Testing Independence Against Endpoint Dependence

Why Test Endpoint Independence?

A joint PK/PD model can represent several endpoints together while allowing their outcomes to be conditionally independent, or it can include an additional dependence structure. Testing independence asks whether the data provide evidence that the joint endpoint distribution needs that structure, after accounting for covariates and exposure included in the model. This matters because dependence can affect joint predictions, uncertainty, and the interpretation of information borrowing across endpoints.

The null hypothesis is always relative to the conditioning set and model. For example, endpoints may be independent conditional on exposure, treatment, and baseline covariates, yet associated marginally if those covariates are related to both endpoints. Thus, a test does not establish that endpoints are “independent in general”; it assesses a specified conditional model.

Definition: Let $Y_{i1}$ and $Y_{i2}$ be two endpoints for subject $i$, and let $X_i$ denote the covariates and exposure variables conditioned on in the model. Conditional endpoint independence means $$ f(y_{i1},y_{i2}\mid X_i)=f_1(y_{i1}\mid X_i)f_2(y_{i2}\mid X_i). $$ A dependent joint model relaxes this factorization through an association parameter, shared latent effect, residual covariance, copula parameter, or other specified dependence structure.

The independent model and dependent model must be fitted to the same observations and use compatible endpoint distributions, covariates, exposure assumptions, and likelihood contributions. The independence model is then a special case of the dependence model, obtained by fixing the association parameter or parameters to their null values. This nested structure permits a likelihood-ratio test (LRT), provided the null lies in the regular interior of the parameter space. If the null is on a boundary, the usual chi-square calibration may not apply.

Likelihood-Ratio Tests Under Regular Conditions

Write $\ell_I$ for the maximized log likelihood under independence and $\ell_D$ for the maximized log likelihood under dependence. Since the independent model is nested within the dependent model, maximizing over the larger parameter space cannot produce a smaller likelihood. The likelihood-ratio statistic is

\Lambda=2(\ell_D-\ell_I).

Under the null hypothesis, the usual Wilks result gives an approximate chi-square reference distribution with degrees of freedom equal to the number $q$ of independent restrictions, when the null parameter value is interior, identifiable, and the regularity conditions for maximum-likelihood inference hold. A small upper-tail p-value indicates that the observed improvement in fit would be unusual if those restrictions were true. It does not give the probability that independence is true.

Regular LRT conditions: The models are nested and use the same data and likelihood basis; the null parameter is an interior point of the alternative parameter space; parameters are identifiable under the null; and the likelihood is sufficiently smooth with a well-behaved information matrix. Under these conditions, $\Lambda$ is approximately distributed as $\chi^2_q$ under $H_0$.

A brief justification explains why the number of restrictions determines the degrees of freedom. Near the true null value, a smooth log likelihood is locally approximated by a quadratic function of the parameter estimates. The unrestricted fit can move in $q$ additional locally identifiable directions; twice the gain in log likelihood is asymptotically the squared length of a $q$-dimensional standard normal vector. That squared length has a $\chi^2_q$ distribution. If the allowed parameter values are constrained by a boundary, the local parameter space is not an unrestricted $q$-dimensional region, so this argument no longer yields the ordinary chi-square law.

Worked Example: A Regular Two-Parameter Dependence Test

Suppose a hypothetical joint model for two non-continuous PD endpoints is fitted to $n=75$ subjects. The independent model has maximized log likelihood $\ell_I=-84.0$ and 6 estimated parameters. The dependent model has $\ell_D=-80.7$ and 8 estimated parameters. Assume the two added association parameters are identifiable under independence, and that both null values are interior points of their allowed ranges.

The models are nested and differ by $q=2$ restrictions. The likelihood-ratio statistic is $$ \Lambda=2\{\ell_D-\ell_I\}=2\{-80.7-(-84.0)\}=2(3.3)=6.6. $$

Under the stated regularity assumptions, compare this with $\chi^2_2$. For 2 degrees of freedom, the upper-tail probability is $P(\chi^2_2\geq 6.6)=\exp(-6.6/2)=0.0369$. At $\alpha=0.05$, this is evidence against conditional independence in favor of the specified dependence model. It is evidence of association after conditioning on the model covariates, not evidence that one endpoint causes the other.

The assumptions matter: if either added parameter is unidentified under independence, or if a null value is on a boundary, the $\chi^2_2$ calibration is not justified merely because two parameters were added.

