Why a Joint Variance Test Has a Mixture Reference Law
A likelihood-ratio test of several variance components is not an ordinary test of several unrestricted coefficients. Each variance must be nonnegative, and under the joint null all tested variances lie at the edge of their allowable parameter space. Consequently, the likelihood can improve in some directions away from the null but not in others. The asymptotic reference distribution reflects both possibilities: it is a mixture of chi-square distributions, called a chi-bar-square distribution.
The result is useful when the null hypothesis sets multiple variance components to zero, such as testing whether a GLMM needs any of several random-effect sources. The mixture is not automatically an ordinary chi-square distribution with degrees of freedom equal to the number of tested components. Its weights depend on the local geometry of the constraints and the information in the data.
For a joint test of \(q\) variance components, write their variances as \(\tau_1^2,\ldots,\tau_q^2\). The null hypothesis sets every tested component to zero; the alternative permits each to be nonnegative. Other model parameters are nuisance parameters. The likelihood-ratio statistic is \(T=2(\ell_1-\ell_0)\), where \(\ell_1\) and \(\ell_0\) are the maximized log likelihoods under the alternative and null, respectively.
These conditions matter. In particular, the result is an asymptotic approximation, not a finite-sample identity. The simple binomial weights require orthogonality of the efficient score coordinates for the tested parameters. Merely having \(q\) variances does not guarantee those weights. If the local constraint geometry or information structure differs, the chi-bar-square weights can differ as well. The general cone result explains why a mixture appears; calculating non-orthogonal weights is a separate problem.
How the Mixture Components Arise
The local likelihood near the null can be approximated by a quadratic function. After accounting for regular nuisance parameters and scaling the tested parameters at their local \(n^{-1/2}\) rate, the relevant part has the form \(2u^\mathsf{T}Z-u^\mathsf{T}u\). Here \(Z\) is an approximately standard multivariate normal score, and \(u\) is restricted to the local cone of permitted parameter directions. In the orthogonal case, that cone is the nonnegative orthant.
Under the null, the local score vector is approximately \(Z\sim N_q(0,I)\). For an allowed local direction \(u\), the quadratic approximation to twice the log-likelihood gain is \(2u^\mathsf{T}Z-u^\mathsf{T}u\).
Completing the square gives \(2u^\mathsf{T}Z-u^\mathsf{T}u=\lVert Z\rVert^2-\lVert Z-u\rVert^2\). Maximizing over the cone therefore gives the squared length of the projection of \(Z\) onto that cone.
For the nonnegative orthant, the projection has coordinates \(\max(Z_j,0)\). A negative score coordinate is projected to zero; a positive one contributes its squared value.
Each independent standard normal coordinate is positive with probability one half, independently of its squared magnitude. If exactly \(k\) coordinates are positive, the sum of their squares has a \(\chi^2_k\) distribution. Thus the mixture weight is the probability of \(k\) positive coordinates.
In the orthogonal case, the number of positive coordinates is binomial with parameters \(q\) and \(1/2\). This yields the stated weights. For other convex cones, projection can land on faces of different dimensions, and the probabilities of those events determine the mixture weights. This geometric description is the reason a joint boundary test may contain a point mass at zero as well as several positive-degree chi-square components.
Worked Example: Weights for Three Orthogonal Variance Components
Suppose a hypothetical GLMM is fitted to binary responses from 50 clinics, with 20 patients per clinic. The model includes three tested nonnegative variance components, measured in log-odds squared. For this illustration, assume their efficient score coordinates are orthogonal and the other nuisance parameters meet the regularity conditions. The fitted component estimates are \(0.12\), \(0.05\), and \(0.00\) log-odds squared. These estimates alone do not determine the test reference law; the stipulated orthogonality and the number of tested components do.
Here \(q=3\), so \(w_k=2^{-3}\binom{3}{k}\). The weights are
They sum to \(1/8+3/8+3/8+1/8=1\). The asymptotic reference law is therefore \(\frac18\chi^2_0+\frac38\chi^2_1+\frac38\chi^2_2+\frac18\chi^2_3\). The \(\chi^2_0\) part represents realizations in which all three local score directions point outside the permitted orthant, so the constrained fit gains nothing to first order over the null.
Calculating a Mixture-Based P-Value
For an observed statistic \(T>0\), the upper-tail p-value under a chi-bar-square reference law is the weighted sum of the component upper-tail probabilities:
When \(T>0\), the \(\chi^2_0\) component contributes zero to this upper tail, because it places all its probability at zero. At \(T=0\), the inclusive upper-tail probability is one. For positive observed statistics, each remaining component contributes its own tail probability multiplied by its mixture weight. This is not the same as selecting one degrees-of-freedom value from the mixture.
