Grouped Data And Information-Criterion Comparisons

Why Grouped Observations Need Their Own Likelihood

Actuarial data may be available as counts in bands rather than as a record of every individual loss. For example, a hypothetical insurer might retain how many reported losses fell below a threshold, in an intermediate band, or above a limit, while no longer retaining the individual amounts. A model must then be assessed against the information actually observed: the probability it assigns to each band.

This changes the likelihood calculation. For raw continuous observations, likelihood contributions involve density values at the observed losses. For grouped observations, the contribution for an observation in a category is the probability of that category. Substituting a band midpoint into a density does not reproduce the grouped-data likelihood.

Definition: Suppose $n$ independent observations are classified into $m$ mutually exclusive and exhaustive categories. Let $n_j$ be the observed count in category $j$, so $\sum_{j=1}^{m}n_j=n$, and let $p_j(\theta)$ be the model probability of that category under parameter vector $\theta$. The grouped-data likelihood is the multinomial likelihood $$ L(\theta)=\frac{n!}{\prod_{j=1}^{m}n_j!}\prod_{j=1}^{m}p_j(\theta)^{n_j}. $$

The probabilities must satisfy $p_j(\theta)\geq 0$ and $\sum_j p_j(\theta)=1$. The categories must describe the same partition under every candidate model being compared. If an observation can belong to more than one category, or some possible observations are omitted without accounting for them, this multinomial formulation does not apply as stated.

Grouped Likelihood, AIC, and BIC

Taking logarithms separates the likelihood into a data-only term and a model-dependent term:

\ell(\theta)=\log\left(\frac{n!}{\prod_{j=1}^{m}n_j!}\right)+\sum_{j=1}^{m}n_j\log p_j(\theta).

For a fitted model, use its fitted category probabilities $\widehat p_j$. The full log-likelihood includes the multinomial coefficient. Because that coefficient depends only on the observed counts, it is identical for models fitted to the same grouped data. It is therefore common to report reduced information criteria, omitting this shared constant. This changes the numerical values, but not the ranking of those candidates on that data.

Definition: For a fitted model with $k$ estimated free parameters and $n$ independent observations, the full criteria are $\mathrm{AIC}=-2\ell(\widehat\theta)+2k$ and $\mathrm{BIC}=-2\ell(\widehat\theta)+k\log n$. The corresponding reduced criteria replace $\ell(\widehat\theta)$ by the grouped log-likelihood kernel $\sum_jn_j\log\widehat p_j$. State which convention is being used when reporting numerical values.

Here $k$ is the dimension of the estimated parameter vector, not automatically the number of categories. Probability constraints matter: a freely specified probability vector for $m$ categories has $m-1$ free probabilities, while a parametric model may use fewer parameters. Include every parameter estimated for the candidate model under the stated convention. For BIC, $n$ is the number of independent observations represented by the counts, not the number of categories.

The reason category probabilities enter is direct. For any particular assignment of the $n$ observations to categories, independence gives a probability equal to the product of the corresponding $p_j$ values. There are $n!/\prod_j n_j!$ assignments that produce the same category counts. Multiplying the assignment probability by this number gives the multinomial likelihood above. Taking logs then gives $n_j\log p_j$ for each category, repeated $n_j$ times.

Worked Calculations

Worked Example: Two Categories and the Reduced Criteria

A hypothetical portfolio has 20 independent losses classified into two exhaustive bands, with counts $8$ and $12$. A candidate model has fitted category probabilities $0.4$ and $0.6$, and $k=1$ free parameter. Calculate its reduced AIC and reduced BIC.

Method. Use the grouped log-likelihood kernel and the reduced-criterion convention. The common multinomial coefficient is omitted. The sample size for BIC is $n=8+12=20$.

The kernel is $$ \ell_{\mathrm{red}}=8\log(0.4)+12\log(0.6). $$

The two terms are $8(-0.9162907)=-7.3303256$ and $12(-0.5108256)=-6.1299072$. Thus $\ell_{\mathrm{red}}=-13.4602328$, and $-2\ell_{\mathrm{red}}=26.9204656$.

Therefore, $\mathrm{AIC}_{\mathrm{red}}=26.9204656+2(1)=28.9204656$. For BIC, $\log(20)=2.9957323$, so $\mathrm{BIC}_{\mathrm{red}}=26.9204656+1(2.9957323)=29.9161979$. To three decimals, the results are 28.920 and 29.916, respectively.

Check and trap avoided. The probabilities sum to $1$, and the penalty uses $k=1$, not $2$. Adding the penalties gives a BIC larger than the unpenalized value, as expected. The common mistake is to use the wrong log-likelihood kernel or a wrong sample size; here the kernel is $-13.4602$, and $n=20$, not two categories.

Worked Example: AIC Comparison with Different Parameter Counts

For 40 hypothetical losses, the counts in three exhaustive bands are $(10,10,20)$. Candidate A assigns probabilities $(0.25,0.25,0.50)$ and has $k_A=2$. Candidate B assigns probabilities $(0.30,0.20,0.50)$ and has $k_B=1$. Using reduced AIC, determine which candidate is preferred.

Method. Calculate each model’s category log-likelihood kernel, multiply by $-2$, and add its own parameter penalty.

For A, the kernel is $$ 10\log(0.25)+10\log(0.25)+20\log(0.50) =20(-1.3862944)+20(-0.6931472)=-41.5888320. $$ Thus $-2\ell_{\mathrm{red},A}=83.1776640$, and $\mathrm{AIC}_{\mathrm{red},A}=83.1776640+2(2)=87.1776640$, or 87.178 to three decimals.

