Why the Information Matrix Changes the Weights
When two variance components are tested jointly at zero, the alternative is not an unrestricted plane: both variances must remain nonnegative. The null therefore lies at the corner of the admissible parameter region. The likelihood-ratio statistic can have a chi-bar-square reference distribution, a weighted mixture of chi-square distributions with different degrees of freedom. For two components, the weights depend on the angle of the admissible region after the information matrix has accounted for the relative precision and dependence of the component estimates.
The word correlated in this setting refers to the geometry induced by the information matrix for the two variance parameters. It does not mean that the random effects themselves have a freely estimated correlation. The test considered here has two nonnegative variance parameters and a local admissible region that is a quadrant.
Assumptions and the Relevant Information
The two-component formula applies under a local asymptotic regime in which the likelihood has a quadratic approximation around the null, the nuisance parameters are identifiable and interior under \(H_0\), and the efficient information for \(\tau\) is finite and positive definite. The local admissible set for \(\tau\) must be the nonnegative quadrant. These conditions are important: if a nuisance parameter is itself unidentified under the null, if the information is singular, or if the parameter constraints make a different local region, the formula below need not apply.
Write the expected information at the null in blocks for \(\tau\) and \(\gamma\). Profiling out the regular nuisance parameters leaves the efficient information matrix \(I_{\mathrm{eff}}=I_{\tau\tau}-I_{\tau\gamma}I_{\gamma\gamma}^{-1}I_{\gamma\tau}\). This Schur complement, rather than the unadjusted \(I_{\tau\tau}\), determines the local geometry. If there are no nuisance parameters, \(I_{\mathrm{eff}}=I_{\tau\tau}\).
A quadratic likelihood approximation measures local displacement using the information metric. Whitening that metric maps the nonnegative quadrant to a wedge in ordinary Euclidean space. Let \(I_{\mathrm{eff}}^{1/2}\) be a symmetric positive-definite square root. The two boundary rays of the transformed wedge are \(I_{\mathrm{eff}}^{1/2}(1,0)^\mathsf{T}\) and \(I_{\mathrm{eff}}^{1/2}(0,1)^\mathsf{T}\). Their angle \(\theta\) satisfies \(\cos(\theta)=I_{\mathrm{eff},12}/\sqrt{I_{\mathrm{eff},11}I_{\mathrm{eff},22}}\).
Geometric Derivation of the Weights
Under the quadratic approximation, the likelihood-ratio statistic is the squared length of the projection of a standard bivariate normal vector onto the whitened admissible wedge. The projection is the point in the wedge nearest to that normal vector. Its possible locations are the wedge interior, either boundary ray, or the vertex. The chi-bar-square degree reflects the dimension of the face on which the projection falls: two in the interior, one on a ray, and zero at the vertex.
For an isotropic standard normal vector, its direction is uniform around the circle. The probability that its direction falls inside a wedge of angle \(\theta\) is \(\theta/(2\pi)\); this is the weight on the two-dimensional component. The polar wedge, consisting of directions whose projection is the vertex, has angle \(\pi-\theta\), so its probability is \((\pi-\theta)/(2\pi)\); this is the weight on the point mass at zero. The remaining probability is \(1-\theta/(2\pi)-(\pi-\theta)/(2\pi)=1/2\), the weight on the one-dimensional component. The conic projection result for a Gaussian vector gives the corresponding chi-square component for each projection dimension. Thus the angular probabilities yield all three mixture weights.
The weights sum to one: \((\pi-\theta)/(2\pi)+1/2+\theta/(2\pi)=1\). Positive definiteness ensures \(0<\theta<\pi\), so both endpoint weights are nonnegative. This derivation also explains why simply counting the two variance parameters does not determine the reference law: the angle, and hence the chance of projection onto the vertex or interior, depends on the information geometry.
Worked Example: Adjusting for a Regular Nuisance Parameter
Suppose hypothetical expected-information blocks at the null are \(I_{\tau\tau}=\begin{pmatrix}5&2\\2&6\end{pmatrix}\), \(I_{\tau\gamma}=\begin{pmatrix}1\\2\end{pmatrix}\), and \(I_{\gamma\gamma}=2\), with one regular nuisance parameter. Find the mixture weights.
First remove the nuisance contribution: \[ I_{\mathrm{eff}}= \begin{pmatrix}5&2\\2&6\end{pmatrix} - \begin{pmatrix}1\\2\end{pmatrix} (2^{-1}) \begin{pmatrix}1&2\end{pmatrix} = \begin{pmatrix}5&2\\2&6\end{pmatrix} - \begin{pmatrix}0.5&1\\1&2\end{pmatrix} = \begin{pmatrix}4.5&1\\1&4\end{pmatrix}. \] Its determinant is \(4.5(4)-1^2=17>0\), and its diagonal entries are positive, so it is positive definite.
The information correlation is \(\rho_I=1/\sqrt{4.5(4)}=1/\sqrt{18}\approx0.2357\). Therefore \(\theta=\arccos(0.2357)\approx1.333\) radians. The weights are \(w_2=1.333/(2\pi)\approx0.2121\), \(w_0=(\pi-1.333)/(2\pi)\approx0.2879\), and \(w_1=0.5000\). They sum to \(1.0000\), as required. Using \(I_{\tau\tau}\) alone would ignore the nuisance adjustment and give a different angle.
