Boundary Targets and One-Sided Profile Intervals

Why a Boundary Changes the Construction

A scalar target may be restricted to an interval such as \([a,\infty)\), or to a probability range such as \([0,1]\). When the maximum likelihood estimate is at an endpoint, a confidence interval cannot extend beyond that endpoint. The resulting confidence set may therefore be one-sided, for example \([a,U]\). Its endpoint is found from a profile likelihood, but its interpretation and likelihood-ratio calibration depend on whether the candidate target value is an interior point or a boundary point.

Let \(\theta\) be the scalar target and let \(\lambda\) denote nuisance parameters. Write \(\ell_p(\theta)=\sup_{\lambda}\ell(\theta,\lambda)\) for the profile log-likelihood, and let \(\hat\theta\) maximize it over the allowed parameter space. The profile likelihood-ratio statistic at a candidate value is

W(\theta)=2\{\ell_p(\hat\theta)-\ell_p(\theta)\}.

A one-sided upper confidence limit \(U\) is commonly obtained by solving \(W(U)=c\), with the confidence set on the side containing \(\hat\theta\). If \(\hat\theta=a\), the set has the form \([a,U]\), provided the profile likelihood falls away from its maximum without turning back to produce other accepted regions. The cutoff \(c\) is not automatically the usual two-sided cutoff.

Definition: A one-sided profile likelihood confidence set for a scalar target is formed by retaining candidate values whose profile likelihood-ratio statistic does not exceed a chosen cutoff \(c\), and reporting the relevant one-sided component within the parameter space. Its endpoint is the solution of \(W(\theta)=c\), unless the set reaches a parameter-space boundary first.

One-Sided Cutoffs and Boundary Calibration

For a regular scalar target at an interior candidate value, the signed root of the profile likelihood-ratio statistic is asymptotically standard normal. A one-sided confidence level \(1-\alpha\) therefore uses \(c=z_{1-\alpha}^2\), where \(z_{1-\alpha}\) is the \(1-\alpha\) standard normal quantile. At 95% confidence, \(z_{.95}\approx1.645\), so \(c\approx2.706\). In contrast, a two-sided 95% interval uses \(\chi^2_{1,.95}\approx3.841\).

There is a related, but distinct, boundary result. Suppose the null target is \(\theta=0\) and the parameter space locally requires \(\theta\geq0\). In a regular model with an identifiable scalar target, nuisance parameters in the interior, differentiable likelihood, and nonsingular efficient information, the likelihood-ratio statistic for testing \(\theta=0\) has a limiting mixture distribution. The nuisance parameters must be handled by profiling, and the local likelihood must admit the usual quadratic approximation. Under these conditions, the limit is one half point mass at zero and one half a \(\chi^2_1\) distribution.

Theorem: Under the stated regularity conditions for a single inequality constraint \(\theta\geq0\), the likelihood-ratio statistic for testing \(\theta=0\) against \(\theta\geq0\) converges in distribution to \(\tfrac12\delta_0+\tfrac12\chi^2_1\), where \(\delta_0\) is point mass at zero.

To see why, use the local quadratic likelihood after profiling out the nuisance parameters. After scaling by the efficient information, the local unconstrained estimate is asymptotically \(Z\sim N(0,1)\). Maximizing the quadratic likelihood subject to a nonnegative local target replaces \(Z\) by \(\max(0,Z)\). The likelihood-ratio statistic is consequently asymptotic to \(\{\max(0,Z)\}^2\). With probability \(1/2\), \(Z\leq0\) and this value is zero; with probability \(1/2\), \(Z>0\) and its conditional squared value has the \(\chi^2_1\) distribution. This gives the stated mixture.

For a level-\(\alpha\) test, the upper-tail probability of a positive cutoff \(c\) under this mixture is \(\tfrac12P(\chi^2_1>c)\). Setting that probability equal to \(\alpha\) gives the boundary cutoff:

c=\chi^2_{1,\,1-2\alpha}=z_{1-\alpha}^2,\qquad 0<\alpha<\tfrac12.

At \(\alpha=.05\), this is approximately \(2.706\), rather than the two-sided 95% cutoff \(3.841\). The numerical equality with the regular one-sided cutoff has two related interpretations: one is a one-sided normal tail, and the other is the upper tail of the boundary mixture. The mixture result is not universal; it requires the regularity conditions above.

Worked Profile Limits

Worked Example: A Constrained Normal Mean at the Boundary

Suppose \(Y_1,\ldots,Y_{100}\) are independent \(N(\theta,1)\) observations, with \(\theta\geq0\). The observed sample mean is \(\bar y=-0.20\). Find the 95% one-sided profile likelihood upper limit using \(c=2.706\).

The unconstrained maximum is \(\bar y=-0.20\), but it is outside the parameter space. The constrained maximum is therefore \(\hat\theta=0\). Apart from a constant, the log-likelihood is \(-100(\bar y-\theta)^2/2\). Thus

W(\theta)=100\{(\bar y-\theta)^2-(\bar y-0)^2\} =100\{(\theta+0.20)^2-0.20^2\}.

Set this equal to \(2.706\). The endpoint equation is \(100(\theta^2+0.40\theta)=2.706\), or \(\theta^2+0.40\theta-0.02706=0\). The positive root is

U=\frac{-0.40+\sqrt{0.40^2+4(0.02706)}}{2} \approx 0.059.

