Limited Fluctuation Credibility

Study guide: Exam MAS-II

Limited Fluctuation Credibility (also called Classical Credibility) assigns a credibility factor $Z \in [0,1]$ to observed data based on whether the volume of data is sufficient to meet a specified accuracy standard. The estimate is then a blend of observed experience and an external prior.

$\text{Estimate} = Z \cdot \bar{X} + (1-Z) \cdot \mu_0$

$Z = \min\!\left(1,\, \sqrt{\frac{n}{n_{\text{full}}}}\right)$

Full credibility standard — the number of observations $n_{\text{full}}$ required so that the observed mean is within $100r\%$ of the true mean with probability $1-\alpha$ :

$n_{\text{full}} = \left(\frac{z_{\alpha/2}}{r}\right)^2 \cdot \frac{\text{Var}(X)}{[\text{E}(X)]^2}$

For claim frequency (Poisson): $\text{Var}(X)/[E(X)]^2 = 1/\lambda$ , giving $n_{\text{full}} = (z_{\alpha/2}/r)^2 / \lambda$ ; typically 1,082 claims at $r=5\%$ , $1-\alpha=90\%$
For claim severity: $n_{\text{full}}$ depends on the coefficient of variation $\text{CV} = \sigma/\mu$
For aggregate losses: $n_{\text{full}}$ combines frequency and severity CVs: $\text{CV}^2_S = \text{CV}^2_N + \text{CV}^2_X \cdot E[N]/E[N]$
Partial credibility uses the square root rule: $Z = \sqrt{n/n_{\text{full}}}$ when $n < n_{\text{full}}$

Credibility Factor for a Small Class The full credibility standard for frequency is 1,082 claims. A class has 400 observed claims. Calculate the credibility factor.

Answer $Z = \sqrt{\frac{400}{1{,}082}} = \sqrt{0.3697} \approx 0.608$ The class has about 60.8% credibility.

Credibility-Weighted Rate Indication A territory has $Z = 0.70$ . The observed loss ratio is 0.85 and the complement (prior) is 0.75. Calculate the credibility-weighted indicated loss ratio.

Answer $\text{Estimate} = 0.70(0.85) + 0.30(0.75) = 0.595 + 0.225 = 0.820$