Worked Frequency Model Selection from Claim Counts

Why Frequency Model Selection Matters

A frequency model describes how many claims arise in a specified exposure period. Choosing a model affects expected claim counts and the uncertainty around them, so a fit should not be selected just because its mean matches the observed average. This tutorial compares two common models: Poisson, which has equal mean and variance, and negative binomial, which can represent greater variability across exposure periods.

The calculations below assume that counts are observed over equal, fully exposed periods, and that periods are independent and governed by the same frequency distribution. If exposures differ, counts should not be treated as identically distributed without accounting for the exposure differences. Dependence between periods or other important departures from the stated assumptions can also make either model unsuitable.

Definition: Under a Poisson frequency model, \(N\sim\operatorname{Poisson}(\lambda)\), where \(\lambda>0\) is both the mean and variance of the count per exposure period. Its probability mass function is \(P(N=k)=e^{-\lambda}\lambda^k/k!\) for \(k=0,1,2,\ldots\). Under a negative binomial model with mean \(\mu>0\) and size parameter \(r>0\), \(E[N]=\mu\) and \(\operatorname{Var}(N)=\mu+\mu^2/r\). One standard parameterization has \(P(N=k)=\frac{\Gamma(k+r)}{\Gamma(r)k!}\left(\frac{r}{r+\mu}\right)^r\left(\frac{\mu}{r+\mu}\right)^k\).

The negative binomial is more variable than a Poisson distribution with the same mean: \(\mu+\mu^2/r>\mu\) for finite \(r\). As \(r\) tends to infinity, its variance approaches \(\mu\) and its distribution approaches the Poisson distribution. This nesting at the limiting boundary is useful for interpretation, but it also means that a finite negative binomial estimate should not be assumed to exist.

Estimation and Initial Dispersion Checks

For independent observations \(N_1,\ldots,N_n\) from one Poisson distribution, the likelihood is the product of their probability mass functions. Maximizing its log likelihood gives the sample mean as the maximum likelihood estimate of \(\lambda\). A quick diagnostic is therefore to compare the sample variance with the sample mean. A variance appreciably larger than the mean is evidence of overdispersion relative to Poisson, but it is not by itself proof that a negative binomial model is appropriate.

\widehat{\lambda}=\bar N=\frac{1}{n}\sum_{j=1}^n N_j,\qquad s^2=\frac{1}{n-1}\sum_{j=1}^n(N_j-\bar N)^2.

For a negative binomial model, a method-of-moments estimate follows by equating its variance to the observed sample variance. When \(s^2>\bar N\), solving \(s^2=\bar N+\bar N^2/r\) gives \(\widehat r_{\mathrm{MOM}}=\bar N^2/(s^2-\bar N)\). This is a diagnostic estimate, not generally the maximum likelihood estimate. If \(s^2\leq\bar N\), the equation gives no positive finite size estimate; that does not establish that the Poisson model is correct, but it gives no method-of-moments support for finite negative binomial overdispersion.

Derivation: The Poisson log likelihood for observed counts is \(\ell(\lambda)=\sum_{j=1}^n[N_j\ln\lambda-\lambda-\ln(N_j!)]\). Differentiating gives \(\ell'(\lambda)=\sum_jN_j/\lambda-n\). Setting this equal to zero yields \(\widehat{\lambda}=\sum_jN_j/n=\bar N\). The second derivative is \(-\sum_jN_j/\lambda^2<0\) when the total count is positive, so this stationary point is a maximum. If every observed count is zero, the likelihood instead increases toward its boundary maximum at \(\lambda=0\).

Worked Example: Diagnosing Dispersion Before Fitting

A hypothetical insurer records the counts \(0,0,0,1,1,2,3,5\) over eight equal and independent exposure periods. Calculate the Poisson estimate, the unbiased sample variance, and the negative binomial method-of-moments size estimate. State what this diagnostic does and does not establish.

Method and assumptions. Treat the eight periods as identically distributed and independent. Estimate the Poisson mean with the sample mean, then compare the sample variance with that mean.

The total count is \(0+0+0+1+1+2+3+5=12\), so \(n=8\) and \(\bar N=12/8=1.5\). The sum of squared counts is \(0+0+0+1+1+4+9+25=40\). Thus \(s^2=[40-12^2/8]/7=[40-18]/7=22/7\approx3.1429\).

Since \(3.1429>1.5\), the data show sample overdispersion relative to the Poisson mean. The method-of-moments estimate is \(r=1.5^2/(22/7-1.5)=2.25/(22/7-3/2)=2.25/(23/14)=31.5/23\approx1.3696\).

