Why Test a Secondary Signal?
A variant can be associated with a phenotype when tested alone because it is correlated with another associated variant. The relevant question for a proposed secondary signal is narrower: does adding this variant improve a model that already includes the lead variant or variants and the other specified covariates? This is a conditional association question. Its answer depends on the variables already in the model.
The earlier material on joint regression provides the setting: several variant dosages can appear together as predictors. Here the focus is how to test whether one additional dosage contributes evidence beyond those already included, and how three standard tests—likelihood-ratio, Wald, and score—formulate that comparison.
Reduced and Full Models
Let \(Y_i\) be the phenotype for person \(i\), \(G_{i1}\) the dosage at a lead variant, \(G_{i2}\) the dosage at a candidate secondary variant, and \(C_i\) a vector of specified covariates such as age, sex, and ancestry components. For an additive dosage model, each genotype dosage is coded as the expected number of copies of a designated allele, conventionally from 0 to 2. A linear model for a quantitative phenotype can be written as:
The reduced model omits the candidate variant \(G_{i2}\), while the full model includes it. Both models must use the same individuals and the same covariates and existing variants for a straightforward nested-model comparison.
The model requires a meaningful outcome distribution and link or mean structure, an appropriate specification of the covariates and variants, and observations whose dependence is handled by the model or its standard errors. For the normal linear model used in the examples, errors are assumed independent, normally distributed with mean zero and common variance, conditional on the predictors. The predictors must also contain enough independent information to estimate the coefficients: exact collinearity between the candidate dosage and existing predictors prevents a separate conditional effect from being identified.
Three Tests, One Nested Comparison
All three tests address the same stated null hypothesis, but they use different information from the fitted models. Under the null and standard regularity conditions, each statistic has an approximate chi-square distribution with degrees of freedom equal to the number of added parameters. With one added dosage, \(df=1\). The approximation is generally most reliable with adequate sample size and a well-behaved, identifiable model.
Likelihood-ratio test
The likelihood-ratio test compares the maximum likelihoods of the full and reduced models. Write \(\ell_{\mathrm{full}}\) and \(\ell_{\mathrm{reduced}}\) for their maximized log-likelihoods. The full model cannot have a lower maximized likelihood because it contains the reduced model as a special case. The statistic is:
Under \(H_0\), \(\Lambda\) is approximately distributed as \(\chi^2_q\), where \(q\) is the number of added parameters. A large value indicates that allowing the added coefficient or coefficients to vary improves the fit more than would usually be expected from sampling variation under the null. For a normal linear model with an unknown variance estimated by maximum likelihood, the likelihood-ratio statistic can also be expressed using the residual sums of squares (RSS):
Wald test
The Wald test uses the full-model estimate and its estimated covariance matrix. For one added coefficient, let \(\hat{\beta}_2\) be its estimate and \(SE(\hat{\beta}_2)\) its standard error. The statistic is:
Under \(H_0\), \(W\) is approximately \(\chi^2_1\). Equivalently, the signed ratio \(z=\hat{\beta}_2/SE(\hat{\beta}_2)\) is approximately standard normal under the null, and the two-sided p-value is \(2P(Z\geq |z|)\) for \(Z\sim N(0,1)\). The coefficient itself is an estimated conditional effect; the Wald statistic measures how far that estimate is from zero relative to its uncertainty.
Score test
The score test evaluates whether the likelihood would increase locally if the omitted coefficient were allowed to move away from its null value. It uses the reduced-model fit, where the added coefficient is fixed at zero, rather than requiring a full-model estimate. In a model with nuisance parameters, the score and information for the added coefficient must account for estimation of those nuisance parameters. Denote these adjusted quantities by \(U\) and \(I\). For one added coefficient:
Under \(H_0\), \(S\) is approximately \(\chi^2_1\). For \(q\) added parameters, the corresponding quadratic form is \(S=\mathbf{U}^\mathsf{T}\mathbf{I}^{-1}\mathbf{U}\), with \(df=q\), when the information matrix is invertible. The score test can be useful when many candidate variants must be screened because each test can be based on a reduced-model fit. It still tests the specified conditional null; it does not establish that a variant is causal.
Worked Examples
Worked Example: Likelihood-Ratio Test From Residual Sums of Squares
Suppose a hypothetical analysis of a quantitative phenotype includes \(n=200\) individuals. Both models contain an intercept, the same covariates, and the same lead-variant dosage. The reduced model has \(\mathrm{RSS}_{\mathrm{reduced}}=1000\), and the full model additionally includes a candidate dosage and has \(\mathrm{RSS}_{\mathrm{full}}=960\). Assume the normal linear model is appropriate and its variance is estimated by maximum likelihood. Test \(H_0:\beta_2=0\) against \(H_a:\beta_2\ne 0\).
Because one coefficient is added, the likelihood-ratio statistic is compared with \(\chi^2_1\). Substituting the values into the RSS expression gives:
The upper-tail probability for \(\chi^2_1\) at 8.164 is approximately 0.0043. At significance level \(\alpha=0.05\), this is evidence against \(H_0\): in this specified model, adding the candidate dosage improves fit conditional on the lead variant and covariates. The result is evidence of conditional association, not proof that the candidate variant itself causes the phenotype.
