Full Credibility Standard

Study guide: Exam MAS-II

The full credibility standard is the volume of experience at which classical credibility assigns $Z = 1$ . It is the smallest expected claim count $n_F$ for which the observed mean stays within $\pm k$ of the true mean with probability $p$ .

$n_F = \left(\frac{z_{(1+p)/2}}{k}\right)^{2}\left(\frac{\sigma_S^{2}}{\mu_S^{2}}\right)$

$\text{where } z_{(1+p)/2} \text{ is the standard normal quantile and } k \text{ the tolerance}$

$p$ is the probability level (how often the estimate must land inside the range) and $k$ the range parameter (how wide "inside" is) — the pair $(p, k)$ defines the standard, not the data
Frequency, Poisson: $\sigma_N^2/\mu_N^2 = 1/\lambda$ , so the standard reduces to $n_F = (z/k)^2$ expected claims
Severity: $n_F = (z/k)^2 \cdot \mathrm{CV}_X^2$ claims, where $\mathrm{CV}_X = \sigma_X/\mu_X$ is the Coefficient of Variation of claim size
Aggregate loss with Poisson frequency: $n_F = (z/k)^2\left(1 + \mathrm{CV}_X^2\right)$ claims
A standard stated in claims converts to exposures by dividing by the expected claim frequency per exposure
Below the standard the estimate is not discarded — it is blended by Partial Credibility

The 1,082-Claim Standard An actuary wants the observed frequency to be within $\pm 5\%$ of the true frequency $90\%$ of the time. Claim counts are Poisson. How many claims are needed for full credibility?

Answer $p = 0.90 \Rightarrow z_{0.95} = 1.645$ , and $k = 0.05$ : $n_F = \left(\frac{1.645}{0.05}\right)^{2} = (32.9)^{2} = 1{,}082.4$ 1,082 expected claims — the standard quoted throughout classical credibility. Tightening to $k = 0.025$ quadruples it to 4,330 claims.

Adding Severity Variation Claim severity has mean 3,000 and standard deviation 6,000, and frequency is Poisson. Find the full-credibility standard for aggregate losses at $p = 0.90$ , $k = 0.05$ .

Answer $\mathrm{CV}_X = 6{,}000/3{,}000 = 2$ , so $\mathrm{CV}_X^2 = 4$ : $n_F = 1{,}082.4\,(1 + 4) = 5{,}412$ 5,412 claims. Severity variation multiplies the frequency-only standard by $1 + \mathrm{CV}_X^2$ — heavy-tailed lines need far more data to be fully credible.