Why Test Genotype and Local Ancestry Together?
At an admixed locus, a phenotype may be associated with an individual's genotype, with the local ancestry inherited at that locus, or with both. These terms can be correlated: ancestry affects which alleles are more likely to be observed. A joint test asks whether the model needs either term, while separate conditional tests ask whether one term contributes after accounting for the other. Those are different questions, and a significant joint test alone cannot say which term carries the evidence.
The model below treats genotype dosage and local ancestry as predictors of an outcome. It is an association model: even a well-supported coefficient does not establish that the predictor causes the outcome. The introductory material on local ancestry association distinguishes the SNP-based and admixture-mapping questions; here the emphasis is on testing them together.
Model and Hypotheses
For a quantitative outcome, a basic additive model is
where \(\mathbf{C}_i\) contains prespecified adjustment covariates and \(\varepsilon_i\) is the residual. For the classical linear-model Wald and likelihood calculations, assume independent individuals, a correctly specified conditional mean, a full-rank design matrix, and normally distributed errors with common variance \(\sigma^2\). If a suitable mixed model or other dependence-aware model is required, the coefficient covariance and likelihood must instead come from that model; treating related observations as independent can invalidate the test.
The joint null hypothesis is that neither predictor contributes conditionally. The alternative is that at least one does:
The usual test has 2 degrees of freedom because it tests two independent restrictions. This requires the two columns to be estimable after adjustment for the intercept and covariates. If genotype and local ancestry are perfectly collinear in the analyzed sample, their separate coefficients cannot be identified, and a two-parameter test is not available.
The conditional, one-degree-of-freedom hypotheses are \(H_0:\beta_G=0\) and \(H_0:\beta_L=0\), each with the other term retained in the model. These tests help locate evidence, but they are not substitutes for the joint null. In particular, failure to reject one conditional null does not prove that its coefficient is zero.
Wald and Likelihood-Ratio Tests
Let \(\hat{\boldsymbol{\beta}}=(\hat{\beta}_G,\hat{\beta}_L)^{\mathsf T}\), and let \(\widehat{\mathbf{V}}\) be its estimated covariance matrix from the model that includes both terms. The joint Wald statistic measures the estimated coefficient vector relative to its covariance:
Under \(H_0\), a large-sample approximation is \(W\sim\chi^2_2\). The covariance between estimates matters: adding two separate squared \(z\)-statistics is only valid when the estimates are uncorrelated. The Wald test is convenient because it uses the fitted model's estimates and covariance, but can be unreliable when estimates are imprecise, strongly correlated, or far from the null in a small sample.
The likelihood-ratio (LR) test compares the maximized likelihood of the full model with that of the nested null model, which retains the same covariates but omits both \(G_i\) and \(L_i\). If their maximized log likelihoods are \(\ell_{\mathrm{full}}\) and \(\ell_{\mathrm{null}}\), then
Under the null and regularity conditions, \(LR\) is approximately \(\chi^2_2\). The two models must use the same observations and likelihood specification. For a binary outcome, for example, one can use logistic regression and compare the nested logistic likelihoods; the same two restrictions give 2 degrees of freedom. Wald and LR tests are asymptotically equivalent under the null, but their finite-sample results can differ.
Worked Calculations
Worked Example: A Two-Parameter Wald Test
Suppose a hypothetical study fits the quantitative-outcome model to 600 unrelated participants, adjusting for age and sex. The additive model and residual diagnostics are considered adequate for this illustration, and the genotype and local-ancestry columns are not collinear with each other or the covariates. The estimated genotype coefficient is 0.20 outcome units per additional alternate allele, with \(SE=0.08\). The local-ancestry coefficient is 0.30 outcome units per additional copy of the specified ancestry, with \(SE=0.15\). Their estimated correlation is 0.30.
The covariance matrix has diagonal entries equal to the squared standard errors and off-diagonal entry \(0.30(0.08)(0.15)=0.0036\):
Equivalently, standardize each coefficient by its standard error. The resulting \(z\)-values are \(0.20/0.08=2.50\) and \(0.30/0.15=2.00\). For two estimates with correlation \(\rho=0.30\), the joint Wald statistic is
Comparing with a \(\chi^2_2\) distribution gives \(p\approx 0.0186\). At a prespecified \(\alpha=0.05\), reject the joint null: the data provide evidence that at least one of the two conditional terms is associated with the outcome. This result does not by itself identify which term is responsible. The coefficients describe conditional associations in outcome units per allele and per ancestry copy, respectively.
Worked Example: Joint Evidence Driven by Genotype
In a second hypothetical study, suppose the fitted model gives \(\hat{\beta}_G=0.24\) outcome units per allele with \(SE=0.08\), and \(\hat{\beta}_L=0.03\) outcome units per ancestry copy with \(SE=0.15\). Assume their estimated covariance is zero, and that the same model assumptions and full-rank condition hold.
The joint Wald statistic is the sum of squared standardized coefficients:
For 2 degrees of freedom, \(p=\exp(-9.04/2)\approx 0.0109\), so the joint null is rejected at the 0.05 level. The one-degree-of-freedom genotype statistic is \(3.00^2=9.00\), with \(p\approx0.0027\); the local-ancestry statistic is \(0.20^2=0.040\), with \(p\approx0.841\). This pattern is evidence for a genotype association conditional on local ancestry, but not evidence for a local-ancestry association conditional on genotype. The non-significant ancestry result is not proof of no ancestry association.
