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Estimate the two variances behind the credibility factor from a claims table and compute a credibility premium two ways
The problem, which is a real one
You are pricing next year's cover for one factory. You have four years of that factory's own claims, averaging ₹12 lakh a year. You also have the whole book of similar factories, averaging ₹21 lakh a year.
Charge the book average and you ignore four years of evidence that this factory is better than average — and it will leave for a competitor who does not. Charge the factory's own average and you are betting a premium on four numbers, which may simply have been four quiet years.
The right answer is somewhere in between, and the only interesting question is where. "Use judgement" is not an answer; "blend them 50:50" is not either. Credibility theory gives a number, derived rather than chosen, and that number is the whole content of the topic.
The form of the answer
Every credibility estimate looks like this:
with 0 ≤ Z ≤ 1. Z is the credibility factor: the weight the risk's own experience has earned. Z = 1 means trust the risk entirely, Z = 0 means ignore it.
It is worth noticing immediately that the same thing can be written
which is the more revealing form: start from the book and move a fraction Z of the way towards what this risk is telling you. The two expressions are algebraically identical, and computing both is a free check on the arithmetic. The worked example does exactly that.
The Bühlmann answer, and why it has that shape
Bühlmann's model supposes each risk has its own underlying expected claim amount, written m(θ), which you cannot see; what you see is n years of claims scattered around it. Two sources of variation are therefore in play, and the whole of the theory is about telling them apart.
E[s2(θ)] is the average within-risk variance — how much one risk's claims bounce around from year to year even though nothing about the risk has changed. Call this the noise.
Var[m(θ)] is the between-risk variance — how much the underlying expected claim amount genuinely differs from one risk to another. Call this the signal.
The credibility factor is then
In words: weight the risk's own data by the number of years you have, against a penalty equal to the ratio of noise to signal.
Read the formula three times, once for each thing it says, because an exam question asking you to "comment" wants these and not the algebra.
More data raises Z. Z increases with n and tends to 1 as n grows. Enough years and the risk's own experience is all you need. It never quite reaches 1 with finite data, which is the formula declining to be certain.
More noise lowers Z. If a single risk's claims swing wildly year to year, its four-year average says little about its underlying level, and the formula discounts it.
More signal raises Z. This is the one people find counter-intuitive and it is the heart of the method. If the risks in the book are genuinely very different from one another, then the book average is a poor description of any one of them, so the risk's own data — however noisy — is the better guide. The book average is only worth leaning on when the risks are alike.
Notice what the formula does not contain: any judgement, any chosen blend, any target. The quantity E[s2]/Var[m] has units of years and is often written K, so that Z = n/(n + K). K is the number of years of data at which the risk's own experience earns exactly half the weight, which is a useful way to feel what a fitted K means.
Empirical Bayes: estimating the two variances from the data
The formula needs two numbers nobody gave you. Estimating them from the claims table itself is empirical Bayes credibility theory, and in its simplest form — claim amounts only, equal volumes of business, N risks observed over n years each — it is three sample calculations.
The overall mean, which estimates E[m(θ)]:
The noise, estimated by computing each risk's own sample variance across its n years and averaging those:
Each risk's variance is computed about its own mean, which is what makes it a within-risk quantity. This is the step that gets confused with the one below.
The signal, which needs a correction:
The correction term is the point of the whole formula and is where the marks are. The risk means scatter for two reasons: because the risks really are different, and because each mean is only an average of n noisy years. The sample variance of the risk means picks up both. Subtracting E[s2]/n — which is exactly the sampling variance of a mean of n observations — removes the part that is noise and leaves the part that is signal. Omitting it overstates the signal and overstates Z.
One honest consequence: this estimator can come out negative, when the between-risk scatter is no bigger than sampling noise would produce on its own. The convention is to set it to zero, which sets Z to zero and charges every risk the book average. That is not a failure of the method; it is the method reporting that the data contains no evidence the risks differ.
Bühlmann-Straub, in one paragraph
Real risks are not the same size. Bühlmann-Straub extends the model to risks with different volumes — a factory with 500 employees against one with 50 — by letting each observation carry a volume measure Pij. The credibility factor becomes
with Pi the risk's total volume rather than its number of years. The structure is identical and the reading is identical: more exposure earns more weight. The arithmetic of the estimators is heavier, which is why the equal-volume case is the one to be fluent in first.
Questions students ask
Why is this called credibility rather than just weighted averaging?
Because the weight is derived rather than chosen. The name is historical, from American casualty practice, where the question was literally how much credibility to give a risk's own record. What makes it a method rather than a rule of thumb is that Z comes out of the variances, so two actuaries with the same data get the same answer.
Is Bühlmann credibility Bayesian?
It is the best linear approximation to the Bayesian answer, and that is the precise statement. A full Bayesian treatment needs a prior distribution for θ and gives a posterior mean that is generally not linear in the data. Bühlmann restricts attention to estimates of the form a + bX̄ and finds the best one, which needs only the two variances and no distributional assumption at all. That is why it is used: it asks for much less and gives up little.
What is the difference from classical credibility?
Classical, or limited fluctuation, credibility starts from a different question — how much data would be needed for the risk's own mean to be within a chosen tolerance with a chosen probability — and sets Z from that. It is simpler and it requires choosing the tolerance and the probability, which is judgement re-entering by the back door. Bühlmann needs no such choice.
Does credibility theory tell me the risk is actually better?
No, and this is worth being clear about. It tells you the best linear estimate of the risk's expected claims given what you have seen. A low credibility premium is not a finding that the factory is safer; it is a statement that the evidence points that way to a degree the variances quantify. Underwriting judgement about why a risk differs is a separate activity and the method does not replace it.
