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Competing Risks and Recurrent Events · Tutorial 424 of 697

Comparing Cause-Specific Association Models

Compare cause-specific association structures with appropriate likelihood criteria, and judge the results only when models use the same observed event and censoring information.

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What You'll Learn

  • Define nested restrictions on cause-specific association parameters.
  • Use a likelihood-ratio test when nested-model regularity conditions hold.
  • Compare nonnested structures with AIC or BIC, not a routine likelihood-ratio test.
  • Interpret estimates and uncertainty for cause-specific association contrasts.
  • Check that candidate models use the same event, censoring, and subject information.
  • Recognize convergence, integration, and post-selection limitations.

Why Compare Association Structures?

A competing-risks joint model can allow a longitudinal marker or its latent trajectory to be associated with each cause-specific hazard. The scientific and statistical question is often not simply whether an association exists, but whether the structure of association should be shared across causes or allowed to differ. A common association parameter is more parsimonious; separate parameters can represent cause-specific differences. Other candidate structures may link risk to different summaries of the longitudinal process.

A comparison is meaningful only when it contrasts models for the same observed information. In particular, the likelihoods must account for the same subjects, longitudinal records, event times, event-cause information, and censoring. This requirement is especially important when event causes are uncertain or misclassified: changing how those records enter the likelihood can change the data being modeled, not merely the association structure.

Define the Candidate Structures and Likelihood

For clarity, consider \(K\) competing event causes. For subject \(i\), let \(T_i\) be the observed event or censoring time, let \(\delta_i=1\) indicate an observed event and \(\delta_i=0\) indicate right censoring, and let \(J_i\) denote the observed cause when \(\delta_i=1\). Let \(b_i\) represent the subject-specific random effects in the longitudinal and event model. A familiar cause-specific hazard form is

$$ \lambda_{ik}(t\mid b_i) = \lambda_{0k}(t)\exp\{x_i(t)^\mathsf{T}\beta_k+\alpha_k m_i(t;b_i)\}, \qquad k=1,\ldots,K. $$

Here \(m_i(t;b_i)\) is the modeled longitudinal quantity entering the hazard, \(\alpha_k\) is its association parameter for cause \(k\), and \(\lambda_{0k}(t)\) is the cause-specific baseline hazard. The comparison in this tutorial concerns how the association parameters or association terms are structured; other components should be held fixed unless their variation is explicitly part of the candidate model.

Definition: A cause-specific association structure specifies which longitudinal quantities enter each cause-specific hazard and which association parameters are shared, constrained, or estimated separately across causes. Two models are nested when every parameter value allowed by the smaller model is obtained from the larger model by imposing restrictions.

For example, the common-association model imposes \(\alpha_1=\cdots=\alpha_K=\alpha\), while the separate-association model estimates \(\alpha_1,\ldots,\alpha_K\). The former is nested in the latter. By contrast, a model using current marker value and a model using marker slope may be nonnested: neither is generally obtained merely by fixing or equating parameters in the other.

With right-censored event times and known causes for observed events, a subject's event contribution conditional on \(b_i\) is

$$ \left\{\lambda_{iJ_i}(T_i\mid b_i)\right\}^{\delta_i} \exp\left\{-\sum_{k=1}^{K}\int_0^{T_i}\lambda_{ik}(u\mid b_i)\,du\right\}. $$

The observed-data likelihood combines this contribution with the longitudinal-data contribution and integrates over the random-effects distribution. Denote the resulting log likelihood by \(\ell\). If the event-cause information is uncertain, its likelihood contribution must represent that uncertainty consistently in every candidate model. The same principle applies to delayed entry or interval censoring: use the appropriate observed-data likelihood for the actual observation scheme in every model being compared.

Nested Models: Likelihood-Ratio Comparison

Let \(M_R\) be a restricted model and \(M_F\) a larger model containing it. Write their maximized log likelihoods as \(\ell_R\) and \(\ell_F\). Under the null hypothesis that the restrictions hold, a likelihood-ratio statistic is

$$ D=2(\ell_F-\ell_R). $$

Under standard regularity conditions, \(D\) is approximately distributed as a chi-square random variable with \(q\) degrees of freedom, where \(q\) is the number of independent restrictions. These conditions include correctly specified likelihoods, a true parameter in the interior of the larger model's parameter space, identifiable parameters, and sufficiently regular large-sample behavior. A test of equal association parameters across two causes usually imposes one restriction, \(\alpha_1-\alpha_2=0\), so \(q=1\).

Theorem: For regular nested models fitted by maximum likelihood to the same observed data, \(D=2(\ell_F-\ell_R)\) provides an asymptotic likelihood-ratio test of the \(q\) restrictions, using a \(\chi^2_q\) reference distribution. The test does not have this routine reference distribution when relevant regularity conditions fail, for example when a tested parameter lies on a boundary or is not identifiable under the null.

