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CAS Exam 8 · Model Selection and Rate Indications · Tutorial 270 of 989

AIC as a Predictive Model Selection Criterion

Learn to calculate AIC from a fitted model’s full log likelihood and parameter count, interpret its predictive tradeoff, and avoid invalid comparisons.

Advanced 12 min read

What You'll Learn

  • Derive the AIC penalty from the optimism of fitted training likelihood.
  • Calculate AIC from a GLM’s maximized log likelihood and number of estimated parameters.
  • Compare candidate models using AIC differences rather than raw fit alone.
  • Count estimated dispersion parameters and distinguish them from fixed offsets.
  • Identify the assumptions and data-comparability conditions behind AIC.
  • Recognize why a lower AIC is a relative predictive criterion, not proof that a model is true.

Why Penalize In-Sample Fit?

A more flexible ratemaking model can usually fit its observed data at least as well as a simpler model. For example, adding a territory interaction to a frequency GLM cannot reduce the maximized training likelihood when the simpler model is nested within it. But some of that improved fit may reflect random features of the training data rather than patterns that will recur in future policies. A predictive selection criterion should therefore reward fit while accounting for the optimism created by estimating parameters.

The Akaike information criterion, or AIC, makes this adjustment using the number of freely estimated parameters. It is a relative criterion: among comparable candidate models, the model with the smaller AIC is preferred as an estimate of predictive performance. The criterion does not establish that a model is correct, and it does not replace validation on future or held-out data.

Definition: For a model with maximized log likelihood \(\ell(\hat{\theta})\) and \(k\) freely estimated parameters, the Akaike information criterion is
$$ \operatorname{AIC}=-2\ell(\hat{\theta})+2k. $$
Here, \(\hat{\theta}\) is the maximum-likelihood estimate. Lower AIC indicates a better estimated predictive tradeoff among the candidate models being compared. Count all parameters estimated by the likelihood, including an estimated dispersion parameter when applicable. A fixed offset is not an estimated parameter.

The \(-2\ell(\hat{\theta})\) term rewards fit: a larger maximized log likelihood gives a smaller fit term. The \(2k\) term penalizes complexity. A model with more parameters can still have a lower AIC, but only if its improvement in likelihood is large enough to offset that penalty. When both models use the same observations and the same likelihood definition, AIC differences can be calculated directly; common constants in the likelihood then cancel in the difference.

Why the Penalty Is \(2k\)

AIC is motivated by estimating expected predictive information loss, commonly expressed through expected log likelihood or, equivalently, expected deviance. The fitted model has an advantage on the data used to estimate it: its parameters were chosen to fit that particular sample. The derivation below explains why the usual correction is \(2k\), rather than \(k\).

Assume a regular, correctly specified likelihood model with \(k\) identifiable free parameters, an interior true parameter value \(\theta_0\), and finite, nonsingular Fisher information. Assume independent observations and a sufficiently large sample for a local quadratic approximation to the log likelihood. Write \(\ell(\theta)\) for the training log likelihood, \(U\) for its score at \(\theta_0\), and \(I\) for its expected information matrix. Near \(\theta_0\), the log likelihood is approximately

$$ \ell(\theta_0+h)\approx \ell(\theta_0)+U^\mathsf{T}h-\frac{1}{2}h^\mathsf{T}Ih. $$

Maximizing this quadratic approximation gives \(\hat{h}\approx I^{-1}U\). Substituting it into the approximation, the training log-likelihood gain over the true parameter is

$$ \ell(\hat{\theta})-\ell(\theta_0) \approx \frac{1}{2}U^\mathsf{T}I^{-1}U. $$

Under the stated conditions, the score has mean zero and covariance approximately \(I\). Thus the expected quadratic form is approximately its dimension, \(k\), and the expected training log-likelihood gain is approximately \(k/2\):

$$ E\!\left[U^\mathsf{T}I^{-1}U\right]\approx k, \qquad E\!\left[\ell(\hat{\theta})-\ell(\theta_0)\right]\approx \frac{k}{2}. $$

Now consider an independent test sample of the same size, drawn from the same process. Conditional on the fitted training estimate, its expected log likelihood is maximized at \(\theta_0\). The local expected curvature of the test log likelihood is approximately \(-I\), so using \(\hat{\theta}\) rather than \(\theta_0\) reduces expected test log likelihood by approximately \(\frac{1}{2}E[\hat{h}^\mathsf{T}I\hat{h}]\). Since \(\hat{h}\approx I^{-1}U\), that expected reduction is also approximately \(k/2\). The training fit gains about \(k/2\), while its expected test log likelihood loses about \(k/2\), relative to the true-parameter benchmark. Therefore, for independent test data of the same size, the expected difference between training and test log likelihood is approximately \(k\).

