Confidence Sets for Variance Components at Zero

Why Confidence Sets Need to Respect the Boundary

A variance component in a generalized linear mixed model (GLMM) cannot be negative. Yet an unconstrained, symmetric estimate-plus-or-minus-standard-error interval can extend below zero, and truncating that interval at zero does not correct its calibration. The boundary also changes the reference distribution for a likelihood-ratio test of the null value zero. A confidence set for a variance component should account for both facts: it must lie in the nonnegative parameter space, and its construction should use the appropriate test calibration at each candidate value.

The method here inverts likelihood-ratio tests. This connects confidence-set construction to the earlier material on profile likelihood, boundary tests, and interpreting p-values from mixture distributions. The key practical detail is that the test for a candidate value of zero has a different asymptotic cutoff from the test for a positive candidate value.

Definition: Let $\theta\geq 0$ be a GLMM variance component, such as the variance of a random intercept, and let $\ell_p(\theta)$ be the profile log likelihood: the log likelihood at $\theta$, maximized over the nuisance parameters. Let $\hat\theta$ be the constrained maximum likelihood estimate over $\theta\geq0$. The profile likelihood-ratio statistic for a candidate value $\theta_0$ is $$ D(\theta_0)=2\{\ell_p(\hat\theta)-\ell_p(\theta_0)\}. $$ A likelihood-ratio confidence set contains the candidate values whose tests are not rejected at the chosen significance level. In particular, its values are restricted to $\theta_0\geq0$.

Which Cutoff Applies?

For a positive candidate value $\theta_0>0$, the null value is in the interior of the parameter space. Under the usual regularity conditions, the likelihood-ratio statistic has an asymptotic chi-square distribution with one degree of freedom under $H_0:\theta=\theta_0$. Thus, a level-$\alpha$ test rejects when $D(\theta_0)$ exceeds the $1-\alpha$ quantile of $\chi^2_1$.

At $\theta_0=0$, the null is on the boundary. For a single variance component, under the conditions that give the half-at-zero, half-chi-square reference law, the statistic has asymptotic distribution $0.5\delta_0+0.5\chi^2_1$. Here $\delta_0$ denotes a point mass at zero. For a positive cutoff $c$, the upper-tail probability under this mixture is half the upper-tail probability of $\chi^2_1$. Consequently, its level-$\alpha$ cutoff is the $1-2\alpha$ quantile of $\chi^2_1$, when $\alpha<0.5$.

P(D(0)>c)=\tfrac12 P(\chi^2_1>c)=\alpha \quad\Longrightarrow\quad c=F^{-1}_{\chi^2_1}(1-2\alpha).

At the conventional $\alpha=0.05$ level, the positive-candidate cutoff is $F^{-1}_{\chi^2_1}(0.95)\approx3.841$, while the zero-candidate cutoff is $F^{-1}_{\chi^2_1}(0.90)\approx2.706$. These are not interchangeable. The smaller boundary cutoff follows from the point mass at zero: for a given positive $c$, the mixture assigns only half as much upper-tail probability as $\chi^2_1$.

Definition: For the single-component setting above, the asymptotic level-$1-\alpha$ likelihood-ratio confidence set is $$ C_{1-\alpha} = \{\theta_0>0:D(\theta_0)\leq F^{-1}_{\chi^2_1}(1-\alpha)\} \;\cup\; \begin{cases} \{0\},&D(0)\leq F^{-1}_{\chi^2_1}(1-2\alpha),\\ \varnothing,&D(0)>F^{-1}_{\chi^2_1}(1-2\alpha). \end{cases} $$ The boundary value is tested separately using its boundary reference law.

Why Inverting Tests Produces a Confidence Set

For each fixed candidate $\theta_0$, consider a level-$\alpha$ test of $H_0:\theta=\theta_0$. Its acceptance rule is $D(\theta_0)\leq c(\theta_0)$, where the cutoff is chosen for that candidate's null distribution. The confidence set contains exactly the candidates accepted by these tests. If the true value is $\theta$, the event that the set covers it is therefore the event that the test of $H_0:\theta=\theta$ does not reject. Under the assumed calibration, that event has probability approximately $1-\alpha$.

The word “approximately” matters: the chi-square and mixture reference laws are asymptotic. They rely on suitable identifiability and smoothness conditions and on enough independent information, often supplied by a sufficiently large number of clusters. If these approximations are poor, the resulting confidence set may not attain its nominal coverage in a finite sample.

