Regularity Conditions For Likelihood-Ratio Tests

Why Regularity Conditions Matter

A likelihood-ratio statistic can be calculated whenever restricted and unrestricted likelihoods can be compared. But the familiar chi-square reference distribution is not automatic. It is an asymptotic result that depends on how the model behaves near the parameter value specified by the null hypothesis. Before using that calibration, check whether the model is identifiable and smooth, whether the null value is interior, and whether the information matrix is nonsingular.

These checks matter in actuarial models because likelihoods may have parameter-dependent supports, redundant parameters, or parameter values on the edge of the allowed range. In such settings, a numerical likelihood ratio can still be computed, but the usual chi-square conclusion may not follow. This tutorial is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: For nested models with parameter spaces $\Theta_0\subset\Theta$, the likelihood-ratio statistic is $$ \Lambda=2\left\{\sup_{\theta\in\Theta}\ell(\theta)-\sup_{\theta\in\Theta_0}\ell(\theta)\right\}, $$ where $\ell(\theta)$ is the log likelihood. Under the usual regularity conditions, if the null model is true and imposes $q$ independent smooth restrictions, then $\Lambda$ converges in distribution to a chi-square random variable with $q$ degrees of freedom as the sample size increases.

Conditions Behind the Usual Chi-Square Limit

The conditions are properties of the model near the true parameter, not a checklist that can be verified from a single likelihood-ratio value. The precise technical versions vary with the model, but the following practical formulation captures the main requirements for the standard result.

Identifiability. Distinct parameter values must correspond to distinct probability distributions. If different parameter vectors produce the same distribution, the model contains parameters the data cannot distinguish.
An interior null value. The true parameter must lie in the interior of the full model’s parameter space. The parameter space must allow small changes in every relevant direction around that value. A null hypothesis may impose equality restrictions and therefore be a lower-dimensional subset of the full model; what matters is that its true value is not at the edge of the full model’s allowed parameter range.
Smooth likelihood. The log likelihood must be sufficiently differentiable in a neighborhood of the true value. The support of the observations should not change with the parameter, and regularity must permit the relevant derivatives to be taken through expectations or integrals.
Finite, nonsingular information. The Fisher information matrix must be finite and positive definite at the true value. This ensures that the parameters have distinguishable local effects and that the usual local quadratic approximation is well behaved.
Regular null restrictions. The $q$ restrictions must be locally independent and smooth, so that they remove $q$ distinct parameter directions. The comparison must be between nested models fitted to the same data under the likelihood setup assumed by the result.

For an independent and identically distributed model with density $f(x;\theta)$, the score is the vector of first derivatives of $\log f(x;\theta)$. Under regularity conditions, its expectation is zero, and the Fisher information per observation can be written as the expected outer product of the score or as the negative expected Hessian of the log density. These expressions require the differentiability and interchange conditions just described; they should not be used merely because a derivative can be written down.

Why the Degrees of Freedom Count the Restrictions

The chi-square limit follows from the local shape of the likelihood. Near the true parameter, a regular log likelihood is approximately a concave quadratic: its first derivative gives the local slope, and its information matrix gives the curvature. Maximizing the unrestricted likelihood allows movement in every parameter direction. Maximizing under the null removes movement in the $q$ directions excluded by the restrictions.

After scaling by the information matrix, the score is asymptotically normal with mean zero and identity covariance. The improvement in twice the log likelihood is therefore asymptotically the squared length of the component of this normal vector in the $q$ directions available only in the unrestricted model. The sum of squares of $q$ independent standard normal components has a chi-square distribution with $q$ degrees of freedom.

\Lambda \ \overset{d}{\longrightarrow}\ \sum_{j=1}^{q} Z_j^2 \ \sim\ \chi^2_q, \qquad Z_j\ \text{independent with}\ Z_j\sim N(0,1).

This argument explains both the degrees of freedom and the role of regularity. A singular information matrix prevents the usual scaling, while a non-smooth or parameter-dependent support can invalidate the quadratic approximation. A statistic that looks numerically like an ordinary likelihood ratio does not repair those failures.

Worked Examples: Auditing Regularity in Practice

Worked Example: A Regular Exponential Severity Comparison

Suppose three hypothetical independent positive losses $x_1=1$, $x_2=2$, and $x_3=3$ are modeled using an exponential density $f(x;\lambda)=\lambda e^{-\lambda x}$, for $x>0$ and $\lambda>0$. Test $H_0:\lambda=1$ against the unrestricted alternative $\lambda>0$. Calculate the likelihood-ratio statistic and assess whether the usual chi-square calibration is structurally appropriate.

