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CAS Exam 8 · Frequency Model Specification · Tutorial 113 of 989

How Overdispersion Changes Relativity Uncertainty

Learn to adjust coefficient uncertainty for overdispersion and translate the resulting confidence intervals to rating-factor scale.

Intermediate 11 min read

What You'll Learn

  • Distinguish a Poisson model-based standard error from a dispersion-adjusted standard error.
  • Apply the square-root dispersion adjustment when its assumptions hold.
  • Construct and interpret confidence intervals for log-link rating factors.
  • Account for covariance when estimating uncertainty in a contrast between classes.
  • Recognize when a simple scalar adjustment is inappropriate.

Why Overdispersion Changes Rating-Factor Uncertainty

A frequency model can describe the expected claim count well while the observed counts vary more than a Poisson model allows. This distinction matters: if the mean structure is retained but the variance is larger, a Poisson model-based standard error can make a rating factor look more precisely estimated than it is. This tutorial quantifies that effect for fitted coefficients and multiplicative relativities, aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: In a Poisson frequency model, observations have conditional mean \(\mu_j\) and conditional variance \(\mu_j\). A common overdispersed specification retains the same mean model but uses \(\operatorname{Var}(N_j\mid\mathbf{x}_j)=\phi\mu_j\), where \(N_j\) is the observed count, \(\mathbf{x}_j\) is its rating information, and \(\phi>0\) is the dispersion parameter. Values \(\phi>1\) indicate variance beyond the Poisson assumption; \(\phi=1\) gives the Poisson variance.

For example, with exposure \(e_j\) and a log link, the expected count may be written \(\mu_j=e_j\exp(\mathbf{x}_j^\mathsf{T}\boldsymbol{\beta})\). The model-based Poisson standard errors use the Poisson variance assumption. Under the overdispersed specification, the coefficient estimates can remain the same as the Poisson estimates under the usual quasi-Poisson estimating equations, but their estimated covariance is scaled by \(\phi\). Consequently, when this scalar adjustment applies, standard errors are scaled by \(\sqrt{\phi}\), not by \(\phi\).

From a Coefficient to a Relativity Interval

With a log link, a coefficient for a binary rating indicator represents a log relativity. If \(\beta_k\) is the coefficient for a class compared with its reference class, the fitted relativity is \(\widehat{R}_k=\exp(\widehat{\beta}_k)\). Uncertainty is first assessed on the coefficient scale, where an approximate normal interval is formed, and then transformed to the positive relativity scale.

$$ \widehat{\operatorname{SE}}_{\phi}(\widehat{\beta}_k) =\sqrt{\widehat{\phi}}\, \widehat{\operatorname{SE}}_{\text{Pois}}(\widehat{\beta}_k), \qquad \widehat{\beta}_k\ \pm\ z_{1-\alpha/2} \widehat{\operatorname{SE}}_{\phi}(\widehat{\beta}_k) $$

For a 95% interval, \(z_{0.975}\) is approximately 1.96. Exponentiate both endpoints to obtain the interval for the relativity. Because the exponential function is increasing, the lower coefficient limit remains the lower relativity limit. The resulting interval is generally asymmetric around \(\widehat{R}_k\), even though the coefficient interval is symmetric around \(\widehat{\beta}_k\).

Why the Standard Error Is Multiplied by the Square Root

The scaling follows from the variance of the coefficient estimator. Let \(A\) denote the information-like matrix in the Poisson estimating equations. Under the Poisson variance assumption, the model-based covariance is approximately \(A^{-1}\). If the actual conditional variance is instead \(\phi\mu_j\), the variance of the estimating equations is multiplied by \(\phi\). The sandwich covariance then has the form below, when the same mean structure and scalar dispersion assumption apply.

$$ \operatorname{Cov}(\widehat{\boldsymbol{\beta}}) \approx A^{-1}(\phi A)A^{-1} =\phi A^{-1} $$

Thus each coefficient variance is multiplied by \(\phi\), and each standard error, being the square root of a variance, is multiplied by \(\sqrt{\phi}\). This derivation also explains why multiplying a standard error directly by \(\phi\) is incorrect. In practice, \(\phi\) is usually estimated, for example from a Pearson statistic divided by residual degrees of freedom. The resulting interval is a plug-in interval: it uses the estimated dispersion but does not separately account for uncertainty in that estimate.

