Likelihood-Ratio Tests for Nested Mechanisms

Why Compare Nested Mechanisms?

A mechanistic model may describe a biological system with a core pathway, while a second model adds a candidate pathway or interaction. If the core model is contained within the expanded model, the models are nested: setting the added mechanism’s parameter or parameters to particular null values recovers the core model. The likelihood-ratio test (LRT) asks whether the best fit available in the expanded model is sufficiently better to be inconsistent with those null values, given the assumed observation model.

The test compares maximized likelihoods, not the biological plausibility of the mechanisms. Its interpretation depends on the likelihood being appropriate for the data and on regularity conditions that justify the reference distribution. The earlier material on nested and non-nested QSP model relationships distinguishes this comparison from comparisons between models that cannot be written as restrictions of one another.

Definition: Let $M_0$ be a reduced model and $M_1$ an expanded model, with $M_0$ obtained by fixing $q$ parameters of $M_1$ to their null values. For data $y$, let $\ell_0$ and $\ell_1$ be the log likelihoods maximized over the respective parameter spaces. The likelihood-ratio statistic is $$ \Lambda = 2(\ell_1-\ell_0). $$ A large $\Lambda$ indicates that the expanded model fits better than the reduced model by more than would typically be expected from the added flexibility under $H_0$.

Reference Distribution and Assumptions

Under $H_0$, the models differ only because the $q$ added parameters are fixed at their null values. If the models satisfy the regularity conditions for Wilks’ theorem, then, in large samples, $\Lambda$ has an approximate chi-square distribution with $q$ degrees of freedom. The p-value is the probability, under that reference distribution, of a statistic at least as large as the observed value.

The usual result requires a correctly specified likelihood for the same observations in both fits; a genuinely nested reduced model; identifiable parameters; a sufficiently large sample; and a smooth likelihood with nonsingular information near the null parameter values. The null values must be interior points of the parameter space. The added parameters must be estimable under the expanded model, rather than being parameters whose values cannot be distinguished from other model quantities under $H_0$. When measurements are correlated or repeated within experimental units, the likelihood must represent that joint data structure rather than treating dependent measurements as independent.

The degrees of freedom are the number of independent, identifiable restrictions imposed by the null, not automatically the number of parameters printed in a model file. For a single freely varying pathway-strength parameter set to zero, $q=1$. For multiple added parameters, count the independent restrictions. If parameters are constrained or unidentified under the null, the ordinary chi-square result may not apply; in particular, a null value on the boundary violates the interior-point condition. The appropriate treatment of boundary parameters is a distinct issue.

Why the Chi-Square Approximation Appears

A local quadratic approximation explains the result. Write the expanded parameter as $(\psi,\lambda)$, where $\psi$ contains nuisance parameters shared with the reduced model and $\lambda$ is the $q$-vector of added parameters. Under $H_0$, $\lambda=0$. Near the true parameter, the log likelihood is approximately a quadratic function of the parameter displacement. After accounting for estimation of $\psi$, the part of the likelihood gain due to $\lambda$ has the form below.

\Lambda \ \approx\ Z^\mathsf{T}Z,\qquad Z\sim N(0,I_q)\ \text{under }H_0.

The sum of squares of $q$ independent standard normal variables has a chi-square distribution with $q$ degrees of freedom. This gives the large-sample reference distribution. The approximation relies on the likelihood behaving smoothly enough that the local quadratic expansion is valid and the estimated parameters behave approximately normally. It is not a universal property of every model comparison.

Worked Example: Interpreting a Likelihood Gain with Three Added Parameters

Suppose a hypothetical calibration fits a reduced mechanism and an expanded, nested mechanism to the same observations using the same likelihood. The expanded mechanism adds three identifiable, unconstrained parameters. The maximized log likelihoods are $\ell_0=-52.4$ and $\ell_1=-47.9$. Assume the regularity conditions above are reasonable.

