Three Ways to Test a Genetic Effect
In a covariate-adjusted association analysis, the question is whether a genetic coefficient contributes information about the phenotype after accounting for specified covariates. The score, Wald, and likelihood-ratio tests address the same null hypothesis, but use different features of the likelihood. Their distinction matters in practice: one test can be computed from a null model, another requires a full-model estimate, and finite-sample results need not agree exactly.
We will use logistic regression for a binary phenotype. Let \(Y_i\) be the case indicator for individual \(i\), \(G_i\) a genotype dosage under a specified genetic coding, and \(C_i\) a vector of nuisance covariates such as age, sex, or ancestry variables. “Nuisance” means that the covariate effects are included to adjust the genetic-effect estimate, but are not the target of this test.
The tests below use the same specified model, genotype coding, and covariate set. Their usual reference distribution is a chi-square distribution with one degree of freedom. This approximation requires a correctly specified likelihood, identifiable parameters, finite nonzero information, and sufficiently large samples for the maximum-likelihood estimates and likelihood approximations to behave regularly. In particular, the null and alternative parameters should be interior points of the parameter space, and the model should not be on a boundary such as complete separation.
Definitions of the Three Tests
Write \(\ell(\beta,\eta)\) for the log-likelihood, where \(\eta\) collects all nuisance parameters, including the intercept and covariate coefficients. Let \(\hat{\beta}\) and \(\hat{\eta}\) be the maximum-likelihood estimates in the full model. Let \(\tilde{\eta}\) be the nuisance-parameter estimate under \(H_0\), where \(\beta\) is fixed at zero.
For the score test, partition the score vector and expected information matrix into the genetic parameter and nuisance parameters, evaluating them at the null fit \((0,\tilde{\eta})\): $$ U= \begin{pmatrix}U_\beta\\U_\eta\end{pmatrix}, \qquad I= \begin{pmatrix}I_{\beta\beta}&I_{\beta\eta}\\I_{\eta\beta}&I_{\eta\eta}\end{pmatrix}. $$ At the null maximum-likelihood estimate, \(U_\eta=0\). The information for \(\beta\) after allowing for nuisance parameters is the Schur complement \(I_{\beta\beta\cdot\eta}=I_{\beta\beta}-I_{\beta\eta}I_{\eta\eta}^{-1}I_{\eta\beta}\).
Worked Example: Computing All Three Tests
Worked Example: Computing All Three Tests
Suppose a hypothetical, sufficiently large study fits the logistic model above using a prespecified genotype dosage and covariates. The fitted genetic coefficient is \(\hat{\beta}=0.30\), with model-based \(SE(\hat{\beta})=0.075\). The score calculation at the null fit gives \(U_{\beta\cdot\eta}=18.0\) and \(I_{\beta\beta\cdot\eta}=20.0\). The fitted log-likelihoods are \(-414.95\) for the full model and \(-419.95\) for the null model. Calculate each test.
Wald. Substituting the estimate and its standard error gives $$ T_{\mathrm{Wald}}=\left(\frac{0.30}{0.075}\right)^2=4^2=16.0. $$
Score. Using the null-model score and adjusted information gives $$ T_{\mathrm{Score}}=\frac{18.0^2}{20.0}=\frac{324}{20.0}=16.2. $$
Likelihood ratio. The difference in maximized log-likelihoods is \(-414.95-(-419.95)=5.00\), so $$ T_{\mathrm{LR}}=2(5.00)=10.0. $$
Each statistic is compared with \(\chi^2_1\). The corresponding approximate p-values are \(6.3\times10^{-5}\), \(5.7\times10^{-5}\), and \(0.0016\), respectively, for the Wald, score, and likelihood-ratio tests. Each provides strong evidence against no genetic association under the specified model. The coefficient corresponds to an estimated conditional odds ratio of \(\exp(0.30)\approx1.35\) per dosage unit; the tests assess evidence against \(\beta=0\), not whether the association is causal.
