depends on \(x^\mathsf{T}Rx\); relatedness is most consequential when the genotype pattern aligns with correlated residual directions.

Worked Examples

Worked Example: Correlated Residuals Within Similar-Genotype Pairs

Suppose a hypothetical sample contains four related pairs, with genotype dosages \((0,0),(1,1),(1,1),(2,2)\). The mean dosage is 1, so the centered genotype vector is \(x=(-1,-1,0,0,0,0,1,1)^\mathsf{T}\). Assume residual variance \(\sigma^2=4\) phenotype-units squared, residual correlation \(\rho=0.5\) within each pair, and zero residual correlation between different pairs. These assumptions define a valid pairwise covariance structure: each pair has correlation 0.5, which lies between -1 and 1, and different pairs are independent.

The genotype sum of squares is \(S_{xx}=1+1+1+1=4\). With pairwise correlation \(\rho\), the quadratic form is the sum of the diagonal contributions plus the within-pair cross-products:

$$ x^\mathsf{T}Rx =S_{xx}+2\rho\sum_{\text{pairs }(i,j)}x_i x_j =4+2(0.5)\{(-1)(-1)+0+0+(1)(1)\} =6. $$

Therefore, the covariance-aware variance is \(4(6)/4^2=1.5\) phenotype-units squared, and its standard error is \(\sqrt{1.5}=1.225\) phenotype units per allele. The naive variance is \(4/4=1\), giving a standard error of 1.000 phenotype unit per allele. Here the naive standard error is too small because the members of each pair have matching centered genotype values and positively correlated residuals.

Worked Example: Consequence for a Null Test

Continue with the hypothetical design above, and suppose the estimated association effect is \(\hat{\beta}=2.400\) phenotype units per allele. For a known residual variance, the ordinary standardized statistic is \(Z_{\text{naive}}=\hat{\beta}/SE_{\text{naive}}=2.400/1.000=2.400\). Its two-sided standard-normal p-value is approximately 0.0164.

Under the stated covariance model, the standard error is 1.225, so the covariance-aware standardized statistic is \(Z_{\text{cov}}=2.400/1.225=1.960\). Its two-sided standard-normal p-value is approximately 0.0500. This comparison assumes the linear mean model, the stated covariance matrix, and a known \(\sigma^2\); it illustrates the calibration change rather than providing a general recipe for estimating a covariance matrix.

Under \(H_0\), with the Gaussian residual model and known \(\sigma^2\), the naive statistic has variance 1.5 rather than 1 in this design. Thus it does not follow the standard normal distribution assumed for the naive p-value. The smaller naive p-value overstates the evidence against no association for this hypothetical sample structure.

Worked Example: Positive Relatedness Does Not Always Inflate Variance

Consider two hypothetical related pairs with centered genotype values \((-1,1)\) in each pair. This pattern could occur when the two members of each pair have different dosages. Let the residual correlation within each pair be \(\rho=0.5\), with zero correlation between pairs. The genotype sum of squares is \(S_{xx}=4\), and the two within-pair products are both \(-1\).

The variance ratio is

$$ \frac{x^\mathsf{T}Rx}{S_{xx}} =\frac{4+2(0.5)(-1-1)}{4} =\frac{2}{4} =0.5. $$

In this design, the covariance-aware variance is half the naive variance, and the covariance-aware standard error is \(\sqrt{0.5}=0.7071\) times the naive standard error. Positive correlation between relatives does not by itself imply inflation of every association statistic: the sign and size of the genotype cross-products within correlated groups matter.

Justification for the Variance Formula

The formula follows directly from the OLS slope and the covariance of a linear combination. Under the model and centering condition, the intercept contribution disappears from the centered-genotype numerator. The estimation error is therefore a weighted sum of residuals. For any fixed vector \(a\), the variance of \(a^\mathsf{T}\varepsilon\) is \(a^\mathsf{T}\operatorname{Cov}(\varepsilon\mid X)a\). Applying that fact gives the result:

1
Express the slope error.
Since \(\sum_i x_i=0\), \(\hat{\beta}-\beta=\sum_i x_i\varepsilon_i/S_{xx}\).
2
Take its conditional variance.
\(\operatorname{Var}(\hat{\beta}\mid X)=\operatorname{Var}(x^\mathsf{T}\varepsilon\mid X)/S_{xx}^2\).
3
Insert the covariance matrix.
Because \(\operatorname{Cov}(\varepsilon\mid X)=\sigma^2R\), the numerator is \(x^\mathsf{T}(\sigma^2R)x=\sigma^2x^\mathsf{T}Rx\).

