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Mixed-Model GWAS, Relatedness and Inflation Diagnostics · Tutorial 259 of 795

Proximal Contamination in LMM Association Tests

Learn why local genetic signal in an LMM relationship matrix can weaken tests of nearby variants, and how to diagnose and reduce that effect.

Advanced 10 min read

What You'll Learn

  • Define proximal contamination in an LMM association test.
  • Derive how local covariance aligned with a tested genotype reduces its effective information.
  • Distinguish reduced power from bias in the estimated fixed variant effect.
  • Examine how linkage disequilibrium can expose a test to nearby variants in the relationship matrix.
  • Compare genome-wide, leave-one-chromosome-out, and leave-region-out matrices.
  • Diagnose proximal contamination using matched tests and local covariance structure.

Why a Relationship Matrix Can Weaken a Nearby Association Test

A genomic relationship matrix summarizes genetic similarity among individuals and supplies the covariance structure for a linear mixed model (LMM). If the matrix includes the same region as the variant being tested, part of the local genetic pattern can be represented as random-effect covariance. The test must then distinguish a fixed effect of the tested genotype from variation the model has already allowed along a similar direction.

This phenomenon is called proximal contamination. It is most important when the tested variant, or variants in linkage disequilibrium (LD) with it, contribute appreciably to the relationship matrix. The practical consequence is often reduced information and power. It is important to be precise: with a fixed, correctly specified covariance matrix, including a genotype-related covariance direction does not by itself bias the generalized least-squares estimate of the fixed effect. It can nevertheless make that estimate less precise, and variance-component estimation under a null model can further weaken the test.

Definition: Proximal contamination occurs when genetic markers in or near a tested genomic region contribute to the LMM relationship matrix, so that the random-effect covariance includes directions aligned with the tested genotype. This overlap can reduce the effective information for the fixed variant effect and, in some fitting settings, lead to conservative association tests.

Model and the Source of Information Loss

For one tested genotype vector \(x\), write the association model as

$$ y=C\gamma+x\beta+u+\varepsilon,\qquad u\sim N(0,\tau K),\qquad \varepsilon\sim N(0,\sigma_e^2 I). $$

Here \(y\) is the phenotype vector, \(C\) is the fixed-covariate design matrix, \(\gamma\) is its coefficient vector, \(\beta\) is the tested variant effect, \(K\) is the relationship matrix, and \(\tau\) and \(\sigma_e^2\) are variance components. The model assumes that the random effect and residual have mean zero, are independent, and have the stated covariance matrices. Conditional on the genotypes and covariates, the marginal covariance of \(y\) is \(V=\tau K+\sigma_e^2I\). For the derivation below, take \(V\) as known and positive definite.

After accounting for covariates, the generalized least-squares (GLS) information for \(\beta\) is \(x'Px\), where

$$ P=V^{-1}-V^{-1}C(C'V^{-1}C)^{-1}C'V^{-1}. $$

Thus the GLS variance is \(1/(x'Px)\), and the standard error is \(1/\sqrt{x'Px}\). A covariance contribution aligned with \(x\) makes the model treat variation in that direction as less informative about a fixed effect. This raises the standard error. It does not necessarily pull the GLS coefficient estimate toward zero.

Key distinction: For fixed \(V\), the GLS estimator of \(\beta\) is unbiased conditional on the design when the mean model is correct. Proximal contamination is fundamentally an information and testing problem, not an automatic attenuation of the estimated coefficient. Claims of coefficient bias require additional conditions, such as mean-model misspecification or effects of estimating variance components.

A Rank-One Derivation

The simplest calculation isolates the tested genotype's contribution to the covariance. Suppose there are no covariates beyond an intercept, \(x\) is centered, \(x'x=n\), and

$$ V=\sigma_e^2 I+\tau\frac{xx'}{n}. $$

The local covariance term has \(x\) as an eigenvector: \(Vx=(\sigma_e^2+\tau)x\). Therefore \(V^{-1}x=x/(\sigma_e^2+\tau)\), and the information for \(\beta\) is

$$ x'V^{-1}x=\frac{x'x}{\sigma_e^2+\tau} =\frac{n}{\sigma_e^2+\tau}. $$

Without that local covariance term, the information would be \(n/\sigma_e^2\). The information ratio is \(\sigma_e^2/(\sigma_e^2+\tau)\), and the standard error is multiplied by \(\sqrt{(\sigma_e^2+\tau)/\sigma_e^2}\). This comparison assumes the same residual variance and known covariance parameters; real fitted models may also change their variance-component estimates.

