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Exam ASTAM · Likelihood-Based Model Comparison · Tutorial 430 of 979

Nuisance Parameters Absent Under The Null

Identify when a mixture component creates an unidentified nuisance parameter under the null, and use the resulting Gaussian-process limit rather than an automatic chi-square calibration.

Intermediate 12 min read

What You'll Learn

  • Recognize when a mixture component parameter is unidentified under the null hypothesis.
  • Distinguish a single boundary direction from a search over unidentified component parameters.
  • Derive the local likelihood-ratio approximation using an efficient score.
  • Calculate the limiting distribution for a finite set of independent candidate components.
  • Explain how correlation among candidate scores changes the calibration.
  • Avoid applying a standard chi-square or boundary-mixture reference distribution without checking the model.

Why an Unidentified Parameter Changes the Test

A likelihood-ratio test often compares a restricted model with a larger model. When the null hypothesis fixes a parameter at the boundary of its allowable range, the usual chi-square approximation may fail. A further complication arises when another parameter has no meaning under the null. In that case, the test is not simply moving in one fixed direction away from the null: it is searching across possible directions.

A mixture model makes this issue concrete. Suppose hypothetical claim severities \(X_1,\ldots,X_n\) are independent and have density \(f(x;\alpha,\eta,\beta)=(1-\alpha)f_0(x;\beta)+\alpha f_1(x;\eta)\). Here \(f_0\) is the baseline severity density, \(\beta\) is its parameter, \(\alpha\in[0,1]\) is the mixture weight, and \(\eta\) controls the added component. The null hypothesis is \(H_0:\alpha=0\). Under this null, the density is \(f_0(x;\beta)\), regardless of \(\eta\). Thus the data under the null contain no information about \(\eta\): it is unidentified under the null.

This is different from a test in which a single, identifiable parameter is constrained to a boundary. In a mixture test, the added component can enter at many possible values of \(\eta\). The likelihood-ratio statistic can benefit from choosing whichever value fits the observed sample best. Its null distribution must account for that search.

Definition: A nuisance parameter is unidentified under the null when its value does not affect the probability distribution specified by the null hypothesis. In the mixture above, every value of \(\eta\) gives the same null density when \(\alpha=0\). Consequently, the null does not determine \(\eta\), and ordinary likelihood-ratio degrees-of-freedom counting does not apply automatically.

From a Local Score to a Supremum

The likelihood-ratio statistic is \(T_n=2\{\sup_{\alpha,\eta,\beta}\ell_n(\alpha,\eta,\beta)-\sup_{\beta}\ell_n(0,\eta,\beta)\}\), where \(\ell_n\) is the log likelihood. The second supremum does not depend on \(\eta\), because \(\eta\) disappears under \(H_0\). For a fixed candidate value of \(\eta\), the mixture weight is constrained by \(\alpha\geq 0\), so its local fitted value cannot move in a negative direction.

The following derivation describes the standard local approximation. Assume independent observations from the null baseline; a smooth, identifiable baseline parameter \(\beta\); finite, positive score variances; and sufficient regularity for a uniform Gaussian limit over the allowed values of \(\eta\). Also assume the mixture model has no additional singularities that invalidate the local expansion. These qualifications matter: mixture models can violate regularity in other ways, so the formula is not universal for every mixture.

For each fixed \(\eta\), the score for increasing \(\alpha\) at zero is based on \(s_\eta(x)=f_1(x;\eta)/f_0(x;\beta)-1\), when this density ratio is well-defined and has finite second moment under the null. If \(\beta\) is estimated, the score must be adjusted for the baseline parameter. Write \(\widetilde{s}_\eta\) for the efficient score: the component of \(s_\eta\) remaining after its linear projection onto the baseline score has been removed. Let \(J_\eta=E_0[\widetilde{s}_\eta^2]>0\), and define the standardized efficient score \(Z_n(\eta)=\sum_{i=1}^n\widetilde{s}_\eta(X_i)/\sqrt{nJ_\eta}\).

