Likelihood-Ratio Tests for Joint-Model Parameters

Why Compare Nested Joint Models?

A joint model links repeated longitudinal measurements and an event-time outcome, often through shared subject-specific random effects. A common inferential question is whether a specified part of that model is needed: for example, whether the longitudinal process is associated with event risk after accounting for the other parameters. A likelihood-ratio test (LRT) answers this by comparing the best fit available under the null hypothesis with the best fit under a broader alternative.

The comparison is meaningful only when the null model is genuinely contained in the alternative. It also relies on regularity conditions for the large-sample chi-squared approximation. Joint models require special care because the observed-data likelihood integrates over latent random effects, parameters can lie on boundaries, and the likelihood may be numerically difficult to optimize.

Observed Likelihood and Nested Hypotheses

Suppose there are $n$ independent subjects. For subject $i$, let $\mathbf y_i$ denote the observed longitudinal measurements, $T_i$ the observed event or censoring time, and $\delta_i$ the event indicator. Let $\mathbf b_i$ be the subject's latent random effects and $\boldsymbol\theta$ the model parameters. The observed-data contribution integrates out the random effects:

L_i(\boldsymbol\theta) = \int p(\mathbf y_i,T_i,\delta_i\mid \mathbf b_i;\boldsymbol\theta) \,p(\mathbf b_i;\boldsymbol\theta)\,d\mathbf b_i, \qquad \ell(\boldsymbol\theta)=\sum_{i=1}^{n}\log L_i(\boldsymbol\theta).

The event contribution in this expression must reflect the event and censoring information actually observed, and the joint model must specify the dependence between the longitudinal and event processes. The LRT uses this observed-data likelihood, not a likelihood that treats the unobserved random effects as if they were observed.

Definition: Two models are nested when every distribution allowed by the null model is also allowed by the alternative model. If $\boldsymbol\theta\in\Theta$ indexes the alternative and the null imposes $q$ independent restrictions, write $H_0:\boldsymbol\theta\in\Theta_0\subset\Theta$. The likelihood-ratio statistic is $$ \Lambda = 2\left\{ \ell(\widehat{\boldsymbol\theta})- \ell(\widehat{\boldsymbol\theta}_0) \right\}, $$ where each estimate maximizes the observed log likelihood over its respective parameter space.

The unrestricted fit cannot have a smaller maximized log likelihood than the restricted fit, because it has at least all the null model's candidate parameter values. Thus $\Lambda\geq 0$, apart from small numerical discrepancies. For a regular test, the degrees of freedom are $q$, the number of independent restrictions—not the number of parameters in either model.

When the Chi-Squared Reference Is Appropriate

Under standard regularity conditions, if $H_0$ is true and the sample size grows, $\Lambda$ has an asymptotic chi-squared distribution with $q$ degrees of freedom. In this setting, “asymptotic” means that the approximation is justified in the limit of many independent subjects; it is not an exact finite-sample result.

Correct nesting and regular restrictions. The null is represented within the alternative, and its $q$ restrictions are locally independent. The null value is an interior point of the alternative parameter space.
Identifiability. Distinct parameter values correspond locally to distinguishable observed-data distributions. Parameters that disappear or become undefined under the null can break this condition.
Smoothness and information. The observed log likelihood is sufficiently smooth near the true parameter, and the information matrix for the identifiable parameters is finite and nonsingular.
Independent sampling and a suitable model. Subjects contribute independent observations under the stated sampling scheme, and the likelihood correctly represents the relevant data and censoring process.
Reliable maximization. Both fits reach their appropriate maxima, and numerical integration and optimization are accurate enough that the log-likelihood difference is trustworthy.

The chi-squared reference can fail when a variance is tested at zero, when a correlation or association parameter is unidentified under the null, or when constraints place the null on the boundary. It can also be unreliable in small samples or when the likelihood surface is poorly behaved. In those situations, the statistic can still be calculated, but the usual chi-squared p-value is not automatically justified.