Method. First check the model conditions, then find the restricted and unrestricted maxima. The log likelihood is $$ \ell(\lambda)=3\log\lambda-\lambda(1+2+3) =3\log\lambda-6\lambda. $$

The support $x>0$ does not depend on $\lambda$. Distinct positive rates give distinct exponential distributions; $\lambda=1$ is interior to $(0,\infty)$; and the log density is smooth there. For one observation, the Fisher information is $1/\lambda^2$, finite and positive at $\lambda=1$. Thus the model is regular for this comparison.

Unrestricted maximum. Differentiate: $$ \frac{d\ell}{d\lambda}=\frac{3}{\lambda}-6. $$ Setting this equal to zero gives $\hat{\lambda}=3/6=0.5$. The second derivative is $-3/\lambda^2<0$, so this is a maximum.

Evaluate both log likelihoods. $$ \ell(0.5)=3\log(0.5)-6(0.5) =-2.0794-3=-5.0794, \qquad \ell(1)=3\log(1)-6=-6. $$ Therefore, $$ \Lambda=2\{-5.0794-(-6)\}=1.8411. $$

The null fixes one parameter, so the difference in parameter dimensions is one and the usual asymptotic reference is $\chi^2_1$. The method avoids the mistake of counting the null parameter as an additional restriction beyond the one parameter fixed. The conditions are structurally appropriate, but there are only three observations: the chi-square result is asymptotic, and this calculation alone does not establish that its finite-sample approximation is accurate.

Worked Example: A Shared Poisson Rate Across Two Groups

In a hypothetical portfolio, two independent groups have observed claim counts $y_A=8$ and $y_B=18$ over exposures $t_A=4$ and $t_B=6$ years. Assume constant Poisson rates. Compare a model with separate rates $\lambda_A,\lambda_B>0$ against $H_0:\lambda_A=\lambda_B$. Calculate the likelihood-ratio statistic and check the conditions relevant to its usual calibration.

Method. For a group with count $y$ and exposure $t$, the likelihood is proportional to $e^{-t\lambda}(t\lambda)^y$. The separate-rate estimates are $$ \hat{\lambda}_A=\frac{8}{4}=2, \qquad \hat{\lambda}_B=\frac{18}{6}=3. $$

Under the shared-rate restriction, the total count is $8+18=26$ and total exposure is $4+6=10$. Hence $$ \hat{\lambda}=\frac{26}{10}=2.6. $$

The factorial terms and exposure-only terms cancel in the log-likelihood difference. Thus $$ \begin{aligned} \Lambda &=2\left\{8\log\left(\frac{2}{2.6}\right) +18\log\left(\frac{3}{2.6}\right)\right\}\\ &=2\left\{8\log\left(\frac{10}{13}\right) +18\log\left(\frac{15}{13}\right)\right\}\\ &=2\{-2.0989+2.5758\}=0.9538. \end{aligned} $$

The unrestricted model has two rate parameters and the shared-rate model has one, so there is one independent restriction. Positive rates give identifiable Poisson distributions; the null rate $2.6$ is interior to the positive-rate parameter space; and the log likelihood is smooth. The information for the separate rates is diagonal, with entries $t_A/\lambda_A$ and $t_B/\lambda_B$, which are finite and positive at positive rates. The usual asymptotic reference is therefore $\chi^2_1$. A common mistake is to use $26$, the total count, as the common rate; the correct estimate divides by total exposure, giving $26/10=2.6$.

Worked Example: Parameter-Dependent Support and a Supremum

Suppose three hypothetical losses $2,5,8$ are modeled as independent observations from a uniform distribution on $(0,\theta)$, with density $f(x;\theta)=1/\theta$ for $0<x<\theta$ and zero otherwise. Compare $H_0:\theta=10$ with the unrestricted model. Calculate the likelihood-ratio expression and decide whether the usual chi-square limit is justified.

Method. The likelihood is $$ L(\theta)= \begin{cases} \theta^{-3},&\theta>8,\\ 0,&\theta\leq 8. \end{cases} $$ The inequality is strict because the density’s support is open at $\theta$, and the largest observed loss equals $8$. For $\theta>8$, $\theta^{-3}$ decreases as $\theta$ increases. Therefore the unrestricted likelihood has supremum $8^{-3}=1/512$ as $\theta$ decreases toward $8$, but it does not attain that value. In particular, $L(8)=0$, not $1/512$.

At the null value, $L(10)=10^{-3}=1/1000$. Using the supremum-based likelihood-ratio expression gives $$ \Lambda=2\log\left(\frac{1/512}{1/1000}\right) =2\log\left(\frac{125}{64}\right) \approx 1.3381. $$

The calculation is a likelihood comparison, but the usual chi-square calibration is not justified: the support depends on $\theta$, the unrestricted supremum is not attained, and the regular local quadratic argument does not apply. The exam trap is to write $L(8)=8^{-3}$ by treating “below $\theta$” as inclusive. Under the stated open-support density, observations must satisfy $x<\theta$, so the correct statement is a supremum as $\theta$ approaches $8$ from above.

