Three Tests for One SNP Effect
In a single-variant case-control analysis, the association test often asks whether a SNP coefficient is zero after adjusting for covariates. Wald, score, and likelihood-ratio tests address this same question, but they use different information from the fitted logistic model. In a large, well-behaved sample, their test statistics are usually close. In finite samples, they can differ enough to produce different p-values, so it matters which model each test fits and what its statistic measures.
Let \(Y_i\) indicate case status for person \(i\), let \(G_i\) be the chosen genotype coding (for example, an allele dosage), and let \(Z_i\) denote the covariates, including an intercept. The additive logistic regression model is
Here \(\operatorname{logit}(\pi)=\log\{\pi/(1-\pi)\}\), \(\gamma\) contains nuisance coefficients, and \(\beta\) is the SNP coefficient. For a dosage coded 0, 1, or 2, \(\exp(\beta)\) is the modelled odds ratio for a one-allele increase, conditional on the covariates.
Assumptions and the Fits Each Test Uses
The standard calculations assume independent observations conditional on the model covariates, a correctly specified logistic mean model, and an identifiable design. For the usual Wald and likelihood-ratio procedures, the full model must also have a finite, estimable maximum-likelihood estimate. The chi-square approximations rely on regularity conditions, including adequate information and a parameter value not at a boundary. These are approximations, not guarantees that a small data set is well behaved.
Write \(\ell(\gamma,\beta)\) for the logistic log-likelihood. The null model sets \(\beta=0\) and estimates \(\gamma\). The full model estimates both \(\gamma\) and \(\beta\). Each test uses these fits differently:
| Test | Fits required | What is measured |
|---|---|---|
| Wald | Full model | Distance of the estimated coefficient from zero, in estimated standard-error units |
| Score | Null model | Slope of the likelihood away from the null, scaled by its efficient information |
| Likelihood-ratio | Null and full models | Improvement in maximized log-likelihood from adding the SNP coefficient |
In the Wald test, let \(\hat{\beta}\) be the full-model maximum-likelihood estimate and \(SE(\hat{\beta})\) its estimated standard error. The statistic is
The score test evaluates the derivative of the log-likelihood with respect to \(\beta\) at \(\beta=0\), using nuisance estimates from the null model. Because nuisance coefficients are estimated, its information is the information for \(\beta\) after accounting for those coefficients. Denote the resulting score by \(U\) and efficient information by \(I\). Then \(S=U^2/I\). The Logistic Score Test Derivation develops this calculation; the essential point here is that the score test does not need to estimate the full-model SNP coefficient.
For the likelihood-ratio test, let \(\hat{\ell}_{\mathrm{full}}\) and \(\hat{\ell}_{\mathrm{null}}\) be the maximized log-likelihoods. The statistic is
The full model includes the null model as the special case \(\beta=0\), so its maximized log-likelihood cannot be smaller. Consequently, \(L\geq 0\). A large value of any of the three statistics is evidence against \(H_0\), but its p-value is a tail probability calculated under the relevant chi-square reference distribution; it is not the probability that \(H_0\) is true.
Worked Example: Comparing All Three Tests in a Two-Group Table
Consider hypothetical independent participants with a binary genotype indicator \(G\), where \(G=0\) denotes the reference group and \(G=1\) the other group. There are 20 cases and 80 controls for \(G=0\), and 35 cases and 65 controls for \(G=1\). Fit a logistic model with an intercept and \(G\). The groups have nonzero case and control counts, so the two group-specific fitted probabilities are finite; this simple model has an identifiable intercept and genotype effect.
The null model has a common case proportion \(\hat p=(20+35)/200=0.275\). Under the null, the expected number of cases in the \(G=1\) group is \(100(0.275)=27.5\), so its score contribution is \(U=35-27.5=7.5\). Each group has null-model weight \(100(0.275)(0.725)=19.9375\). After adjusting for the fitted intercept, the efficient information is \(I=19.9375(19.9375)/(19.9375+19.9375)=9.96875\). Thus
For the full model, the estimated case proportions are \(20/100=0.20\) and \(35/100=0.35\). The coefficient is the difference in their log odds: \( \hat{\beta}=\log\{(35/65)/(20/80)\}=\log(2.15385)=0.7673\). Its standard error in this two-group logistic model is \(SE(\hat{\beta})=\sqrt{1/35+1/65+1/20+1/80}=0.3263\). Therefore
For the likelihood-ratio statistic, the null fit uses the pooled proportion \(0.275\), while the full fit uses \(0.20\) and \(0.35\). The binomial log-likelihood difference is \(20\log(0.20/0.275)+80\log(0.80/0.725)+35\log(0.35/0.275)+65\log(0.65/0.725)\). The individual contributions sum to approximately \(2.849\), giving \(L=2(2.849)=5.698\). The three statistics are close, but not identical. With one degree of freedom, all give p-values around 0.02, and all indicate evidence of association at \(\alpha=0.05\) under the stated model and approximation.
Why the Tests Are Close Asymptotically
The connection between the tests can be seen by expanding the log-likelihood near the null. For a regular model, after accounting for nuisance parameters, the log-likelihood as a function of \(\beta\) is locally approximated by a quadratic. If \(U\) is the score and \(I\) the efficient information at the null, the approximation for a small change \(b\) is
The maximum of this quadratic occurs at \(b=U/I\). Substituting that value gives a log-likelihood increase of approximately \(U^2/(2I)\), so twice the increase is \(U^2/I\), the score statistic. The quadratic approximation also implies that the full-model estimate is approximately \(U/I\), with variance approximately \(1/I\). Its squared Wald statistic is therefore approximately \((U/I)^2/(1/I)=U^2/I\). Thus the three procedures agree to first order near the null when the likelihood is well approximated by a quadratic and the estimates behave regularly.