Worked Example: A Local Covariance Component Doubles the Variance

Suppose a hypothetical study has a centered genotype vector with \(x'x=100\). Let \(\sigma_e^2=1\), and suppose the relationship matrix contributes the local term \(xx'/100\) with \(\tau=1\). Assume there are no other covariance components aligned with \(x\), and treat the variances as known.

Without the local term, the information is \(100/1=100\), so \(SE(\hat\beta)=1/\sqrt{100}=0.1000\), in phenotype units per genotype unit. With the local term, \(\sigma_e^2+\tau=2\), giving information \(100/2=50\) and \(SE(\hat\beta)=1/\sqrt{50}=0.1414\).

For a hypothetical phenotype vector with \(x'y=20\), the GLS estimate in either model is \(x'y/(x'x)=20/100=0.2000\) phenotype units per genotype unit. The corresponding Wald statistics are \(0.2000/0.1000=2.000\) without the local term and \(0.2000/0.1414=1.414\) with it. The two-sided standard-normal p-values are approximately 0.0455 and 0.1573, respectively. The example shows weaker evidence because of lower information, not a smaller estimated effect. These p-values use known variances and the stated normal model.

How Nearby Markers Can Contaminate a Test

The tested variant need not appear explicitly in the matrix for the problem to arise. A nearby marker in LD can have a genotype vector strongly aligned with the tested genotype. For a simple illustration, suppose the relationship contribution is \(K_R=zz'/n\), where \(z'z=n\) is a centered nearby-marker vector, and the tested vector also satisfies \(x'x=n\). Let their sample correlation be \(r=x'z/n\). With \(V=I+\tau K_R\), the Sherman–Morrison identity gives

$$ x'V^{-1}x =x'x-\frac{\tau}{n+\tau z'z}(x'z)^2 =n\left(1-\frac{\tau r^2}{1+\tau}\right). $$

The identity applies because \(I\) is invertible and \(1+\tau>0\), as required for this covariance model. The formula shows how LD transfers the effect of a nearby marker's covariance onto the tested variant's information. The size of the reduction depends on both the alignment \(r\) and the covariance contribution \(\tau\); proximity in physical distance alone does not determine it.

Worked Example: A Nearby Marker in Strong LD

Consider hypothetical centered genotype vectors with \(n=100\), \(x'x=z'z=100\), and correlation \(r=0.90\). Suppose only the nearby marker contributes the covariance term \(K_R=zz'/100\), with \(\tau=4\), and residual variance equal to 1.

The information for testing \(x\) is

$$ 100\left(1-\frac{4(0.90)^2}{1+4}\right) =100(1-0.648)=35.2. $$

If this covariance contribution were absent, the information would be 100. The standard error therefore increases from \(1/\sqrt{100}=0.1000\) to \(1/\sqrt{35.2}=0.1685\) phenotype units per genotype unit, a factor of \(0.1685/0.1000=1.685\). This calculation isolates the effect of one aligned covariance component; it does not assert that a real relationship matrix consists of only one marker.

Exclusion Strategies and Their Trade-Offs

The standard remedy is to construct the relationship matrix without markers from the tested region, then use that matrix in the association test. The leave-one-chromosome-out (LOCO) approach excludes the tested chromosome. A leave-region-out approach excludes a smaller window around the variant. Both aim to reduce local alignment between the tested genotype and the random-effect covariance while retaining information about genome-wide relatedness.

The appropriate exclusion depends on the genomic structure and analysis goal. A narrow window may leave correlated markers outside the window; a whole-chromosome exclusion removes more local contribution but can make the covariance estimate less representative of genome-wide relatedness. Recomputing the matrix can also change estimated variance components and standard errors. Therefore, differences between approaches should be assessed as matched analyses, not interpreted as a simple correction factor.

ApproachWhat it changesMain consideration
All-marker relationship matrixUses the full marker set, including the tested regionCan retain local covariance aligned with the test
LOCO matrixExcludes the tested chromosomeReduces chromosome-wide proximal contribution, but removes more markers
Leave-region-out matrixExcludes a defined local windowPreserves other markers, but the window must account for LD beyond its boundaries
Alternative association test using the same covarianceChanges the test statistic, not the local covariance sourceWald, score, and likelihood-ratio tests do not automatically remove proximal contamination

Worked Example: Comparing Matched All-Marker and LOCO Tests

Suppose a hypothetical analysis tests the same variant under the same fixed covariates, with the same genotype coding and phenotype. Under the all-marker matrix, the fitted effect is \(0.200\) phenotype units per genotype unit with \(SE=0.1414\). Under a LOCO matrix, it is \(0.200\) with \(SE=0.1000\). Assume the reported Wald statistics use a standard normal reference distribution.

