AIC and BIC for Frequency and Severity Model Choice

Why Use Information Criteria?

A fitted model can describe observed data well simply because it has many parameters. When choosing between candidate frequency or severity models, fit alone is therefore not enough: the comparison should also account for model complexity. The Akaike information criterion (AIC) and Bayesian information criterion (BIC) are two commonly used summaries that combine a model’s maximized likelihood with a penalty for its number of estimated parameters.

Both criteria are calculated from the same basic inputs, but their penalties differ. AIC uses a fixed penalty for each parameter; BIC’s penalty grows with the sample size. A smaller value indicates the preferred model among the candidates being compared. Neither criterion establishes that a model is true, nor does either replace checking that the model and likelihood are appropriate for the data.

Definition: For a model with maximized log-likelihood $\hat{\ell}$, $k$ estimated free parameters, and $n$ independent observational units, define $$ \mathrm{AIC}=-2\hat{\ell}+2k, \qquad \mathrm{BIC}=-2\hat{\ell}+k\ln(n). $$ The likelihood must be maximized over the model’s parameters. The parameter count $k$ includes all parameters estimated from the data that contribute to that likelihood. The sample size $n$ for BIC is the number of independent observations represented by the likelihood, not the number of parameters or the number of candidate models.

What Must Be Consistent in a Comparison?

AIC and BIC are relative criteria. Compare models fitted to the same data using likelihoods that describe the same observation process. For frequency models, the observational units might be policy periods, with one count observed per period. For complete severity data, they are usually the individual observed losses. If the records are censored, truncated, grouped, or rounded, use the likelihood contributions for that data structure rather than pretending every record is an exact complete observation.

The log-likelihood values must also use a consistent convention. For example, a continuous severity likelihood is built from densities with respect to the same measurement scale. Changing the units of every loss changes all the density log-likelihoods by the same additive amount when the models use the same observations and scale; that changes the reported criteria but not their ranking. Comparing likelihoods calculated on different scales or from different records can produce a meaningless ranking.

The value of $\hat{\ell}$ is a log-likelihood, not a likelihood. Since larger maximized log-likelihood means better fit, the term $-2\hat{\ell}$ gets smaller as fit improves. The added term then penalizes the number of estimated parameters. For AIC the penalty is $2$ per parameter. For BIC it is $\ln(n)$ per parameter, which exceeds $2$ when $n>e^2$, approximately $7.39$.

Interpretation: When comparing candidates, select the model with the smaller criterion value. AIC and BIC can disagree because they apply different complexity penalties. BIC is often more demanding of additional parameters as $n$ grows, but neither criterion is a probability that a model is correct.

Why the Formulas Balance Fit and Complexity

The fit contribution in either criterion is $-2\hat{\ell}$. The factor of $2$ puts the log-likelihood on the conventional information-criterion scale. The remaining term is a penalty: it increases when the model uses more estimated parameters, even if the more complex model fits at least as well.

The practical comparison follows directly by subtracting the criteria for a more complex model from those for a simpler one. Suppose model 1 has $k_1$ parameters and log-likelihood $\hat{\ell}_1$, while model 2 has $k_2>k_1$ parameters and log-likelihood $\hat{\ell}_2$. Then

\mathrm{AIC}_2-\mathrm{AIC}_1 =-2(\hat{\ell}_2-\hat{\ell}_1)+2(k_2-k_1),

so AIC favors model 2 precisely when its increase in maximized log-likelihood exceeds $k_2-k_1$. For BIC, the corresponding difference is

\mathrm{BIC}_2-\mathrm{BIC}_1 =-2(\hat{\ell}_2-\hat{\ell}_1)+(k_2-k_1)\ln(n).

Thus BIC favors the more complex model only when its fit improvement exceeds $\tfrac12(k_2-k_1)\ln(n)$. This subtraction is a useful check: a better likelihood does not automatically imply a better information criterion.

Worked Comparisons

Worked Example: Selecting a Frequency Model

A hypothetical portfolio has 15 independent policy-period claim counts. A Poisson model with fitted mean count $\hat{\lambda}=2.0$ has maximized log-likelihood $-22.0$. A negative binomial model, parameterized by fitted mean $\hat{\mu}=2.0$ and fitted size parameter $\hat{r}=4.0$, has maximized log-likelihood $-20.5$. Treat these as the fitting results for the same 15 counts. Compare the models by AIC and BIC.

Method. Use $n=15$, count one estimated parameter for the Poisson model and two for the negative binomial model, and substitute each model’s own maximized log-likelihood. The negative binomial fit is better because $-20.5>-22.0$, but its extra parameter is penalized.

For the Poisson model, $k=1$:

\mathrm{AIC}_{\mathrm{Pois}} =-2(-22.0)+2(1) =44+2 =46.0000.

