Diagnosing a Portfolio in Sequence
A portfolio can have a reasonable average claim frequency and still be poorly described by a Poisson model. The Poisson distribution requires conditional variance to equal conditional mean. If policies with similar modeled means show substantially more variation than that assumption permits, Poisson-based uncertainty measures may be too small. A useful diagnosis therefore looks at both where observed counts differ from expected counts and how much variation occurs within risk classes.
This tutorial follows a practical sequence: calculate policy-level Poisson residuals, review class-level observed-to-expected ratios, and then estimate a Negative Binomial dispersion parameter. These diagnostics answer related but different questions. A class-level ratio can point to a pattern in the fitted mean structure; it cannot, by itself, reveal all within-class variation.
What the Diagnostics Measure
The Pearson statistic summarizes departures from the Poisson mean-variance assumption. Under a correctly specified Poisson model, \(E[Y_i]=\operatorname{Var}(Y_i)=\mu_i\). Thus each squared Pearson residual has expected value 1, and their sum is approximately on the scale of the residual degrees of freedom. A large statistic relative to those degrees of freedom is evidence that the Poisson model does not capture the observed variation. It is a diagnostic, not a proof of one particular cause.
For a class \(j\), define its observed-to-expected ratio as \(O_j/E_j\), where \(O_j\) is the observed claim total and \(E_j\) is the total expected under the model. A ratio above 1 means the class produced more claims than expected; a ratio below 1 means fewer. When expected counts are fixed, a class-level Pearson contribution is \((O_j-E_j)^2/E_j\). But aggregation can hide differences among policies within a class: positive and negative policy residuals can cancel in the class total.
A common Negative Binomial (NB2) specification keeps the mean structure and allows variance to exceed the mean:
The parameter \(\alpha=0\) recovers the Poisson variance. A positive \(\alpha\) makes the variance increase faster than the mean. One transparent way to estimate \(\alpha\) for a diagnostic refit is the method of moments, when the \(\mu_i\) are treated as fixed reference means:
If the unconstrained estimate is negative, it is not evidence for a negative NB2 dispersion parameter: the model requires \(\alpha\geq0\), so a constrained estimate is set to zero. In an applied analysis, one should also consider whether the mean structure or dependence assumptions need attention; extra variation need not have a single cause.
Why the Moment Estimate Has This Form
For fixed \(\mu_i\), the NB2 variance assumption implies the following expectation for each centered squared count:
Summing over independent observations gives an expected numerator of \(\alpha\sum_i\mu_i^2\). Dividing the observed numerator by \(\sum_i\mu_i^2\) therefore gives the method-of-moments estimate. The derivation depends on the specified means being appropriate and treated as fixed for this calculation. If means are estimated from the same observations, fitting affects residual variation; the simple fixed-mean calculation should not be presented as an exact finite-sample adjustment. A likelihood fit or an appropriate adjustment is preferable when required by the analysis.
Worked Example: A Portfolio with Excess Within-Class Variation
A hypothetical insurer evaluates 20 policies using a Poisson frequency model calibrated on separate experience. The expected count is 1.5 for each of 10 policies in Class A and 3 for each of 10 policies in Class B. In Class A, five policies have zero claims and five have four claims. In Class B, five have zero claims and five have six claims. Calculate class-level observed-to-expected ratios, the policy-level Poisson Pearson statistic, and a fixed-mean NB2 method-of-moments estimate.
Method and assumptions. Treat the stated Poisson means as fixed reference means, and treat policy counts as independent. Because the means were calibrated on separate experience, no parameters are subtracted from the diagnostic sample to obtain its reference degrees of freedom.
Class-level ratios. Class A has \(5(0)+5(4)=20\) observed claims and \(10(1.5)=15\) expected claims, so \(O_A/E_A=20/15=1.3333\). Class B has \(5(0)+5(6)=30\) observed and \(10(3)=30\) expected, so \(O_B/E_B=30/30=1.0000\). The portfolio totals are 50 observed and 45 expected.
