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CAS Exam 8 · Frequency Model Specification · Tutorial 107 of 989

Recognizing Overdispersion In Claim Counts

Distinguish genuine evidence of excess conditional claim-count variation from the variability expected when policies have different fitted means.

Intermediate 10 min read

What You'll Learn

  • State the conditional mean and equidispersion assumptions of a Poisson frequency model
  • Calculate Pearson dispersion from policy counts and fitted means
  • Interpret a dispersion estimate while accounting for fitted parameters and sample size
  • Explain why a portfolio’s raw variance can exceed its average mean without conditional overdispersion
  • Recognize common misinterpretations of dispersion evidence

Why Overdispersion Matters

A Poisson frequency model is useful for pricing because it connects each policy’s expected claim count to its exposure and rating characteristics. But a good fit to average counts does not guarantee that the model describes how variable counts are around those averages. If counts vary more than the fitted Poisson model permits, the model may understate uncertainty and its fit should be examined before relying on its conclusions.

The central comparison is conditional: for policies with the same modeled characteristics and exposure, does observed claim-count variability exceed what their fitted means imply? This distinction matters because policies can have different fitted means even when a Poisson model is correct for each policy.

Definition: Let \(Y_i\) be the claim count for policy \(i\), and let \(\mu_i\) be its fitted expected claim count for the policy’s exposure period. Under the Poisson frequency model, conditional on the modeled characteristics and exposure, \(Y_i\sim\operatorname{Poisson}(\mu_i)\), so \(E[Y_i\mid\text{modeled information}]=\mu_i\) and \(\operatorname{Var}(Y_i\mid\text{modeled information})=\mu_i\). Overdispersion means that the conditional variation in the counts is greater than the fitted Poisson model allows.

In a frequency model with a log link and an exposure offset, a typical fitted mean has the form \(\mu_i=e_i\exp(x_i^\mathsf{T}\widehat\beta)\), where \(e_i\) is exposure and \(x_i\) represents the policy’s modeled rating characteristics. The offset scales the expected count for exposure; it does not change the Poisson model’s conditional equality of mean and variance. The question here is whether that equality is adequate after conditioning on the fitted means.

Why the Poisson Mean Equals Its Variance

The equidispersion condition is a property of the Poisson distribution, not an empirical rule that every small group must satisfy exactly. For \(Y\sim\operatorname{Poisson}(\mu)\), the probability of count \(y\) is \(P(Y=y)=e^{-\mu}\mu^y/y!\), for \(y=0,1,2,\ldots\). The first two factorial moments are \(E[Y]=\mu\) and \(E[Y(Y-1)]=\mu^2\). Since \(Y^2=Y(Y-1)+Y\), the variance follows:

$$ \operatorname{Var}(Y) =E[Y^2]-E[Y]^2 =E[Y(Y-1)]+E[Y]-\mu^2 =\mu^2+\mu-\mu^2 =\mu. $$

This gives the model’s benchmark: a policy with fitted mean \(0.6\) has conditional variance \(0.6\), while a policy with fitted mean \(2.4\) has conditional variance \(2.4\). A count is discrete and will not equal its expected value in general. Overdispersion is a claim about variability across observations relative to the fitted benchmark, not about whether an individual policy has a count above or below its mean.

Assessing Excess Variation With Pearson Residuals

For policy \(i\), the Pearson residual is the observed count minus its fitted mean, scaled by the square root of the fitted mean. Squaring and adding these residuals gives the Pearson statistic \(X_P^2\). Dividing by the residual degrees of freedom gives a familiar summary of excess variation.

$$ r_{Pi}=\frac{Y_i-\widehat\mu_i}{\sqrt{\widehat\mu_i}}, \qquad X_P^2=\sum_{i=1}^{n}\frac{(Y_i-\widehat\mu_i)^2}{\widehat\mu_i}, \qquad \widehat\phi=\frac{X_P^2}{n-p}. $$

Here \(n\) is the number of policy observations and \(p\) is the number of fitted model parameters, including an intercept if one is fitted. The quantity \(n-p\) is the residual degrees of freedom in the usual large-sample approximation. A value of \(\widehat\phi\) near 1 is broadly consistent with Poisson-scale variation. A value materially above 1 is a warning that the counts may be overdispersed relative to the fitted means. A value below 1 indicates less residual variation than the Poisson benchmark, not overdispersion.

This ratio is a diagnostic, not a universal pass-or-fail rule. Its interpretation depends on the amount of data, whether the mean structure is reasonable, and whether the model’s assumptions are appropriate. Small samples can produce substantial variation in the ratio. Also, fitted parameters use information from the same counts being checked, which is why the residual degrees of freedom account approximately for the parameters estimated.