Why the Statistic Has a Chi-Squared Limit

A short local argument explains the usual result. Write the parameter as $(\boldsymbol\eta,\boldsymbol\psi)$, where $\boldsymbol\eta$ contains nuisance parameters and $H_0$ fixes the $q$-dimensional vector $\boldsymbol\psi$ at an interior value. Near the true parameter, a regular log likelihood is approximately quadratic. After accounting for estimation of $\boldsymbol\eta$, the relevant score for $\boldsymbol\psi$ is its efficient score, $U_{\mathrm{eff}}$, and its corresponding information is $I_{\mathrm{eff}}$.

2\{\ell(\widehat{\boldsymbol\theta})-\ell(\widehat{\boldsymbol\theta}_0)\} = U_{\mathrm{eff}}^{\mathsf T}I_{\mathrm{eff}}^{-1}U_{\mathrm{eff}}+o_p(1).

Under $H_0$, the efficient score divided by the square root of its information converges to a $q$-dimensional standard normal vector. If $Z\sim N(\mathbf 0,I_q)$, then $Z^{\mathsf T}Z\sim\chi^2_q$, since it is the sum of squares of $q$ independent standard normal variables. The displayed approximation therefore gives $\Lambda\overset{d}{\longrightarrow}\chi^2_q$. The argument depends on the regularity conditions: boundary points or nonidentifiable parameters do not generally have this same quadratic, unconstrained limit.

Worked Example: Testing a Single Association Parameter

Consider a hypothetical joint model whose alternative includes an association parameter $\alpha$, while the null sets $\alpha=0$. Assume this is an interior value, all other parameters remain identifiable under the null, and the required regularity conditions hold. The maximized observed log likelihoods are $-809.89$ under the alternative and $-812.40$ under the null.

There is one restriction, so $q=1$. The statistic is

\Lambda=2\{-809.89-(-812.40)\} =2(2.51)=5.02.

Using the $\chi^2_1$ reference distribution gives the upper-tail p-value $P(\chi^2_1\geq5.02)\approx0.0251$. At significance level $\alpha=0.05$, reject $H_0:\alpha=0$. In this hypothetical analysis, the data provide evidence against no association under the specified model and its assumptions. The p-value is the probability, assuming the null and the reference conditions, of a statistic at least this large; it is not the probability that the null is true.

Worked Example: Jointly Testing Two Restrictions

Suppose a hypothetical alternative model estimates two association parameters, $\alpha_1$ and $\alpha_2$, and the null sets both to zero. All remaining parameters are identifiable under the null, and the null point is interior. The maximized log likelihoods are $-598.595$ for the alternative and $-603.200$ for the null.

The null imposes two independent restrictions, so $q=2$. The calculation is

\Lambda=2\{-598.595-(-603.200)\} =2(4.605)=9.210.

For two degrees of freedom, the upper-tail probability is $P(\chi^2_2\geq x)=\exp(-x/2)$. Hence

p=\exp(-9.210/2)=\exp(-4.605)\approx0.0100.

At the 0.05 significance level, reject the joint null. This result tests the two restrictions together; it does not establish that each parameter is individually nonzero. The chi-squared calibration here relies on the stated interior, identifiability, and regularity assumptions.

Boundary and Numerical Failure Modes

A common difficulty is testing whether a random-effects variance is zero. Variances cannot be negative, so zero is on the boundary of the parameter space. The usual interior-point argument does not apply. In some simple variance-component problems, a particular mixture of chi-squared distributions is justified under additional assumptions; it is not a universal correction for joint models. If other parameters, such as a random-effects correlation or an association parameter, are undefined when the variance is zero, the null distribution may require still different treatment.

