Limited Fluctuation Credibility Calculations

Your company insures 3,000 taxicabs. Last year: 412 claims. Is that enough to let the fleet's experience stand on its own, or must you blend with the class rate? The answer is the full credibility standard, and it depends on whether you are measuring claim counts, average severity, or aggregate losses. Get the target wrong and your standard is off by a factor of 2 or more.

To perform limited fluctuation calculations, start by specifying the tolerance and required probability. Then match the full credibility standard to the statistic being estimated: claim count, average severity, or aggregate loss.

HIGH-FREQUENCY: The Poisson full credibility formula is the most commonly tested credibility calculation.

Tolerance criterion: observed claims within $100r\%$ of expected with probability $p$:

$$n_0 = \left(\frac{z_p}{r}\right)^2 \cdot \frac{\text{Var}(N)}{E[N]}$$

Here $z_p$ is the two-sided critical value for probability $p$, not the $p$th percentile of the standard normal: $z_p = \Phi^{-1}\left(\frac{1+p}{2}\right)$. A 90% probability level therefore uses the 95th percentile, 1.645, and a 95% level uses the 97.5th percentile, 1.960.

Common mistakes

• Forgetting the frequency term for aggregate losses. With Poisson frequency, aggregate uses $1+CV^2$, not just $CV^2$. For exponential severity the standard is twice the unrounded frequency requirement. Other count models use their variance-to-mean ratio in place of 1.
• Using the wrong $z_p$. $z_p$ is the two-sided critical value $\Phi^{-1}((1+p)/2)$, so $p = 0.90$ takes the 95th percentile and $p = 0.95$ takes the 97.5th percentile. In percentile subscripts that is $z_{0.95} = 1.645$ and $z_{0.975} = 1.960$; the 90th percentile value $z_{0.90} = 1.282$ is not used by either standard. Carrying 1.645 into a $p = 0.95$ problem gives 1,083 instead of 1,537.
• Computing $CV^2$ for the wrong distribution. Pareto($\alpha = 3$): $CV^2 = 3$. Gamma($\alpha = 3$): $CV^2 = 1/3$. A factor of 9 difference. Trap: using the gamma formula for a Pareto or vice versa.

Bottom line

• Poisson frequency standard: $n_0 = (z_p/r)^2$; memorize 1,083 ($p = 0.90$) and 1,537 ($p = 0.95$) at $r = 0.05$.
• Severity-only standard: $n_0 = (z_p/r)^2 \cdot CV^2$, no "+1".
• With Poisson frequency, aggregate standard is $n_0=(z_p/r)^2(1+CV^2)$. Otherwise replace 1 with the count variance-to-mean ratio.
• Partial credibility: $Z = \sqrt{n/n_0}$, capped at 1.

Exam shortcut

Before computing, identify the target, tolerance, and required probability. For Poisson frequency, use factors 1 for counts, $CV^2$ for severity, and $1+CV^2$ for aggregate losses. Compare the actual squared coefficients of variation: gamma needs less data than exponential only when its shape exceeds one, the same at shape one, and more below one. Finite-variance Pareto has squared coefficient of variation above one.

The full lesson (about 2,323 words, 15 min read) adds 3 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• 4b

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