The statistic is $$ \Lambda=2\{\ell_1-\ell_0\} =2\{-47.9-(-52.4)\} =2(4.5)=9.00. $$

The null reference distribution is approximately $\chi^2_3$, since three independent parameters are fixed under $H_0$. Its upper-tail probability at 9.00 is approximately $0.0293$. At significance level $\alpha=0.05$, this is evidence against the reduced mechanism under the specified likelihood and regularity assumptions. It does not establish that the expanded mechanism is biologically true, nor does it show that every added parameter is individually supported.

Worked Pathway-Addition Example

Consider a small hypothetical experiment measuring a response at five conditions. A baseline QSP mechanism predicts a common response level $a$. The expanded mechanism adds a pathway contribution represented by a prespecified activity profile $z_i$ and a freely varying strength $\beta$. Assume independent normal measurement errors with known standard deviation $1$ response unit. The profile is centered, so its values sum to zero.

Worked Example: Testing an Added Pathway

The hypothetical observations are $y=(-1.8,-0.9,0.2,1.1,1.9)$ response units, and the corresponding pathway-profile values are $z=(-2,-1,0,1,2)$. The reduced model is $y_i=a+\epsilon_i$. The expanded model is $y_i=a+\beta z_i+\epsilon_i$, with $\epsilon_i\sim N(0,1)$. Thus $H_0:\beta=0$, and the alternative allows $\beta$ to vary freely. Because the error variance is specified and the models are linear Gaussian models, the likelihood calculations below are exact for these assumptions.

Under the reduced model, the maximum-likelihood estimate of $a$ is the sample mean: $$ \hat{a}_0=\bar{x}=\frac{-1.8-0.9+0.2+1.1+1.9}{5}=0.10. $$ The residual sum of squares is $$ RSS_0=(-1.9)^2+(-1.0)^2+(0.1)^2+(1.0)^2+(1.8)^2=8.86. $$

In the expanded model, because $\sum_i z_i=0$, the estimates are $\hat{a}_1=0.10$ and $$ \hat{\beta}=\frac{\sum_i z_i(y_i-\bar{x})}{\sum_i z_i^2} =\frac{9.4}{10}=0.94. $$ The fitted values are $(-1.78,-0.84,0.10,1.04,1.98)$, giving residuals $(-0.02,-0.06,0.10,0.06,-0.08)$. Therefore $$ RSS_1=(-0.02)^2+(-0.06)^2+(0.10)^2+(0.06)^2+(-0.08)^2=0.024. $$

For independent normal errors with known variance $\sigma^2=1$, the log likelihood differs between models only through the residual sum of squares. Consequently, $$ \Lambda=\frac{RSS_0-RSS_1}{\sigma^2} =\frac{8.86-0.024}{1}=8.836. $$ There is one added parameter, so the null reference distribution is $\chi^2_1$. The upper-tail p-value is approximately $0.00295$. Under this model and its assumptions, the data provide evidence against $\beta=0$; the estimated pathway contribution is $0.94$ response units per unit of pathway profile. This test does not establish that the pathway causes the response, and the conclusion is conditional on the assumed profile and error model.

When the Error Variance Is Estimated

If the same normal error variance is unknown and estimated separately within each model, the likelihood gain is not simply the reduction in residual sum of squares divided by a known variance. For Gaussian models with $n$ independent observations, the maximum-likelihood variance estimate is $RSS/n$. Substituting these estimates into the likelihood gives the following LRT statistic:

\Lambda=n\log\left(\frac{RSS_0}{RSS_1}\right).

This form illustrates why the observation model must be stated before calculating a likelihood ratio. It also illustrates that a reduction in residual sum of squares is not itself the LRT statistic when variance is estimated. The chi-square reference distribution remains an asymptotic result under regularity conditions; the formula does not make that approximation exact.