Check and common mistake. The sample variance uses divisor \(n-1=7\), not \(n=8\); using \(n\) would calculate a different quantity. The positive finite estimate is consistent with overdispersion, but it does not establish that the negative binomial is the best-fitting model. A likelihood comparison and distributional diagnostics are still needed.

Worked Model Selection: Poisson and Negative Binomial

The next example carries the comparison through estimation, a grouped goodness-of-fit check, and AIC. For the negative binomial, the mean is estimated by the sample mean, and the size parameter is then found by maximizing the likelihood over \(r>0\). Numerical maximization is appropriate because the score equation for \(r\) does not usually have a closed-form solution.

Worked Example: Comparing Models for a Claim-Count Portfolio

A hypothetical portfolio has 100 independent, equal exposure periods. Its observed count frequencies are: 40 periods with zero claims, 30 with one, 15 with two, 10 with three, and 5 with four. Fit Poisson and negative binomial models, check grouped Poisson fit, and compare AIC.

Step 1: Calculate the sample mean and variance. The total count is \(30+30+30+20=110\), so \(\bar N=110/100=1.1\). The sum of squared counts is \(30+60+90+80=260\). Therefore, \(s^2=[260-110^2/100]/99=(260-121)/99=139/99\approx1.4040\). The sample variance exceeds the mean, indicating overdispersion relative to Poisson. The method-of-moments negative binomial size estimate is \(1.1^2/(139/99-1.1)\approx3.9797\); this is a useful check, not the likelihood estimate.

Step 2: Fit the Poisson model. The maximum likelihood estimate is \(\widehat{\lambda}=1.1\). The aggregated Poisson log likelihood is \(\ell_{\mathrm{P}}=110\ln(1.1)-100(1.1)-[15\ln(2!)+10\ln(3!)+5\ln(4!)]\). The factorial contribution is \(15\ln(2!)+10\ln(3!)+5\ln(4!)\approx44.205\). Also, \(110\ln(1.1)\approx10.4841\), so \(\ell_{\mathrm{P}}\approx10.4841-110-44.2051=-143.7210\). There is one estimated parameter, giving \(\mathrm{AIC}_{\mathrm{P}}=-2\ell_{\mathrm{P}}+2(1)\approx289.44\).

Step 3: Fit the negative binomial model. With \(\widehat{\mu}=1.1\), the score equation for \(r\), using these observed frequencies, is \(100\ln[r/(r+1.1)]+60/r+30/(r+1)+15/(r+2)+5/(r+3)=0\). Numerically solving this equation gives \(\widehat r\approx3.369\). Substitution into the negative binomial log likelihood gives \(\ell_{\mathrm{NB}}\approx-142.019\). This model has two estimated parameters, \(\mu\) and \(r\), so \(\mathrm{AIC}_{\mathrm{NB}}=-2(-142.019)+2(2)\approx288.04\). The negative binomial AIC is about \(289.44-288.04=1.40\) lower.

Step 4: Check grouped Poisson fit. With \(\widehat{\lambda}=1.1\), the fitted probabilities are \(P(N=0)=e^{-1.1}\approx0.3329\), \(P(N=1)=1.1e^{-1.1}\approx0.3662\), and \(P(N\geq2)\approx0.3010\), calculated from unrounded probabilities. The corresponding expected counts are approximately 33.287, 36.616, and 30.097. Grouping the observations into \(0\), \(1\), and \(\geq2\), the Pearson statistic is \((40-33.287)^2/33.287+(30-36.616)^2/36.616+(30-30.097)^2/30.097\approx2.55\). There are three groups and one estimated parameter, so the usual reference degrees of freedom are \(3-1-1=1\). The approximate chi-square upper-tail probability is 0.11.

Conclusion and check. The variance and AIC favor the negative binomial, but the AIC difference of about 1.40 is modest rather than decisive. The grouped Poisson diagnostic does not show a strong lack of fit at conventional thresholds, and grouping into only three categories discards detail. The evidence supports considering the negative binomial, not declaring it unquestionably correct.

Common mistake avoided. Using the displayed rounded probabilities, \(1-0.3329-0.3662=0.3009\), not \(0.3010\). The tail probability reported above is approximately \(0.3010\) when calculated from unrounded probabilities; rounded intermediate values should not be used to claim an exact equality.