Worked Example: A Two-Component Test
In a hypothetical study of binary outcomes from 40 clinics, investigators test two nonnegative variance components. Assume orthogonal efficient score coordinates and otherwise suitable asymptotic conditions. The null-model log likelihood is \(\ell_0=-126.40\), and the alternative-model log likelihood is \(\ell_1=-124.15\). Both models use the same observations and fixed-effects structure.
The likelihood-ratio statistic is
For \(q=2\), the orthogonal-case weights are \(w_0=1/4\), \(w_1=1/2\), and \(w_2=1/4\). Using \(P(\chi^2_1\geq4.50)\approx0.0339\) and \(P(\chi^2_2\geq4.50)=e^{-4.50/2}\approx0.1054\), the mixture p-value is
At a significance level of \(0.05\), this asymptotic calculation provides evidence against the joint null that both variance components are zero. It does not give the probability that the null is true. For comparison, using \(\chi^2_2\) alone would give approximately \(0.1054\), while using \(\chi^2_1\) alone would give approximately \(0.0339\). Neither is the stipulated mixture reference law.
Worked Example: A Three-Component Tail Probability
Consider a second hypothetical binary-outcome GLMM with three tested variance components and an observed likelihood-ratio statistic \(T=5.60\). Assume again that the efficient score coordinates are orthogonal and the asymptotic conditions are plausible. The weights are \(1/8,3/8,3/8,1/8\), for degrees of freedom \(0,1,2,3\), respectively.
The component upper-tail probabilities at \(5.60\) are approximately \(0\) for \(\chi^2_0\), \(0.0180\) for \(\chi^2_1\), \(e^{-2.8}=0.0608\) for \(\chi^2_2\), and \(0.1328\) for \(\chi^2_3\). Therefore,
At \(\alpha=0.05\), the result is just below the significance threshold under these assumptions. The conclusion is sensitive to the reference distribution: for instance, the \(\chi^2_3\) tail alone is about \(0.1328\), not \(0.0462\). The mixture calculation incorporates the boundary constraints rather than treating all three variances as unrestricted parameters.
Assumptions, Diagnostics, and Limitations
Before using this reference law, check that the models being compared are nested, that the alternative permits the tested variances to be nonnegative, and that the null sets exactly those components to zero. Also check that the likelihood comparison is valid for the fitted models and that the regular nuisance parameters are handled consistently. The derivation assumes a locally quadratic likelihood and an appropriate large-sample approximation. Orthogonal-case binomial weights require the additional orthogonality condition; if that condition is not justified, those weights should not be assumed.
Numerical diagnostics matter because a likelihood-ratio statistic is only useful if the fits are reliable. Check convergence reports, whether the maximized log likelihoods are reproducible from sensible starting values, and whether estimated variance components or other parameters sit at their allowable boundaries. A singular or nearly singular fit, a very small number of independent clusters, or flat likelihood profiles can make the asymptotic approximation unreliable. A boundary estimate is not by itself proof that the test is invalid; it is a reason to check whether the model and approximation behave as assumed.
The chi-bar-square result does not supply a universal set of weights for every GLMM. In particular, one must not infer binomial weights merely by counting the tested components. Nor does an asymptotic mixture guarantee accurate finite-sample calibration. When the design is small or the likelihood is poorly behaved, conclusions should reflect that uncertainty and may require a suitable finite-sample assessment.
- Wrong: “There are three tested variances, so the test uses \(\chi^2_3\).” Correction: The nonnegative constraints put the null on a boundary; under orthogonal conditions the reference law is a mixture with degrees of freedom from zero through three.
- Wrong: “All tests of three variance components have weights \(1/8,3/8,3/8,1/8\).” Correction: Those binomial weights require orthogonal efficient score coordinates. The general mixture weights depend on the local constraint geometry and information.
- Wrong: “A p-value of \(0.043\) means there is a 4.3% probability that the joint null is true.” Correction: It is the probability, assuming the null and reference conditions, of obtaining a statistic at least as large as the observed one.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- For \(q=2\) orthogonal nonnegative variance components, what are the mixture weights, including the weight at zero?
- Why does a \(\chi^2_0\) component appear in a boundary-test reference distribution?
- With three orthogonal components, calculate the weight on \(\chi^2_2\).
- For an observed statistic \(T>0\), what contribution does the \(\chi^2_0\) component make to the upper-tail p-value?
- Is the binomial weight formula justified solely because a test includes \(q\) variance components? Explain.
Solutions. (1) The weights are \(1/4,1/2,1/4\) for degrees of freedom \(0,1,2\). (2) The constrained score can point entirely outside the permitted cone, leaving a projection of length zero and hence a likelihood-ratio limit of zero. (3) \(w_2=2^{-3}\binom{3}{2}=3/8\). (4) Zero, because the point mass is at zero and \(T\) is positive. (5) No; the binomial weights require orthogonal efficient score coordinates, as well as the regularity conditions for the asymptotic result.