What Is Conditional in a Mixed-Model Association Test?
A genome-wide association analysis tests one candidate variant at a time while accounting for covariates and correlation among individuals. In a linear mixed model (LMM), that correlation is represented by a covariance structure, often informed by a genomic relationship matrix. The null-model fitting step estimates the covariate effects and covariance parameters without the candidate variant. A null-fitted conditional association test then uses those fitted values as fixed inputs when testing the variant’s effect.
“Conditional” here describes the analysis after plugging in estimates from the fitted null model. It does not mean that estimates from the phenotype have magically become known constants, nor does it mean that the test conditions on the observed phenotype in a probability-theory sense. The distinction matters: a test that freezes nuisance estimates is not identical to one that estimates them again after adding the variant.
The Model and the Hypothesis
Let \(y\) be the \(n\)-vector of quantitative phenotypes, \(X\) the \(n\)-by-\(q\) matrix of fixed covariates (including an intercept if appropriate), and \(g\) the \(n\)-vector of dosages for one variant. A dosage commonly records the number of copies of a designated allele, from 0 to 2. The LMM for this candidate variant is
Here \(\gamma\) is the vector of covariate coefficients, \(\beta\) is the per-dosage association effect, \(K\) is a specified relationship matrix, \(\tau\) is the random-effect variance component, and \(\sigma_e^2\) is the residual variance. Under these assumptions, the marginal distribution of \(y\), after accounting for the random effect, is normal with mean \(X\gamma+g\beta\) and covariance
The single-variant null hypothesis is \(H_0:\beta=0\). A two-sided alternative is \(H_a:\beta\ne 0\); a one-sided alternative is appropriate only when a directional effect was specified in advance. For a quantitative phenotype, \(\beta\) represents the change in the modeled phenotype mean per additional dosage unit, conditional on the modeled covariates and the assumed covariance structure. It is an association parameter, not by itself a causal effect.
The null model omits \(g\beta\). Suppose its fitted parameters are denoted by \(\hat{\gamma}_0\), \(\hat{\tau}_0\), and \(\hat{\sigma}_{e,0}^2\). The conditional formulation substitutes these values into the candidate-variant model:
In this formulation, \(\hat{\gamma}_0\), \(\hat{\tau}_0\), \(\hat{\sigma}_{e,0}^2\), and the specified \(K\) are fixed while the candidate effect is tested. The null sets \(\beta=0\); the alternative permits \(\beta\ne0\). This description defines what is held fixed, but does not select a particular test statistic or its calibration.
Why the Conditional Model Has This Form
The justification follows directly from the LMM assumptions. Since \(u\) and \(\varepsilon\) are independent normal vectors, their sum is normal, with mean zero and covariance equal to the sum of their covariances. Therefore \(u+\varepsilon\sim N(0,\tau K+\sigma_e^2I)\). Adding the fixed mean \(X\gamma+g\beta\) gives the stated marginal distribution for \(y\). Substituting the null-model estimates gives the plug-in conditional model.
The cross-covariance terms are zero because the random effect and residual are assumed independent. Under \(H_0\), the mean reduces to \(X\gamma\); after plugging in the null estimates, it is \(X\hat{\gamma}_0\). This argument establishes the model distribution under the stated assumptions. It does not establish that the plug-in test has exact finite-sample calibration: the fitted values were estimated from data, so their uncertainty has been set aside in this conditional calculation.
Worked Example: Writing the Null-Fitted Model
Consider a hypothetical study with \(n=4\) individuals, an intercept and one covariate \(x\), and a candidate genotype vector \(g=(0,1,1,2)^\mathsf{T}\). Suppose the fitted null model gives \(\hat{\gamma}_{0,1}=5.0\) phenotype units for the intercept and \(\hat{\gamma}_{0,2}=0.4\) phenotype units per unit of \(x\). The fitted covariance parameters are \(\hat{\tau}_0=0.30\) and \(\hat{\sigma}_{e,0}^2=0.70\). Assume the specified relationship matrix has diagonal entries 1 and off-diagonal entries 0.2.
