Deriving A Boundary Mixture Distribution

Why a Boundary Changes the Likelihood-Ratio Reference Distribution

A likelihood-ratio test usually compares a null model with a larger alternative model. Under the regularity conditions for Wilks’ theorem, the test statistic has an asymptotic chi-square distribution, with degrees of freedom determined by the number of restrictions. A different calibration is needed when the null value lies at the edge of the allowed parameter space. The maximum-likelihood estimate cannot move past that edge, even when the data would favor doing so in an unconstrained model.

For a single parameter constrained to be nonnegative, testing whether it is zero gives a particularly useful result. In the regular one-parameter setting, the limiting likelihood-ratio distribution is a mixture: probability one-half is concentrated at zero, and probability one-half follows a chi-square distribution with one degree of freedom. The derivation explains both parts of this result and how to use it for testing.

Definition: Let $\theta\geq 0$ be one parameter, and consider $H_0:\theta=0$ against $H_1:\theta>0$. Write $\ell(\theta)$ for the log-likelihood and define the likelihood-ratio statistic as $$ T=2\left\{\sup_{\theta\geq 0}\ell(\theta)-\ell(0)\right\}. $$ Under suitable regularity conditions, when $H_0$ is true, $T$ converges in distribution to a mixture with probability one-half at zero and probability one-half from $\chi^2_1$.

The relevant conditions include a well-behaved likelihood near $\theta=0$, finite positive information for $\theta$, and a single constrained parameter whose null value is the boundary. Any other parameters must be identifiable and satisfy the regularity conditions needed for the local likelihood approximation. This result is not a general replacement for Wilks’ theorem in every boundary problem: different constraints, multiple boundary parameters, or failures of regularity can lead to other limits.

Deriving the Half-and-Half Mixture

Near the true parameter value, a regular log-likelihood can be approximated by a quadratic. Let $U$ be the score at $\theta=0$, and let $I>0$ denote the local information. The local form is

\ell(\theta)-\ell(0)\ \approx\ U\theta-\frac{1}{2}I\theta^2,\qquad \theta\geq 0.

The unconstrained maximum of this quadratic occurs at $U/I$. Enforcing $\theta\geq 0$ replaces a negative unconstrained maximum by the boundary value zero. Thus the constrained maximizer is $\widehat{\theta}=\max(0,U/I)$ in this approximation. Substituting it into the likelihood-ratio statistic gives

T\ \approx\ \frac{\{\max(0,U)\}^2}{I} =\{\max(0,Z)\}^2,\qquad Z=\frac{U}{\sqrt{I}}.

Under the null, the standardized score $Z$ converges in distribution to a standard normal random variable. A standard normal variable is positive with probability one-half and negative with probability one-half. If $Z\leq 0$, the constrained maximum is at $\theta=0$, so $T=0$. If $Z>0$, then $T=Z^2$. Squaring a standard normal variable gives a $\chi^2_1$ variable; conditioning on $Z>0$ does not change the distribution of its square, because the positive and negative halves are symmetric.

Consequently, half the limiting probability is an atom at zero, while the remaining half has the $\chi^2_1$ distribution. In mixture notation,

T\ \xrightarrow{d}\ \frac{1}{2}\delta_0+\frac{1}{2}\chi^2_1,

where $\delta_0$ denotes a point mass at zero. This is an asymptotic result in general. In some simple models, such as the known-variance normal-mean examples below, the same distribution holds exactly under the null.

Calculating p-Values and Critical Values

For an observed statistic $t>0$, the mixture’s upper-tail probability is half the upper-tail probability of $\chi^2_1$. Since a $\chi^2_1$ variable has the distribution of the square of a standard normal variable, the p-value can also be written using the standard normal cumulative distribution function $\Phi$:

p=\frac{1}{2}P(\chi^2_1\geq t) =1-\Phi(\sqrt{t}),\qquad t>0.

If $t=0$, the usual inclusive upper-tail p-value is $P(T\geq 0)=1$, because the entire distribution is at least zero. For a significance level $\alpha<0.5$, the upper-tail critical value $c_\alpha$ satisfies

\frac{1}{2}P(\chi^2_1\geq c_\alpha)=\alpha, \qquad c_\alpha=\chi^2_{1,\,1-2\alpha}.