1
Expand locally.
For a fixed \(\eta\), consider a mixture weight of local size \(\alpha=h/\sqrt{nJ_\eta}\), with \(h\geq 0\). After profiling the regular baseline parameter, the log-likelihood difference has the local form \(hZ_n(\eta)-h^2/2+o_p(1)\).
2
Maximize over the allowed direction.
The quadratic expression is maximized over \(h\geq0\) at \(h=\max(0,Z_n(\eta))\). Its maximum is \(\tfrac12\{\max(0,Z_n(\eta))\}^2\).
3
Allow the component parameter to vary.
The alternative also chooses \(\eta\). Multiplying the maximized log-likelihood difference by 2 gives a supremum of squared positive score parts, not the square of one fixed score.
$$ T_n \ \Rightarrow\ \sup_{\eta\in\mathcal{H}}\{\max(0,G(\eta))\}^2 $$

Here \(\mathcal{H}\) is the allowed set of component parameters, and \(G(\eta)\) is a mean-zero, unit-variance Gaussian process. Its correlation function is the correlation between the efficient scores at different values of \(\eta\): \(\operatorname{Corr}(G(\eta_1),G(\eta_2))= E_0[\widetilde{s}_{\eta_1}\widetilde{s}_{\eta_2}]/\sqrt{J_{\eta_1}J_{\eta_2}}\). The supremum appears because the alternative selects the best-fitting \(\eta\); the positive part appears because \(\alpha\) cannot be negative. This explains why the limiting distribution depends on the range of \(\eta\) and the dependence among its score directions.

Theorem (local mixture likelihood-ratio limit): Under the stated local regularity and Gaussian-process convergence conditions, the likelihood-ratio statistic for adding a component with \(\alpha\geq0\), when \(\eta\) is unidentified at \(\alpha=0\), converges to \(\sup_{\eta\in\mathcal{H}}\{\max(0,G(\eta))\}^2\). The limit is generally not a chi-square distribution and is not determined by counting the added parameters.

Finite Candidate Components

A useful special case restricts \(\eta\) to \(K\) candidate values. If the corresponding limiting scores \(Z_1,\ldots,Z_K\) are independent standard normal variables, then the limiting statistic is \(T=\max_{1\leq j\leq K}\{\max(0,Z_j)\}^2\). For \(t\geq0\), the event \(T\leq t\) occurs exactly when every score is at most \(\sqrt{t}\). Independence therefore gives

$$ P(T\leq t)=\Phi(\sqrt{t})^K,\qquad P(T>t)=1-\Phi(\sqrt{t})^K, $$

where \(\Phi\) is the standard normal cumulative distribution function. In particular, \(P(T=0)=2^{-K}\): all \(K\) scores must be nonpositive. If the scores are correlated, the product formula does not apply. The joint Gaussian distribution, including its correlations, determines the calibration.

Worked Example: A Three-Component Candidate Set

A hypothetical severity-model comparison considers three possible added-component parameter values. The fitted, standardized efficient scores are \(1.4\), \(-0.3\), and \(0.9\). Assume that under the null their limiting scores are independent standard normal variables. Find the observed limiting statistic and its approximate upper-tail p-value. No interest or mortality assumptions enter this likelihood comparison; the stated distributional assumption is the independent standard normal score limit.

Method. For each candidate component, retain only the positive score, square it, and take the maximum. Thus \(T=\max\{1.4^2,0,0.9^2\}=\max\{1.96,0,0.81\}=1.96\). The negative score cannot improve the fit because the mixture weight is constrained to be nonnegative.

The threshold is \(t=1.96\), so \(\sqrt{t}=1.4\). With the unrounded standard normal value \(\Phi(1.4)\approx0.9192433\), the independent-candidate tail probability is \(1-\Phi(1.4)^3\approx1-(0.9192433)^3\approx0.2232\). The approximate p-value is therefore \(0.2232\), to four decimal places.

Check and trap avoided. The statistic is \(1.96\), not \(1.4^2+0.9^2\): the alternative chooses one candidate value of \(\eta\), rather than adding evidence from mutually exclusive candidates. The cube in the tail formula accounts for all three possible directions.

Worked Example: Correlation Changes the Probability of a Zero Statistic

A hypothetical model has two candidate values of \(\eta\). Their limiting efficient scores are jointly standard normal with correlation \(0.5\). In one sample, the two standardized scores are \(0.7\) and \(-0.4\). Find the statistic and the probability, under the limiting null, that the statistic is exactly zero. No discounting or mortality assumptions are involved.

Observed statistic. The positive parts are \(0.7\) and \(0\). Therefore \(T=\max\{0.7^2,0\}=0.49\).