The resulting one-sided profile set is \([0,0.059]\). The lower endpoint is fixed by the parameter space, not obtained by solving another likelihood equation. The exact normal likelihood has been used for the profile curve; the choice of \(2.706\) gives the stated one-sided asymptotic calibration.

Worked Example: A Binomial Estimate at Zero

Let \(X\sim\operatorname{Binomial}(20,p)\), and suppose \(x=0\). The likelihood, omitting a factor that does not depend on \(p\), is \(L(p)=(1-p)^{20}\), for \(0\leq p\leq1\). Its maximum occurs at \(\hat p=0\). Find the upper profile endpoint using the 95% one-sided cutoff \(2.706\).

For a candidate \(p>0\), the profile likelihood-ratio statistic is

W(p)=2\{\ell(0)-\ell(p)\} =-40\log(1-p).

Solving \(W(p)=2.706\) gives \(-40\log(1-U)=2.706\), so \(1-U=\exp(-2.706/40)\). Therefore

U=1-\exp(-2.706/40)\approx 0.0654.

The likelihood-based set is \([0,0.0654]\) for this cutoff. For comparison, using the two-sided 95% cutoff \(3.841\) would give \(U=1-\exp(-3.841/40)\approx0.0916\). These computations show the effect of the cutoff on the endpoint; they do not establish exact finite-sample coverage.

In particular, do not apply the boundary-mixture theorem mechanically to \(p=0\) in this binomial model. If the true probability is zero, \(X=0\) with probability one, and the likelihood-ratio statistic comparing \(p=0\) with the unrestricted binomial model is exactly zero with probability one. It does not have the half-point-mass, half-\(\chi^2_1\) limit. The binomial endpoint is nonregular for that theorem, so a mixture cutoff cannot be justified there by the regular boundary argument.

Worked Example: An Interior Binomial Candidate

Suppose \(X\sim\operatorname{Binomial}(30,p)\) and \(x=6\). The maximum likelihood estimate is \(\hat p=6/30=0.20\). Find the one-sided 95% upper profile endpoint using \(c=2.706\).

For a candidate \(p>0.20\), the log-likelihood difference gives

W(p)=2\left\{6\log\left(\frac{0.20}{p}\right) +24\log\left(\frac{0.80}{1-p}\right)\right\}.

The upper endpoint solves \(W(p)=2.706\), or equivalently

0.20\log\left(\frac{0.20}{p}\right) +0.80\log\left(\frac{0.80}{1-p}\right) =\frac{2.706}{60}=0.04510.

The left side is zero at \(p=0.20\) and increases for \(p>0.20\): its derivative is \((p-0.20)/\{p(1-p)\}\), which is positive there. Thus there is a unique upper solution. Numerically it is \(U\approx0.3356\); substitution gives a left side of approximately \(0.0451\), and hence \(W(U)\approx60(0.0451)=2.706\). This is an interior-candidate, one-sided asymptotic profile limit, not an exact binomial bound.

Using the Boundary Correctly

A boundary may affect the reported interval in two ways: it can constrain the set of allowable target values, and it can alter the reference distribution for a test whose null value is itself on the boundary. Those are separate issues. A one-sided interval can have a fixed endpoint at the boundary even when the likelihood-ratio calibration for its other endpoint is justified at interior candidate values. Conversely, a boundary null test needs its own calibration, and its validity depends on the local likelihood structure.

The binomial calculation illustrates why this distinction matters. The formula for the likelihood profile is exact, but a likelihood-ratio cutoff is a calibration choice. In small samples, and especially when the estimate lies at a boundary, asymptotic profile limits can have coverage unlike their nominal level. If finite-sample guarantees are required, use a method with an appropriate finite-sample justification rather than interpreting a profile likelihood contour as automatically exact.

Common pitfall: Using the two-sided 95% cutoff \(3.841\) for every interval, or assuming every boundary null produces the half-\(\chi^2_1\) mixture. For a regular one-sided 95% likelihood-ratio limit the cutoff is \(2.706\); the mixture result also gives \(2.706\) for a regular scalar boundary test. But the binomial probability endpoint \(p=0\) is nonregular, so the mixture argument does not apply there.

Identify the allowed target space.
Record whether the target has a finite lower or upper boundary and whether the estimate lies at that boundary.

Form the profile likelihood.
For each candidate target, maximize over nuisance parameters and calculate \(W(\theta)=2\{\ell_p(\hat\theta)-\ell_p(\theta)\}\).

Choose a justified cutoff.
Use a one-sided cutoff for a one-sided regular limit; use a boundary-mixture cutoff only when its regularity assumptions hold.

Solve and respect the parameter space.
Find the relevant likelihood contour endpoint, report the interval only within the allowed space, and state when the calibration is asymptotic.

Check Your Understanding

Answer these questions by referring to both the profile curve and its calibration.

Why can a one-sided 95% likelihood-ratio limit use \(2.706\) rather than the two-sided cutoff \(3.841\)?
State the regularity conditions that support the half-point-mass, half-\(\chi^2_1\) boundary limit.
For the binomial example with \(n=20\) and \(x=0\), write the equation that determines the upper endpoint for a general cutoff \(c\).
Why does the likelihood-ratio statistic at \(p=0\) in that binomial model not have the regular boundary-mixture distribution?
What two different roles can a parameter-space boundary play in constructing and calibrating a one-sided profile interval?

Boundary Targets and One-Sided Profile Intervals

What You'll Learn