Empirical Bayes Credibility

Study guides: Exam ASTAM, Exam MAS-II

Empirical Bayes credibility estimates the structural parameters $\mu$ , $v$ and $a$ of a Bühlmann Credibility or Bühlmann-Straub Credibility model from the observed experience itself, rather than assuming them known. It is how credibility is actually applied when no prior is given.

$\hat{v} = \frac{1}{r(n-1)}\sum_{i=1}^{r}\sum_{j=1}^{n}\left(X_{ij} - \bar{X}_i\right)^{2}$

$\hat{a} = \frac{1}{r-1}\sum_{i=1}^{r}\left(\bar{X}_i - \bar{X}\right)^{2} - \frac{\hat{v}}{n}$

Non-parametric: no distributional assumption at all — $\hat v$ is the pooled within-risk variance, $\hat a$ the between-risk variance corrected for the process noise that inflates it
Semi-parametric: assume a family for the conditional distribution to shortcut one estimate — for Poisson frequency $\hat v = \bar{X}$ , leaving only $\hat a$ to compute
The correction term $\hat v / n$ matters: the raw spread of the group means already contains process variance, and subtracting it is what makes $\hat a$ unbiased
$\hat a$ can come out negative when the groups look no more different than noise; it is then set to 0, giving $Z = 0$ and full weight to the overall mean
With unequal exposures the same idea carries over with exposure weights — the Bühlmann-Straub form

Estimating the Structural Parameters Two risks are observed for 2 years each: risk 1 reports 4 and 6 claims, risk 2 reports 10 and 12. Estimate $\mu$ , $v$ , $a$ and the credibility for one risk's 2 years.

Answer $\bar{X}_1 = 5$ , $\bar{X}_2 = 11$ , $\bar{X} = 8$ . $\hat v = \frac{(4-5)^2 + (6-5)^2 + (10-11)^2 + (12-11)^2}{2(2-1)} = \frac{4}{2} = 2$ $\hat a = \frac{(5-8)^2 + (11-8)^2}{2-1} - \frac{2}{2} = 18 - 1 = 17$ $\hat k = \frac{2}{17} = 0.1176 \qquad Z = \frac{2}{2 + 0.1176} = 0.944$ The risks differ far more than the year-to-year noise, so each risk's own experience carries 94% of the weight: risk 1 is estimated at $0.944(5) + 0.056(8) = 5.17$ .