Worked Example: Calculating the Weights from a Correlated Information Matrix
Consider a hypothetical efficient information matrix \(I_{\mathrm{eff}}=\begin{pmatrix}2.75&0.75\\0.75&1.75\end{pmatrix}\). Assume the model satisfies the regularity and quadrant conditions stated above. Calculate the angle and the weights.
The determinant is \(2.75(1.75)-0.75^2=4.8125-0.5625=4.25>0\), and both diagonal entries are positive, so the matrix is positive definite. Its information correlation is \(\rho_I=0.75/\sqrt{2.75(1.75)}=0.75/\sqrt{4.8125}\approx0.3419\). Thus \(\theta=\arccos(0.3419)\approx1.2219\) radians.
Consequently, \(w_2=1.2219/(2\pi)\approx0.1945\), \(w_0=(\pi-1.2219)/(2\pi)\approx0.3055\), and \(w_1=0.5000\). The asymptotic reference law is therefore approximately \(0.3055\chi_0^2+0.5000\chi_1^2+0.1945\chi_2^2\). The positive off-diagonal information makes the wedge angle smaller than a right angle, shifting weight from the two-dimensional component toward the point mass at zero.
Worked Example: Negative Information Association
For another hypothetical test, let \(I_{\mathrm{eff}}=\begin{pmatrix}1&-0.6\\-0.6&1\end{pmatrix}\). Its determinant is \(1-0.36=0.64>0\), with positive diagonal entries, so it is positive definite. Find the weights.
Here \(\rho_I=-0.6\), so \(\theta=\arccos(-0.6)\approx2.2143\) radians, an angle greater than \(\pi/2\). Therefore \(w_2=2.2143/(2\pi)\approx0.3524\), \(w_0=(\pi-2.2143)/(2\pi)\approx0.1476\), and \(w_1=0.5000\). The weights sum to one. Compared with an orthogonal wedge, the wider transformed wedge gives more weight to the two-dimensional component and less to the point mass.
Interpreting the Geometry and Checking the Calculation
If the two efficient score directions are orthogonal, then \(I_{\mathrm{eff},12}=0\), \(\theta=\pi/2\), and the weights are \(1/4,1/2,1/4\) on \(\chi_0^2,\chi_1^2,\chi_2^2\), respectively. This is a useful special case, not a default for every two-component test. With positive information correlation, \(0<\theta<\pi/2\); with negative information correlation, \(\pi/2<\theta<\pi\). The endpoint weights shift accordingly, while \(w_1=1/2\) remains fixed for this two-dimensional quadrant problem.
The correlation in the angle formula is dimensionless. If either variance parameter is rescaled by a positive constant, the corresponding row and column of the information matrix are rescaled, but the ratio \(I_{\mathrm{eff},12}/\sqrt{I_{\mathrm{eff},11}I_{\mathrm{eff},22}}\) does not change. More generally, a valid reparameterization must carry along the transformed parameter region as well as the information matrix; changing coordinates while retaining the old quadrant can give the wrong geometry.
In practice, check that the nuisance parameters are regular under the null, that the local alternative really is a quadrant, and that \(I_{\mathrm{eff}}\) is positive definite. A computed information correlation outside \([-1,1]\), a nonpositive determinant, or a nonfinite inverse for \(I_{\gamma\gamma}\) signals a calculation or identification problem. If the smallest eigenvalue is close to zero, the angle and weights may be numerically unstable; report that limitation rather than treating a rounded matrix as reliable.
Common Mistakes and Limitations
- Using equal endpoint weights automatically. The weights \(1/4,1/2,1/4\) require a right-angle wedge, equivalently zero off-diagonal efficient information. Calculate the angle from \(I_{\mathrm{eff}}\) before using them.
- Using the raw information when nuisance parameters are present. The geometry for the profiled likelihood uses the Schur complement. Ignoring it generally changes the angle.
- Confusing information correlation with random-effect correlation. The off-diagonal entry here describes local information coupling between variance-component estimates; it is not itself a covariance parameter of the random effects.
- Applying the result to a different constraint geometry. A test with one component already positive, a non-quadrant admissible set, or additional boundary constraints has a different local cone and may have different weights.
The formula is an asymptotic result. It does not guarantee accurate finite-sample calibration, particularly when there are few independent clusters or the information is nearly singular. The chi-bar-square weights describe the limiting reference distribution under the stated assumptions; they do not resolve a failure of those assumptions.
Check Your Understanding
Use the stated quadrant assumptions unless a question asks you to identify a failure.
- If the efficient information matrix is diagonal and positive definite, what are the three mixture weights?
- For \(I_{\mathrm{eff}}=\begin{pmatrix}1&0.8\\0.8&1\end{pmatrix}\), determine the information correlation and whether the two-dimensional weight is above or below \(1/4\).
- Why must the Schur complement be used when regular nuisance parameters are estimated?
- A computed efficient information matrix has determinant zero. Can the two-component angle formula be applied as stated?
Solutions. (1) A diagonal matrix has \(I_{\mathrm{eff},12}=0\), so \(\theta=\pi/2\). Thus \(w_0=1/4\), \(w_1=1/2\), and \(w_2=1/4\). (2) The information correlation is \(0.8/\sqrt{1(1)}=0.8\). Since \(\arccos(0.8)<\pi/2\), \(w_2=\theta/(2\pi)<1/4\). (3) Profiling the nuisance parameters changes the curvature of the likelihood in the variance-component directions; the Schur complement is the efficient information for those directions after that adjustment. (4) No. A zero determinant means the matrix is not positive definite, so the whitening and nondegenerate wedge assumptions fail. Investigate identification or numerical problems before selecting a reference law.