Why the Score Test Adjusts for Nuisance Covariates
The score test asks whether the likelihood is increasing in the genetic direction at the null fit, while allowing for the fact that changes in the nuisance parameters can also change the likelihood. The adjustment is therefore not optional: testing the raw genetic score against the unadjusted information generally fails to account for correlation between the genetic predictor and covariates.
Set \(\beta=0\) and estimate the intercept and covariate coefficients, giving \(\tilde{\eta}\). At this fit, the nuisance score satisfies \(U_\eta=0\).
Evaluate \(U_\beta\), the likelihood's derivative in the genetic direction, at \((0,\tilde{\eta})\).
Subtract the part of the genetic information explained by its overlap with nuisance directions: \(I_{\beta\beta\cdot\eta}=I_{\beta\beta}-I_{\beta\eta}I_{\eta\eta}^{-1}I_{\eta\beta}\).
Compute \(U_\beta^2/I_{\beta\beta\cdot\eta}\), then compare it with \(\chi^2_1\) under the null and regularity conditions.
The adjustment can also be justified from a local quadratic approximation to the log-likelihood. Let \(h\) be a small change in \(\beta\) from zero and \(k\) a small change in \(\eta\) from \(\tilde{\eta}\). Near the null fit, the log-likelihood change is approximately $$ \Delta\ell(h,k)=U_\beta h-\frac{1}{2}\left(I_{\beta\beta}h^2+2hI_{\beta\eta}k+k^\mathsf{T}I_{\eta\eta}k\right), $$ because \(U_\eta=0\). For a fixed \(h\), maximizing this expression over \(k\) gives \(k=-I_{\eta\eta}^{-1}I_{\eta\beta}h\). Substitution yields $$ \Delta\ell_{\mathrm{profile}}(h)\approx U_\beta h-\frac{1}{2}I_{\beta\beta\cdot\eta}h^2. $$ Thus the curvature relevant to a genetic change after nuisance parameters are allowed to adjust is \(I_{\beta\beta\cdot\eta}\), not \(I_{\beta\beta}\) alone. Maximizing this quadratic over \(h\) gives an increase of \(U_\beta^2/(2I_{\beta\beta\cdot\eta})\); twice that increase is the score statistic. This approximation explains both the nuisance adjustment and the statistic's form.
Worked Example: The Effect of Nuisance Adjustment
For a hypothetical null-model fit with one nuisance coefficient, suppose the score and information components are \(U_\beta=6.0\), \(U_\eta=0\), \(I_{\beta\beta}=10.0\), \(I_{\beta\eta}=4.0\), and \(I_{\eta\eta}=8.0\). Calculate the properly adjusted score statistic.
The nuisance-adjusted information is $$ I_{\beta\beta\cdot\eta}=10.0-\frac{4.0^2}{8.0}=10.0-2.0=8.0. $$ Since the nuisance score is zero at its null maximum-likelihood estimate, \(U_{\beta\cdot\eta}=6.0\). Therefore, $$ T_{\mathrm{Score}}=\frac{6.0^2}{8.0}=\frac{36.0}{8.0}=4.50. $$ Against \(\chi^2_1\), this has an approximate p-value of \(0.0339\). Using \(I_{\beta\beta}=10.0\) without adjustment would instead give \(36.0/10.0=3.60\), which is not the nuisance-adjusted score test. The lower adjusted information reflects the overlap between the genetic and nuisance directions in this hypothetical model.