An estimate of expected test log likelihood is consequently \(\ell(\hat{\theta})-k\). Multiplying its negative by two gives AIC:

$$ -2\{\ell(\hat{\theta})-k\}=-2\ell(\hat{\theta})+2k. $$

This is an asymptotic justification, not an exact finite-sample identity. The standard correction can be unreliable for small samples, boundary estimates, nonidentifiable or singular models, or settings where the likelihood assumptions do not hold. AIC is also tied to the likelihood and predictive target being used; it should not be treated as a universal score for any model output.

Calculating and Comparing AIC

For each candidate, obtain the maximized log likelihood from the fitted likelihood model and count its estimated parameters. Use the full log likelihood, including terms that may be constant with respect to the parameters, when comparing different likelihood families for the same response data. For a Poisson count model, for example, the likelihood includes the \(-\log(y!)\) terms. Omitting such terms may not change parameter estimates within one model, but can make AIC comparisons across likelihood specifications invalid.

In a GLM, a fitted intercept and each independently estimated regression coefficient count toward \(k\). An exposure offset is specified rather than estimated, so it does not count. A dispersion parameter estimated as part of a likelihood does count. A model’s coefficient count is determined by its actual fitted design matrix and constraints, not simply by the number of variable names in its formula.

Worked Example: Deciding Whether an Added Rating Factor Is Worthwhile

A hypothetical insurer fits two Poisson frequency GLMs to the same policy-level claim counts, using the same exposure offsets and complete records. The simpler model has maximized log likelihood \(-184.2\) and \(k=5\) estimated regression parameters. The expanded model adds a rating factor and has maximized log likelihood \(-179.8\) with \(k=8\). Compare the models by AIC.

Method. Calculate \(-2\ell(\hat{\theta})+2k\) for each model. Since the data and likelihood definition are the same, the resulting AIC values are comparable.

For the simpler model, the fit term is \(-2(-184.2)=368.4\), and its penalty is \(2(5)=10\). Therefore,

$$ \operatorname{AIC}_{\text{simple}}=368.4+10=378.4. $$

For the expanded model, the fit term is \(-2(-179.8)=359.6\), and its penalty is \(2(8)=16\). Therefore,

$$ \operatorname{AIC}_{\text{expanded}}=359.6+16=375.6. $$

The expanded model has the lower AIC by \(378.4-375.6=2.8\), so AIC favors it among these two candidates. Equivalently, its log likelihood improved by \(4.4\); twice that gain is \(8.8\), while the three added parameters increase the penalty by \(2(3)=6\). The net AIC change is \(-8.8+6=-2.8\). The calculation checks: both methods give the same difference. The common mistake avoided is to regard the parameter penalty as \(k\) rather than \(2k\).

Worked Example: Comparing Poisson and Negative Binomial Likelihoods

A hypothetical analyst models claim counts per policy with exposure offsets. Both models use the same policy records and response definition. The Poisson GLM has maximized full log likelihood \(-92.6\) and four estimated regression coefficients. A negative binomial model has maximized full log likelihood \(-89.1\), five estimated regression coefficients, and one estimated dispersion parameter. Calculate and compare AIC.

Method. Count the negative binomial dispersion estimate because it is part of the fitted likelihood. Its total parameter count is \(k=5+1=6\). The Poisson model has \(k=4\).

For the Poisson model,

$$ \operatorname{AIC}_{\text{Poisson}} =-2(-92.6)+2(4) =185.2+8 =193.2. $$

For the negative binomial model,

$$ \operatorname{AIC}_{\text{NB}} =-2(-89.1)+2(6) =178.2+12 =190.2. $$

The negative binomial AIC is lower by \(193.2-190.2=3.0\), so it is preferred by AIC for this comparison. Its log-likelihood gain is \(3.5\), which gives a fit-term improvement of \(7.0\); its two additional parameters incur a penalty increase of \(4\). The net AIC improvement is \(7.0-4=3.0\). The calculation checks. The common mistake avoided is failing to count the estimated dispersion parameter, which would understate the negative binomial penalty. This comparison is meaningful because both models describe the same response observations with full likelihoods.

Worked Example: A Higher Likelihood Can Still Lose on AIC

A hypothetical insurer fits a base severity model and a model adding a three-parameter interaction. Both use the same observed losses and the same likelihood specification. The base model has log likelihood \(-310.0\) and six estimated parameters. The interaction model has log likelihood \(-307.2\) and nine estimated parameters. Which model has the lower AIC?

