Why Nesting Matters
A likelihood-ratio statistic compares the best fit allowed by a null hypothesis with the best fit allowed by a larger model. The word larger matters: if every distribution allowed by the null is also allowed by the alternative, the two maximizations have a fixed ordering. That ordering gives the statistic its usual nonnegative form and is part of the setting in which Wilks' theorem supplies a chi-square calibration.
For two models that are not nested, one may still calculate and compare their maximized likelihoods. But the comparison is not automatically the usual likelihood-ratio test. There is no general containment-based ordering, and subtracting the two maxima does not, on its own, produce a statistic with a chi-square distribution. The issue is not whether the arithmetic can be done; it is whether the hypotheses and assumptions justify the proposed reference distribution.
Nested Models and the Likelihood-Ratio Statistic
In a regular parametric model, the null is often expressed as a restriction on a parameter vector: \(\theta\in\Theta_0\subseteq\Theta_1\). For example, an unrestricted model may estimate a probability \(p\), while the null fixes it at a specified value. The restricted fit and unrestricted fit use the same observed data and likelihood, but maximize over different parameter sets.
Proof. Every parameter value permitted by the null is also permitted by the alternative. Thus the null maximizes over a subset of the alternative's allowable values. A supremum over a subset cannot exceed the supremum over the full set, so \(L_0\leq L_1\). Dividing by the positive \(L_1\) gives \(\Lambda\leq1\), while likelihoods are nonnegative, so \(\Lambda\geq0\). For \(L_0>0\), taking logarithms gives \(\log\Lambda\leq0\), hence \(T=-2\log\Lambda\geq0\). If \(L_0=0\), the statistic is interpreted as infinite. \(\square\)
This ordering is a structural consequence of nesting; it is not a feature that must be checked after observing the data. For regular nested hypotheses, Wilks' theorem states that under the null, \(T\) converges in distribution to a chi-square random variable with degrees of freedom equal to the number of independent restrictions. The standard result requires, among other conditions, an identifiable model, an interior true parameter value, suitable differentiability and information conditions, and a smooth restriction of the stated rank. Boundary points, nonidentifiability, and other failures of regularity can invalidate that calibration.
Worked Example: A Bernoulli Probability Restricted to One Value
Suppose \(X_1,\ldots,X_{10}\) are independent Bernoulli observations, with \(7\) successes and \(3\) failures. Test \(H_0:p=0.5\) against the alternative that \(p\) is unrestricted in \((0,1)\).
The null permits just \(p=0.5\), and the alternative permits every value in \((0,1)\), so the models are nested. The restricted likelihood is \(L_0=(0.5)^7(0.5)^3=(0.5)^{10}=0.0009765625\).
The unrestricted maximum-likelihood estimate is \(\hat p=7/10=0.7\). The maximized likelihood is \(L_1=(0.7)^7(0.3)^3=0.0022235661\) (rounded). It exceeds \(L_0\), as nesting requires.
Therefore, \(T=2\log(L_1/L_0)=2\{7\log(1.4)+3\log(0.6)\}\approx1.6457\). This is a one-restriction test, so under the regular null its large-sample reference distribution is \(\chi^2_1\). The reference distribution is justified by the nested restriction and regularity conditions, not merely by the fact that two likelihoods were calculated.
What Changes for Nonnested Models
If neither model contains the other, their maximized likelihoods have no guaranteed order. One model may fit the observed data better than the other, but reversing which model is subtracted reverses the sign of the log-likelihood difference. As a result, there is no automatically defined nonnegative deviance with the same null-versus-expanded-model interpretation.
This does not mean that likelihoods cannot be compared, or that no test can compare nonnested hypotheses. It means that a test and its calibration require a specified testing problem and a distributional argument of their own. For two simple hypotheses, for instance, a likelihood ratio can be used in a Neyman–Pearson test with a designated null and alternative. That fact does not make the statistic a Wilks chi-square statistic.
Worked Example: Two Fixed Bernoulli Probabilities
Again observe \(7\) successes and \(3\) failures in \(10\) independent Bernoulli trials. Compare the two simple models \(A:p=0.5\) and \(B:p=0.8\). Neither model contains the other.
The likelihoods are \(L_A=(0.5)^{10}=0.0009765625\) and \(L_B=(0.8)^7(0.2)^3=0.0016777216\). With \(B\) in the numerator, the directed log-likelihood comparison is \(2\log(L_B/L_A)=2\log(0.0016777216/0.0009765625)\approx1.082306\), or \(1.082\) to three decimal places.
Reversing the order gives \(2\log(L_A/L_B)\approx-1.082306\). The change of sign is not a numerical problem: it reflects that neither model is an expanded version of the other. The positive comparison says only that \(B\) assigns greater likelihood to this observed sample than \(A\) does. It does not justify treating the number as a \(\chi^2\) statistic with one degree of freedom.