Worked Example: A Variance Test at the Boundary

In a hypothetical joint-model fit, suppose the null fixes a random-intercept variance $\tau^2$ to zero and the alternative estimates $\tau^2\geq0$. The maximized log likelihoods are $-421.70$ under the null and $-419.90$ under the alternative. The numerical statistic is

\Lambda=2\{-419.90-(-421.70)\} =2(1.80)=3.60.

This calculation is valid, but assigning $3.60$ a $\chi^2_1$ p-value just because one variance was fixed is not justified by the ordinary Wilks result: the null value $\tau^2=0$ is on the boundary. One must derive a reference distribution appropriate to this model and test, or use a carefully designed parametric bootstrap. If additional parameters cease to be identifiable under the null, even a simple one-half-$\chi^2_0$, one-half-$\chi^2_1$ mixture may not apply. The statistic alone does not settle the calibration.

Numerical errors can cause a different failure. The alternative fit should not have a worse maximized likelihood than the restricted fit. If it does, check the optimization, parameter constraints, and integration settings before interpreting the result. In joint models, numerical integration approximates each subject's likelihood contribution; inaccurate integration can shift the difference between two large log likelihoods enough to distort $\Lambda$.

Fit both models with the same data, likelihood definition, and treatment of censoring.
Check that the restricted model can be obtained by imposing the stated null restrictions on the alternative.
Use multiple starting values or other convergence checks when the likelihood may have local maxima.
Inspect convergence information and estimates near parameter boundaries; a successful algorithm termination alone does not establish regularity.
Reassess the likelihood difference when increasing numerical-integration accuracy, especially if the two fitted log likelihoods are close.

How the LRT Relates to Other Tests

The Wald test uses an estimate and its estimated standard error to assess distance from the null. The score test evaluates the likelihood's slope at the null fit, without requiring a full alternative fit. The LRT compares the maximized likelihoods directly. Under regularity and large samples, these tests often agree approximately, but their finite-sample results can differ. Wald tests can behave poorly when estimates are near a boundary; score tests can be sensitive to the null fit; and an LRT requires two reliable optimizations. None of them removes the need to check identifiability, model assumptions, or the validity of the reference distribution.

A profile likelihood is useful for examining how the likelihood changes as a parameter of interest varies while nuisance parameters are re-optimized. An LRT can be viewed as a comparison of the unrestricted maximum with the maximum subject to the null restriction. A likelihood-ratio confidence set may use the same likelihood-drop principle, but its coverage also depends on suitable calibration. These methods describe evidence under the model; they do not make an association causal.

Key takeaway: For nested joint models, compute twice the difference between the unrestricted and restricted maximized observed log likelihoods. Use a $\chi^2_q$ reference only when the null imposes $q$ regular, interior, identifiable restrictions and the likelihood fits are reliable.

Check Your Understanding

Answer each question, then compare with the solutions below.

The alternative model has maximized log likelihood $-250.0$; the nested null model has maximized log likelihood $-253.1$ and imposes one regular restriction. Calculate the statistic and state its reference degrees of freedom.
A null hypothesis fixes three independently varying association parameters at specified interior values. How many degrees of freedom are used under the usual regularity conditions?
Why is a chi-squared reference not automatically valid when testing a variance equal to zero?
The restricted model has a larger maximized log likelihood than the alternative fit. Name two checks to make before reporting an LRT.
A joint test of two parameters is significant. Does that show that each parameter is individually nonzero?

Solutions. (1) $\Lambda=2\{-250.0-(-253.1)\}=2(3.1)=6.20$, with $q=1$ degree of freedom. (2) The usual reference is $\chi^2_3$, provided the restrictions are independent and the regularity conditions hold. (3) Zero is a boundary value for a nonnegative variance, so the interior-point conditions behind the ordinary chi-squared result fail; the appropriate null distribution must be justified separately. (4) Check that the models are nested and use the same likelihood and data, and check convergence, optimization starts, constraints, and integration accuracy. (5) No. The joint test assesses the restrictions together; individual conclusions require appropriate individual tests or intervals.