The null mean for individual \(i\) is \(5.0+0.4x_i\). For example, if \(x_3=2\), the null fitted mean for individual 3 is \(5.0+0.4(2)=5.8\) phenotype units. The conditional alternative adds the candidate effect: for that individual its mean is \(5.8+1\beta\), since \(g_3=1\).
For each diagonal covariance entry, \(0.30(1)+0.70=1.00\) phenotype-units squared. For each off-diagonal entry, \(0.30(0.2)+0.70(0)=0.060\) phenotype-units squared. Thus, with \(I\) the four-dimensional identity matrix,
The test is \(H_0:\beta=0\) against, for example, \(H_a:\beta\ne0\). In the fully frozen conditional formulation, both null covariate coefficients and both covariance components remain at the values above. The genotype coefficient is the parameter being tested. The covariance entries have units of phenotype-units squared; the genotype dosage is a copy count, so \(\beta\) has units of phenotype units per dosage.
Which Parameters Stay Fixed?
It is useful to name the distinction explicitly. The candidate effect \(\beta\) is the parameter of interest; under the null it equals zero, while under the alternative it is allowed to differ from zero. The nuisance parameters describe the baseline mean and covariance rather than this variant’s effect. In the fully frozen conditional formulation, these nuisance parameters are replaced by their null-model estimates and are not updated during the variant test.
| Quantity | Role in the conditional test | What happens after null fitting? |
|---|---|---|
| \(\beta\) | Candidate variant effect | Tested; fixed to zero under \(H_0\) |
| \(\gamma\) | Covariate coefficients | Held at \(\hat{\gamma}_0\) in the fully frozen formulation |
| \(\tau,\sigma_e^2\) | Random-effect and residual variance components | Held at \(\hat{\tau}_0,\hat{\sigma}_{e,0}^2\) |
| \(K\) | Relationship structure | Treated as specified and unchanged for this test |
This is a definition of one conditional testing setup, not a claim that every mixed-model software implementation freezes every nuisance parameter. Some procedures keep the covariance parameters at their null estimates but re-estimate covariate coefficients when fitting the alternative. Others use a null-model calculation that does not fit the alternative mean model at all. Reports should therefore identify the actual procedure and its parameter treatment. The phrase “conditional on the fitted null model” is not enough to infer every implementation detail.
Worked Example: Freezing Is Not Refitting
Suppose a hypothetical null fit gives \(\hat{\gamma}_{0,1}=10.0\), \(\hat{\gamma}_{0,2}=1.5\), \(\hat{\tau}_0=0.40\), and \(\hat{\sigma}_{e,0}^2=0.60\). An analyst’s fully frozen conditional test of a new genotype vector \(g\) uses the mean \(10.0+1.5x_i+g_i\beta\) and covariance \(0.40K+0.60I\). These numbers do not change as \(\beta\) is varied.
Imagine instead that a separate alternative-model fit returns a covariate slope of 1.4, a random-effect variance of 0.35, and a residual variance of 0.65. That fit uses the mean \(10.0+1.4x_i+g_i\beta\) and covariance \(0.35K+0.65I\), if its intercept remains 10.0. Those are not the same fixed inputs: the covariate slope and both variance components have been re-estimated. The difference is not a rounding detail; it changes the model comparison or test being conducted.
The two sets of values may be numerically close, but closeness does not make the procedures identical. To describe the fully frozen conditional test, state that the null estimates remain fixed and only the variant effect is the target. If an implementation re-estimates any nuisance parameter, name that exception rather than calling all parameters fixed.
Assumptions, Diagnostics, and Failure Modes
The formulation depends on assumptions about both the mean and covariance. The covariates and genotype coding must represent the intended comparison; the LMM’s residual and random-effect distributions must be suitable for the phenotype; and \(K\) must be an appropriate representation of the modeled relatedness. The covariance matrix \(\tau K+\sigma_e^2I\) must be valid (positive semidefinite, and positive definite for an ordinary nonsingular multivariate normal density). The random effect and residual must be independent under the specified model. A test’s reference distribution also depends on its chosen statistic and calibration; merely writing the conditional model does not settle that issue.