If the task is specifically to test \(A\) against \(B\), the likelihood ratio can still be part of a valid simple-hypothesis test. Its rejection threshold and error probabilities must be justified for that testing problem; the nested-model Wilks argument does not supply them.
A useful way to understand why a likelihood comparison can favor either model is to examine its expectation under a specified data-generating distribution. The following inequality gives that expectation a precise interpretation.
Proof. For any \(t>0\), \(\log t\leq t-1\). Apply this with \(t=q(x)/p(x)\) wherever \(p(x)>0\). Multiplying by \(p(x)\) and integrating gives \(\int p(x)\log\{q(x)/p(x)\}\,dx\leq\int q(x)\,dx-\int p(x)\,dx=0\). Changing the sign yields \(\int p(x)\log\{p(x)/q(x)\}\,dx\geq0\). Equality in \(\log t\leq t-1\) holds only at \(t=1\); therefore equality in the integral requires \(p=q\) almost everywhere. For independent observations \(X_1,\ldots,X_n\), the log likelihood ratio is the sum of the \(n\) individual log ratios. Linearity of expectation then gives \(nE_P[\log\{p(X)/q(X)\}]\). If \(P\) assigns positive probability to a region where \(q=0\), the expected log ratio is infinite rather than a finite negative value. \(\square\)
This result explains why the direction of a nonnested comparison depends on which distribution generates the data: the expected log likelihood ratio favors the model closer to that distribution in this particular divergence sense. But it does not give a chi-square null distribution for the observed statistic. That distribution depends on the hypotheses, how their parameters are estimated, and the sampling behavior under the null.
Worked Example: Comparing a Poisson and a Geometric Model
Observe one count \(Y=1\). Compare a Poisson model with fixed rate \(1\) to a geometric model with fixed success probability \(1/2\), where the geometric variable takes values \(1,2,\ldots\). These model families are not nested: the Poisson distribution assigns positive probability to \(0\), while the geometric distribution does not, and the Poisson model is not a member of the geometric family.
Under the Poisson model, \(P(Y=1)=e^{-1}1^1/1!=e^{-1}\). Under the geometric model, \(P(Y=1)=\tfrac12(1-\tfrac12)^{1-1}=\tfrac12\). Thus the directed comparison, with Poisson in the numerator, is \(2\log\{e^{-1}/(1/2)\}=2(\log 2-1)\approx-0.613706\).
In the reverse direction the value is \(2(1-\log 2)\approx0.613706\). Both calculations are correct, but neither direction inherits the usual nested likelihood-ratio statistic's nonnegative ordering and chi-square calibration. The example is a likelihood comparison for this observation, not by itself a standard Wilks test.
How to Use the Comparison Responsibly
Before assigning a chi-square reference distribution to a likelihood difference, check the model relationship rather than relying on the statistic's name. The relevant questions are:
- Is the null distribution set contained in the alternative distribution set?
- Are both likelihoods based on the same observed data and compatible probability models?
- If the models are nested, are the regularity conditions for Wilks' theorem satisfied?
- If they are not nested, what testing procedure and null distribution justify the chosen statistic?
A common mistake is to reason, “Both models have a maximized likelihood, so twice their log-likelihood difference must be chi-square.” This skips the key hypotheses of the usual result. For the nested test, the alternative maximizes over a set containing the null set, and the null imposes a specified number of regular restrictions. For nonnested models, neither this containment nor the corresponding degrees-of-freedom argument is available. A chi-square approximation must not be borrowed solely because the expression resembles \(2(\ell_1-\ell_0)\).
Nesting is therefore more than a convenient way to organize model fits. It fixes the direction of the usual likelihood-ratio statistic, gives a meaningful null-versus-expanded-model comparison, and—together with regularity conditions—connects that statistic to Wilks' chi-square limit. Without nesting, likelihood values remain informative descriptive quantities, but a test requires its own justification.
Check Your Understanding
Try these questions using the distinction between a likelihood comparison and a calibrated test.
- If \(\mathcal{P}_0\subseteq\mathcal{P}_1\), why must the unrestricted maximized likelihood be at least as large as the restricted maximum?
- What additional assumptions, beyond nesting, are needed for the usual Wilks chi-square calibration?
- Two fixed Bernoulli models have probabilities \(0.5\) and \(0.8\). Why does comparing their likelihoods not by itself produce a Wilks chi-square test?
- Can a likelihood ratio be useful for testing two simple nonnested hypotheses? What does that not imply about its reference distribution?
- Under a specified true density \(p\), what does the expected log likelihood ratio against a density \(q\) measure, and why is it nonnegative under the stated conditions?