Justification: If the Poisson means were known and the model were correct, each standardized residual would have expected square approximately 1, since \(E[(Y_i-\mu_i)^2/\mu_i]=\operatorname{Var}(Y_i)/\mu_i=1\). Thus the sum of squared residuals would be roughly the number of observations. Estimating \(p\) parameters uses some of the observed variation, so the corresponding benchmark for the residual sum is approximately \(n-p\). This motivates \(\widehat\phi=X_P^2/(n-p)\); it is an approximation, especially in small samples or when fitted means are very small.

A large Pearson ratio signals a mismatch between observed residual variation and the Poisson benchmark, but does not by itself establish why the mismatch occurred. Potential explanations include an inadequate mean specification, heterogeneous risk not represented by the rating variables, dependence between observations, or problems in the data. Determining which explanation is relevant requires further investigation; the dispersion calculation alone does not identify a cause.

Worked Example: Equal Fitted Means and a Large Dispersion Ratio

A hypothetical insurer models eight policies with one fitted intercept and equal exposure. The observed claim counts are \(0,0,0,0,2,2,4,4\). Calculate the fitted mean under an intercept-only Poisson model and assess the Pearson dispersion. State the method and a common mistake to avoid.

Method. For an intercept-only Poisson model with equal exposure, the fitted mean is the sample average. First add the counts:

\(0+0+0+0+2+2+4+4=12\). Therefore \(\widehat\mu=12/8=1.5\) for each policy.

For counts \(0,2,4\), the squared deviations from \(1.5\) are respectively \(2.25,0.25,6.25\). Their frequencies are four, two, and two, so the total squared deviation is \(4(2.25)+2(0.25)+2(6.25)=9+0.5+12.5=22\).

The Pearson statistic is \(X_P^2=22/1.5=14.6667\). One parameter was estimated, so the residual degrees of freedom are \(8-1=7\), and \(\widehat\phi=14.6667/7=2.0952\). To four decimal places, the estimate is \(2.0952\), which is a warning of excess variation relative to the fitted Poisson model.

Check. The Pearson statistic is the total squared deviation divided by the common fitted mean: \(22/1.5=14.6667\). The ratio times the residual degrees of freedom returns the same statistic: \(2.0952(7)\approx14.6664\), with the small difference due to rounding.

Common mistake avoided. Do not divide the Pearson statistic by \(n\) here. One fitted parameter reduces the residual degrees of freedom to \(7\), so the relevant ratio is \(14.6667/7\), not \(14.6667/8\). The result is evidence to investigate, not proof that a particular alternative model is correct.

Unequal Means Require a Conditional Comparison

When policies have different exposures or rating characteristics, their fitted Poisson means differ. A policy with a larger fitted mean is expected to have a larger variance under the model. Consequently, the raw variance of all counts pooled together is not, by itself, a valid test of conditional Poisson equidispersion.

The reason can be shown with the law of total variance. Let \(\mu\) denote a policy’s fitted mean, which varies across the portfolio, and suppose counts are Poisson conditional on that mean. Then:

$$ \operatorname{Var}(Y) =E[\operatorname{Var}(Y\mid\mu)] +\operatorname{Var}(E[Y\mid\mu]) =E[\mu]+\operatorname{Var}(\mu). $$

The first term, \(E[\mu]\), is the average conditional Poisson variance. The second term is the extra variation that arises simply because policy means differ. Therefore, the portfolio-wide variance can exceed the portfolio-wide average fitted mean even if the conditional Poisson model is exactly right. Pearson residuals address this by scaling each policy’s deviation using its own fitted mean.

Worked Example: Heterogeneous Means Without Conditional Overdispersion

Consider a hypothetical portfolio with four equally represented policy types. Their fitted means are \(0.5,1.0,1.5,\) and \(2.0\) claims per policy-year. Suppose each type follows its own conditional Poisson distribution exactly. Calculate the portfolio’s average fitted mean and its theoretical marginal variance. Explain whether the variance difference demonstrates overdispersion.

Method. The average fitted mean is \((0.5+1.0+1.5+2.0)/4=5.0/4=1.25\).

The mean of the four squared deviations of the fitted means from \(1.25\) is \((0.75^2+0.25^2+0.25^2+0.75^2)/4=(0.5625+0.0625+0.0625+0.5625)/4=1.25/4=0.3125\). Thus the variance of the type-specific means is \(0.3125\).

By the law of total variance, the marginal variance is the average conditional Poisson variance plus the variance of the conditional means: \(1.25+0.3125=1.5625\). The marginal variance is \(1.5625\), greater than the average fitted mean of \(1.25\).

Conclusion and check. This difference does not demonstrate conditional overdispersion: every type was assumed to follow its fitted Poisson distribution. The additional \(0.3125\) is entirely due to differences between the policy-type means. The calculation checks because the marginal variance equals \(E[\mu]+\operatorname{Var}(\mu)=1.25+0.3125=1.5625\).