Poisson Pearson statistic. In Class A, the five zero-count policies each contribute \((0-1.5)^2/1.5=2.25/1.5=1.5\), for \(5(1.5)=7.5\). The five four-count policies each contribute \((4-1.5)^2/1.5=6.25/1.5=4.1667\), for \(5(4.1667)=20.8333\). Class A contributes \(28.3333\).
In Class B, each zero-count policy contributes \((0-3)^2/3=3\), and each six-count policy contributes \((6-3)^2/3=3\). Its contribution is \(5(3)+5(3)=30\). Therefore the total statistic is \(28.3333+30=58.3333\), with 20 reference degrees of freedom. The statistic is substantially larger than its reference degrees of freedom, indicating more policy-level variation than the Poisson model predicts.
NB2 moment estimate. For Class A, the sum of squared residuals before division by \(\mu_i\) is \(5(1.5^2)+5(2.5^2)=11.25+31.25=42.5\). For Class B it is \(5(3^2)+5(3^2)=90\). Thus \(\sum_i(Y_i-\mu_i)^2=132.5\), while \(\sum_i\mu_i=10(1.5)+10(3)=45\). The numerator of the moment estimate is \(132.5-45=87.5\). Its denominator is \(\sum_i\mu_i^2=10(1.5^2)+10(3^2)=22.5+90=112.5\). Therefore \(\widehat{\alpha}_{\mathrm{MOM}}=87.5/112.5=0.7778\), rounded to four decimals.
At this estimate, an individual Class A count has fitted variance \(1.5+0.7778(1.5^2)=3.25\); a Class B count has fitted variance \(3+0.7778(3^2)\approx10.00\). The method confirms that excess variation is material in this example. The class ratios alone would not show its full extent: the Class B total matches expectation, despite substantial variation among its policies. The mistake to avoid is treating a class total as a substitute for policy-level residuals.
Class-Level Departures and a Second Diagnostic
Class-level ratios remain valuable because a pattern across classes can reveal that expected frequencies are systematically high or low for particular groups. They should be read alongside the residual statistic, not instead of it. The next example uses fixed benchmark expectations so the calculation has a direct interpretation without a same-sample parameter adjustment.
Worked Example: Finding a Class-Level Pattern
A hypothetical portfolio is divided into three vehicle-use classes. A model calibrated independently predicts 80, 120, and 200 claims, respectively. The observed counts are 72, 150, and 178. Calculate each class’s observed-to-expected ratio and the total class-level Pearson statistic. State what the results do and do not establish.
Method. Compare each observed class total with its fixed expected total. Because the expectations are independent benchmarks, the three class contributions can be summed directly; the reference degrees of freedom are 3 for this fixed-expectation diagnostic.
The ratios are \(72/80=0.9000\), \(150/120=1.2500\), and \(178/200=0.8900\). The class contributions are \((72-80)^2/80=64/80=0.8000\), \((150-120)^2/120=900/120=7.5000\), and \((178-200)^2/200=484/200=2.4200\). The total is \(0.8000+7.5000+2.4200=10.7200\).
Class 2 is notably above expectation, while Classes 1 and 3 are below. The statistic signals departure from the fixed class expectations, but does not by itself distinguish a misspecified class-level mean pattern from excess count variation. If the expectations instead came from parameters estimated on these same observations, the reference degrees of freedom would need to account for those fitted parameters. Do not apply the fixed-expectation degrees of freedom automatically to an in-sample fit.
Deciding Whether the NB2 Variance Matters
A positive estimate of \(\alpha\) is not, by itself, a decision rule. Assess its size in relation to the portfolio’s fitted means. Under NB2, the ratio of variance to mean for a risk with expected count \(\mu_i\) is \(1+\alpha\mu_i\). Thus the same \(\alpha\) has a larger variance impact at higher expected frequencies. Consider the practical effect on uncertainty, the evidence in the residual diagnostics, and whether the mean structure is credible.