What the Diagnostics Can and Cannot Tell You

A Pearson goodness-of-fit statistic compares observed counts with expected counts in categories. If the categories have adequate expected counts and the usual large-sample approximation is reasonable, the statistic is compared with a chi-square distribution. When \(k\) categories are used and \(p\) parameters have been estimated from the data, the customary degrees of freedom are \(k-1-p\). Categories may need to be combined when expected counts are too small; the category definitions and resulting degrees of freedom must then match the calculation.

A grouped test is only one diagnostic. It can miss discrepancies within a broad category, such as \(N\geq2\). Always ensure the expected counts correspond to the model and parameter estimates being tested. In particular, do not interpret a nonsignificant test as proof that the model is true: it may reflect limited data or a coarse grouping.

AIC compares fitted models using \(\mathrm{AIC}=-2\ell_{\max}+2p\), where \(\ell_{\max}\) is the maximized log likelihood and \(p\) is the number of estimated parameters. It balances fit against parameter count. A lower AIC is preferred among the models compared, but the size of the difference matters: a small difference is weak comparative evidence, not a decisive selection. The likelihoods must be based on the same observations and the same likelihood convention.

Worked Example: Turning Fitted Likelihoods into an AIC Comparison

For a separate hypothetical portfolio, a fitting procedure reports maximized log likelihoods of \(-50.2\) for a one-parameter Poisson model and \(-47.9\) for a two-parameter negative binomial model, both fitted to the same observations. Calculate both AIC values and interpret the comparison.

Method. Apply the AIC formula to each reported maximized log likelihood, counting every estimated model parameter.

For Poisson, \(p=1\), so \(\mathrm{AIC}_{\mathrm{P}}=-2(-50.2)+2(1)=100.4+2=102.4\). For negative binomial, \(p=2\), so \(\mathrm{AIC}_{\mathrm{NB}}=-2(-47.9)+2(2)=95.8+4=99.8\).

The negative binomial AIC is lower by \(102.4-99.8=2.6\). It is preferred by this comparison, but the difference is not overwhelming. The negative binomial improves the maximized log likelihood by \(2.3\), while its extra parameter incurs a penalty of 2 AIC units.

Check and common mistake. The Poisson log likelihood is less favorable, but AIC uses minus twice that log likelihood, then adds the parameter penalty. Comparing the raw log likelihoods alone would omit the cost of the extra negative binomial parameter.

A Practical Selection Sequence

Check the data structure.
Confirm that periods, exposure amounts, and independence assumptions are compatible with the models being considered.

Estimate the mean and inspect dispersion.
Compare the sample variance with the sample mean. Treat this as evidence about variability, not as a final model-selection rule.

Assess fitted distributions.
Compare observed and expected frequencies, using sensible grouping and a goodness-of-fit statistic where appropriate.

Compare likelihood-based criteria.
Use AIC with the correct number of parameters and the same observations for each candidate model.

State a qualified conclusion.
Explain which model is favored and whether the evidence is strong, modest, or inconclusive; do not treat one diagnostic as proof.

A frequent exam trap is to infer “negative binomial” automatically whenever the sample variance exceeds the mean. Sampling variation can produce that pattern under a Poisson model, and a negative binomial can still fit poorly for other reasons. The reverse trap is to select Poisson merely because its grouped goodness-of-fit test is not significant. A defensible choice uses the assumptions, dispersion, fitted frequencies, and likelihood evidence together. This model-selection framework is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

Work each problem using the stated assumptions, then compare your result with the answer.

A hypothetical portfolio records counts \(0,1,1,2,3,3\) in six independent, equal exposure periods. Find the Poisson maximum likelihood estimate of the mean. (Answer: \(\widehat{\lambda}=10/6\approx1.6667\))
For counts \(0,0,1,2,4\), calculate the sample mean, unbiased sample variance, and negative binomial method-of-moments size estimate. (Answer: \(\bar N=1.4,\ s^2=2.8,\ \widehat r_{\mathrm{MOM}}=1.4\))
Two models fitted to the same hypothetical counts have log likelihoods \(-50.2\) for Poisson and \(-47.9\) for negative binomial, with one and two parameters respectively. Which has lower AIC, and by how much? (Answer: Negative binomial; its AIC is 99.8 versus 102.4, lower by 2.6)
Under a Poisson model with \(\lambda=0.8\), calculate the probabilities of zero claims, one claim, and at least two claims, to four decimals. (Answer: \(P(N=0)=0.4493,\ P(N=1)=0.3595,\ P(N\geq2)=0.1912\))
A grouped goodness-of-fit test uses four categories and estimates two model parameters from the data. What are the usual chi-square reference degrees of freedom, assuming the expected counts are adequate? (Answer: \(4-1-2=1\) degree of freedom)

What You'll Learn