Worked Example: A Likelihood-Ratio Test
For a hypothetical binary outcome, suppose a logistic model includes genotype dosage, local ancestry, age, and sex. The same 800 participants are used in both fits; the null model retains age and sex and omits the two locus terms. Assume the logistic mean model is appropriate, the estimates are finite, and the full model is identifiable. Suppose the null-model log likelihood is \(-210.4\), and the full-model log likelihood is \(-205.7\).
The LR statistic is
Against \(\chi^2_2\), the p-value is \(\exp(-9.40/2)\approx0.00910\). The joint null is rejected at \(\alpha=0.05\): adding genotype and local ancestry together improves fit beyond the covariates-only model. This test alone does not distinguish a SNP association from an admixture association; inspect the conditional coefficients and their uncertainty, or conduct the corresponding prespecified one-term tests.
Why the Joint Test Has Two Degrees of Freedom
The Wald reference distribution follows from the approximate sampling distribution of the two coefficient estimates. Under standard large-sample regularity conditions, under \(H_0\),
The Wald statistic is the squared length of this standardized vector. If its two coordinates are independent standard normal variables \(Z_1\) and \(Z_2\), that squared length is \(Z_1^2+Z_2^2\), which has a \(\chi^2_2\) distribution. The LR reference distribution follows from the standard likelihood-ratio result: under regularity conditions, twice the log-likelihood gain for \(r\) independent restrictions is asymptotically \(\chi^2_r\). Here \(r=2\). In a model with multiple ancestry indicators, count the independent ancestry parameters: testing genotype plus \(q\) independent ancestry terms jointly has \(1+q\) restrictions.
Interpretation, Diagnostics, and Alternatives
A joint p-value answers whether the data are inconsistent with both conditional coefficients being zero, under the fitted model. It does not give the probability that the null is true, measure the size of an effect, or attribute the signal to one predictor. For attribution, report the conditional estimates, standard errors or intervals, and the relevant one-term tests alongside the joint result. Interpret each coefficient using its units and coding. Neither a genotype association nor a local-ancestry association is automatically causal.
Before interpreting either test, check that the data support estimating both terms. Strong genotype–ancestry correlation can make \(\widehat{\mathbf{V}}\) nearly singular: coefficients may have large standard errors, change substantially with small model changes, or have wide intervals even when the joint test shows evidence. Check model convergence, the rank and conditioning of the design matrix, residual behavior for a quantitative model, and whether the fitted outcome model is credible. Also confirm that both nested models use the same participants and covariates. If local ancestry is uncertain, treating a hard call as error-free can affect inference; this tutorial's calculations assume the supplied local-ancestry values are adequate for the analysis.
| Approach | Question answered | Practical consideration |
|---|---|---|
| Joint Wald test | Do the genotype and ancestry terms jointly contribute? | Uses coefficient estimates and their covariance; can be sensitive to unstable estimates. |
| Joint likelihood-ratio test | Does adding both terms improve likelihood? | Requires a correctly specified nested likelihood and regularity conditions. |
| Separate conditional tests | Does one term contribute when the other remains in the model? | Can help locate evidence, but does not replace the joint test. |
A common mistake is to see a significant joint test and conclude that both genotype and ancestry are associated. The joint alternative is “at least one coefficient is nonzero,” not “both coefficients are nonzero.” The correction is to report the joint test as an omnibus result and evaluate each conditional term separately, with appropriate uncertainty. Conversely, testing each term alone while omitting the other answers a different question and can confound their associations.
If the scientific model includes a genotype-by-ancestry interaction, a joint test of the two main-effect coefficients alone does not test whether all locus terms are absent. The interaction term must be included in the model and in an omnibus null if the question is whether any genotype, ancestry, or interaction term contributes. Main-effect interpretations also depend on the interaction model's coding. The comparison here is limited to the additive model with the two specified terms.
Check Your Understanding
Answer each question using the model and testing principles above. Solutions follow.
- In the additive model, state the joint null and the degrees of freedom for a two-ancestry setting.
- A joint test is significant, but the conditional genotype test is not and the conditional ancestry test is significant. What conclusion is justified?
- For estimates with standard errors \(0.10\) and \(0.20\), and coefficient estimates \(0.10\) and \(0.40\), respectively, assume zero covariance. Calculate the joint Wald statistic.
- What must be held constant when comparing full and null models with a likelihood-ratio test?
Solutions. (1) The null is \(H_0:\beta_G=0,\ \beta_L=0\); there are two restrictions and thus 2 degrees of freedom, provided both terms are estimable. (2) The result supports a conditional ancestry association, but does not establish that the genotype coefficient is zero; the genotype test may lack power. The joint test says at least one term contributes. (3) The standardized estimates are \(0.10/0.10=1\) and \(0.40/0.20=2\). With zero covariance, \(W=1^2+2^2=5.00\), compared with \(\chi^2_2\). (4) Use the same participants, covariates, and likelihood specification; the null model must be nested in the full model and differ by the terms being tested.