For BIC, $\ln(15)=2.7080502$, so

\mathrm{BIC}_{\mathrm{Pois}} =44+1(2.7080502) =46.7081.

For the negative binomial model, $k=2$:

\mathrm{AIC}_{\mathrm{NB}} =-2(-20.5)+2(2) =41+4 =45.0000, $$ $$ \mathrm{BIC}_{\mathrm{NB}} =41+2(2.7080502) =41+5.4161004 =46.4161.

Decision and check. AIC prefers the negative binomial model because $45.0000<46.0000$. BIC also prefers it because $46.4161<46.7081$. The extra parameter improves the fit enough to overcome both penalties, although the BIC difference is small. This calculation avoids a common error: counting the negative binomial’s two parameters in AIC but forgetting to multiply $\ln(n)$ by two in BIC.

Worked Example: Comparing Exponential and Lognormal Severity Fits

Four hypothetical complete losses, measured in dollars, are $1,2,4,8$. Fit an exponential severity model with rate $\lambda$ and a lognormal severity model whose logarithm has a normal distribution with mean $\mu$ and variance $\sigma^2$. Use maximum likelihood, count the exponential rate as one parameter and the lognormal mean and variance as two parameters, and calculate AIC and BIC. Use natural logarithms.

Exponential fit. The sample mean is $\bar{x}=(1+2+4+8)/4=15/4$. The exponential maximum-likelihood estimate is $\hat{\lambda}=1/\bar{x}=4/15$. The maximized log-likelihood for complete positive observations is $\hat{\ell}_{\mathrm{Exp}}=n\ln(\hat{\lambda})-\hat{\lambda}\sum x_i$. Therefore,

\hat{\ell}_{\mathrm{Exp}} =4\ln(4/15)-(4/15)(15) =4(-1.3217558)-4 =-9.2870234.

With $k=1$, the criteria are

\mathrm{AIC}_{\mathrm{Exp}} =-2(-9.2870234)+2 =18.5740468+2 =20.5740, $$ $$ \mathrm{BIC}_{\mathrm{Exp}} =18.5740468+\ln(4) =18.5740468+1.3862944 =19.9603.

Lognormal fit. The log observations are $0,\ln 2,2\ln 2,3\ln 2$. Their mean is $\hat{\mu}=1.5\ln 2$. The squared deviations from this mean are $2.25(\ln 2)^2,0.25(\ln 2)^2,0.25(\ln 2)^2,2.25(\ln 2)^2$, with sum $5(\ln 2)^2$. The maximum-likelihood variance divides this sum by $n=4$, not by $n-1$:

\hat{\sigma}^2 =\frac{5(\ln 2)^2}{4} =1.25(\ln 2)^2 \approx 0.600566.

For a lognormal density, the complete-data log-likelihood is $-\frac{n}{2}\ln(2\pi)-\frac{n}{2}\ln(\hat{\sigma}^2)-\frac{\sum(\ln x_i-\hat{\mu})^2}{2\hat{\sigma}^2}-\sum\ln x_i$. Here the four contributions are $-3.675754$, $+1.019765$, $-2$, and $-4.158883$, respectively. In particular, the squared-deviation term is $-5(\ln2)^2/[2(1.25(\ln2)^2)]=-2$. Hence

\hat{\ell}_{\mathrm{LN}} =-3.675754+1.019765-2-4.158883 =-8.8148726.

With $k=2$,

\mathrm{AIC}_{\mathrm{LN}} =-2(-8.8148726)+2(2) =17.6297452+4 =21.6297, $$ $$ \mathrm{BIC}_{\mathrm{LN}} =17.6297452+2\ln(4) =17.6297452+2.7725887 =20.4023.

Decision and check. AIC prefers the exponential model, $20.5740<21.6297$, and BIC also prefers the exponential model, $19.9603<20.4023$. Both fitted log-likelihoods and criteria are shown to four decimal places at the final comparison. The common mistake this example avoids is halving the squared-deviation sum incorrectly: it is $5(\ln2)^2$, so the lognormal variance estimate is $1.25(\ln2)^2$, and the quadratic term in the log-likelihood is $-2$.

Worked Example: When AIC and BIC Disagree

A hypothetical set of 100 independent, complete severity observations, measured in thousands of dollars, is fitted by two positive-loss models. The exponential fit has estimated rate $\hat{\lambda}=0.010$ per thousand dollars, one estimated parameter, and maximized log-likelihood $-210.0$. A two-parameter Weibull fit has estimated shape $\hat{\alpha}=1.25$, estimated scale $\hat{\theta}=90$ thousand dollars, and maximized log-likelihood $-208.5$. Treat the reported maximized log-likelihoods as fitting outputs for these same observations. Compare the criteria.