Worked Example: Adjusting a Single Rating-Factor Interval

A hypothetical commercial property frequency model uses a log link. The fitted coefficient for a protection class relative to the reference class is \(\widehat{\beta}=\ln(1.25)\), and its model-based Poisson standard error is 0.10. The estimated dispersion is \(\widehat{\phi}=2.25\). Calculate the fitted relativity and its approximate 95% interval. Assume the scalar quasi-Poisson adjustment is appropriate.

Method and adjustment. The fitted factor is \(\exp(\ln(1.25))=1.25\). Since \(\sqrt{2.25}=1.5\), the adjusted standard error is \(0.10(1.5)=0.15\).

Coefficient-scale limits. The 95% margin is \(1.96(0.15)=0.294\). The coefficient estimate is \(\ln(1.25)=0.22314\), so the limits are \(0.22314-0.294=-0.07086\) and \(0.22314+0.294=0.51714\).

Relativity-scale limits. Exponentiating gives \(\exp(-0.07086)=0.9316\) and \(\exp(0.51714)=1.6772\), to four decimal places. The fitted factor and interval are therefore \(1.25\) and \([0.9316,\ 1.6772]\).

Check and trap avoided. The interval contains the fitted factor, and its asymmetry is expected after exponentiation. Multiplying 0.10 by 2.25 would incorrectly give a standard error of 0.225; the correct multiplier for a standard error is \(\sqrt{2.25}\).

Intervals for Contrasts Between Rating Classes

A comparison of two non-reference classes is a contrast of coefficients. Its uncertainty depends on both coefficient variances and their covariance. If the Poisson model-based covariance matrix is available and the scalar dispersion assumption applies, multiply the variance of the contrast by \(\widehat{\phi}\), or equivalently multiply its standard error by \(\sqrt{\widehat{\phi}}\). Do not calculate the contrast uncertainty by adding standard errors.

$$ \widehat{\operatorname{Var}}_{\text{Pois}}(\widehat{\beta}_b-\widehat{\beta}_a) = \widehat{\operatorname{Var}}_{\text{Pois}}(\widehat{\beta}_b) +\widehat{\operatorname{Var}}_{\text{Pois}}(\widehat{\beta}_a) -2\widehat{\operatorname{Cov}}_{\text{Pois}}(\widehat{\beta}_a,\widehat{\beta}_b) $$

The contrast \(\widehat{\beta}_b-\widehat{\beta}_a\) is the log of the factor for class \(b\) relative to class \(a\). Exponentiating its confidence limits gives an interval for that ratio of rating factors.

Worked Example: Comparing Two Classes With Covariance

In a hypothetical frequency model, two class coefficients have estimates \(\widehat{\beta}_a=0.30\) and \(\widehat{\beta}_b=0.45\). Their Poisson model-based variances are 0.0064 and 0.0049, respectively, and their covariance is \(-0.0010\). The estimated dispersion is 2. Find a 95% interval for the factor for class \(b\) relative to class \(a\).

Method and contrast estimate. The log contrast is \(0.45-0.30=0.15\), so its fitted ratio is \(\exp(0.15)=1.1618\).

Poisson contrast variance. Including the covariance gives \(0.0049+0.0064-2(-0.0010)=0.0133\). The unadjusted contrast standard error is \(\sqrt{0.0133}=0.1153\).

Dispersion adjustment. The adjusted variance is \(2(0.0133)=0.0266\), and the adjusted standard error is \(\sqrt{0.0266}=0.1631\). The 95% margin is \(1.96(0.1631)=0.3197\), so the log limits are \(0.15-0.3197=-0.1697\) and \(0.15+0.3197=0.4697\).

Ratio interval and check. Exponentiating the limits gives approximately \(0.844\) and \(1.599\), to three decimals. The interval \([0.844,\ 1.599]\) contains 1.1618. Ignoring the negative covariance, or omitting it from the variance formula, would give the wrong uncertainty; the covariance term here increases the variance because the covariance is negative.

Estimating Dispersion and Knowing the Adjustment’s Limits

A common estimate of dispersion for count data is the Pearson chi-square statistic divided by its residual degrees of freedom. In a fitted model, Pearson residuals compare observed counts with fitted means while accounting for the Poisson variance. This estimate is a diagnostic and a scale estimate; it does not by itself identify why the data are overdispersed.

$$ \widehat{\phi} = \frac{\sum_j (N_j-\widehat{\mu}_j)^2/\widehat{\mu}_j} {\text{residual degrees of freedom}} $$

Worked Example: Using a Pearson Dispersion Estimate

A hypothetical fitted frequency model has a Pearson chi-square statistic of 30 on 15 residual degrees of freedom. A rating coefficient is estimated as 0.20 with a Poisson model-based standard error of 0.05. Calculate the dispersion-adjusted 95% interval for its multiplicative factor.