Worked Example: Comparing Fits with an Estimated Common Variance

Suppose two hypothetical, nested Gaussian models are fit to $n=40$ independent observations with a common variance estimated within each model. The reduced model has $RSS_0=64$, and the expanded model has $RSS_1=52$. Suppose one identifiable parameter is added and the regularity conditions are plausible.

The statistic is $$ \Lambda=40\log\left(\frac{64}{52}\right) =40\log(1.23077) =40(0.20764)=8.306. $$ The reference distribution is approximately $\chi^2_1$, giving an upper-tail p-value of about $0.00395$. This is evidence against the reduced model at $\alpha=0.05$, conditional on the Gaussian likelihood and the assumptions used to fit both models. Since the variance is estimated, using $(64-52)/52$, or treating the known-variance formula as applicable, would give the wrong LRT statistic.

Diagnostics, Alternatives, and Limitations

Before interpreting an LRT, verify that the reduced model is obtainable from the expanded one by the stated restrictions, and that both fits use the same data and observation model. Confirm that each optimizer has converged to a credible maximum: inspect convergence messages, compare multiple starting values where appropriate, and check that the expanded model’s maximized log likelihood is not lower than the reduced model’s. For nested models, the expanded parameter space contains the reduced one, so a lower reported maximum usually signals an optimization failure or inconsistent fitting.

Also examine whether the assumed likelihood is defensible. A poor residual pattern, an implausible variance structure, or unmodelled dependence can undermine the p-value even if the arithmetic is correct. The LRT measures relative fit within the specified model family; it cannot diagnose all forms of model misspecification or guarantee useful predictions outside the observed conditions.

Wald tests and score tests are other likelihood-based tests of parameter restrictions. A Wald test evaluates an estimated parameter relative to its estimated uncertainty; a score test evaluates the likelihood’s local slope at the null fit. Under regularity conditions, these tests and the LRT have the same large-sample chi-square calibration, but their finite-sample behavior and sensitivity to parameterization can differ. The LRT directly compares optimized fits and is often convenient for nested mechanisms. For non-nested models, this ordinary LRT is not appropriate; a different comparison strategy is needed.

A small p-value is not a measure of the size or biological importance of the pathway effect. A large p-value is not proof that the reduced mechanism is correct: it may reflect limited information, weakly discriminating predictions, or parameter identifiability problems. Report the statistic, degrees of freedom, p-value, model restriction, and effect estimates in context. Treat the result as evidence conditional on the models and assumptions, rather than as a stand-alone mechanism-selection rule.

Common Pitfall

Common mistake: “The expanded model has a higher likelihood, so the added pathway is supported.” Maximized likelihood cannot decrease when a nested model is expanded, because the expanded model can reproduce the reduced fit. The relevant question is whether the likelihood gain is large relative to the gain expected under $H_0$, using a valid reference distribution. Calculate $2(\ell_1-\ell_0)$, use the correct degrees of freedom, and check the regularity conditions before interpreting a p-value.

Check Your Understanding

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

A nested expanded model adds two identifiable, unconstrained parameters. Under regularity conditions, what is the LRT reference distribution under $H_0$?
For maximized log likelihoods $\ell_0=-30.6$ and $\ell_1=-27.6$, calculate the LRT statistic. If one parameter was added, what is the approximate p-value?
Why can an ordinary chi-square LRT calibration fail when a null parameter value is on the boundary of its allowed range?
In a Gaussian model with known measurement standard deviation $\sigma$, what is the LRT statistic in terms of the two residual sums of squares?

Solutions. (1) The reference distribution is approximately $\chi^2_2$, with degrees of freedom equal to the two independent restrictions under the null. (2) $\Lambda=2\{-27.6-(-30.6)\}=6.00$. With one degree of freedom, the upper-tail p-value is approximately $0.0143$. (3) Wilks’ usual result assumes the null values are interior points of a smooth parameter space; a boundary violates that condition, so the standard chi-square calibration is not justified. (4) For independent Gaussian errors with known variance, $\Lambda=(RSS_0-RSS_1)/\sigma^2$.