For B, the kernel is $$ 10\log(0.30)+10\log(0.20)+20\log(0.50) =-12.0397280-16.0943791-13.8629436=-41.9970507. $$ Thus $-2\ell_{\mathrm{red},B}=83.9941014$, and $\mathrm{AIC}_{\mathrm{red},B}=83.9941014+2(1)=85.9941014$, or 85.994 to three decimals.

Candidate B has the lower reduced AIC by $87.1776640-85.9941014=1.1835626$, approximately $1.184$. It is preferred by this criterion for these candidates and these data.

Check and trap avoided. Both probability vectors sum to $1$, and each uses its stated $k$. Candidate A has a better fit kernel because $-41.5888320$ is larger than $-41.9970507$, but its extra parameter penalty is $2$, rather than $1$. Comparing fit alone would miss the penalty and give the wrong ranking.

Worked Example: Binned Losses Are Not Raw Loss Values

A hypothetical severity study retains only counts in three bands: below \$100, from \$100 to below \$300, and at least \$300. The counts are $(6,9,5)$, so $n=20$. Candidate C assigns probabilities $(0.30,0.45,0.25)$ and has $k_C=2$. Candidate D assigns probabilities $(0.25,0.50,0.25)$ and has $k_D=1$. Calculate reduced AIC and identify the lower value.

Method. Apply probabilities to the observed bands, not densities to invented individual losses. For C, $$ \ell_{\mathrm{red},C}=6\log(0.30)+9\log(0.45)+5\log(0.25). $$

The terms are $6(-1.2039728)=-7.2238368$, $9(-0.7985077)=-7.1865693$, and $5(-1.3862944)=-6.9314720$. Hence $\ell_{\mathrm{red},C}=-21.3418781$, so $-2\ell_{\mathrm{red},C}=42.6837562$. Adding $2k_C=4$ gives reduced AIC $46.6837562$, or 46.684.

For D, $$ \ell_{\mathrm{red},D}=6\log(0.25)+9\log(0.50)+5\log(0.25). $$

The terms are $6(-1.3862944)=-8.3177664$, $9(-0.6931472)=-6.2383248$, and $5(-1.3862944)=-6.9314720$. Hence $\ell_{\mathrm{red},D}=-21.4875632$, so $-2\ell_{\mathrm{red},D}=42.9751264$. Adding $2k_D=2$ gives reduced AIC $44.9751264$, or 44.975.

D has the lower reduced AIC by $46.6837562-44.9751264=1.7086298$, approximately $1.709$. The counts do not tell us the exact losses within each band. Consequently, they cannot support a raw-data likelihood calculation that requires individual loss amounts.

Check and trap avoided. Each candidate’s probabilities sum to $1$. Replacing each loss by its band midpoint and evaluating a continuous density there would create data that were never observed, so it is not a valid substitute for this grouped likelihood.

Using the Comparison Correctly

Information criteria are meaningful comparisons only when candidates are evaluated against the same observed data and the same category definitions. If one model is evaluated on a different sample or a different grouping, the omitted multinomial constants or likelihood terms may no longer be common. A smaller reported reduced AIC or BIC is not a license to compare incompatible likelihoods.

The full and reduced criteria have identical differences when the same counts are used. Specifically, if $C=\log(n!/\prod_jn_j!)$, then $\ell_{\mathrm{full}}=C+\ell_{\mathrm{red}}$, so $$ \mathrm{AIC}_{\mathrm{full}}=\mathrm{AIC}_{\mathrm{red}}-2C,\qquad \mathrm{BIC}_{\mathrm{full}}=\mathrm{BIC}_{\mathrm{red}}-2C. $$

This is why rankings agree under either convention for a fixed grouped data set, even though the numerical criterion values differ. If a task requests the full likelihood-based criterion, retain the multinomial coefficient. If it requests or clearly adopts reduced criteria for candidate comparison, omit that shared term and label the result accordingly.

A further trap is to confuse the number of occupied bands with the BIC sample size. Twenty losses recorded in three bands give $n=20$, not $n=3$. Also, $k$ counts estimated model parameters, not observed counts and not automatically all $m$ category probabilities. State the fitted probability vector, the parameter count, the likelihood convention, and $n$ before finalizing a comparison.

Key takeaway: For grouped independent observations, the model-dependent log-likelihood is $\sum_j n_j\log\widehat p_j$. Use the fitted probability of each observed category, apply the correct parameter penalty, and distinguish the full multinomial likelihood from a reduced version that omits a common data-only constant.

Check Your Understanding

For each question, use the stated reduced criterion unless the question asks otherwise. Round numerical answers to three decimals.

A grouped sample has counts $(5,15)$, fitted probabilities $(0.25,0.75)$, and $k=1$. Find reduced AIC. (Answer: 24.493)
Counts in three categories are $(4,8,8)$. A model has probabilities $(0.20,0.40,0.40)$, $k=2$, and $n=20$. Find reduced BIC. (Answer: 48.188)
Counts are $(12,6,2)$. Candidate A has probabilities $(0.50,0.30,0.20)$, $k=2$; candidate B has probabilities $(0.60,0.25,0.15)$, $k=2$. Which has lower reduced AIC, and what is the difference? (Answer: B, by 1.037)
Fifty observations are grouped into two bands with counts $(30,20)$. A model has fitted probabilities $(0.60,0.40)$ and $k=1$. Find reduced BIC and state the $n$ used. (Answer: 71.213, using $n=50$)
Counts are $(5,10,5)$. Candidate A has probabilities $(0.25,0.50,0.25)$, $k=2$; candidate B has probabilities $(0.20,0.60,0.20)$, $k=2$. Which has lower reduced AIC? (Answer: A; AIC values are 45.589 for A and 46.405 for B)

What You'll Learn