Likelihood Ratio Test

Likelihood Ratio Test for Nested Models in Health Economics

The likelihood ratio test asks whether a richer statistical model fits the data better than a simpler one that sits inside it, by more than chance would explain. Health economists use it on the regressions that feed decision models, for example to decide whether a dispersion or survival shape parameter earns its place. The statistic is twice the gain in maximised log-likelihood, and in large samples it is read against a chi-square distribution whose degrees of freedom count the extra parameters. The test says nothing about models that are not nested, and its usual reference fails for some common restrictions. The model comparison test is distinct from the diagnostic accuracy measure with a similar name, described on the likelihood ratio page.

Two maximised likelihoods and one statistic

Both models are fitted to the same observations by maximum likelihood estimation. The restricted model is the larger model with some parameters fixed, usually at zero, and the test compares the best likelihood each can reach.

$$\Lambda = \frac{\max_{\theta \in \Theta_0} L(\theta)}{\max_{\theta \in \Theta} L(\theta)}$$

where $\Lambda$ is the likelihood ratio, $L(\theta)$ is the likelihood of the data at parameter values $\theta$, $\Theta$ is the set of values allowed by the larger model and $\Theta_0$ is the smaller set allowed by the restricted model. Because $\Theta_0$ lies inside $\Theta$, the ratio cannot exceed 1.

$$LR = -2 \ln \Lambda = 2\left(\ell_1 - \ell_0\right)$$

where $LR$ is the likelihood ratio statistic, $\ell_1$ is the maximised log-likelihood of the larger model and $\ell_0$ is that of the restricted model.

For a generalised linear model the same quantity appears as a change in deviance. The R documentation for analysis of deviance tables reports the reduction in residual deviance as terms are added and notes that comparing several fits only makes statistical sense if the models are nested. For models with dispersion estimated by moments, such as quasi-Poisson fits, it judges an F test more appropriate.

Wilks' theorem and the chi-square reference

The reference distribution comes from large-sample theory known as Wilks' theorem, after a 1938 paper by Wilks. As Chen, Moustaki and Zhang summarise it, when two nested models satisfy certain regularity conditions, the p-value comes from a chi-square distribution with degrees of freedom equal to the difference in the number of free parameters.

$$LR \sim \chi^2_{q}, \qquad q = p_1 - p_0$$

where $\chi^2_q$ is a chi-square distribution with $q$ degrees of freedom, $p_1$ is the number of free parameters in the larger model and $p_0$ is the number in the restricted model. The 5% critical values are 3.841 on one degree of freedom and 5.991 on two, and the p-value is the right-tail area beyond the observed statistic.

Chen and colleagues single out two regularity conditions that their examples breach. The true parameter value must be an interior point of the parameter space, and the Fisher information matrix at that value must be strictly positive definite.

What counts as a nested comparison

Two models are nested when fixing some parameters of one gives exactly the other. The NICE Decision Support Unit's Technical Support Document 14 (TSD 14) gives as its example one model that adds a covariate to another. Several familiar pairs are nested.

1. Exponential within Weibull. A Weibull model with its shape parameter fixed at 1 is the exponential model.
2. Weibull within generalised gamma. TSD 14 notes that the generalised gamma model includes the Weibull, exponential and log normal as special cases, and equals the Weibull when one parameter is 1.
3. Poisson within negative binomial. The Stata manual for negative binomial regression describes the Poisson model as the special case with dispersion parameter zero.

Different distribution families, such as a Weibull and a log-logistic curve, are not nested. TSD 14 states that the negative 2 log likelihood is only suitable for comparing nested models, that it has been used erroneously in past NICE technology appraisals, and that AIC and BIC allow comparison of models that do not have to be nested. It describes one appraisal, TA172, in which model choice rested partly on the log likelihood of different parametric families, and calls the approach flawed for that reason.

Nesting also requires the same data. The R documentation warns that the comparison is only valid for fits to the same dataset, which can fail when a new covariate has missing values and those rows are dropped.

