Why AIC Balances Fit and Complexity
A more flexible NLME model can often fit the observed data at least as well as a simpler one. But a model that fits its own data unusually well may predict new data less well: some of its apparent fit reflects the particular sample. The Akaike Information Criterion (AIC) addresses this trade-off by subtracting a complexity penalty from the reward for better likelihood fit.
For NLME models, the likelihood ordinarily describes the observed data after the subject-specific random effects have been integrated out. AIC compares maximized likelihoods on a common basis and penalizes the number of estimated model parameters. Its purpose is to compare estimated predictive information loss, not to test whether one model is statistically significantly better.
In an NLME model, population fixed effects and estimated random-effects distribution parameters contribute to \(K\). In the usual marginal likelihood, the individual random effects have been integrated out; their estimated subject-specific values are not treated as additional free parameters in this AIC formula. The details of counting parameters in particular NLME models require care, but the central principle here is that \(K\) counts the free parameters estimated in the likelihood being used.
Why the Penalty Is \(2K\)
AIC's penalty corrects, approximately, for the optimism created by fitting parameters to the same data used to assess fit. The following derivation gives the standard justification under a correctly specified, regular model. It is an asymptotic argument: it describes the expected difference over repeated datasets, rather than an exact correction for every finite sample.
Suppose independent observational units have a correctly specified parametric density \(f(y\mid\theta)\), with true parameter \(\theta_0\) in the interior of the parameter space. Assume the model is identifiable and sufficiently smooth, the maximum likelihood estimator is consistent and asymptotically normal, and the Fisher information is finite and nonsingular. For NLME data, independent units are typically subjects; dependence among observations within a subject is represented in that subject's likelihood. Let \(U\) be the score vector at \(\theta_0\), and let \(I\) be the expected information matrix for the full dataset, that is, the negative expected Hessian of the log likelihood at \(\theta_0\).
A second-order expansion of the log likelihood around \(\theta_0\), together with the likelihood approximation \(\hat{\theta}-\theta_0\approx I^{-1}U\), shows that fitting the parameters raises the training log likelihood above its value at the true parameter by approximately \(\frac{1}{2}U^\mathsf{T}I^{-1}U\). Under correct specification, the information identity gives \(\operatorname{Cov}(U)=I\). Therefore,
The fitted model also loses approximately \(K/2\) units of expected log likelihood when evaluated on independent new data rather than at the true parameter. To see the latter term, average the new-data log likelihood over data generated at \(\theta_0\). Its expected curvature near \(\theta_0\) is \(-I\), so the expected change at \(\hat{\theta}\) is approximately \(-\frac{1}{2}E[(\hat{\theta}-\theta_0)^\mathsf{T}I(\hat{\theta}-\theta_0)]\), which is \(-K/2\). Thus the expected training-to-new-data optimism is approximately \(K/2-(-K/2)=K\) in log-likelihood units. Multiplying the log-likelihood scale by \(-2\) gives the penalty \(2K\) in AIC.
Correct specification matters in this derivation. If the candidate model is misspecified, the score covariance need not equal the expected information. In that case, with \(J=\operatorname{Cov}(U)\), the corresponding asymptotic penalty generally involves \(2\operatorname{tr}(I^{-1}J)\), not \(2K\). The ordinary AIC formula remains widely used for model comparison, including when candidates are approximations, but its standard \(2K\) penalty does not have the same unbiased-optimism justification under misspecification.
Worked Example: Calculating AIC for Two NLME Models
Consider hypothetical concentration data from the same subjects and sampling times, analyzed with the same marginal likelihood. Model A has maximized log likelihood \(-812.4\) and \(K=7\). Model B has maximized log likelihood \(-808.9\) and \(K=10\). Calculate both AIC values and compare them.
For Model A, $$ \mathrm{AIC}_A=-2(-812.4)+2(7)=1624.8+14=1638.8. $$ For Model B, $$ \mathrm{AIC}_B=-2(-808.9)+2(10)=1617.8+20=1637.8. $$
Model B has the lower AIC by \(1638.8-1637.8=1.0\). Its log likelihood improves by \(3.5\), but its three additional parameters add a penalty of \(2(3)=6\); the likelihood improvement reduces the deviance term by \(2(3.5)=7\), leaving a net AIC improvement of 1.0. On this criterion, Model B has slightly less estimated relative information loss. The one-point difference is not evidence from a significance test and does not establish that Model B is the true model.
Interpreting AIC Differences
AIC values are most useful as differences within a set of candidate models. If \(A_{\min}\) is the smallest AIC in that set, define \(\Delta_i=\mathrm{AIC}_i-A_{\min}\). The model with \(\Delta_i=0\) has the smallest AIC in the set; other models have larger estimated relative information loss according to AIC. Because AIC is on a deviance-like scale, a difference of 6 means a difference of 6 AIC units, not a 6% change in predictive accuracy.
The information-loss interpretation comes from estimating expected Kullback–Leibler discrepancy between a candidate model's predictive distribution and the data-generating distribution. For a given set of candidates, AIC estimates that discrepancy up to terms common to the candidates. It does not directly report an absolute prediction error, and the difference should be interpreted as approximate and relative. AIC alone also does not provide the probability that a particular model is true.