The justification follows from the local quadratic approximation to the log likelihood around its maximum. Under the null, the \(q\) directions released by the larger model behave asymptotically like \(q\) independent standard-normal components after appropriate scaling. The increase in maximized log likelihood is approximately half the sum of their squares; doubling it gives a sum of \(q\) squared standard-normal variables, which has a \(\chi^2_q\) distribution. This is an asymptotic argument, not a guarantee for a small sample or a poorly identified model.

Worked Example: Testing a Shared Association Against Two Cause-Specific Associations

Suppose a hypothetical study has two causes and the same event and censoring information is used in both fits. The shared-association model has maximized log likelihood \(\ell_R=-214.6\). The model with separate associations has \(\ell_F=-212.8\). It adds one free association parameter, and assume the fits are regular and adequately converged.

The null restriction is \(\alpha_1=\alpha_2\), so \(q=1\). The statistic is

$$ D=2\{(-212.8)-(-214.6)\}=2(1.8)=3.60. $$

For a \(\chi^2_1\) reference distribution, the upper-tail p-value is approximately \(0.058\). At significance level \(\alpha=0.05\), this test does not reject the equality restriction. This is not proof that the associations are identical: the result says that this dataset does not provide sufficient evidence against equality at that threshold.

Suppose the separate-model estimates are \(\hat{\alpha}_1=0.80\) and \(\hat{\alpha}_2=0.42\), so the estimated contrast is \(0.80-0.42=0.38\). If its standard error is \(0.20\), a Wald statistic is \(0.38/0.20=1.90\), with a two-sided normal-reference p-value of about \(0.057\). The close agreement with the likelihood-ratio result is reassuring, though the two procedures need not agree exactly in finite samples. The association parameters are on the hazard's log-relative-risk scale per unit of \(m_i(t;b_i)\); their numerical interpretation therefore depends on the units of that longitudinal quantity.

Nonnested Models: Information Criteria

A routine likelihood-ratio test is not justified for nonnested models. A likelihood difference alone does not have the chi-square reference distribution used for a nested-model test. A common likelihood-based comparison uses the Akaike information criterion (AIC) or Bayesian information criterion (BIC):

$$ \mathrm{AIC}=-2\ell+2p, \qquad \mathrm{BIC}=-2\ell+p\log(n), $$

where \(p\) is the number of estimated parameters in the likelihood being compared and \(n\) is the number of independent subjects for the usual subject-level BIC convention. Smaller values indicate a better tradeoff, under that criterion, between fit and complexity. AIC targets expected relative predictive performance under its assumptions; BIC applies a stronger complexity penalty as sample size grows and has a large-sample model-selection interpretation under specific conditions. Neither criterion supplies a test that a model is true.

Worked Example: Comparing Two Nonnested Association Forms by AIC

Consider two hypothetical models fitted to the same 180 subjects and the same longitudinal and competing-event records. Model A links the hazard to current marker value and has \(p_A=12\) estimated parameters with \(\ell_A=-305.0\). Model B uses marker slope instead and has \(p_B=13\) parameters with \(\ell_B=-301.5\). Neither model is a restricted version of the other.

For Model A,

$$ \mathrm{AIC}_A=-2(-305.0)+2(12)=610+24=634.0. $$

For Model B,

$$ \mathrm{AIC}_B=-2(-301.5)+2(13)=603+26=629.0. $$

Model B has the lower AIC by \(634.0-629.0=5.0\), so it is preferred among these two candidates by AIC. This is a relative ranking, not evidence that slope is the uniquely correct association or that the difference is statistically significant. The interpretation also depends on the models representing the same observed data and using comparable likelihoods.

Uncertainty, Comparability, and Diagnostics

Model rankings and parameter uncertainty answer different questions. For an estimated association \(\hat{\alpha}_k\), report its standard error and, where appropriate, a confidence interval such as \(\hat{\alpha}_k\pm1.96\,SE(\hat{\alpha}_k)\). A confidence interval describes uncertainty under the fitted model and its regularity assumptions; it does not account automatically for searching across many association structures. For a cause contrast, report the contrast itself and its standard error, using the covariance between estimates:

$$ SE(\hat{\alpha}_1-\hat{\alpha}_2) = \sqrt{\operatorname{Var}(\hat{\alpha}_1)+\operatorname{Var}(\hat{\alpha}_2) -2\operatorname{Cov}(\hat{\alpha}_1,\hat{\alpha}_2)}. $$

Do not calculate this standard error by treating estimates from a joint model as independent unless their covariance is actually zero. If the likelihood is asymmetric or a variance estimate is near a boundary, a profile-likelihood interval or a carefully designed subject-level bootstrap may be more reliable than a Wald interval. Resampling should preserve the independent sampling unit and refit the models using the same fitting procedure.