Method. Calculate the full AIC for each candidate; do not select solely by the larger raw log likelihood.

For the base model, \(-2(-310.0)=620.0\), and its penalty is \(2(6)=12\). Thus,

$$ \operatorname{AIC}_{\text{base}}=620.0+12=632.0. $$

For the interaction model, \(-2(-307.2)=614.4\), and its penalty is \(2(9)=18\). Thus,

$$ \operatorname{AIC}_{\text{interaction}}=614.4+18=632.4. $$

The base model has the lower AIC by \(0.4\). The interaction increases log likelihood by \(2.8\), improving the fit term by \(5.6\), but its three extra parameters add a penalty of \(2(3)=6\). Its AIC therefore increases by \(6-5.6=0.4\). The result checks. The common mistake avoided is choosing the interaction simply because it has the higher training likelihood. AIC’s preference is close and relative; the \(0.4\) difference is not proof that the base model is true or that the interaction has no predictive value.

Interpreting AIC Differences Carefully

AIC is useful because it puts the fit-complexity tradeoff on a single scale. It is not a test of whether an added term is statistically significant, a measure of calibration, or a guarantee of stable indicated rates. A low AIC says that a model has a better estimated predictive information tradeoff than the particular alternatives considered. It does not say that predictions are well calibrated in every segment or that the chosen specification is acceptable for implementation.

The penalty depends on how many parameters are estimated, not on how many variables appear in a model description. A categorical factor with several levels may require multiple coefficients, while an offset contributes none if fixed. Constraints such as reference levels affect the number of free coefficients. When fitting software reports likelihood and parameter counts, confirm that the count includes every estimated likelihood parameter and that the likelihoods are on a comparable basis.

AIC comparisons require the same observed response data and a common likelihood basis. If one model is fit to a different set of records, a different response definition, or a different aggregation without a likelihood that supports a valid comparison, the numerical AIC ranking may not answer the intended question. In particular, likelihood constants can matter when comparing different distributions, so use full likelihoods rather than arbitrary software summaries. A quasi-likelihood fit without a fully specified likelihood does not automatically supply an ordinary AIC.

AIC is designed to estimate relative predictive information loss, not to reward simplicity for its own sake. A more complex candidate can be preferable when it produces enough improvement in likelihood. Conversely, a modestly higher likelihood does not necessarily justify extra coefficients. Since the selection itself can add uncertainty, use AIC alongside sound model diagnostics and genuinely out-of-sample assessment where available. This interpretation is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Key takeaway: Compare \(-2\ell(\hat{\theta})+2k\) for candidate models fit to the same data on a comparable likelihood basis. The \(2k\) term estimates likelihood optimism under regularity assumptions; lower AIC favors a better estimated predictive tradeoff, not a proven true model.

Check Your Understanding

For each problem, calculate the relevant AIC quantity before deciding what it implies.

  1. A hypothetical frequency GLM has log likelihood \(-74.3\) and three estimated coefficients. A second model fitted to the same records has log likelihood \(-71.9\) and five estimated coefficients. Calculate both AIC values and identify the preferred candidate. (Answer: AICs are 154.6 and 153.8, respectively; the second model is preferred.)
  2. Two severity models use the same observations and full likelihood basis. Model X has log likelihood \(-126.0\) and \(k=6\); Model Y has log likelihood \(-123.5\) and \(k=8\). Calculate their AIC values and select by AIC. (Answer: Model X has AIC 264.0; Model Y has AIC 263.0; Model Y is preferred.)
  3. An expanded model adds two coefficients to a base model, and its maximized log likelihood is higher by \(1.4\). Without knowing the absolute log likelihoods, determine the AIC change and which model has lower AIC. (Answer: The AIC change is \(-2(1.4)+2(2)=+1.2\); the base model has lower AIC by 1.2.)
  4. A Poisson model for policy claim counts has log likelihood \(-51.7\) and three regression coefficients. A negative binomial model on the same records has log likelihood \(-48.8\), four regression coefficients, and one estimated dispersion parameter. Calculate both AICs and identify the preferred model. (Answer: The Poisson AIC is 109.4; the negative binomial AIC is 107.6; the negative binomial model is preferred by 1.8.)
  5. For two regular, correctly specified models fit to samples of equal size, state the approximate training log-likelihood gain for a \(k\)-parameter fit relative to the true parameter, and the approximate training-versus-independent-test log-likelihood optimism. (Answer: The expected training gain is approximately \(k/2\); for an independent test sample of the same size, expected test log likelihood is reduced by approximately \(k/2\), giving training-test optimism of approximately \(k\).)
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