The exponential model has $k=1$, and the Weibull model has $k=2$. For the exponential fit,

\mathrm{AIC}_{\mathrm{Exp}} =-2(-210.0)+2(1) =420+2 =422.0000, $$ $$ \mathrm{BIC}_{\mathrm{Exp}} =420+\ln(100) =420+4.6051702 =424.6052.

For the Weibull fit,

\mathrm{AIC}_{\mathrm{Wei}} =-2(-208.5)+2(2) =417+4 =421.0000, $$ $$ \mathrm{BIC}_{\mathrm{Wei}} =417+2\ln(100) =417+9.2103404 =426.2103.

Decision and check. AIC prefers the Weibull model, $421.0000<422.0000$; BIC prefers the exponential model, $424.6052<426.2103$. The Weibull model improves the maximized log-likelihood by $1.5$, so $-2\hat{\ell}$ falls by $3$. That is enough to overcome AIC’s extra-parameter penalty of $2$, but not BIC’s extra-parameter penalty of $\ln(100)=4.6052$. This is a genuine disagreement, not an arithmetic inconsistency.

Common Errors and Practical Checks

Using the wrong sample size. For 15 independent policy-period counts, use $n=15$, not the total number of claims or the number of fitted parameters. For complete severity observations, $n$ is the number of observations represented in the likelihood.
Forgetting the parameter multiplier in BIC. If $k=2$, the penalty is $2\ln(n)$, not just $\ln(n)$. For example, with $n=15$, the penalty is $2(2.7080502)=5.4161004$.
Using a parameter count inconsistently. Count every free estimated parameter in the likelihood. Do not count a parameter as fixed if it was estimated from the data; do not count a known fixed value as an estimated parameter.
Treating a larger log-likelihood as the final decision. A more complex model can have a larger $\hat{\ell}$ and still have a larger AIC or BIC. Calculate the full criteria before selecting.
Mixing likelihood definitions. Use likelihood contributions appropriate to the observations and the same data for every candidate. A censored record, for example, contributes a probability of exceeding the censoring point rather than a density evaluated as if the exact loss were observed.

Key takeaway: For every candidate, record the same four items: the data and likelihood convention, the maximized log-likelihood $\hat{\ell}$, the number of estimated parameters $k$, and the BIC sample size $n$. Calculate both criteria from those inputs, then compare values within each criterion. Lower is preferred, but the result is a model-selection aid rather than proof that a candidate is correct.

Check Your Understanding

For each problem, identify the relevant parameter count and sample size before comparing criteria. Give criterion values to four decimal places where needed.

Two frequency models are fitted to the same 20 independent policy-period counts. Model A has $k=2$ and $\hat{\ell}=-18.0$; model B has $k=3$ and $\hat{\ell}=-16.8$. Calculate both criteria and state which model each criterion prefers. (Answer: AIC A $=40.0000$, B $=39.6000$, so AIC prefers B; BIC A $=41.9915$, B $=42.5872$, so BIC prefers A.)
For 8 complete severity observations, a one-parameter exponential fit has $\hat{\ell}=-12.0$, and a two-parameter alternative has $\hat{\ell}=-11.5$. Calculate both criteria. (Answer: AIC exponential $=26.0000$, alternative $=27.0000$; BIC exponential $=26.0794$, alternative $=27.1589$; both prefer the exponential model.)
Two frequency fits use 50 independent periods. Model C has $k=1$, $\hat{\ell}=-60.0$; model D has $k=2$, $\hat{\ell}=-57.0$. Which model is preferred by each criterion? (Answer: AIC C $=122.0000$, D $=118.0000$, so AIC prefers D; BIC C $=123.9120$, D $=121.8240$, so BIC prefers D.)
A severity model is fitted to 30 records, of which 5 are right-censored at a policy limit. A candidate model has two estimated parameters. What sample size belongs in its BIC penalty, and what likelihood contribution should be used for a censored record? (Answer: use $n=30$, if the 30 records are the independent observational units represented in the likelihood; use the model’s probability of exceeding the applicable censoring threshold for each right-censored observation, not a density at an unobserved exact loss.)
For $n=100$, a model with one parameter has $\hat{\ell}=-210$, while a model with two parameters has $\hat{\ell}=-208.5$. Explain the differing choices without recalculating every term from scratch. (Answer: the improvement in log-likelihood is $1.5$, so $-2\hat{\ell}$ improves by $3$; the additional AIC penalty is $2$, so AIC favors the two-parameter model, while the additional BIC penalty is $\ln(100)=4.6052$, so BIC favors the one-parameter model.)

This treatment of model comparison is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

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