What Each Test Requires and When They Agree
| Test | Fits or estimates required | What it uses | Practical consideration |
|---|---|---|---|
| Score | Null model and its score/information | Likelihood slope at the null and nuisance-adjusted information | Useful when many genetic effects are tested against a common null model; it does not require estimating each full alternative model. |
| Wald | Full model and standard error of \(\hat{\beta}\) | Distance of the estimated coefficient from zero in standard-error units | Convenient when the full model has already been fitted; it depends on the estimate and its local uncertainty. |
| Likelihood ratio | Null and full models | Difference in maximized log-likelihoods | Requires both fits but directly compares their optimized likelihoods. |
Under the regularity conditions above, the three statistics have the same first-order large-sample null distribution. They also have the same first-order behavior under local alternatives that approach the null at rate \(1/\sqrt{n}\). Informally, with abundant information and a likelihood well approximated by a quadratic near its maximum, the slope at the null, the distance to the full-model estimate, and the increase in maximized likelihood encode nearly the same evidence.
That similarity is not an exact finite-sample identity. The score test evaluates the slope and information at the null; Wald uses the full-model estimate and its standard error; the likelihood-ratio test uses the change in optimized likelihood. When the likelihood is asymmetric, the estimate is far from zero, or information is limited, their numerical values and p-values can differ. A Wald test can also depend on the parameter scale: a test formed by treating a nonlinear re-expression as approximately normal need not match the test on the original coefficient. The likelihood-ratio test is invariant to a one-to-one reparameterization of the same model and null hypothesis.
Worked Example: A Finite-Sample Difference Near a Threshold
Consider another hypothetical covariate-adjusted logistic fit with \(\hat{\beta}=0.650\) and \(SE(\hat{\beta})=0.330\). Suppose the null-model score calculation gives \(T_{\mathrm{Score}}=3.72\), while the fitted log-likelihood difference gives \(T_{\mathrm{LR}}=4.03\). Compare the three tests at \(\alpha=0.05\).
The Wald statistic, calculated from the unrounded ratio, is $$ T_{\mathrm{Wald}}=\left(\frac{0.650}{0.330}\right)^2\approx3.880. $$ For one degree of freedom, the approximate p-values are \(0.0489\) for Wald, \(0.0538\) for score, and \(0.0447\) for the likelihood-ratio test. Thus Wald and likelihood ratio reject \(H_0\) at the 0.05 level, while the score test does not. These close, differing results are possible in finite samples; the statistics do not have to coincide. The data and estimates here are hypothetical, and the example is not a claim that any one test is universally preferable.
Model-Based Inference and Common Pitfalls
The formulas described here are the conventional likelihood-based tests, with standard errors and information calculated under the specified model. A sandwich standard error changes the Wald calculation, but simply inserting a robust standard error into a conventional likelihood-ratio formula does not make that likelihood-ratio test robust. Score tests also require an appropriate variance and information calculation if the model-based variance assumptions are not used. The test's assumptions and variance estimator should therefore be stated alongside its result.
- Ignoring nuisance adjustment. Using genetic information \(I_{\beta\beta}\) alone can misstate the score statistic when genotype and covariates are associated. Use the information after projecting out the nuisance directions.
- Assuming the tests are identical. Their large-sample agreement is approximate, not an algebraic equality for every data set. Report which test was used and its statistic.
- Using a Wald test when the full estimate is unstable. Wald relies on a useful full-model estimate and local standard-error approximation. If regularity fails, the usual chi-square calibration may fail as well.
- Interpreting a p-value as the probability that there is no effect. A p-value is calculated under \(H_0\); it is the probability, under that null model and its assumptions, of a statistic at least as extreme as observed. It is not the probability that \(H_0\) is true.
Check Your Understanding
Answer these questions using the model and definitions above.
- What does \(\beta=0\) mean in the covariate-adjusted logistic model, and what does \(\exp(\beta)\) represent?
- Which model fits are needed for the score, Wald, and likelihood-ratio tests?
- Why does the score test use \(I_{\beta\beta\cdot\eta}\) rather than \(I_{\beta\beta}\) alone?
- State two regularity conditions that support the usual \(\chi^2_1\) calibration.
- Why can score, Wald, and likelihood-ratio p-values differ in a finite sample even when they test the same null?