When the Null Is on a Boundary

A common dependence structure uses a shared random effect with variance $\tau^2\geq 0$. Independence corresponds to $\tau^2=0$. Because a variance cannot be negative, the null value is at the boundary of the allowable parameter space. In a simple case with one variance component and otherwise regular, identifiable nuisance parameters, the asymptotic null distribution of the LRT is a 50:50 mixture: it has probability one-half at zero and probability one-half according to $\chi^2_1$. This is often written $\tfrac12\chi^2_0+\tfrac12\chi^2_1$, where $\chi^2_0$ denotes a point mass at zero.

For a positive observed statistic $\Lambda$, the mixture upper-tail p-value is one-half the ordinary $\chi^2_1$ upper-tail probability. This result is not a universal correction for every dependence model. Multiple variance components, correlations undefined under the null, or additional constraints can produce other limiting distributions. In such cases, derive an appropriate reference distribution or use a carefully designed parametric bootstrap.

Worked Example: Testing a Shared Random-Effect Variance

In a hypothetical study, a joint model includes a shared subject-level random effect with variance $\tau^2\geq0$. The independence fit has $\ell_I=-52.000$, and the dependent fit has $\ell_D=-50.645$. Suppose this is the simple one-variance-component case described above, with the other parameters identifiable and regular under $H_0:\tau^2=0$.

The statistic is $$ \Lambda=2\{-50.645-(-52.000)\}=2(1.355)=2.710. $$ For the boundary-mixture reference distribution, the p-value is $$ P(\Lambda_{\mathrm{mix}}\geq2.710)=\tfrac12P(\chi^2_1\geq2.710) \approx\tfrac12(0.0998)=0.0499. $$

At the 0.05 level, this is just below the threshold and provides modest evidence against the zero-variance independence model under the stated mixture approximation. The result is sensitive to calibration and model assumptions; reporting the statistic and boundary reference distribution is more informative than reporting only “significant” or “not significant.”

Information Criteria: Fit Balanced Against Complexity

Information criteria compare relative fit while penalizing additional parameters. For maximized log likelihood $\ell$, parameter count $k$, and sample size $n$, the Akaike information criterion and Bayesian information criterion are

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

Lower values indicate a preferred model among the candidates being compared; neither criterion is a hypothesis test or a p-value. AIC and BIC can rank models differently because their complexity penalties differ. BIC’s penalty depends on sample size and is often stronger than AIC’s for larger $n$. Both require comparable likelihoods fitted to the same data; using different observations or incompatible likelihood definitions can make the comparison misleading.

Worked Example: AIC and BIC Give Different Rankings

Use the hypothetical fits from the regular LRT example: $-2\ell_I=168.0$, $k_I=6$, $-2\ell_D=161.4$, and $k_D=8$, with $n=75$. The AIC values are $$ \mathrm{AIC}_I=168.0+2(6)=180.0,\qquad \mathrm{AIC}_D=161.4+2(8)=177.4. $$ The dependent model has AIC lower by $180.0-177.4=2.6$, so AIC favors it among these two candidates.

For BIC, $\log(75)=4.31749$, giving $$ \mathrm{BIC}_I=168.0+6(4.31749)=193.905, $$ $$ \mathrm{BIC}_D=161.4+8(4.31749)=195.940. $$ Thus $\mathrm{BIC}_D-\mathrm{BIC}_I=195.940-193.905=2.035$, so BIC favors the independent model by 2.035 points. The two criteria differ in their complexity penalties; neither result overrides the LRT’s null-hypothesis interpretation. In particular, these calculations use the deviance terms $-2\ell$, not AIC values as though they were deviance.

A Direct Table-Based Test and Its Interpretation

For two binary endpoints, a simple contingency table illustrates the independence hypothesis directly. Under independence, the expected count in each cell is its row total multiplied by its column total, divided by the grand total. The likelihood-ratio statistic for independence in a $2\times2$ table is $G^2=2\sum O\log(O/E)$, where $O$ and $E$ are the observed and expected cell counts. When expected counts are adequate and subjects contribute independent observations, this has an approximate $\chi^2_1$ reference distribution. In a PK/PD analysis, however, covariates, repeated observations, and subject-level dependence may require a joint likelihood model rather than a raw table test.