If the residuals are jointly Gaussian, this linear combination is normally distributed. Under \(H_0\), the naive statistic formed using the independent-residual standard error has variance \(x^\mathsf{T}Rx/S_{xx}\), rather than 1, when \(\sigma^2\) is known. A variance ratio above 1 makes the null statistic more dispersed than the reference distribution and can increase the frequency of extreme test statistics. A ratio below 1 makes it less dispersed. When \(\sigma^2\) is estimated, the usual residual-based standard error also reflects that estimation; correlation can affect that estimate as well, so the known-variance calculation isolates rather than exhausts the potential miscalibration.

Distinguishing Covariance from Mean-Model Problems

Relatedness can raise two different concerns. This tutorial’s calculation concerns covariance: the residuals have mean zero under the specified model, but are not independent. Separately, ancestry, family environment, or other omitted factors may make the conditional mean wrong—for example, genotype and phenotype may both differ across subgroups. In that case, an apparent association can arise from mean-model misspecification, not just an underestimated standard error. Adjusting standard errors alone does not generally remove such bias.

The earlier material on calibration checks and quantile-quantile plots provides useful ways to notice that a scan’s test statistics depart from a reference distribution. Excess dispersion across many variants can be consistent with covariance-related miscalibration, but it does not identify its cause. True associations, population structure, technical artifacts, and model misspecification can also affect the distribution. A quantile-quantile plot or an inflation summary is a diagnostic clue, not proof that relatedness is responsible or a replacement for a suitable association model.

Possible responses depend on the sampling structure and the inferential target. A covariance-aware association model represents dependence directly; a robust variance estimator can adjust uncertainty under appropriate conditions. These approaches are not interchangeable in all settings, and neither repairs a misspecified mean automatically. In particular, a cluster-robust estimate generally needs enough independent clusters for its approximation to work well. A sample containing only a few large families is not made safe merely by labeling the standard error “robust.” The subsequent material on random genetic effects and mixed-model association develops one model-based approach; the point here is to understand why the independence calculation can fail.

Diagnostics, Limitations, and Common Mistakes

  • Ignoring the sampling structure. Check whether participants include relatives or other groups likely to have correlated outcomes. If so, state what covariance assumption supports the reported standard errors.
  • Assuming every related sample is inflated by the same amount. The variance ratio \(x^\mathsf{T}Rx/S_{xx}\) depends on genotype alignment with covariance. It can exceed, equal, or fall below 1.
  • Equating a large test statistic with a causal effect. This calculation concerns uncertainty in an association estimate. It does not establish causation, and covariance correction alone does not address confounding.
  • Treating a genomic inflation signal as a diagnosis. Broad test-statistic inflation can motivate investigation, but several mechanisms can produce it. Inspect study design and model assumptions as well as summary diagnostics.
  • Assuming a known covariance structure in practice. The examples specify the correlations to make the arithmetic transparent. Real analyses must justify and estimate or otherwise account for dependence; a convenient assumed correlation is not automatically correct.

The central limitation of the simple pair calculations is that they use a deliberately specified covariance matrix and, for the test illustration, known residual variance. Real GWAS may involve uneven relationships, covariates, binary phenotypes, estimated variance parameters, and population structure. The simple result remains valuable because it shows the basic mechanism: the OLS coefficient is a weighted sum of outcomes, and covariance among outcomes changes the variance of that sum.

Key takeaway: Relatedness can make naive GWAS evidence anti-conservative when correlated residuals align with genotype patterns, but it does not guarantee inflation. Valid inference requires attention to both the mean model and the covariance assumptions behind the standard error.

Check Your Understanding

Answer each question using the covariance argument above. Brief solutions follow the questions.

  1. In a centered-genotype linear model, what quantity replaces \(S_{xx}\) in the numerator of the OLS coefficient variance when residual covariance is \( \sigma^2R \)?
  2. A design has \(S_{xx}=10\) and \(x^\mathsf{T}Rx=15\). By what factor is the covariance-aware standard error larger than the naive standard error?
  3. Why can positive within-pair residual correlation reduce, rather than increase, the slope variance in some genotype patterns?
  4. Does a quantile-quantile plot showing excess dispersion prove that relatedness caused the inflation? Give one reason.
  5. Why is correcting a standard error not necessarily enough when genotype and phenotype means differ across ancestry groups?

Solutions. (1) The variance is \(\sigma^2x^\mathsf{T}Rx/S_{xx}^2\); the covariance quadratic form \(x^\mathsf{T}Rx\) replaces the independent-residual numerator \(S_{xx}\). (2) The variance ratio is \(15/10=1.5\), so the standard-error ratio is \(\sqrt{1.5}=1.225\). (3) When related individuals have oppositely signed centered genotype values, their within-pair cross-products are negative, so positive residual covariance contributes negative cross-terms to \(x^\mathsf{T}Rx\). (4) No. Population structure, technical artifacts, true associations, and other model problems can also affect dispersion. (5) A standard-error correction changes uncertainty, not the fitted mean; omitted structure can bias the estimated association itself.

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