Worked Example: Redundant Parameters and Singular Information

Let hypothetical observations be independent with $X_i\sim N(\alpha+\beta,1)$, where the variance is known to be $1$ and $\alpha,\beta$ are unrestricted real parameters. For observed values $1,2,3,4$, check identifiability and calculate the information matrix. Explain why treating $\alpha$ and $\beta$ as two separately estimable parameters is invalid.

Method. The density depends on $\alpha$ and $\beta$ only through $\mu=\alpha+\beta$. The data therefore identify $\mu$, but not the individual values of $\alpha$ and $\beta$. For example, every pair satisfying $\alpha+\beta=2.5$ gives the same fitted distribution when the sample mean is $2.5$: $$ \bar{x}=\frac{1+2+3+4}{4}=\frac{10}{4}=2.5. $$ There are infinitely many such pairs, including $(\alpha,\beta)=(1,1.5)$ and $(0,2.5)$.

For one observation, the score vector is $$ \begin{pmatrix}X_i-(\alpha+\beta)\\X_i-(\alpha+\beta)\end{pmatrix}. $$ Since $X_i$ has variance $1$, the information per observation is $$ \begin{pmatrix}1&1\\1&1\end{pmatrix}. $$ For four observations it is $$ I_4= \begin{pmatrix}4&4\\4&4\end{pmatrix}, \qquad \det(I_4)=16-16=0. $$

The parameters are not identifiable and the information matrix is singular. Thus the standard chi-square result cannot be applied by simply counting $\alpha$ and $\beta$ as two distinct estimable directions. The mistake is to infer regularity from the smoothness of the normal density alone; smoothness does not compensate for redundant parameters.

A Practical Regularity Audit

Before interpreting a likelihood-ratio statistic using a chi-square distribution, make the checks in a deliberate order. If a check fails, do not proceed as if the usual limit were guaranteed. A failed condition does not mean that the likelihood comparison is meaningless; it means that its reference distribution needs separate justification.

Write the parameter spaces.
Specify the full model and the null model, including positivity or other parameter restrictions. Identify the true null value being assessed.

Check identifiability and restriction count.
Ask whether distinct parameter values give distinct distributions, and whether the null imposes the stated number of independent restrictions.

Check local regularity.
Verify that the null value is interior to the full model, that the support is parameter-independent, and that the log likelihood is sufficiently smooth nearby.

Check information and conclude cautiously.
Confirm that the Fisher information is finite and nonsingular. Only then use the usual chi-square limit, while remembering that it is an asymptotic result.

A common error is to check only that the likelihood can be differentiated at the fitted value. That is not enough: parameters may be unidentifiable, the support may move with the parameter, or the null value may fail to be interior. Another error is to treat the parameter-count difference as sufficient evidence for the degrees of freedom. It gives the usual degrees of freedom only when the restrictions are locally independent and the regularity conditions hold.

Key takeaway: The usual likelihood-ratio chi-square limit is justified by a regular, locally identifiable, smooth model with finite nonsingular information and an interior true null value. Check those properties before interpreting the statistic; a computed likelihood ratio alone does not establish them.

Check Your Understanding

For each problem, decide what regularity conclusion is justified. Where requested, calculate the stated quantity.

A sample is modeled as exponential with rate $\lambda>0$, and the null is $\lambda=1.4$. The density has support $x>0$, independent of $\lambda$. Is the null rate interior, and is parameter-dependent support a concern? (Answer: The rate is interior to $(0,\infty)$; the support is parameter-independent.)
A model uses two real parameters $\gamma,\eta$, but its density depends only on $\gamma-\eta$. Which regularity issue is present? (Answer: The parameters are not separately identifiable; the information matrix is singular in the redundant direction.)
For four independent Poisson counts $y_1=3,y_2=5,y_3=4,y_4=8$ with equal exposure of one year, compare separate positive rates with a common positive rate. What is the common-rate estimate and how many restrictions are imposed? (Answer: $\hat{\lambda}=20/4=5$ claims per year; there are $4-1=3$ restrictions.)
A density is defined by $f(x;\theta)=1/\theta$ for $0<x<\theta$, and the observed maximum is $m$. Is the unrestricted likelihood maximized at $\theta=m$? (Answer: No. It is positive only for $\theta>m$ and has a supremum $m^{-n}$ approached as $\theta\downarrow m$, not an attained maximum.)
A regular identifiable model has four parameters, while a smooth null model imposes two independent restrictions at an interior null value. What is the usual asymptotic reference distribution, assuming the information matrix is finite and nonsingular? (Answer: $\chi^2_2$.)

What You'll Learn