This argument explains asymptotic agreement; it does not say the finite-sample statistics must match. The score statistic uses curvature and slope at the null, the Wald statistic uses the estimate and uncertainty at the full-model fit, and the likelihood-ratio statistic compares the two maximized likelihoods over the path between them. When the likelihood is asymmetric or poorly approximated by a quadratic, these summaries can differ.
Worked Example: A Finite-Sample Difference in Decisions
Now consider a second hypothetical two-group data set: among 40 participants with \(G=0\), 2 are cases and 38 are controls; among 40 with \(G=1\), 8 are cases and 32 are controls. Both groups have cases and controls, so the intercept-plus-indicator model has finite group-specific estimates. The small counts nevertheless make the large-sample calibration something to check rather than take for granted.
The pooled null proportion is \(\hat p=10/80=0.125\). The \(G=1\) group has 5 expected cases under the null, so \(U=8-5=3\). Each group has null weight \(40(0.125)(0.875)=4.375\), giving efficient information \(I=4.375^2/(4.375+4.375)=2.1875\). Hence
The full-model coefficient is \(\hat{\beta}=\log\{(8/32)/(2/38)\}=\log(4.75)=1.558\). Its estimated standard error is \(SE(\hat{\beta})=\sqrt{1/8+1/32+1/2+1/38}=0.8262\), so
For the likelihood ratio, the log-likelihood difference from the pooled fit is \(2\log(0.05/0.125)+38\log(0.95/0.875)+8\log(0.20/0.125)+32\log(0.80/0.875)=2.185\). Thus \(L=2(2.185)=4.371\). The corresponding one-degree-of-freedom chi-square p-values are approximately 0.059 for Wald, 0.043 for score, and 0.037 for likelihood ratio. At \(\alpha=0.05\), the score and likelihood-ratio tests reject \(H_0\), while the Wald test does not. This disagreement is possible because the estimates and likelihood are not perfectly quadratic in this finite sample. The p-values also rely on an asymptotic approximation, which may be less reliable with small counts.
Worked Example: Choosing a Test in an Adjusted Scan
Suppose a hypothetical scan includes 500 unrelated participants, an intercept, age, two ancestry covariates, and one SNP dosage \(G\). Assume the logistic model is identifiable, the null and full optimizations converge, and the full-model coefficient is finite. For one variant, suppose the full fit reports \(\hat{\beta}=0.40\) and \(SE(\hat{\beta})=0.18\); the null fit gives efficient score \(U=12.0\) and information \(I=30.0\); and the two maximized log-likelihoods differ by \(2.44\). These are hypothetical fitted outputs, not results from a real study.
The statistics are \(W=(0.40/0.18)^2=4.938\), \(S=12.0^2/30.0=4.800\), and \(L=2(2.44)=4.880\). Each is compared with \(\chi^2_1\), giving p-values of roughly 0.026, 0.028, and 0.027, respectively. Here the adjusted tests give similar evidence. The Wald calculation needs the full fit; the score calculation can be obtained from the null fit; and the likelihood ratio requires both. A scan may therefore use the score test to reduce the computational burden of fitting a full model at every variant, while selected variants can be examined with additional fits and diagnostics.
Diagnostics, Pitfalls, and Alternatives
Before interpreting any of the three tests, check that the genotype coding and covariates are as intended, the null model converges, and the full model converges when a full fit is required. For Wald and likelihood-ratio tests, inspect whether the SNP estimate and its standard error are finite and plausible. A very large coefficient or standard error, a convergence warning, or a fitted probability near zero or one can indicate that the likelihood is poorly behaved. The score test avoids fitting the SNP effect under the alternative, but it does not repair a misspecified model, a problematic null fit, or inadequate information.
The score test is often convenient for a large variant scan because the covariate-only null model can be fitted once and reused. The Wald test is straightforward when reliable full-model estimates are already available. The likelihood-ratio test directly measures the improvement in likelihood and is often a useful alternative when the likelihood is asymmetric, though it needs both fits. None is universally best: the choice should account for computational demands, fit stability, and whether the chi-square approximation is credible. For especially small or sparse data, a standard asymptotic test may not provide dependable calibration; specialized approaches require separate consideration.
A common mistake is to treat the three statistics as algebraically identical, or to choose whichever gives the smallest p-value. They are not identical in finite samples, and selecting a test after seeing which one is most significant changes the inferential procedure. A better practice is to specify the test in advance, report the model and test used, and investigate material disagreement rather than presenting it as proof that one result is correct.
Check Your Understanding
Answer each question, then compare with the solutions below.
- Which fitted models are required for the Wald, score, and likelihood-ratio tests?
- In a one-degree-of-freedom test, if a score statistic is \(U=4\) and efficient information is \(I=8\), what is the statistic?
- Why can the three tests yield different p-values in a finite sample even though they test the same null hypothesis?
- A full logistic fit reports a very large SNP coefficient and a convergence warning. Which test statistics require that full fit, and what should an analyst avoid concluding?
Solutions. (1) Wald requires the full model; score requires the null model; likelihood ratio requires both. (2) \(S=U^2/I=4^2/8=2\), compared with a one-degree-of-freedom chi-square reference distribution under the usual approximation. (3) They use different parts of the likelihood: the Wald test uses the full-fit estimate and standard error, the score test uses the null-fit slope and information, and the likelihood-ratio test uses the change in maximized log-likelihood. Their equality is an asymptotic approximation, not a finite-sample identity. (4) Wald and likelihood ratio require the full fit, and both may be unreliable when it is unstable; do not interpret a resulting large statistic or small p-value as decisive evidence without investigating the fitting problem and the model conditions.