Here $\chi^2_{1,\,1-2\alpha}$ is the $1-2\alpha$ quantile of the chi-square distribution with one degree of freedom. For a 5% test, this critical value is approximately $2.7055$. The ordinary 5% $\chi^2_1$ critical value, approximately $3.8415$, is not the correct cutoff for this one-parameter boundary case.

Observed statistic	Boundary-mixture upper-tail p-value	How to calculate it
$t=0$	1	Include the probability mass at zero in $P(T\geq t)$
$t>0$	$\frac{1}{2}P(\chi^2_1\geq t)$	Equivalently, $1-\Phi(\sqrt{t})$
5% critical value	Approximately 2.7055	Use the 90th percentile of $\chi^2_1$

Worked Example: Applying a Local Quadratic Likelihood

A hypothetical one-parameter model has $\theta\geq 0$, and testing is based on the local log-likelihood $\ell(\theta)-\ell(0)=2.4\theta-2\theta^2$, with $\ell(0)=0$. Use the boundary-mixture approximation to calculate the likelihood-ratio statistic and p-value. State the decision at the 5% level.

Method. Match the quadratic to $U\theta-\frac{1}{2}I\theta^2$. Here $U=2.4$, and $\frac{1}{2}I=2$, so $I=4$. The unconstrained maximizer is $U/I=2.4/4=0.6$, which satisfies the constraint $\theta\geq 0$. Therefore the constrained estimate is $\widehat{\theta}=0.6$.

The maximized log-likelihood difference is $2.4(0.6)-2(0.6)^2=1.44-0.72=0.72$. Thus $T=2(0.72)=1.44$. Equivalently, $T=U^2/I=2.4^2/4=5.76/4=1.44$.

For a positive statistic, the mixture p-value is $1-\Phi(\sqrt{T})$. Since $\sqrt{1.44}=1.2$, the p-value is $1-\Phi(1.2)\approx 1-0.8849=0.1151$. Because $0.1151>0.05$, do not reject $H_0$ at the 5% level.

Check and common mistake. The 5% boundary-mixture cutoff is about $2.7055$, and $1.44<2.7055$, agreeing with the p-value decision. Using the ordinary $\chi^2_1$ tail without halving it would give approximately $0.2301$, not the boundary-mixture p-value.

Worked Examples From a Normal Likelihood

The next examples use a hypothetical statistical model that makes the constraint and its effect especially clear. Suppose $X_1,\ldots,X_n$ are independent observations from $N(\theta,1)$, with the variance known to be 1 and $\theta\geq 0$. The hypotheses are $H_0:\theta=0$ and $H_1:\theta>0$. No interest or mortality assumptions are involved in this model.

Result: Under this normal model, the unconstrained maximum-likelihood estimate is $\bar{X}$, and the constrained estimate is $\widehat{\theta}=\max(0,\bar{X})$. The likelihood-ratio statistic is $T=n\{\max(0,\bar{X})\}^2$. Under $H_0$, $\sqrt{n}\bar{X}\sim N(0,1)$, so the half-at-zero, half-$\chi^2_1$ distribution is exact in this example.

Worked Example: A Negative Score Leaves the Estimate at the Boundary

Suppose three independent observations follow $N(\theta,1)$, where the variance is known to be 1 and $\theta\geq 0$. The observed values are $-1,-2,-1$. Test $H_0:\theta=0$ against $H_1:\theta>0$ using the likelihood-ratio statistic and the boundary-mixture calibration.

Method. Calculate the sample mean: $\bar{x}=(-1-2-1)/3=-4/3$. The unconstrained estimate is $-4/3$, but it is not allowed because $\theta\geq 0$. Hence the constrained estimate is $\widehat{\theta}=0$.

For this model, $T=3\{\max(0,\bar{x})\}^2$. Since $\max(0,-4/3)=0$, $T=3(0)^2=0$. The inclusive upper-tail p-value is $P(T\geq 0)=1$, so the data provide no evidence in favor of a positive value of $\theta$; do not reject the null.

Check and common mistake. The constrained estimate equals the null estimate, so the two maximized log-likelihoods are equal and $T=0$. Squaring the negative unconstrained estimate and reporting a positive statistic would ignore the parameter constraint and produce an invalid likelihood-ratio calculation.