Before interpreting comparisons, check the following:

  • Same observed information. Confirm that subjects, longitudinal measurements, event times, censoring indicators, and event-cause treatment are the same. Differences in missing-data handling or cause classification can invalidate a direct comparison.
  • Same likelihood target. Compare maximized observed-data likelihoods, not a marginal likelihood for one model with a conditional likelihood for another. Keep the censoring and entry-time treatment consistent.
  • Convergence and integration. Review convergence messages, gradients, Hessian behavior, and sensitivity to numerical integration settings. A lower log likelihood can reflect failed optimization rather than a worse model.
  • Identification and sparsity. Cause-specific parameters may be weakly identified when few events occur for a cause. Very large standard errors, unstable estimates across starting values, or nearly singular Hessians are warnings that an apparent ranking may be fragile.
  • Model adequacy. AIC or BIC selects only among the candidates considered. Check whether the baseline hazards, covariate effects, event-cause assumptions, and longitudinal model are plausible; a preferred model can still fit poorly.

When fitting uses numerical integration over random effects, use settings accurate enough that small likelihood differences are not numerical noise. A practical check is to refit leading candidates with tighter integration settings and alternative reasonable starting values. If their rankings or estimates shift materially, report that instability rather than treating the smallest criterion value as decisive.

What the Comparison Can and Cannot Establish

The nested likelihood-ratio test asks whether the data are inconsistent with a specified set of restrictions, subject to its assumptions. AIC and BIC rank candidate models according to different fit-complexity criteria. Neither procedure establishes a causal effect, and a cause-specific association parameter remains a model-based association, not a causal claim. Likewise, a selected association structure does not establish that the underlying biological mechanism is shared or cause-specific.

Selection also adds uncertainty. If a structure is chosen after comparing many alternatives, the usual standard errors and confidence intervals computed as though that structure had been fixed in advance can understate total uncertainty. Present the candidate structures, criterion values, parameter estimates, and relevant uncertainty summaries together. If scientific conclusions change across plausible structures, that sensitivity is itself an important result.

Key takeaway: Use a likelihood-ratio test for a regular nested comparison, and use criteria such as AIC or BIC for nonnested candidates. In either case, first verify that the fits use the same observed event and censoring information and a comparable likelihood.

Worked Example: Why Different Event Records Cannot Be Compared Directly

Suppose two hypothetical fits report \(\ell_A=-180.0\) and \(\ell_B=-176.0\), but Model A excludes 6 subjects whose event causes are uncertain while Model B includes them through a cause-misclassification likelihood. The arithmetic difference would be \(2\{(-176.0)-(-180.0)\}=8.0\), but it is not a valid likelihood-ratio statistic for comparing association structures: the likelihoods do not describe the same observed information.

To make the comparison meaningful, refit both association structures using the same set of subjects and the same treatment of uncertain causes. If the shared-data refits yield \(\ell_A=-180.0\) and \(\ell_B=-178.7\), then the log-likelihood gain is \(1.3\), not \(4.0\). If these models are nested with one restriction and regularity conditions hold, the LR statistic is \(2(1.3)=2.60\), with a \(\chi^2_1\) p-value of about \(0.107\). If they are nonnested, do not use that statistic as a routine test; calculate AIC or BIC from the comparable fits instead.

Common Mistakes

  • Using a likelihood-ratio test for nonnested models. A difference in maximized log likelihood does not by itself have a chi-square reference distribution. Use an appropriate comparison method, such as AIC or BIC, for nonnested candidates.
  • Comparing likelihoods from different records. A larger likelihood can arise because the subjects or event-cause contributions differ. Refit using the same observed information before comparing.
  • Calling the lower AIC a significant winner. AIC is a relative criterion, not a hypothesis test or probability that a model is true. Describe the size of the difference and the candidate set.
  • Ignoring uncertainty in a cause contrast. Subtracting two estimates is not enough; its standard error depends on their covariance.
  • Interpreting an association as causal. A fitted cause-specific association describes the model's conditional relationship; causal interpretation requires additional design and assumptions.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. A shared-association model has \(\ell_R=-90.0\), and a two-parameter extension has \(\ell_F=-87.4\). What is the LR statistic and its degrees of freedom, assuming the extension releases two regular restrictions?
  2. Two nonnested models have the same likelihood and subject count. Model A has 10 parameters and Model B has 12. Which has lower AIC, and by how much?
  3. Why is a comparison invalid if one cause-specific model omits uncertain-cause subjects while the other includes them?
  4. Estimates for two cause associations are \(0.6\) and \(0.2\), with variances \(0.04\) and \(0.09\), and covariance \(0.03\). Find the standard error of their difference.

Solutions. (1) \(D=2\{(-87.4)-(-90.0)\}=5.2\), with \(df=2\). Its p-value would be obtained from a \(\chi^2_2\) distribution only if the models are nested, use the same data and likelihood, and meet regularity conditions. (2) Because the log likelihoods are equal, the AIC difference is the penalty difference: \(2(12-10)=4\). Model A has lower AIC by 4. (3) The fits do not describe the same observed information, so their likelihood difference can reflect the changed records rather than the association structure. Refit both under the same cause-uncertainty treatment. (4) The estimated difference is \(0.6-0.2=0.4\), and its variance is \(0.04+0.09-2(0.03)=0.07\), so its standard error is \(\sqrt{0.07}=0.265\) to three significant figures.

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