advanced32 min studyLesson 708 in Mathematics

Likelihood-ratio asymptotics and boundaries

Check the null geometry before assigning chi-square degrees of freedom.

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What you’ll understand

  • State regular interior conditions behind Wilks-type chi-square limits.
  • Derive a boundary likelihood-ratio mixture in a normal model.
  • Calculate a one-sided critical value and handle the atom at zero.

A likelihood-ratio statistic compares the best fit under a null model with the best fit in a larger model. Its familiar chi-square approximation relies on local geometric and probabilistic assumptions, not simply a count of free parameters. At a boundary, the likelihood cannot move in every direction, so the limiting optimization changes. A constrained normal mean gives an exact example: the statistic has a point mass at zero and a chi-square component. Deriving that mixture clarifies both critical values and the p-value convention at the atom.

The regular interior theorem has a geometric hypothesis

Consider a correctly specified, fixed-dimensional, dominated model at an identifiable interior point with nonsingular Fisher information. Assume the smoothness, local uniform quadratic likelihood expansion, and consistent full and restricted maximizers needed for regular likelihood asymptotics. If the null is locally a smooth embedded submanifold of codimension rr, the twice-log-likelihood difference converges to chi-square with rr degrees of freedom. These hypotheses exclude a boundary point, unidentified nuisance parameters, singular information, and parameter dimensions growing with sample size.

In information-normalized coordinates, the quadratic likelihood is optimized over a linear tangent space for the null and the full Euclidean space for the alternative. Their difference is a squared Gaussian projection onto the orthogonal complement of the null tangent space. This gives the chi-square law and explains why codimension is the relevant count. The geometry must actually be linear at the point; writing down a nominal number of restrictions does not establish the projection argument.

Wn=2{ℓn(θ^)−ℓn(θ~)}⇒χr2under the regular smooth interior null.W_n=2\{\ell_n(\widehat\theta)-\ell_n(\widetilde\theta)\}\Rightarrow\chi_r^2\quad\text{under the regular smooth interior null}.
The full and null-constrained maximizers require the stated likelihood and tangent-space conditions.↔ Scroll wide equations horizontally.

A one-sided cone changes the distribution

For a normal mean constrained to be nonnegative, the maximum-likelihood estimator is the positive part of the sample mean. At a null mean of zero, a negative unconstrained estimate cannot improve the fit over the null. The likelihood-ratio statistic is then zero. A positive unconstrained estimate contributes its usual standardized square. Since a centered normal has equal mass on either side of zero, the resulting distribution mixes a point mass and a one-degree chi-square law with equal weights.

Other boundary problems can have different mixture weights, several chi-square components, or entirely different limits. The answer depends on the parameter constraints, information metric, and whether nuisance parameters remain identified. The half-and-half mixture is a result for the specific one-sided scalar geometry here. In a general cone problem, one derives the limiting optimization or invokes a theorem whose assumptions fit that cone; the presence of any boundary does not by itself identify this distribution.

Z=n X‾σ∼N(0,1),Wn=(max⁡{Z,0})2∼12δ0+12χ12.Z=\frac{\sqrt n\,\overline X}{\sigma}\sim N(0,1),\qquad W_n=(\max\{Z,0\})^2\sim\tfrac12\delta_0+\tfrac12\chi_1^2.
For known-variance normal observations and a nonnegative mean, the boundary mixture is exact under a zero mean.↔ Scroll wide equations horizontally.

Tail probabilities must respect the atom

For a positive observed statistic, the upper-tail probability is one half of the corresponding one-degree chi-square tail, equivalently the upper standard-normal tail at its square root. At an observed value of zero, the conventional nonrandomized upper-tail p-value using a greater-than-or-equal event is one: the statistic is always nonnegative. Extending the positive-value formula to zero would ignore the atom and report one half instead. Different randomized conventions must be stated explicitly rather than silently substituted.

For a five-percent test, the correct positive critical value is the square of the ninety-fifth standard-normal percentile. Using the usual ninety-fifth percentile of a one-degree chi-square distribution instead gives a conservative two-and-a-half-percent test in this example. More elaborate boundary models often require specialized asymptotic theory or a carefully validated parametric simulation. A bootstrap is not automatically valid just because the ordinary chi-square rule fails; resampling must reproduce the boundary experiment and its nuisance behavior.

A nonnegative mean with known variance

Observe 6464 independent N(θ,4)N(\theta,4) variables with θ≥0\theta\geq0. Test θ=0\theta=0. If the sample mean is 0.50.5, calculate the likelihood-ratio statistic and its boundary p-value, then compare the ordinary chi-square tail.

  1. The constrained estimator is max⁡(X‾,0)=0.5\max(\overline X,0)=0.5. The standard deviation is 22, so Z=64(0.5)/2=2Z=\sqrt{64}(0.5)/2=2.

  2. The twice-log-likelihood improvement is W=(max⁡(Z,0))2=4W=(\max(Z,0))^2=4. Under the null, negative standardized means instead produce W=0W=0.

  3. For this positive statistic, the mixture upper tail is 1−Φ(2)≈0.0227501-\Phi(2)\approx0.022750. The ordinary one-degree chi-square upper tail is twice as large, approximately 0.0455000.045500.

The result

The boundary calculation yields a one-sided p-value. If the observed mean had been negative, the statistic would be zero and the conventional nonrandomized p-value would be one.

Explore the model

Interactive lab

Interactive lab · function

Squaring alone does not implement a constraint

Pause/Reset and set BOTH controls to a=1a=1, x=−1x=-1: predict value 11 and slope −2-2. For a fresh comparison Pause/Reset and set BOTH to a=1a=1, x=1x=1: value remains 11 and slope is 22. Manually apply the positive-part rule to these positions: boundary statistics would be 00 and 11. Changes retain the paused clock.

Squaring alone does not implement a constraintThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe instrument plots an ordinary quadratic. The manual clipping calculation reveals the boundary step that an unconstrained square would miss.

Check your understanding

In this exact boundary model, what is the conventional nonrandomized upper-tail p-value when the observed likelihood-ratio statistic is zero?

Try it yourself

Find the five-percent critical value for the nonnegative-normal-mean likelihood-ratio test. What is the actual null rejection probability if one instead uses the usual one-degree chi-square five-percent critical value?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

For a positive threshold cc, the boundary tail is 1−Φ(c)1-\Phi(\sqrt c).

Compare with the solution

Solve 1−Φ(c)=0.051-\Phi(\sqrt c)=0.05, giving c={Φ−1(0.95)}2≈2.705543c=\{\Phi^{-1}(0.95)\}^2\approx2.705543. The usual one-degree chi-square threshold is {Φ−1(0.975)}2≈3.841459\{\Phi^{-1}(0.975)\}^2\approx3.841459. Under the boundary mixture, its upper tail is 1−Φ(Φ−1(0.975))=0.0251-\Phi(\Phi^{-1}(0.975))=0.025. Thus that familiar threshold halves the intended size in this specific boundary model.

Take it with you

  • Wilks-type limits require smooth interior null geometry and regular likelihood behavior.
  • A scalar one-sided normal boundary produces a specific atom-plus-chi-square mixture.
  • Tail calculations at zero must include the atom and state their convention.

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