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Covariate Model Building Strategies · Tutorial 466 of 797

Degrees of Freedom in Covariate Likelihood-Ratio Tests

You will learn to count the independent parameter restrictions in a covariate likelihood-ratio test and use that count to interpret its chi-square statistic.

Intermediate 10 min read

What You'll Learn

  • Define the likelihood-ratio statistic for comparing nested NLME covariate models.
  • Calculate degrees of freedom from the difference in free parameter counts.
  • Count the parameters for categorical effects and joint covariate hypotheses.
  • Recognize the assumptions needed for a chi-square reference distribution.
  • Diagnose common problems with model comparability, estimation, and parameter identifiability.

Why Degrees of Freedom Matter

When a covariate is added to a nonlinear mixed-effects model, the likelihood generally improves, even if the covariate has no meaningful contribution. A likelihood-ratio test (LRT) evaluates whether the improvement is large relative to what would be expected from the additional parameters. Its degrees of freedom determine the reference chi-square distribution, and therefore affect the p-value. Counting them correctly is essential: the number is not automatically the number of covariates, nor the number of subjects.

For the comparison to have its usual interpretation, the models must be nested: the reduced model must be obtainable from the full model by imposing restrictions on its parameters. Both models must be fit to the same observations using comparable likelihoods. In NLME covariate testing, this normally means comparing the marginal likelihoods under maximum likelihood (ML), with the random effects integrated out. Restricted maximum likelihood (REML) likelihoods are generally not comparable when the fixed-effects design changes.

Definition: For a reduced model and a nested full model, let \(\ell_R\) and \(\ell_F\) be their maximized log likelihoods. The likelihood-ratio statistic is \(LR=2(\ell_F-\ell_R)\). The degrees of freedom \(q\) are the number of independent parameter restrictions imposed by the reduced model on the full model. Under the regularity conditions for the test, \(LR\) is compared with a chi-square distribution with \(q\) degrees of freedom.

Equivalently, when the two models differ only in the parameters being tested, \(q\) is the number of additional identifiable, freely estimated parameters in the full model. It is the difference in model dimension, not the total parameter count in either model. If other features also change—such as the random-effects structure or residual-error model—the parameter-count difference may not describe a test of the covariate alone.

Counting the Restrictions

A single continuous covariate with a linear effect usually adds one coefficient. For example, if clearance is modeled as \(\log(CL_i)=\theta_{CL}+\beta x_i+\eta_{CL,i}\), testing whether this covariate has an effect means comparing \(H_0:\beta=0\) with a model in which \(\beta\) is free. There is one restriction, so \(q=1\).

For a categorical covariate with \(K\) levels, a standard reference-level parameterization uses \(K-1\) coefficients. Testing the categorical effect as a whole sets all \(K-1\) coefficients to zero, so \(q=K-1\), provided those coefficients are identifiable. The reference level is represented by the intercept or baseline parameter; it does not require an additional effect coefficient. Replacing reference coding with another full-rank coding changes the coefficient labels, but not the dimension of the overall effect.

The same principle applies to a joint hypothesis. If a model includes two coefficients for a covariate’s effects on two different model parameters, testing that covariate has no effect on either parameter imposes two restrictions. If the scientific question concerns only one of those effects, the comparison should constrain only its coefficient and has one degree of freedom. The null hypothesis being tested—not the informal name of a covariate—determines the count.

Worked Example: One Continuous Covariate

Suppose hypothetical concentration data from 120 participants are fit with the same structural and random-effects model in both cases. The reduced model has no body-weight effect on clearance and has maximized log likelihood \(\ell_R=-512.8\). The full model adds one body-weight coefficient and has \(\ell_F=-509.6\). Both fits use ML, converge successfully, and estimate the added coefficient away from a boundary.

The models are nested because setting the added coefficient to zero recovers the reduced model. The full model adds one free coefficient, so \(q=1\). The statistic is

$$ LR=2(\ell_F-\ell_R) =2[-509.6-(-512.8)] =2(3.2) =6.4. $$

Against a chi-square distribution with 1 degree of freedom, the p-value is approximately 0.0114. At a 0.05 significance level, this is evidence against the null hypothesis of no body-weight effect in this model comparison. The test does not establish that body weight causes a change in clearance, nor does it by itself quantify whether the estimated effect is clinically important.

Worked Example: A Four-Level Categorical Covariate

In a hypothetical analysis, a four-level formulation group is added to a model for relative bioavailability. The reference group is coded in the intercept, and the other three groups each have a coefficient. The reduced and full models have log likelihoods \(\ell_R=-430.2\) and \(\ell_F=-424.1\), respectively. Assume that all four groups are represented in the data, the three coefficients are identifiable, and both models use the same observations and ML likelihood.

The joint null hypothesis is that all three group coefficients are zero. It imposes three independent restrictions, so \(q=3\), not 1 simply because there is one named covariate. The statistic is

$$ LR=2[-424.1-(-430.2)] =2(6.1) =12.2. $$

For a chi-square distribution with 3 degrees of freedom, the p-value is approximately 0.00672. This provides evidence against the joint null that formulation group has no effect, under the assumptions of the comparison. It is an overall test: it does not say that every non-reference group differs from the reference group, or identify which contrasts are responsible for the improvement.