Method and dispersion estimate. The estimated dispersion is \(30/15=2\). The factor estimate is \(\exp(0.20)=1.2214\).

Adjusted uncertainty. The adjusted standard error is \(0.05\sqrt{2}=0.07071\). The 95% margin is \(1.96(0.07071)=0.13859\).

Limits and check. The coefficient limits are \(0.20-0.13859=0.06141\) and \(0.20+0.13859=0.33859\). Exponentiating gives \([1.0633,\ 1.4031]\), to four decimals. The point estimate 1.2214 lies between these limits. Rounding the dispersion-adjusted standard error before calculating the margin can shift the final endpoints, so intermediate values were retained here.

The scalar adjustment has a defined scope. It is appropriate when the mean model is retained and the variance is treated as a common multiple of the Poisson variance. It is not a universal replacement for refitting a different count distribution. For example, a negative binomial model specifies a particular mean-variance relationship and generally has its own fitted covariance matrix; one should use that model’s uncertainty estimates rather than automatically applying a Pearson-based multiplier to Poisson standard errors. Likewise, robust sandwich standard errors need not equal a model-based covariance multiplied by one scalar, especially when the variance inflation differs across observations or the mean model is misspecified.

Exam Traps and Practical Interpretation

  • Scaling the wrong quantity. If \(\widehat{\phi}=2.25\), the standard error multiplier is 1.5, while the variance multiplier is 2.25.
  • Leaving the interval on the wrong scale. A coefficient interval describes log relativities. Exponentiate both limits to report a multiplicative-factor interval.
  • Dropping the covariance in a contrast. Variances of differences include \(-2\operatorname{Cov}\); separate standard errors cannot simply be added or subtracted.
  • Assuming all overdispersion has the same structure. A scalar quasi-Poisson adjustment is not automatically the correct uncertainty calculation for every alternative count model or for every robust covariance estimate.
  • Confusing uncertainty with the fitted indication. The adjustment changes standard errors and intervals, not the coefficient estimate or the fitted factor under the stated quasi-Poisson setup.
Key takeaway: When a common dispersion factor scales the Poisson coefficient covariance, multiply a model-based standard error by \(\sqrt{\widehat{\phi}}\), form the coefficient-scale interval, and exponentiate its endpoints to obtain the rating-factor interval. For contrasts, first calculate the full variance, including covariance.

Check Your Understanding

For each problem, use a 95% normal interval with \(z=1.96\). Unless stated otherwise, assume the scalar dispersion adjustment is appropriate.

  1. A coefficient estimate is 0.08, its Poisson model-based standard error is 0.03, and \(\widehat{\phi}=3\). Find the adjusted standard error and the interval for its multiplicative factor, to three decimals. (Answer: adjusted SE \(=0.0520\); interval \([0.978,\ 1.199]\))
  2. A log-relativity estimate is 0.10 with Poisson model-based standard error 0.08. The estimated dispersion is 1.25. Find the adjusted standard error and factor interval, to three decimals. (Answer: adjusted SE \(=0.0894\); interval \([0.927,\ 1.317]\))
  3. A coefficient estimate is 0.15 with model-based standard error 0.06, and \(\widehat{\phi}=1\). Find the factor interval, to three decimals. Explain why the point estimate need not be the midpoint of the interval. (Answer: interval \([1.033,\ 1.307]\); the interval is symmetric on the coefficient scale, not after exponentiation.)
  4. For two fitted coefficients, the estimates are 0.20 and 0.35; their Poisson model-based variances are 0.0064 and 0.0049, and their covariance is \(-0.0010\). If \(\widehat{\phi}=2\), find the adjusted standard error of the difference and the 95% interval for the factor ratio, to three decimals. (Answer: adjusted SE \(=0.1631\); interval \([0.844,\ 1.599]\))
  5. A model has Pearson chi-square 27 on 18 residual degrees of freedom. A coefficient estimate is 0.10 with Poisson model-based standard error 0.04. Estimate dispersion and calculate the 95% factor interval, to three decimals. (Answer: \(\widehat{\phi}=1.5\); adjusted SE \(=0.0490\); interval \([1.004,\ 1.217]\))
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