Probability of zero. The statistic equals zero precisely when both scores are nonpositive. For a bivariate standard normal pair with correlation \(\rho\), the probability that both coordinates are nonpositive is \(1/4+\arcsin(\rho)/(2\pi)\). Substituting \(\rho=0.5\) gives \(1/4+\arcsin(0.5)/(2\pi)=1/4+(\pi/6)/(2\pi)=1/4+1/12=1/3\). Thus the limiting statistic has probability \(1/3\) at zero.

Check and trap avoided. Treating the scores as independent would give \(1/4\), which is incorrect because their correlation is positive. The joint score dependence affects the null distribution even though each individual score is standard normal.

Worked Example: Calibrating a Two-Candidate Test

For a hypothetical two-candidate mixture test, assume independent standard normal limiting scores. The observed likelihood-ratio statistic is \(2.25\). Calculate its approximate upper-tail probability using the correct limit and compare it with the single-direction boundary-mixture result. The score model is as stated; interest and mortality assumptions are not relevant.

Here \(\sqrt{2.25}=1.5\), and \(q=1-\Phi(1.5)\approx0.0668072\). The statistic exceeds \(2.25\) if at least one of the two independent scores exceeds \(1.5\). Hence \(P(T>2.25)=1-(1-q)^2=2q-q^2\). Substitution gives \(2(0.0668072)-(0.0668072)^2\approx0.1291512\), or \(0.1292\) to four decimal places.

For comparison, a single identifiable boundary direction has the familiar limit with probability \(1/2\) at zero and probability \(1/2\) distributed as \(\chi^2_1\). Its upper tail at \(2.25\) is \(q\approx0.0668\). That is not the calibration here: the two-candidate search raises the upper-tail probability to \(0.1292\).

Check and trap avoided. The independent-candidate calculation can also be checked from the distribution function: \(1-\Phi(1.5)^2=1-(1-0.0668072)^2\approx0.1292\). The candidate directions are maximized over, not averaged.

Common Exam Trap: Reusing a Boundary Mixture Automatically

A frequent error is to see \(\alpha=0\) at the edge of its parameter space and immediately use the single-direction boundary distribution. That calibration is appropriate only when the local alternative has one relevant, identifiable score direction under the null, along with the required regularity conditions. When \(\eta\) is absent under the null, the alternative searches across score directions, and the limiting statistic is a supremum.

For example, at \(T=2.25\), the wrong single-direction boundary calculation gives \(1-\Phi(1.5)\approx0.0668\). With two independent candidate component values, the correct calculation is \(1-\Phi(1.5)^2\approx0.1292\). The first calculation understates the tail probability because it omits the search over candidates. Counting \(\alpha\) and \(\eta\) and using a chi-square distribution with two degrees of freedom is not a general fix; \(\eta\) is not identified by the null, so the regular Wilks degrees-of-freedom argument does not apply.

For a continuous set of candidate values, there is generally no finite-\(K\) product formula. Calibration requires the covariance structure of the Gaussian score process and the allowed parameter set. The key exam decision is therefore made before looking up a reference distribution: determine whether the nuisance parameter is identified under the null and whether the alternative contributes one direction or a search over directions.

Check Your Understanding

For each problem, use the stated limiting score model and give probabilities to four decimal places.

  1. A test has two independent candidate scores, and the observed scores are \(1.1\) and \(0.6\). Find the limiting statistic and p-value. Use \(\Phi(1.1)=0.8643339\). (Answer: \(T=1.21\); p-value \(=1-\Phi(1.1)^2\approx0.2529\).)
  2. There are three independent candidate scores with the standard normal limit. What is the probability that the limiting likelihood-ratio statistic is zero? (Answer: \(1/8=0.1250\).)
  3. There are two independent candidate scores and the observed statistic is \(1\). Find the upper-tail probability using \(\Phi(1)=0.8413447\). (Answer: \(1-\Phi(1)^2\approx0.2921\).)
  4. Two candidate scores have a bivariate standard normal limit with correlation \(-0.5\). Find the probability that the limiting statistic is zero. (Answer: \(1/4+\arcsin(-0.5)/(2\pi)=1/6\approx0.1667\).)
  5. Four independent candidate scores have the standard normal limit. Find the probability that the statistic exceeds \(4\), using \(\Phi(2)=0.9772499\). (Answer: \(1-\Phi(2)^4\approx0.0879\).)
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