A worked example
Three risks in a small portfolio, four years of claims each, in ₹ lakh. Every figure is invented for this example. The volumes are equal, so this is the simplest empirical Bayes case.
| Risk | Year 1 | Year 2 | Year 3 | Year 4 | Mean |
|---|---|---|---|---|---|
| 1 | 10 | 12 | 11 | 15 | 12 |
| 2 | 18 | 22 | 19 | 21 | 20 |
| 3 | 28 | 31 | 35 | 30 | 31 |
So N = 3 risks, n = 4 years.
The overall mean
m̂ = (12 + 20 + 31) / 3 = 63/3 = 21.0000000The noise: within-risk variances
Each risk's deviations are taken about its own mean, and each sum of squares is divided by
n − 1 = 3.Risk 1, about 12: −2, 0, −1, 3 → 4 + 0 + 1 + 9 = 14, so s2 = 14/3 = 4.6666667
Risk 2, about 20: −2, 2, −1, 1 → 4 + 4 + 1 + 1 = 10, so s2 = 10/3 = 3.3333333
Risk 3, about 31: −3, 0, 4, −1 → 9 + 0 + 16 + 1 = 26, so s2 = 26/3 = 8.6666667Each set of deviations sums to zero — −2+0−1+3 = 0, −2+2−1+1 = 0, −3+0+4−1 = 0 — which is the cheapest possible check that the three means are right.
E[s2]̂ = (4.6666667 + 3.3333333 + 8.6666667) / 3 = 16.6666667/3 = 5.5555556The signal: between-risk variance, with the correction
Risk means about 21: −9, −1, 10 → 81 + 1 + 100 = 182
divide by N − 1 = 2: 182/2 = 91.0000000That 91 is the raw scatter of the three risk means, and it contains both signal and noise. Take the noise out:
Var[m(θ)]̂ = 91.0000000 − 5.5555556/4 = 91.0000000 − 1.3888889 = 89.6111111Positive, so there is no truncation to worry about. The correction removed about 1.5% of the raw figure. It is small here precisely because the risks are so different; on a book of similar risks it can remove most of it, and omitting it would then be the difference between a sensible answer and a nonsense one.
The credibility factor
K = E[s2]̂ / Var[m(θ)]̂ = 5.5555556 / 89.6111111 = 0.0619963 years
Z = n/(n + K) = 4 / (4 + 0.0619963) = 4 / 4.0619963 = 0.9847375Z is 0.985 — the risks' own experience takes almost all the weight. That is the data speaking, not a modelling choice. K is 0.062 years, meaning that about three weeks of experience would be enough to earn half the weight: the three risks are so plainly different from one another that the book average is nearly useless as a description of any of them. A book of genuinely similar risks would produce a K of several years and a far smaller Z.
The credibility premiums, computed two ways
Risk Own mean Z×own + (1−Z)×book book + Z×(own − book) 1 12 12.1373626 12.1373626 2 20 20.0152625 20.0152625 3 31 30.8473748 30.8473748 Taking risk 1 in full, so the arithmetic is visible:
route one: 0.9847375 × 12 + 0.0152625 × 21 = 11.8168500 + 0.3205125 = 12.1373625
route two: 21 + 0.9847375 × (12 − 21) = 21 − 8.8626375 = 12.1373625The two routes agree exactly, as they must, since the second expression is the first with the Z multiplied out. Both come out at 12.1373625 against the table's 12.1373626, and that last digit is Z having been rounded to seven places here and carried unrounded into the table — rounding, not a mistake, and worth recognising as such rather than hunting for an error that is not there.
The check that catches a wrong Z
Add the three credibility premiums:
12.1373626 + 20.0152625 + 30.8473748 = 63.0000000
and 3 × 21 = 63.0000000They must be equal. Each credibility premium is
book + Z × (own − book), and the deviationsown − bookare −9, −1 and 10, which sum to zero — so whatever Z is, the adjustments cancel and the premiums average back to the book mean. This is the single best check in the topic: it is quick, it uses every number, and it fails loudly if Z has been computed wrongly or applied with the weights the wrong way round.What a different amount of data would have done
Holding the two variances fixed and varying only the number of years:
Years of data n Z 1 0.9416229 4 0.9847375 10 0.9938386 20 0.9969098 50 0.9987616 Even a single year earns 94% of the weight on this book, and Z climbs towards 1 without reaching it. Both features are the formula behaving correctly: with a tiny K the data dominates immediately, and
n/(n+K)is strictly below 1 for every finite n, which is the method never claiming to be certain.Reading the result
Risk 1's premium moves from its own 12.00 to 12.14 — a shift of about 1.1% towards the book. Risk 3's moves from 31.00 down to 30.85. The adjustments are small because Z is large, and Z is large because these three risks are visibly not interchangeable. A portfolio of three factories whose claims averaged 19, 21 and 23 would produce a small Var[m], a large K, and credibility premiums bunched tightly around 21 — the same formula reporting that there is no real evidence of a difference to charge for.
Where this fits in CS2
This page explains one idea from CS2 Risk Modelling & Survival Analysis. The full CS2 syllabus, with what each part is examined on, sits on the course page, and past-paper style mock exams with full solutions are where you find out whether the idea has stuck under time pressure.
Other CS2 topics explained
- The survival function and the hazard rate, explained
- The Kaplan-Meier estimator, and why censoring needs it
- The Cox proportional hazards model, and what a hazard ratio means
- Markov chains: transition matrices, Chapman-Kolmogorov and the stationary distribution
- Central exposed to risk, and the two mortality rates it produces
- Extreme value theory: block maxima, thresholds and the generalised Pareto distribution
- Copulas and Sklar's theorem: modelling dependence without touching the margins
Stuck on a term rather than the idea? The actuarial glossary defines the notation used above. If something here is wrong or unclear, tell us — these pages get corrected.