The following example isolates that variance comparison. It uses a fixed mean model and an already estimated NB2 dispersion value; it is not another estimate of \(\alpha\).
Worked Example: Comparing Poisson and NB2 Uncertainty
A hypothetical block contains 12 independent policies, each with expected count \(\mu_i=0.5\). A prior diagnostic has produced an NB2 estimate \(\alpha=0.4\). Compare the Poisson and NB2 variances of the total claim count, and state the effect on the standard deviation.
Method. For independent counts, the variance of the total is the sum of individual variances. Under Poisson, each policy’s variance is 0.5, so the total variance is \(12(0.5)=6\), and its standard deviation is \(\sqrt{6}=2.4495\).
Under NB2, the variance for one policy is \(0.5+0.4(0.5^2)=0.5+0.1=0.6\). The total variance is \(12(0.6)=7.2\), and the standard deviation is \(\sqrt{7.2}=2.6833\). Rounded to four decimals, NB2 raises the standard deviation from 2.4495 to 2.6833, an increase of \(2.6833/2.4495-1\approx0.0954\), or 9.54%. The variance increase is \(7.2/6-1=20\%\). This change may matter for uncertainty measures even though the mean remains unchanged.
Common Traps in a Portfolio Diagnosis
The most consequential trap is aggregation. In the first example, a class-total calculation for Class A would give \((20-15)^2/15=1.6667\), and Class B would give \((30-30)^2/30=0\). Those class-level contributions total only 1.6667, while the policy-level Pearson statistic is 58.3333. The two calculations answer different questions: the class-total version discards within-class variation. Use policy-level residuals when the purpose is to assess policy-level Poisson dispersion.
Other checks help keep the diagnosis coherent:
- Use the correct reference degrees of freedom. Subtract parameters estimated from the diagnostic sample; do not subtract parameters that were fixed from independent calibration data.
- Keep the mean and variance questions separate. A class ratio far from 1 may point to a mean-model issue. Excess residual variation may persist even when class totals are close to expected.
- Use the stated NB2 parameterization. Here \(\operatorname{Var}(Y_i)=\mu_i+\alpha\mu_i^2\). Do not transfer a dispersion estimate from another parameterization without converting it.
- Interpret the size of \(\alpha\) at relevant means. Its variance effect depends on \(\alpha\mu_i\), not on \(\alpha\) alone.
A defensible conclusion describes the evidence and its scope: for example, that the Poisson residual statistic is large relative to its reference degrees of freedom, that specific classes are over or under expected, and that an NB2 variance allowance materially changes uncertainty at the portfolio’s fitted means. This diagnostic approach is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.
Check Your Understanding
Work each problem using the stated reference means and assumptions.
- A fixed benchmark predicts 60, 90, and 150 claims in three classes; observed counts are 54, 99, and 147. Find each observed-to-expected ratio and the total class-level Pearson statistic. (Answer: Ratios \(0.9000,\ 1.1000,\ 0.9800\); statistic \(1.5600\), with 3 fixed-expectation degrees of freedom.)
- Four independent policies have fixed means \(\mu_i=2\) and observed counts \(0,1,3,4\). Calculate the Poisson Pearson statistic and the unconstrained fixed-mean NB2 moment estimate. (Answer: Pearson statistic \(5.0000\); \(\widehat{\alpha}_{\mathrm{MOM}}=0.1250\).)
- A fixed class model expects 40, 80, and 120 claims, while observed counts are 50, 72, and 138. Calculate the three observed-to-expected ratios and the total class-level Pearson statistic. (Answer: Ratios \(1.2500,\ 0.9000,\ 1.1500\); statistic \(6.0000\), with 3 fixed-expectation degrees of freedom.)
- Twelve independent policies each have mean count \(0.5\), and an NB2 fit gives \(\alpha=0.4\). Find the total expected count, Poisson variance, NB2 variance, and NB2 standard deviation. (Answer: Expected total \(6\); Poisson variance \(6\); NB2 variance \(7.2\); NB2 standard deviation \(2.6833\).)