Common mistake avoided. Comparing pooled variance directly with pooled average mean would label this portfolio overdispersed. That comparison ignores the variation in fitted means; use policy-level residuals scaled by their fitted means instead.

Worked Example: Pearson Dispersion With Unequal Fitted Means

A hypothetical fitted frequency model has four policy observations with fitted means \(0.5,1.0,1.5,\) and \(2.0\). Their observed counts are \(0,0,4,\) and \(5\), respectively. The model has one estimated parameter. Calculate the Pearson statistic and dispersion ratio, then interpret them. State the method and a common mistake to avoid.

Method. Compute each squared residual divided by its own fitted mean. The four contributions are:

For the first policy, \((0-0.5)^2/0.5=0.25/0.5=0.5\). For the second, \((0-1.0)^2/1.0=1/1=1.0\). For the third, \((4-1.5)^2/1.5=2.5^2/1.5=6.25/1.5=4.1667\). For the fourth, \((5-2.0)^2/2.0=3^2/2=9/2=4.5\).

Adding the contributions gives \(X_P^2=0.5+1.0+4.1667+4.5=10.1667\). With \(n=4\) and \(p=1\), the residual degrees of freedom are \(4-1=3\). Hence \(\widehat\phi=10.1667/3=3.3889\), to four decimal places.

Conclusion and check. A ratio of \(3.3889\) is a substantial warning that the observed counts vary more than the fitted Poisson model allows. Multiplying the rounded ratio by 3 gives \(10.1667\), confirming the calculation up to rounding.

Common mistake avoided. Do not divide every squared residual by one common average fitted mean. The Pearson statistic uses each policy’s own fitted mean in its denominator, as shown in the four separate contributions.

Interpreting the Evidence Carefully

A dispersion estimate well above 1 is a useful signal, but its meaning should be stated precisely: the residual count variation is larger than the fitted Poisson model’s benchmark, after an approximate adjustment for estimated parameters. It is not a direct estimate of a particular physical source of variation, and it does not prove that the fitted means are wrong. A model can describe average frequency reasonably while still failing to describe the spread of counts.

A few practical checks help keep the interpretation sound:

  • Use the fitted mean for each policy in the Pearson denominator; do not treat unequal means as if they were equal.
  • Use residual degrees of freedom that account for estimated parameters, rather than dividing automatically by the number of policies.
  • Interpret the ratio in light of sample size. A value modestly above 1 in a small sample is not conclusive, and no single cutoff applies in every setting.
  • Separate the finding from its explanation. A large ratio identifies a lack of fit in variability; further analysis is needed to determine the reason.
  • Do not assume that a raw portfolio variance larger than the average fitted mean establishes overdispersion. Variation among the fitted means can create that result even under conditional Poisson assumptions.

Recognizing overdispersion is therefore an intermediate but important model-checking step: first establish what the Poisson model predicts conditionally, then compare observed residual variation with that benchmark, and only then investigate what the evidence may imply. This approach is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each problem, assess the conditional Poisson comparison rather than relying on a pooled variance alone.

  1. A hypothetical intercept-only model is fitted to six equal-exposure policies with counts \(0,0,1,1,4,6\). Find the fitted mean, Pearson statistic, residual degrees of freedom, and dispersion ratio. (Answer: \(\widehat\mu=2\), \(X_P^2=15\), \(n-p=5\), and \(\widehat\phi=3.0000\).)
  2. Five policies have fitted means \(0.5,1,1.5,2,2.5\) and observed counts \(0,1,1,5,2\). The model estimates one parameter. Calculate \(X_P^2\) and \(\widehat\phi\). (Answer: \(X_P^2=0.5+0+0.1667+4.5+0.1=5.2667\); \(n-p=4\), so \(\widehat\phi=1.3167\).)
  3. A portfolio has two equally represented types with fitted means \(1\) and \(3\), and each type is conditionally Poisson. Find the average fitted mean and theoretical marginal variance. Does this establish conditional overdispersion? (Answer: average fitted mean \(=2\); variance of means \(=1\); marginal variance \(=2+1=3\); no, the difference is caused by heterogeneous means.)
  4. A model has \(n=20\) observations, \(p=4\) fitted parameters, and \(X_P^2=32\). Calculate the dispersion ratio and interpret it briefly. (Answer: \(n-p=16\), \(\widehat\phi=32/16=2.0000\); this is a warning of excess residual variation relative to the Poisson benchmark, not proof of its cause.)
  5. For three policies, fitted means are \(0.25,1,\) and \(4\), and observed counts are \(1,0,\) and \(7\). A model has one fitted parameter. Calculate the Pearson statistic and dispersion ratio. (Answer: contributions are \(0.75^2/0.25=2.25\), \(1\), and \(3^2/4=2.25\); \(X_P^2=5.5\), \(n-p=2\), and \(\widehat\phi=2.7500\).)
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