Fit the constrained model.
Estimate the variance component subject to $\theta\geq0$, and profile the likelihood over nuisance parameters.

Evaluate candidate values.
For each $\theta_0\geq0$, calculate $D(\theta_0)=2\{\ell_p(\hat\theta)-\ell_p(\theta_0)\}$.

Apply the candidate-specific cutoff.
Use the mixture cutoff for $\theta_0=0$, and the one-degree-of-freedom chi-square cutoff for $\theta_0>0$, under the stated assumptions.

Report the accepted values.
Keep only nonnegative candidates that are not rejected, and describe the result as an approximate confidence set.

Worked Calculations

Worked Example: A Set That Includes Zero

Suppose a hypothetical GLMM has a single random-intercept variance component $\theta$, and a local quadratic approximation to its profile log likelihood is $$ \ell_p(\theta)=K-\frac{(\theta-0.040)^2}{2(0.040)^2},\qquad \theta\geq0, $$ where $K$ is a constant. The constrained estimate is $\hat\theta=0.040$. Construct the asymptotic 95% confidence set using the candidate-specific likelihood-ratio cutoffs.

Likelihood-ratio statistic. Subtracting the log likelihood at a candidate from its maximum and multiplying by two gives $$ D(\theta_0)=\frac{(\theta_0-0.040)^2}{0.0016}. $$ At the boundary, $D(0)=(0.040)^2/0.0016=1.000$. Since $1.000<2.706$, zero is included by the boundary test.

Positive candidates. For $\theta_0>0$, retain values satisfying $D(\theta_0)\leq3.841$: $$ \frac{(\theta_0-0.040)^2}{0.0016}\leq3.841 \quad\Longrightarrow\quad |\theta_0-0.040|\leq\sqrt{3.841(0.0016)}\approx0.07839. $$ The corresponding untruncated endpoints are $0.040-0.07839=-0.03839$ and $0.040+0.07839=0.11839$. Restricting to positive candidates gives values above zero through $0.11839$; the separate boundary calculation confirms that zero itself is included. Thus, the set is approximately $[0,0.1184]$ variance units.

Interpretation. Under the model and asymptotic calibration used here, the procedure produces a 95% confidence set for the random-intercept variance component from zero to about $0.1184$ variance units. This does not mean there is a 95% probability that the fixed, unknown variance lies in this particular set.

Worked Example: A Set That Excludes Zero but Reaches Close to It

For another hypothetical model, suppose the profile likelihood gives $$ D(\theta_0)=\frac{(\theta_0-0.056)^2}{0.001},\qquad \theta_0\geq0, $$ with $\hat\theta=0.056$. Find the 95% likelihood-ratio confidence set using the same asymptotic rules.

At zero, $D(0)=(0.056)^2/0.001=3.136$. This exceeds the boundary cutoff $2.706$, so zero is not included. For positive candidates, use the cutoff $3.841$: $$ |\theta_0-0.056|\leq\sqrt{3.841(0.001)} =\sqrt{0.003841}\approx0.06198. $$ The untruncated endpoints are $0.056-0.06198=-0.00598$ and $0.056+0.06198=0.11798$. Among positive candidates, all values greater than zero and no greater than $0.11798$ are accepted. The resulting set is approximately $(0,0.1180]$ variance units: it excludes zero but contains arbitrarily small positive values.

This open endpoint can arise because the candidate zero and positive candidates use different asymptotic calibrations. It is not an arithmetic error. It also signals a limitation: asymptotic tests calibrated separately at the boundary and in the interior need not behave uniformly for candidate values very close to zero. In small samples, it is prudent to examine calibration rather than assume the nominal coverage is exact.

Worked Example: Reporting the Set on the Standard-Deviation Scale

In the first example, the variance-scale confidence set was approximately $[0,0.11839]$. The corresponding random-intercept standard deviation is $\sigma_b=\sqrt{\theta}$, which is a nondecreasing transformation for $\theta\geq0$. Transform both endpoints: $$ [\sqrt{0},\sqrt{0.11839}]=[0,0.3441]. $$

Thus, the corresponding 95% confidence set for the random-intercept standard deviation is approximately $[0,0.3441]$ on the scale of the linear predictor. The endpoint units are those of the random intercept, whereas the variance endpoints have squared units. The transformation preserves the set because taking square roots maps every nonnegative variance value in the original set to exactly one standard-deviation value. It would be incorrect to take a standard error for the variance estimate and simply add and subtract it on the standard-deviation scale.