Why the Chi-Square Reference Applies

The usual chi-square result is an instance of Wilks’ theorem. Under the null hypothesis, the reduced model is correct within the full model, and the full model has \(q\) additional locally identifiable parameter directions. If the log likelihood is smooth near the true parameter and the maximum is well behaved, it can be approximated locally by a quadratic function. The likelihood gain from allowing the \(q\) directions to vary is then approximately the squared length of a \(q\)-dimensional standard normal vector.

That squared length has a chi-square distribution with \(q\) degrees of freedom. More explicitly, if \(Z_1,\ldots,Z_q\) are independent standard normal random variables, then \(\sum_{j=1}^{q}Z_j^2\sim\chi^2_q\). This explains why the degrees of freedom count the independent restrictions released by the full model.

The approximation relies on regularity conditions. The models must be genuinely nested and correctly fit to the same data; the parameters under test should be identifiable; and, under the null, their values should be in the interior of the parameter space. The likelihood should be sufficiently smooth, and the sample should provide enough information for the large-sample approximation to be useful. These conditions are not guaranteed merely because software reports a likelihood and a p-value.

A covariate coefficient tested against zero is commonly an interior restriction when that coefficient can take positive or negative values. In contrast, testing whether a variance component is zero places the null on the boundary of the allowable parameter space. Its null distribution need not be the ordinary chi-square distribution with the simple parameter-count degrees of freedom. The same caution applies when parameters are unidentified under the null or when the fitted model is singular.

Worked Example: A Joint Linear and Curvature Test

Suppose a hypothetical model represents an age effect on a parameter using both a linear and a quadratic term in centered, scaled age \(z\). The reduced model has neither term; the full model adds coefficients \(\beta_1\) and \(\beta_2\). Their log likelihoods are \(\ell_R=-620.0\) and \(\ell_F=-617.5\). Assume both coefficients are identifiable, the models are fit by ML to the same participants, and the other model components are unchanged.

The null hypothesis for this comparison is \(H_0:\beta_1=0\) and \(\beta_2=0\). It imposes two independent restrictions, so \(q=2\). The calculation is

$$ LR=2[-617.5-(-620.0)] =2(2.5) =5.0. $$

For 2 degrees of freedom, the chi-square upper-tail p-value is approximately 0.0821. At a 0.05 level, the comparison does not provide sufficient evidence to reject the joint null. This does not prove that age has no effect. A different comparison—such as testing curvature while retaining the linear term—would test \(H_0:\beta_2=0\) and have 1 degree of freedom. The hypotheses and nested models must be stated before assigning the degrees of freedom.

Practical Checks and Common Pitfalls

Before interpreting a covariate LRT, verify that the full model can be reduced to the proposed null model by setting the tested coefficients to their null values. Confirm that both fits used the same observations, including the same handling of missing covariate values, and that the likelihoods are calculated on the same basis. A difference in the data or likelihood definition can produce a numerical difference that is not the likelihood gain for the intended nested comparison.

Check convergence and parameter estimates in both fits. A full model that has not reached a reliable maximum may have an inaccurate log likelihood, and a nominal likelihood improvement can then mislead. Sparse categories, strong collinearity, or poor covariate support can make coefficients weakly identified; in those situations, the nominal parameter count may not describe a well-behaved test. These issues call for investigating the fit and the design, not mechanically trusting the reported p-value.

A common mistake is to assign one degree of freedom because only one covariate was added. For a categorical covariate with four levels, the overall test usually frees three coefficients, so the correct reference is \(\chi^2_3\), not \(\chi^2_1\). The correction is to write down the null restrictions explicitly and count the independent restrictions. A different mistake is to count every parameter in the full NLME model; parameters shared by both models do not contribute to the degrees of freedom for this comparison.

An LRT of a multi-parameter effect is an overall test. It differs from inspecting one coefficient at a time, which answers a narrower question and does not substitute for a joint test. A Wald test can also test parameter restrictions, but it uses a different approximation and may behave differently in finite samples. For covariate model comparisons, the LRT’s degrees of freedom still come from the restrictions defining the particular nested comparison.

Key takeaway: State the null hypothesis, identify the nested models, and count the independent restrictions it imposes. Use that count as the chi-square degrees of freedom only when the models are comparable and the regularity conditions make the usual reference distribution reasonable.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. A one-coefficient continuous covariate is added to an NLME model. How many degrees of freedom does the usual LRT have, and what restriction defines its null hypothesis?
  2. A five-level categorical covariate is represented by four identifiable coefficients. What is the degrees of freedom for its overall test?
  3. A full model has log likelihood \(-301.4\), and its nested reduced model has log likelihood \(-304.0\). The full model adds two parameters. Calculate the LRT statistic and state the reference distribution.
  4. Why is a comparison of two models fit to different participant sets generally not the intended covariate LRT?

Solutions. (1) One degree of freedom; the null sets the added coefficient to zero. (2) Four degrees of freedom, because the overall null sets four independent coefficients to zero. (3) \(LR=2[-301.4-(-304.0)]=2(2.6)=5.2\), compared with \(\chi^2_2\) under the regularity conditions. (4) The likelihood difference can reflect both the covariate change and the change in observed data, so it does not isolate the gain from the added covariate in a common-data nested comparison.