For the all-marker fit, \(z=0.200/0.1414=1.414\), giving a two-sided p-value of approximately 0.1573. For the LOCO fit, \(z=0.200/0.1000=2.000\), giving a two-sided p-value of approximately 0.0455. The point estimates agree, while the standard errors differ. This pattern is consistent with increased information after removing local covariance, but it does not prove that the all-marker estimate was biased. In an actual analysis, refitting the covariance parameters can change both the effect estimate and its standard error, so each result must be interpreted under its own fitted model.

Diagnostics, Assumptions, and Failure Modes

A useful diagnostic is to compare association results for the same variants using the all-marker matrix and a matrix that excludes the tested region. Keep phenotype processing, covariates, genotype coding, sample, and test definition fixed. Inspect changes in effect estimates, standard errors, and p-values together. A pattern of similar estimates but larger all-marker standard errors is consistent with an information loss. A substantial estimate change deserves further investigation, because it may reflect variance-component refitting, model instability, or a change in the effective analysis rather than proximal contamination alone.

The derivations assume a correct mean model, a positive-definite covariance matrix, and (for the displayed Wald calculations) known variance parameters and normally distributed errors. In practice, variance components are estimated. Under a null association fit, a strong local signal may be represented partly as random-effect variance, which can make a test conservative. The magnitude and direction of the resulting test distortion are not determined by the rank-one calculation; they depend on the fitting procedure and model. Relatedness, population structure, residual variance, and genotyping quality still need appropriate treatment.

Look for markers with high LD to the tested genotype that remain in the matrix, and check which markers were actually used to construct each matrix. An exclusion window that stops before relevant LD decays may provide little protection. Conversely, unusually large changes after exclusion can signal that the two fitted models differ in more than local covariance. Neither a single changed p-value nor a comparison of significance labels is an adequate diagnostic; effect estimates and standard errors are more informative.

A common mistake is to say, “The relationship matrix contains the tested SNP, so the fixed-effect estimate must be shrunk toward zero.” That conclusion does not follow. With known \(V\), the GLS estimate remains unbiased under the stated model; its variance can increase because \(x'Px\) decreases. A different mistake is to switch from a Wald test to a score test and assume the problem is gone. If both tests use a covariance matrix containing the same local contribution, changing the test statistic does not remove that contribution.

Proximal exclusion also has limits. LOCO or leave-region-out matrices reduce one source of overlap; they do not guarantee a well-calibrated test, correct variance components, or absence of population stratification. The region definition should be reported, and sensitivity to reasonable exclusion choices should be examined when local signal is important.

1
Identify the matrix contribution.
Determine whether the tested marker and nearby markers were included in the relationship matrix, and examine their LD with the tested genotype.
2
Fit a matched exclusion analysis.
Use a LOCO or leave-region-out matrix while holding the phenotype, sample, covariates, and association procedure fixed.
3
Compare estimates and uncertainty.
Review effect estimates, standard errors, and test statistics together; do not treat a changed p-value alone as proof of bias.
4
Check model behavior.
Confirm that the matrices and marker sets are as intended, and inspect variance-component estimates and fitting diagnostics for instability.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. In the rank-one model, how does the information ratio change if \(\tau=0\)?
  2. Why can proximal contamination reduce power without biasing the GLS coefficient estimate?
  3. For \(n=100\), \(\tau=4\), and \(r=0\) in the nearby-marker example, what is the information?
  4. Why does a leave-region-out analysis not guarantee that all local covariance overlap is removed?

Solutions. (1) The ratio is \(\sigma_e^2/(\sigma_e^2+\tau)\), so at \(\tau=0\) it is 1 and there is no reduction. (2) The covariance changes the GLS weighting and lowers \(x'Px\), increasing the variance of the estimate; with the correct mean model and fixed \(V\), the GLS estimate is still unbiased conditional on the design. (3) Substitution gives \(100[1-4(0)^2/(1+4)]=100\), so the unaligned marker contributes no information reduction in this calculation. (4) Markers outside the chosen window can still be correlated with the tested genotype, so the window may fail to capture the relevant LD.

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