Disconnected Profile Confidence Sets

Why a Profile Confidence Set Need Not Be One Interval

A profile confidence set is often reported as an interval, but the likelihood-ratio definition does not require that shape. If the profile log-likelihood has two well-separated high regions with a sufficiently low region between them, a likelihood-ratio cutoff retains both high regions and excludes the middle. The result is a union of intervals.

This possibility matters in models with latent classes or mixture components. The likelihood can have multiple local maxima, and profiling nuisance parameters does not generally restore concavity. An interval that joins two accepted regions would include parameter values that fail the likelihood-ratio criterion.

Definition: Let $\psi$ be a scalar target, let $\lambda$ denote nuisance parameters, and write the profile log-likelihood as $\ell_p(\psi)=\sup_{\lambda}\ell(\psi,\lambda)$, with the supremum taken over the allowed nuisance-parameter values at that $\psi$. If $\ell_{\max}=\sup_{\psi}\ell_p(\psi)$ and $c\geq 0$ is a chosen cutoff, the profile likelihood-ratio confidence set is $$ C=\{\psi:2[\ell_{\max}-\ell_p(\psi)]\leq c\}. $$ Equivalently, it is the set where the profile likelihood is at least $\exp(-c/2)$ times its supremum.

The cutoff is part of the procedure, not a guarantee of coverage by itself. In regular settings, a chi-square approximation may motivate a choice of $c$. Mixture models can violate regularity conditions for that approximation, so a nominal chi-square cutoff should not automatically be interpreted as having its usual coverage.

Geometry of the Accepted Set

The definition makes $C$ a superlevel set of the profile log-likelihood: it contains all target values whose profile likelihood is close enough to the global maximum. A dip below the threshold can separate two accepted regions. Nonconcavity permits such a dip, but it does not force one: a nonconcave profile may still remain above the cutoff throughout the parameter range.

Result: Suppose $\ell_p$ is continuous on a real parameter space. If there are $a<b<d$ with $a,d\in C$ but $b\notin C$, then $C$ is disconnected. If the parameter space is compact, $C$ is also closed and compact, and each of its connected components is an interval (possibly a singleton).

Proof. If $C$ were connected, it would be an interval in the real line: every point between any two of its points would also belong to it. Since $a,d\in C$, that property would imply $b\in C$, contradicting the assumption. Thus $C$ is disconnected. For the second claim, continuity makes $2[\ell_{\max}-\ell_p(\psi)]$ continuous, so its sublevel set $C$ is closed relative to the parameter space. A closed subset of a compact space is compact. Finally, every connected subset of the real line is an interval. $\square$

The intermediate-value argument also provides a practical diagnostic: if two values pass the cutoff but a value between them fails it, the set cannot be represented by one interval. A plot can suggest this shape, but the cutoff inequality must be checked to establish the components and their endpoints.

Worked Example: A Profile With Two Accepted Bands

Consider a profile log-likelihood on $[-1,1]$ given by $\ell_p(t)=-(t^2-1/4)^2$. Its maximum is zero, attained at $t=-1/2$ and $t=1/2$. Use cutoff $c=0.08$ and calculate the confidence set.

The likelihood-ratio statistic at $t$ is $$ 2[\ell_{\max}-\ell_p(t)]=2(t^2-1/4)^2. $$

Acceptance requires $2(t^2-1/4)^2\leq 0.08$, or $|t^2-1/4|\leq\sqrt{0.04}=0.2$. Therefore $0.05\leq t^2\leq 0.45$. Taking square roots and respecting both signs gives $$ C=[-\sqrt{0.45},-\sqrt{0.05}]\ \cup\ [\sqrt{0.05},\sqrt{0.45}] \approx[-0.6708,-0.2236]\cup[0.2236,0.6708]. $$

The middle value $t=0$ has likelihood-ratio statistic $2(1/4)^2=0.125$, which exceeds $0.08$. Thus the accepted bands are separated. This example also shows that two equal maxima do not imply that every point between them is accepted.

A Finite-Mixture-Style Likelihood

For a concrete calculation, suppose each of $n=100$ independent subjects has a binary response. Each subject belongs to one of two latent response components, with mixing probabilities $1/2$ and $1/2$. Conditional on component membership, the success probabilities are specified by $q_1(t)=t$ and $q_2(t)=(1-t)^2$, where the scalar target $t$ lies in $[0,1]$. This is a deliberately constrained mixture response model: its component probabilities are functions of the same target, rather than an ordinary single linear-predictor GLM. Its likelihood illustrates the profile-set geometry that can arise in mixture-style fitting.

The marginal success probability is the mixture average $p(t)=\tfrac12q_1(t)+\tfrac12q_2(t)=\tfrac12(t+(1-t)^2)=\tfrac12(1-t+t^2)$. If $Y=50$ successes are observed, the likelihood, omitting the binomial coefficient that does not depend on $t$, is $L(t)=p(t)^{50}[1-p(t)]^{50}$. There are no free nuisance parameters here, so the profile likelihood is this likelihood itself.

