Calculate classical (limited fluctuation), Buhlmann, Buhlmann-Straub, and Bayesian credibility-weighted estimates for frequency, severity, and aggregate loss.

A territory wrote 8 claims last year against a bookwide expectation of 12. Do you price next year off the territory's own experience, the manual rate, or some blend? Credibility theory tells you the exact blend.

Why credibility exists. A single risk's loss experience is noisy. A bookwide rate is stable but ignores what makes the risk different. Credibility $Z\in[0,1]$ is the weight you place on the risk's own data; $1-Z$ is the weight on the external benchmark.

$$P_c = Z\,\bar{X} + (1-Z)\,\mu$$

The classical idea: the risk's data deserves full credibility when its observed mean is within tolerance $k$ of the true mean with probability $1-\alpha$. By the central limit theorem you need enough observations to shrink the standard error of $\bar{X}$ to $k\mu / z_{\alpha/2}$.

Frequency standard (Poisson). If claim count $N$ is Poisson with mean $\lambda$, then $\text{Var}(N) = \lambda$. Setting $z_{\alpha/2}\sqrt{\lambda}/\lambda \le k$ gives the expected-claim standard:

$$n_F = \left(\frac{z_{\alpha/2}}{k}\right)^{2}$$

At the textbook default of 90% confidence ($z = 1.645$) and 5% tolerance, $n_F = (1.645/0.05)^2 \approx 1{,}082$ expected claims.

Common mistakes

• Using $\sqrt{n/n_F}$ instead of $n/(n+K)$ for Buhlmann. Square-root credibility is classical. Buhlmann is rational, not radical. A common wrong answer of 0.36 appears when test-takers compute $\sqrt{4/30.43}$ instead of $4/34.43$.
• Plugging Buhlmann $K$ into the classical square-root formula. $K$ and $n_F$ measure different things. Mixing them gives nonsense like $Z = \sqrt{4/30.43} \approx 0.36$; the correct Buhlmann answer is 0.116.
• Confusing exposure with claim count in $n_F$. Classical $n_F$ is in expected claims. If you compare to vehicle-years directly without multiplying by $\mu$ you understate $Z$ by a factor of $1/\sqrt{\mu}$. Converting the 1,082.41 expected-claim standard to 9,278 vehicle-years by dividing by $\mu$ is the right move.

Bottom line

• Classical (limited fluctuation): $Z = \min\!\big(\sqrt{n/n_F},\,1\big)$, with Poisson full-credibility standard $n_F = (z_{\alpha/2}/k)^2$ expected claims set by confidence level and tolerance.
• Severity and aggregate raise the standard: severity multiplies by $\text{CV}_X^2$, aggregate by $1+\text{CV}_X^2$, so aggregate always needs the most data.
• Buhlmann: $Z = n/(n+K)$ with $K = \text{EPV}/\text{VHM}$; higher process variance lowers $Z$, higher between-risk variance raises it.
• Buhlmann is the best linear unbiased predictor of $\mu(\Theta)$: no linear function of $\bar{X}$ has smaller MSE.

Exam shortcut

If the problem gives a confidence level and tolerance without naming a method, default to classical limited fluctuation and use $n_F = (z/k)^2$ with the appropriate severity or aggregate multiplier. If the problem gives class means and class probabilities (or process variances per class), it is a Buhlmann problem: compute $\mu$, $v$, $a$, then $K$. If exposures differ by period, switch to Buhlmann-Straub with total exposure $m$.

The full lesson (about 3,841 words, 26 min read) adds 9 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• A1

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