Credibility theory calculator
How much should a risk's own claims experience count against the collective's? Edit the grid and watch the EBCT Model 1 machinery answer live: the variance components, the credibility factor Z, and every risk's blended premium.
Aggregate claims Xij (4 risks × 5 years)
| Risk | Yr 1 | Yr 2 | Yr 3 | Yr 4 | Yr 5 | X̄ᵢ | |
|---|---|---|---|---|---|---|---|
| 1 | 102.4 | ||||||
| 2 | 147.0 | ||||||
| 3 | 79.4 | ||||||
| 4 | 112.6 |
EBCT Model 1 estimates
Overall mean X̄110.35
E[s²(θ)] · within-risk variance82.10
V[m(θ)] · between-risk variance773.38
Credibility factor Z0.9792
Credibility premiums
Risk 1 · Z·102.4 + (1−Z)·110.4102.57
Risk 2 · Z·147.0 + (1−Z)·110.4146.24
Risk 3 · Z·79.4 + (1−Z)·110.480.04
Risk 4 · Z·112.6 + (1−Z)·110.4112.55
Z = n / (n + E[s²(θ)]/V[m(θ)]) with n the number of years. Each premium blends the risk’s own experience with the collective’s: make one risk wildly volatile and E[s²] rises, dragging Z down for everyone; spread the risk means apart and V[m] rises, pushing Z towards 1. More years always push Z up — experience earns credibility.
Test + worked solutions
Test yourself
A fresh claims grid every time: means, the two variance components, Z, and the blended premiums, worked exactly as the exam wants them laid out. Allow small rounding differences.
The two variances that decide everything
Empirical Bayes credibility (Model 1) treats each risk’s true mean as a draw from a collective, and the credibility premium Z·X̄ᵢ + (1−Z)·X̄ is the best linear estimate of it. Everything hinges on two variance components: E[s²(θ)], the noise within a risk’s own experience, and V[m(θ)], the genuine spread between risks. Z is just the ratio of signal to signal-plus-noise, scaled by the number of years. The standard exam traps are all mechanised here: the within variance uses divisor n−1, the between variance subtracts E[s²]/n, and a negative between-variance estimate means Z is set to zero, not negative.
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