Worked Example: A Positive Statistic Near the 5% Cutoff

A hypothetical set of four independent observations follows $N(\theta,1)$, with known variance 1 and $\theta\geq 0$. Each observed value is $0.9$. Test $H_0:\theta=0$ against $H_1:\theta>0$ at the 5% level, using the appropriate boundary distribution.

Method. The sample mean is $\bar{x}=(0.9+0.9+0.9+0.9)/4=0.9$, so the constrained estimate is $\widehat{\theta}=0.9$. The likelihood-ratio statistic is $T=4(0.9)^2=4(0.81)=3.24$.

The mixture p-value is $1-\Phi(\sqrt{3.24})=1-\Phi(1.8)$. Using $\Phi(1.8)\approx 0.9641$, the p-value is approximately $0.0359$. Since $0.0359<0.05$, reject $H_0$ at the 5% level.

Check and common mistake. The mixture critical value is approximately $2.7055$, and $3.24>2.7055$, which confirms the decision. The ordinary $\chi^2_1$ p-value is approximately $0.0719$, which would not reject at 5%. Using that ordinary calibration here misses the probability mass at zero and gives the wrong decision.

What the Mixture Result Does and Does Not Say

The atom at zero has a direct likelihood interpretation: in approximately half of repeated samples under the null, the data favor moving the parameter in the forbidden direction. The constraint prevents the estimate from moving, so the likelihood-ratio statistic is zero. In the other half, the score points into the allowed parameter space, and the statistic behaves like a one-degree-of-freedom chi-square variable.

The mixture does not mean that every boundary test uses this distribution. It applies to the stated one-parameter setting when the local likelihood is regular and the null is at a single boundary. Before using it, identify the parameter space, verify that the null value is its boundary, and check that the constrained parameter is the only source of nonstandard behavior. In particular, do not use this result automatically for a parameter that is not identified under the null or for several constrained parameters.

Key takeaway: For one regular parameter constrained by $\theta\geq 0$ and tested at $H_0:\theta=0$, a positive likelihood-ratio statistic $t$ has asymptotic p-value $1-\Phi(\sqrt{t})$. The 5% critical value is approximately $2.7055$, and an observed statistic of zero has inclusive upper-tail p-value 1.

Check Your Understanding

For each problem, use the one-parameter boundary mixture when its assumptions are stated. Answers use probabilities rounded to four decimal places.

A local log-likelihood has $\ell(\theta)-\ell(0)=-1.5\theta-\theta^2$, with $\theta\geq 0$. Find the constrained estimate and likelihood-ratio statistic. (Answer: $\widehat{\theta}=0,\ T=0$.)
A local log-likelihood has $\ell(\theta)-\ell(0)=3\theta-\frac{9}{2}\theta^2$, with $\theta\geq 0$. Find the constrained estimate, statistic, and boundary-mixture p-value. (Answer: $\widehat{\theta}=1/3,\ T=1,\ p=0.1587$.)
Five independent observations follow $N(\theta,4)$, with known variance 4 and $\theta\geq 0$. The observations are $0,0,0,1,1$. Calculate the likelihood-ratio statistic and p-value for $H_0:\theta=0$ against $H_1:\theta>0$. (Answer: $T=0.2000,\ p=0.3274$.)
A one-parameter boundary test has observed statistic $T=2.95$. Calculate the mixture p-value and decide at the 5% level. (Answer: $p\approx0.0430$; reject $H_0$ at 5%.)
A one-parameter boundary test has observed statistic $T=3.10$. Compare the boundary-mixture p-value with the ordinary $\chi^2_1$ p-value and state the 5% decision under the correct calibration. (Answer: mixture p-value $\approx0.0391$, ordinary $\chi^2_1$ p-value $\approx0.0782$; reject $H_0$ using the boundary mixture.)

Observed statistic	Boundary-mixture upper-tail p-value	How to calculate it
\(t=0\)	1	Include the probability mass at zero in \(P(T\geq t)\)
\(t>0\)	\(\frac{1}{2}P(\chi^2_1\geq t)\)	Equivalently, \(1-\Phi(\sqrt{t})\)
5% critical value	Approximately 2.7055	Use the 90th percentile of \(\chi^2_1\)

What You'll Learn