Worked Example: Binary Endpoints in a Hypothetical Table

Consider 100 hypothetical subjects classified on two binary PD endpoints. The observed counts are 30 subjects with both endpoints positive, 20 with endpoint 1 positive only, 15 with endpoint 2 positive only, and 35 with neither positive. The row totals are 50 and 50; the column totals are 45 and 55. Under independence, expected counts are $50(45)/100=22.5$, $50(55)/100=27.5$, $50(45)/100=22.5$, and $50(55)/100=27.5$. All expected counts exceed 5.

The likelihood-ratio statistic is $$ G^2=2\left[ 30\log\left(\frac{30}{22.5}\right)+ 20\log\left(\frac{20}{27.5}\right)+ 15\log\left(\frac{15}{22.5}\right)+ 35\log\left(\frac{35}{27.5}\right) \right]\approx9.240. $$ With 1 degree of freedom, the approximate p-value is 0.00237. The table therefore provides evidence against independence for these hypothetical subjects. The sample odds ratio is $(30\times35)/(20\times15)=3.50$, describing positive association in this table. Neither the p-value nor the odds ratio establishes that one endpoint causes the other; common exposure, unmeasured factors, or selection could contribute to the association.

Diagnostics, Alternatives, and Common Pitfalls

Before interpreting a test, check that both fits converged to plausible maxima, use the same subjects and likelihood contribution, and differ only in the dependence restrictions being tested. Inspect parameter estimates, standard errors, gradients, and sensitivity to starting values. A suspiciously large likelihood gain, unstable association estimates, or estimates at extreme parameter values may signal weak identifiability or numerical problems. Sparse outcome patterns can make asymptotic tests unreliable. The later material on residual and predictive checks addresses model adequacy beyond the comparison of these nested fits.

The principal alternatives answer related but different questions. AIC and BIC compare relative model support with different complexity penalties; they do not test a null hypothesis. Wald tests assess an estimated association parameter relative to its estimated standard error and can perform poorly when estimates are near a boundary or likelihoods are asymmetric. A score test evaluates local evidence at the null without fitting the full alternative, but still needs a valid null reference distribution. Parametric bootstrap calibration can be useful when asymptotic boundary theory is complicated: simulate data under the fitted independence model, refit both models to each simulated data set, and compare the observed statistic with the simulated null distribution. The simulation must preserve the design and refit procedure, and failed or nonconvergent fits require attention rather than silent omission.

Wrong reasoning: “The dependent model has a larger likelihood, so dependence is proven.” A nested model cannot fit worse, even when its added parameters describe only random variation.
Correction: Quantify the gain with an appropriate test or information criterion, check the assumptions and calibration, and describe the conclusion as evidence for or against the specified conditional independence model.
Wrong reasoning: “Every LRT uses a chi-square distribution with one degree of freedom per added parameter.” This fails when a null parameter is on a boundary or is not identifiable under the null.
Correction: Establish the null parameter geometry first. Use the regular chi-square result only under its conditions; otherwise use boundary-aware theory or justified simulation.

Key takeaway: Testing endpoint independence is a comparison of a specified factorized joint model with a specified dependence model. The likelihood gain, its reference distribution, and the interpretation all depend on nesting, identifiability, boundary conditions, and the conditioning set. Evidence of association is not evidence of causation.

Check Your Understanding

Answer each question, then check the concise solutions below.

What does conditional endpoint independence mean when covariates and exposure are included?
A dependent model improves the maximized log likelihood by 1.8 and adds one regular, identifiable parameter. Calculate the LRT statistic and state its reference distribution.
Why does a shared random-effect variance equal to zero require boundary-aware inference?
In the example with 75 subjects, which model does BIC favor, and by how many points?
Can a small p-value for endpoint dependence establish that one endpoint causes the other? Explain.

Solutions. (1) The conditional joint density factors into the product of the two conditional marginal densities given the specified covariates and exposure. (2) $\Lambda=2(1.8)=3.6$; under the stated regular conditions it is compared with $\chi^2_1$. (3) A variance is constrained to be nonnegative, so zero is a boundary value and the ordinary interior-parameter chi-square derivation does not generally apply; the simple one-component case has a 50:50 point-mass/$\chi^2_1$ mixture. (4) BIC favors independence by 2.035 points. (5) No. The test supports association under the specified model; confounding, shared causes, or other mechanisms can produce association without causation.