Worked Example: Both Components of a Mixture Profile Set

Use the 95% one-degree-of-freedom reference cutoff $c=3.841$ as a chosen likelihood-ratio threshold, and calculate all accepted values of $t$.

Write $u=t(1-t)$. Since $1-t+t^2=1-u$, we have $p(t)=(1-u)/2$ and $1-p(t)=(1+u)/2$. Also $0\leq u\leq 1/4$. The maximum success probability $1/2$, and hence the maximum likelihood, occurs at $t=0$ and $t=1$. The profile likelihood-ratio statistic is $$ 2[\ell_{\max}-\ell_p(t)] =100\log\left(\frac{(1/2)^2}{p(t)[1-p(t)]}\right) =-100\log(1-u^2). $$

The cutoff condition is $-100\log(1-u^2)\leq 3.841$. Exponentiating gives $1-u^2\geq\exp(-3.841/100)$, so $$ u\leq d=\sqrt{1-\exp(-0.03841)}\approx 0.1941. $$

Now solve $t(1-t)\leq d$. The equality $t(1-t)=d$ has roots $$ t=\frac{1\pm\sqrt{1-4d}}{2} \approx \frac{1\pm 0.4728}{2}, $$ which are approximately $0.2636$ and $0.7364$. Because $t(1-t)$ is above $d$ between these roots and below $d$ outside them, the full confidence set is $$ C=[0,0.2636]\ \cup\ [0.7364,1]. $$

As a check, at $t=1/2$ we have $u=1/4$, giving statistic $-100\log(15/16)\approx 6.454$, greater than $3.841$. The center is excluded, while both endpoints have statistic zero. The set therefore has exactly two components. The stated cutoff is used here to show the calculation; the mixture structure means its nominal chi-square interpretation requires separate justification.

The calculation rests on the profile likelihood itself, not on a picture of two peaks. Each component is found by solving the cutoff inequality over the entire parameter space. In a model with additional nuisance parameters, one first obtains $\ell_p(t)$ by constrained maximization at each fixed $t$, then applies the same inequality to that profile.

Checking a Polynomial Endpoint Equation

Numerical profile calculations can reduce a cutoff crossing to a polynomial equation. Root checks need accurate endpoint evaluations and a monotonicity argument when uniqueness is claimed. The next example illustrates both steps; it concerns a crossing equation on one branch, not a substitute for checking the likelihood-ratio inequality on every branch.

Worked Example: Verifying a Cubic Crossing

Suppose an endpoint equation on $0\leq t\leq 1/4$ is $h(t)=0$, where $h(t)=1-9t+18t^2-5t^3$. Determine whether there is exactly one root in that range and bracket it.

Differentiate: $h'(t)=-9+36t-15t^2$. Its derivative is $36-30t>0$ throughout $[0,1/4]$, so $h'$ increases there. Its largest value on this range is $h'(1/4)=-9+9-15/16=-15/16<0$. Thus $h$ is strictly decreasing. At the endpoints, $$ h(0)=1,\qquad h(1/4)=1-\frac94+\frac{18}{16}-\frac{5}{64} =\frac{64-144+72-5}{64} =-\frac{13}{64}<0. $$

Continuity and the sign change imply at least one root, and strict decrease implies at most one. For a numerical bracket, $h(0.1600)=0.00032>0$, while direct substitution gives $h(0.1601)\approx-0.0000422<0$. The unique root is therefore between $0.1600$ and $0.1601$, approximately $0.16009$. The endpoint value is $-13/64$; a different value can lead to incorrect root checks even if the qualitative conclusion happens to survive.

Common Mistakes and Interpretation

Reporting the span between the extreme endpoints. That span includes rejected values when a middle region fails the cutoff. Report each connected component instead.
Assuming nonconcavity guarantees disconnection. It only makes multiple components possible. The likelihood-ratio curve must actually fall below the cutoff between accepted values.
Stopping after finding one local maximum. A local optimizer can miss another high region. Profile calculations must examine the full allowed target range and all relevant constrained maxima.
Treating the cutoff as automatically calibrated. A chi-square reference cutoff may be inappropriate under mixture nonregularity. The geometry calculation and the coverage calibration are distinct questions.

A disconnected set is not a computational defect. It communicates that the observed likelihood supports separated target values while disfavoring values between them, often because different latent-component configurations fit well. Joining the components for presentation may conceal precisely that information. Conversely, a disconnected profile set should prompt checks that each constrained optimization and each cutoff crossing has been found, rather than being accepted uncritically as a numerical artifact.

Check Your Understanding

Use the likelihood-ratio definition and check the full parameter range in each problem.

What feature of the profile log-likelihood can cause a likelihood-ratio confidence set to have more than one component?
Can a nonconcave profile log-likelihood still produce a connected confidence set? Explain.
In the mixture example, why is $t=1/2$ excluded at the stated cutoff?
For a continuous profile on a real parameter space, why do two accepted values separated by a rejected value imply disconnection?
Why does calculating the components not, by itself, establish that a chosen chi-square cutoff has its nominal coverage in a mixture model?