Worked Example: Wald Test and Conditional Effect
In a separate hypothetical study of a quantitative phenotype measured in standard-deviation units, a full model includes a lead dosage, the candidate dosage, and the same prespecified covariates. The estimated candidate-variant coefficient is \(\hat{\beta}_2=0.30\) standard deviations per additional allele, with \(SE(\hat{\beta}_2)=0.12\) standard deviations per allele. Assume the model and standard-error calculation are appropriate. Test the two-sided null at \(\alpha=0.05\).
The Wald statistic and its equivalent signed statistic are:
The two-sided normal p-value is approximately 0.0124 (equivalently, the upper-tail probability for \(\chi^2_1\) at 6.25). Since \(0.0124<0.05\), the estimate is inconsistent with a zero conditional coefficient at the 5% level. A 95% Wald confidence interval for the conditional coefficient is:
Under the fitted model, the estimated mean phenotype is 0.30 standard deviations higher per additional dosage unit, holding the included predictors fixed. The interval describes uncertainty about that model coefficient; it does not give a 95% probability that the fixed coefficient lies in this particular interval.
Worked Example: Score Test From the Reduced Model
Consider a hypothetical study with \(n=500\) individuals and a quantitative phenotype. A reduced normal linear model includes the lead variant and all prespecified covariates, but not the candidate dosage. At the reduced-model fit, suppose the nuisance-adjusted score for the candidate coefficient is \(U=10\), and its nuisance-adjusted information is \(I=25\). These are hypothetical summary quantities from the reduced-model calculation. Assume the model is identifiable and the large-sample approximation is adequate.
The score statistic is:
With one added coefficient, compare this value with \(\chi^2_1\). Its upper-tail p-value is approximately 0.0455. At \(\alpha=0.05\), this provides evidence against \(H_0:\beta_2=0\), conditional on the lead dosage and covariates in the reduced model. The score test reaches this conclusion without estimating the full-model candidate coefficient; it does not mean that the candidate has an independent or causal effect outside this model.
Why the Three Results May Differ
The three statistics share a null and, in large samples under regular conditions, often give similar p-values. They are not numerically identical in every data set. The likelihood-ratio test compares optimized fits, the Wald test measures the full-model estimate against its estimated standard error, and the score test assesses the slope of the likelihood at the reduced-model fit. In small samples, with weak information, or with estimates near parameter boundaries, their approximations can differ appreciably.
The one-parameter Wald calculation also shows why a coefficient and its standard error matter together. A modest estimated effect may be estimated precisely, while a larger estimate may be highly uncertain. Neither effect magnitude nor significance alone captures both features. In a genetic association analysis, report the conditional effect estimate and units, its uncertainty, the test used, and the model terms against which the candidate was adjusted.
Short Justification: Why the Likelihood-Ratio Statistic Uses a Log Ratio
For nested models, the likelihood-ratio statistic compares how well each model can fit the observed data after maximizing its parameters. The likelihood ratio in favor of the full model is \(L_{\mathrm{full}}/L_{\mathrm{reduced}}\), which is at least one. Taking twice its logarithm gives \(\Lambda=2\log(L_{\mathrm{full}}/L_{\mathrm{reduced}})=2(\ell_{\mathrm{full}}-\ell_{\mathrm{reduced}})\), a nonnegative measure of the improvement in fit. Under \(H_0\), the regularity conditions for Wilks' theorem imply that this statistic approaches a chi-square distribution with degrees of freedom equal to the number of added parameters. Thus, adding one dosage gives \(df=1\); jointly adding two freely estimated dosage effects gives \(df=2\).
Common Mistakes and Interpretation
- Calling the null a test of no association at the locus. The null here is that the added coefficient or block is zero conditional on the model's existing predictors. Other included variants may remain associated.
- Comparing models fit to different individuals or covariates. A change in fit could then reflect a changed data set or adjustment set, not just the candidate variant. Use the same observations and shared terms for the nested comparison.
- Treating a small p-value as proof of causality. The tests measure evidence against a model-specific zero-coefficient null. Linkage disequilibrium, confounding, model misspecification, or other explanations may remain.
- Using an ordinary one-degree-of-freedom test for several added parameters. A block of \(q\) terms requires a joint test with \(q\) degrees of freedom, subject to identifiability and regularity conditions.
- Reading a p-value as the probability that the null is true. It is the probability, assuming the null model and test assumptions, of observing a statistic at least as extreme as the one calculated.
Check Your Understanding
Answer these questions using the model-specific meaning of the tests.
- In a model already containing one lead variant and ancestry covariates, what does \(H_0:\beta_2=0\) say about a candidate dosage?
- A full model has a higher maximized log-likelihood than its reduced counterpart. Why must this be true for properly nested models fit to the same data?
- For a single added coefficient, how is the Wald statistic calculated from its estimate and standard error?
- Which model's fit is used directly for the score test, and what quantity does that test evaluate?
- Why does evidence from a conditional association test not by itself establish that the tested variant is causal?