Worked Example: Interpreting a Difference of Six AIC Units
Suppose three hypothetical models fitted to the same observations have AIC values 1470, 1476, and 1483. The first model has the minimum. The differences are $$ \Delta_1=1470-1470=0,\qquad \Delta_2=1476-1470=6,\qquad \Delta_3=1483-1470=13. $$
The first model has the least estimated relative information loss among these candidates. The second has an AIC 6 units higher, and the third 13 units higher; these are relative comparisons within this candidate set. They are not p-values, confidence levels, or probabilities that the models are correct. A model omitted from the candidate set could have lower information loss, and even the minimum-AIC model may be an inadequate description of the data.
Worked Example: When Does an Improved Fit Offset More Parameters?
Suppose two hypothetical models are fitted to the same NLME dataset. Model S has \(\ell(\hat{\theta})=-430.0\) and \(K=5\). Model C has \(\ell(\hat{\theta})=-426.8\) and \(K=7\). Calculate the AIC difference and explain the trade-off.
The values are $$ \mathrm{AIC}_S=-2(-430.0)+2(5)=860.0+10=870.0, $$ and $$ \mathrm{AIC}_C=-2(-426.8)+2(7)=853.6+14=867.6. $$ Thus \(\mathrm{AIC}_C-\mathrm{AIC}_S=867.6-870.0=-2.4\), so Model C has the lower AIC by 2.4 units. Its log likelihood is higher by 3.2, reducing \(-2\ell\) by 6.4; the two extra parameters add 4 penalty units. The net change is \(-6.4+4=-2.4\). AIC therefore favors Model C in this comparison, but this is a model-selection result, not a test that the added parameters are statistically significant.
If the log likelihood had instead improved by only 1.5, the deviance reduction would be \(2(1.5)=3.0\), smaller than the 4-unit penalty. The more complex model would then have an AIC one unit higher and would not be favored by AIC.
Conditions, Comparability, and Diagnostics
The arithmetic in the AIC formula is simple; establishing that the inputs are meaningful is the harder part. Compare models fitted to the same observed data using likelihoods defined on the same basis. In particular, the likelihood must represent the same outcome and observations, and the treatment of the random effects and likelihood approximation must be consistent. When fixed-effect structures differ, use maximum-likelihood fits for an AIC comparison rather than comparing restricted maximum-likelihood likelihoods, which are not generally comparable across different fixed-effect structures.
A maximized likelihood is only useful if it has actually been maximized. Check convergence, gradients, and whether alternative initial values lead to the same solution. Boundary estimates, flat likelihood profiles, or poor parameter identifiability can make the reported likelihood and parameter count difficult to interpret. AIC cannot repair a failed fit or distinguish a useful improvement from an artifact of numerical optimization. If an approximate likelihood is used, the AIC values inherit that approximation and should be compared consistently.
The standard \(2K\) penalty is an asymptotic, regular-model correction. It can be inadequate when information is limited relative to the number of estimated parameters, and the standard derivation does not cover every boundary or nonidentifiable case. AIC is also not a guarantee of good absolute fit: inspect whether the candidate models adequately describe the concentration-time patterns and variability relevant to the analysis. Choosing the smallest AIC only ranks the candidates supplied.
What AIC Is Not
AIC is not a likelihood ratio test. A likelihood ratio test compares nested models using a likelihood-ratio statistic and a reference distribution under specified null hypotheses and conditions. AIC instead combines maximized log likelihood with a parameter penalty to estimate relative information loss. Consequently, an AIC difference does not have a null distribution or a p-value by itself. Nor should a small difference be described as proof that two models are equivalent: it says only that their AIC values are close on this criterion.
A common mistake is to say, “Model B has a lower AIC, so its extra random effect is statistically significant.” That conclusion confuses a selection criterion with a hypothesis test. The correction is to say, “Model B has the lower AIC among the candidates, indicating lower estimated relative information loss under this comparison.” If the scientific question is whether a specified parameter or structure differs from a null value, use an appropriate inferential procedure and its assumptions rather than interpreting AIC as a test.
Check Your Understanding
Work each question, then compare your reasoning with the solutions below.
- A hypothetical model has maximized log likelihood \(-250.0\) and \(K=4\). What is its AIC?
- Model A has AIC 920 and Model B has AIC 926 on the same dataset. What can be concluded, and what cannot be concluded?
- Why does the usual derivation of the \(2K\) penalty require the information identity?
- Two models have different fixed-effect structures and were fitted using REML. Is it appropriate to compare their reported restricted likelihoods directly using AIC? Briefly explain.
Solutions. (1) \(\mathrm{AIC}=-2(-250.0)+2(4)=500+8=508\). (2) Model A has the smaller AIC by 6 units and thus less estimated relative information loss among these candidates. This is not a p-value and does not show that A is true or that B is significantly worse. (3) The derivation uses \(\operatorname{Cov}(U)=I\), which makes \(E[U^\mathsf{T}I^{-1}U]=K\). Under misspecification, the score covariance \(J\) generally differs from expected information \(I\), so the expectation is generally \(\operatorname{tr}(I^{-1}J)\), not \(K\). (4) No. Restricted likelihoods are not generally comparable across different fixed-effect structures; maximum-likelihood fits on a common basis are appropriate for that comparison.