Where the test appears in model building

Most uses in health economics sit in the statistical models that estimate decision model inputs. In logistic regression risk equations, the test checks whether a block of risk factors improves the fit of a model for an event such as hospital admission. In count models for resource use it compares Poisson regression with negative binomial regression. In parametric survival models it tests a treatment covariate or a shape parameter within one family. A worked set of nested survival comparisons is on the chi-square distribution page.

Tests on the boundary of the parameter space

Wilks' theorem fails when the restricted model fixes a parameter at the edge of its allowed range. A variance cannot be negative, so testing whether a random intercept variance is zero in a random-effects model, for example hospitals in a multilevel cost model, places the null on the boundary. Chen and colleagues show by simulation that the plain chi-square reference does not fit in this case. They note that the statistic often follows a mixture of chi-square distributions, and derive for their random intercept example a limit with the distribution of $w^2$ when $w \ge 0$ and zero otherwise, where $w$ is standard normal.

$$P(LR > x) = \tfrac{1}{2} P(\chi^2_1 > x)$$

where $x$ is the observed statistic and the factor of one half reflects the half of the null distribution that sits at zero. The Stata manual applies this to the negative binomial dispersion test, stating that the null distribution is a 50 to 50 mixture of a point mass at zero and a chi-square on one degree of freedom, reported as chibar2(01). The naive p-value is therefore twice the correct one, and the 5% critical value falls from 3.841 to 2.706.

Choosing the number of latent classes or mixture components is a harder case. Chen and colleagues list finite mixture models among those where the chi-square approximation is often found not to hold, and cite work showing that latent class and factor models have points of irregularity that can invalidate the test. This matters for latent class models of health care use and for mixture approaches to a mapping function between utility instruments.

Likelihood ratio, Wald and score tests

Three large-sample tests of the same restriction are asymptotically equivalent under the null hypothesis. The Wald test needs only the larger fit and compares an estimate with its standard error. The score test, which the R documentation calls Rao's efficient score test, needs only the restricted fit. The likelihood ratio test needs both fits.

The likelihood ratio statistic is unchanged when a parameter is re-expressed, for instance as a log hazard ratio in place of a hazard ratio, whereas the Wald statistic changes with the scale. Inverting the test gives a profile likelihood confidence interval, the set of values of a parameter not rejected at the chosen level, which can be asymmetric around the estimate.

How the test relates to information criteria

The Akaike information criterion is minus twice the maximised log-likelihood plus twice the number of parameters. For two nested models the difference in AIC is therefore the likelihood ratio statistic less twice the number of extra parameters.

$$\Delta AIC = AIC_0 - AIC_1 = LR - 2q$$

where $AIC_0$ and $AIC_1$ are the criteria for the restricted and larger models and $q$ is the number of extra parameters. With one extra parameter AIC favours the larger model whenever $LR$ exceeds 2, which matches a test at a significance level of about 0.157 instead of 0.05. The Bayesian information criterion charges $\ln n$ per parameter for $n$ observations, so in large datasets it demands a far bigger likelihood gain than a 5% test. TSD 14 notes that the BIC penalises extra parameters more heavily.

Worked example with Poisson and negative binomial admission models

All figures are illustrative. A cost model needs the annual number of emergency admissions for 400 trial participants. Three models are fitted to the same 400 records, each with an intercept, age and treatment arm.

1. Negative binomial against Poisson. Model B adds the dispersion parameter.

$$LR_{BA} = 2 \times (-650.8 - (-652.4)) = 2 \times 1.6 = 3.2$$

where $LR_{BA}$ is the statistic for model B against model A. Against a chi-square on one degree of freedom the p-value would be 0.074. The null value of zero is on the boundary, so the correct p-value is half of that, 0.037, and 3.2 exceeds the boundary critical value of 2.706. The naive reading would have kept the Poisson model at the 5% level.

2. Adding comorbidity. Model C adds two indicator parameters for the three comorbidity categories.

$$LR_{CB} = 2 \times (-645.3 - (-650.8)) = 2 \times 5.5 = 11.0$$

where $LR_{CB}$ is the statistic for model C against model B. On two degrees of freedom the right tail has a closed form.

$$P(\chi^2_2 > 11.0) = e^{-11.0/2} = e^{-5.5} = 0.0041$$

where $e^{-x/2}$ is the chi-square tail on two degrees of freedom. The statistic is well above 5.991, so comorbidity is retained.