Assumptions, Diagnostics, and Alternatives

The simple half-at-zero calibration is appropriate only when the null concerns one variance component at zero and the relevant asymptotic conditions support that reference law. In particular, the component must be identifiable under the alternative, nuisance parameters must behave regularly enough for the result, and the model must be correctly specified for the likelihood calculation. More complicated boundary structures can have other mixture distributions; a one-component cutoff should not be transferred automatically to a test involving several variance components or additional singular covariance parameters.

Inspect the profile likelihood over the full feasible range, not only at the maximum. A flat or irregular profile can indicate weak information, numerical instability, or poor identification; multiple local maxima can make endpoint searches unreliable. Check that the optimizer respects $\theta\geq0$, that the profile is maximized over nuisance parameters at each candidate, and that the reported endpoints satisfy the appropriate likelihood-ratio inequalities. With few clusters or substantial irregularity, a parametric bootstrap or other finite-sample calibration may be preferable to relying on the asymptotic cutoffs.

Approach	Strength	Limitation for a variance component
Profile likelihood-ratio set	Uses the likelihood shape and can be asymmetric.	Requires suitable calibration and reliable profiling; boundary and interior candidates use different reference laws.
Wald interval	Quick to calculate from an estimate and standard error.	A symmetric interval can extend below zero and may be unreliable near the boundary.
Wald interval truncated at zero	Enforces nonnegativity in the reported endpoints.	Truncation alone does not repair the Wald interval's sampling calibration.
Bootstrap-calibrated set	Can better reflect finite-sample behavior when simulation is appropriate.	Requires a valid simulation scheme and adequate computation; bootstrap calibration is not automatic protection against model misspecification.

The profile likelihood-ratio set is not the only possible confidence procedure. Its advantage over a truncated Wald interval is that it is built by inverting tests with a reference distribution suited to the candidate null, rather than by clipping a symmetric approximation. Its limitations are equally important: asymptotic calibration can be inaccurate, and the resulting set can be irregular. When reporting results, state the variance-component definition, the confidence level, the calibration used at zero and at positive values, and whether the set includes the boundary.

Common Mistake

A frequent error is to use the ordinary $\chi^2_1$ 95% cutoff, $3.841$, for every candidate, including zero. The tempting reasoning is that the likelihood-ratio statistic always has a chi-square reference distribution. At a zero variance component, however, the null is on the boundary, and under the half-at-zero mixture law the 5% cutoff is $2.706$, not $3.841$. Another error is to calculate a Wald interval and truncate its negative endpoint to zero, then call the result a boundary-corrected interval. Nonnegativity of the endpoint does not establish correct coverage. Use the boundary test calibration for zero, the interior calibration for positive candidates under its assumptions, and report the resulting set without implying more precision than the model supports.

Key takeaway: Construct a confidence set by inverting likelihood-ratio tests over $\theta\geq0$. Under the single-component asymptotic setting, test zero with the half-at-zero mixture cutoff and positive candidates with the regular $\chi^2_1$ cutoff. Then report whether the boundary is included and assess the calibration assumptions.

Check Your Understanding

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

At a 5% significance level, what likelihood-ratio cutoff applies to a test of a single variance component at zero under the half-at-zero mixture law?
A candidate positive variance value has $D(\theta_0)=4.10$. Is it retained in the asymptotic 95% set under the regular $\chi^2_1$ calibration?
A profile likelihood gives $D(0)=2.50$. Is zero included in the asymptotic 95% set under the single-component boundary calibration?
A variance confidence set is $[0,0.09]$ in squared units. What is the corresponding set for its standard deviation, and what units should be used?

Solutions. (1) The cutoff is $F^{-1}_{\chi^2_1}(0.90)\approx2.706$, since half the $\chi^2_1$ upper tail must equal 0.05. (2) No. For a positive candidate, compare with 3.841; $4.10>3.841$, so it is rejected and excluded. (3) Yes. Since $2.50<2.706$, the boundary candidate is not rejected and zero is included. (4) Taking square roots of both endpoints gives $[0,\sqrt{0.09}]=[0,0.30]$. The transformed set is in the original standard-deviation units, not squared units.