Before interpreting a variant test, check that the null model converged, that its variance estimates are plausible, and that the relationship matrix and sample ordering match the phenotype and genotype vectors. Inspect residual patterns and the distribution of association evidence across variants for signs that the mean or covariance model is inadequate. These checks cannot prove the assumptions, but can reveal problems such as extreme outliers, an inappropriate trait scale, or residual structure left unexplained by the model.
A further limitation is uncertainty in the fitted nuisance parameters. The plug-in model treats \(\hat{\gamma}_0\) and the variance estimates as fixed for the conditional calculation, even though they were estimated. This can be a useful computational and inferential strategy, especially when the same null model serves many variant tests, but it does not automatically account for all uncertainty from null-model fitting. Whether that matters for a particular analysis depends on the sample, the model, and the test calibration.
A common mistake is to say, “The variance components were estimated under the null, so they are known and cannot affect the variant test.” The correction is that the procedure holds their fitted values fixed; it does not turn estimates into known population parameters. Another mistake is to assume that every alternative model reuses the null covariate coefficients and covariance estimates. That is true only for the explicitly frozen formulation; implementations differ, so verify what is refitted.
How This Formulation Differs from Nearby Choices
The main contrast is between a null-fitted conditional analysis and a test that refits nuisance parameters under the alternative. The conditional approach reuses null estimates and changes only the designated genotype effect in the fully frozen version. A refitted alternative allows some nuisance estimates to change when genotype is added. These approaches answer related questions, but they do not necessarily produce the same test statistic or p-value.
A second contrast is with a naive regression that treats observations as independent. The conditional LMM retains the covariance \(\hat{V}_0\), so correlated phenotypes are not represented as independent measurements. Ignoring that covariance can misstate the uncertainty of an association estimate. Conversely, specifying an LMM does not guarantee protection against every source of misspecification: an unsuitable \(K\), omitted covariates, or a poor trait model can still distort inference.
The conditional model is therefore best understood as the setup for a test, not as a complete testing recipe. The chosen statistic and its null calibration must be stated separately. In particular, a conditional formulation alone does not imply a specific standard error, reference distribution, or p-value.
Worked Example: Interpreting a Conditional Effect
Suppose a hypothetical phenotype is measured in concentration units and the candidate genotype is coded as dosage 0, 1, or 2. Under the stated conditional model, \(\beta\) is the modeled mean difference in concentration units for a one-copy increase in dosage, at the same covariate values and under the fitted null covariance structure.
For illustration, if a fitted candidate effect were \(0.25\) concentration units per dosage, the model-implied mean difference between dosage 0 and dosage 2 would be \(2(0.25)=0.50\) concentration units, holding the modeled covariates and other model inputs fixed. This arithmetic describes the coefficient’s scale; it is not evidence that the effect is statistically distinguishable from zero, because no uncertainty measure or test result has been supplied.
The null hypothesis remains \(H_0:\beta=0\), meaning no linear per-dosage association in this model. Rejecting it would support evidence of association under the specified analysis, not prove that the variant causes the phenotype.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- Write the null and two-sided alternative hypotheses for a single candidate genotype effect \(\beta\).
- In the fully frozen conditional formulation, which parameters are varied for the association test, and which are held at null estimates?
- If \(\hat{\tau}_0=0.25\), \(\hat{\sigma}_{e,0}^2=0.75\), and \(K_{ij}=0.4\) for two distinct individuals, what is their fitted off-diagonal covariance?
- Why does “held fixed at the null estimate” not mean “known without uncertainty”?
- A software method holds variance components fixed but re-estimates covariate coefficients under the alternative. Is it the fully frozen formulation defined here? Explain.
Solutions. (1) \(H_0:\beta=0\) and \(H_a:\beta\ne0\). (2) The candidate effect \(\beta\) is tested; \(\gamma,\tau,\sigma_e^2\), and the specified \(K\) remain at their null values in the fully frozen setup. (3) The off-diagonal covariance is \(0.25(0.4)+0.75(0)=0.10\) phenotype-units squared. (4) The null estimates were calculated from data; the conditional calculation treats them as fixed inputs but does not erase the uncertainty involved in estimating them. (5) No. It freezes the variance components but changes the covariate coefficients, so it is a different conditional parameter treatment.