3. Checking against AIC. With AIC equal to minus twice the log-likelihood plus twice the parameter count, the three models score 1,310.8, 1,309.6 and 1,302.6. The gaps of 1.2 and 7.0 match $LR - 2q$, that is 3.2 minus 2 and 11.0 minus 4, so AIC ranks model C first.

4. Checking against BIC. The penalty per parameter is $\ln 400 = 5.991$, giving 1,322.8, 1,325.6 and 1,326.5. BIC ranks the Poisson model first, because the gain of 3.2 is below one parameter's penalty and the gain of 11.0 is below the 11.98 that two parameters cost.

5. Interpretation. At the 5% level, both tests support the negative binomial model with comorbidity, but only once the boundary correction is applied, and BIC disagrees. Models A and B share a mean structure, so the choice between them mainly changes the variance of admissions and the standard errors of the coefficients that feed probabilistic sensitivity analysis. The Poisson fit makes a natural scenario analysis.

Use in economic evaluation and HTA

The NICE manual (PMG36, section 4.6.18) asks that the underlying assumptions of models for cost-utility analysis be checked statistically whenever possible, and a likelihood ratio test is one such check. Section 4.6.20 adds that when statistical measures of relative fit to observed trial data are used to assess alternative extrapolation models, their limitations should be documented, and section 4.6.25 asks for the clinical plausibility of the hazard functions to be assessed. A test of fit within the trial period cannot show which curve extrapolates well, and TSD 14 points to AIC and BIC for comparing parametric families.

A non-significant test shows only that the data cannot distinguish the models. It does not show that the simpler model is true. Where two plausible structures give different cost-effectiveness results, the choice is structural uncertainty, which PMG36 (section 4.7.5) notes can be brought into a probabilistic model, for example by model averaging.

Misreadings and limits

Most errors with the test come from applying the chi-square reference where its conditions do not hold. Others come from reading the result as more than a statement about fit within the data.

1. Comparing families that are not nested. A log-likelihood gap between a Weibull and a log-logistic curve has no chi-square reference.
2. Comparing fits to different samples. Rows dropped for a missing covariate make the two likelihoods incomparable.
3. Ignoring the boundary. Tests of a variance or dispersion parameter at zero need the mixture reference, and tests of the number of latent classes need other methods.
4. Treating a quasi-likelihood as a likelihood. Generalised estimating equations and quasi-Poisson fits rest on estimating equations without a full likelihood, so the test does not apply.
5. Equating significance with importance. In very large datasets small gains become significant, while small trials lack statistical power.

Sources

• Casella G, Berger RL. Statistical Inference. 2nd ed. Duxbury; 2002. Chapters 8 to 10.
• Chen Y, Moustaki I, Zhang H. A note on likelihood ratio tests for models with latent variables. Psychometrika. 2020;85(4):996-1012. https://doi.org/10.1007/s11336-020-09735-0
• Latimer N. NICE DSU Technical Support Document 14: Survival analysis for economic evaluations alongside clinical trials, extrapolation with patient-level data. Decision Support Unit; June 2011, last updated March 2013. Sections 2.2, 2.6, 3.3 and 4.6. https://www.sheffield.ac.uk/nice-dsu/tsds/survival-analysis
• National Institute for Health and Care Excellence. NICE technology appraisal and highly specialised technologies guidance: the manual (PMG36). Published 31 January 2022, last updated 13 October 2026. Sections 4.6.18, 4.6.20, 4.6.25 and 4.7.5. https://www.nice.org.uk/process/pmg36/chapter/economic-evaluation-2
• R Core Team. anova.glm, Analysis of deviance for generalized linear model fits. R Documentation, package stats. Accessed October 2026. https://stat.ethz.ch/R-manual/R-devel/library/stats/html/anova.glm.html
• StataCorp. nbreg, Negative binomial regression. Stata Base Reference Manual. StataCorp LLC; 2025. https://www.stata.com/manuals/rnbreg.pdf