A Worked Bühlmann Estimation Example

From a Risk Model to a Class Estimate

Bühlmann credibility combines a risk class's observed experience with a collective mean. The purpose is not to replace the experience with the overall average, nor to trust a small sample as if it were definitive. Instead, the estimate gives more weight to experience when it is relatively informative compared with the process variation inherent in observations.

A useful worked calculation begins with the model, not with the credibility formula. Specify how observations vary within a risk, how the underlying expected value varies between risks, and how many observations are available. These ingredients determine the variance components and the credibility assigned to the class.

Definition: In the Bühlmann collective model, each risk has an unobserved risk parameter \(\Theta\). Given \(\Theta\), observations \(X_1,\ldots,X_n\) are conditionally independent and identically distributed, with conditional mean \(m(\Theta)=E[X_j\mid\Theta]\) and conditional variance \(s^2(\Theta)=\operatorname{Var}(X_j\mid\Theta)\). Assume the unconditional means and variances below are finite. The collective mean is \(\mu=E[m(\Theta)]\). The expected process variance (EPV) is \(\text{EPV}=E[s^2(\Theta)]\), and the variance of hypothetical means (VHM) is \(\text{VHM}=\operatorname{Var}(m(\Theta))\).

EPV measures the average variability of observations for a fixed risk. VHM measures how much the underlying expected values differ across risks. For \(n\) observations on a risk, the Bühlmann credibility factor \(Z\) and class estimate are

K=\frac{\text{EPV}}{\text{VHM}},\qquad Z=\frac{n}{n+K} =\frac{n\,\text{VHM}}{\text{EPV}+n\,\text{VHM}},\qquad \widehat{m}=Z\bar X+(1-Z)\mu.

Here, \(\bar X\) is the arithmetic average of the risk's \(n\) observations, \(K\) is the credibility constant, and \(\widehat{m}\) is the credibility estimate of its expected value. This form presumes \(\text{VHM}>0\). If \(\text{VHM}=0\), all risks have the same hypothetical mean under the model, so experience does not distinguish their expected values. If EPV is zero and VHM is positive, observations have no within-risk variation and the formula gives \(Z=1\) for \(n>0\).

Why the Credibility Formula Has This Form

The Bühlmann estimate is the best linear predictor of a risk's hypothetical mean \(m(\Theta)\) using the observations. It is not, without further distributional assumptions, necessarily the exact conditional expectation \(E[m(\Theta)\mid X_1,\ldots,X_n]\). To see where the weight comes from, center the observations at the collective mean and seek a linear predictor of the form \(\mu+c\sum_{j=1}^n(X_j-\mu)\).

Conditional independence gives zero conditional covariance between distinct observations. Taking expectations over risks then gives the unconditional covariances needed for the linear projection:

\operatorname{Cov}(m(\Theta),X_j)=\text{VHM},\qquad \operatorname{Var}(X_j)=\text{EPV}+\text{VHM},\qquad \operatorname{Cov}(X_j,X_k)=\text{VHM}\quad (j\ne k).

For a best linear predictor, the prediction error must be uncorrelated with each observation. By symmetry, all \(n\) observations receive the same coefficient \(c\). Equating the covariance of the target with \(X_j\) to the covariance of the predictor with \(X_j\) yields

\text{VHM} =c\left[\operatorname{Var}(X_j)+\sum_{k\ne j}\operatorname{Cov}(X_k,X_j)\right] =c\left[\text{EPV}+n\,\text{VHM}\right].

Thus \(c=\text{VHM}/(\text{EPV}+n\,\text{VHM})\), and the total coefficient on the sample average is \(nc=n\,\text{VHM}/(\text{EPV}+n\,\text{VHM})=Z\). This gives \(\mu+Z(\bar X-\mu)\), which is the same as \(Z\bar X+(1-Z)\mu\). The derivation also shows why conditional independence and a common conditional distribution across observation periods matter: without the covariance structure used above, this formula need not apply.

Worked Calculations

Worked Example: Calculating Both Variance Components

Consider a hypothetical annual loss-cost measure, with no interest or mortality assumptions involved. A risk's type \(\Theta\) is low, middle, or high, with probabilities \(0.25,\ 0.50,\ 0.25\), respectively. Conditional on its type, each annual observation is independent and has the following distribution: low has values 60 and 100 with equal probabilities; middle has 80 and 120 with equal probabilities; high has 100 and 140 with equal probabilities. A middle-type risk has four observed annual values \(80,120,120,80\). Calculate EPV, VHM, \(K\), \(Z\), and the class estimate.

1. Calculate conditional means and variances. The type-specific means are \(80,\ 100,\ 120\). In each type, the two possible values are 20 below and 20 above the mean, each with probability \(0.5\). Therefore the conditional variance in each type is

$$ 0.5(20^2)+0.5(20^2)=400. $$

2. Calculate EPV. The conditional variance is 400 for every type, so its probability-weighted average is \(0.25(400)+0.50(400)+0.25(400)=400\). Hence \(\text{EPV}=400\).

3. Calculate the collective mean and VHM. The collective mean is \(0.25(80)+0.50(100)+0.25(120)=100\). The type-specific mean deviations from 100 are \(-20,\ 0,\ 20\), so

\text{VHM}=0.25(-20)^2+0.50(0)^2+0.25(20)^2=200.

4. Calculate credibility and the estimate. Here \(K=400/200=2\), and the experience volume is \(n=4\). Thus \(Z=4/(4+2)=2/3\). The sample average is \(\bar X=(80+120+120+80)/4=100\). Therefore \(\widehat m=(2/3)(100)+(1/3)(100)=100\).

The result equals both the class experience average and the collective mean because they happen to agree. As a check, the weights sum to one, and the estimate remains 100. The calculation uses the Bühlmann method and avoids treating the type-specific variance as VHM: 400 is within-risk variance, while 200 is between-risk variance.

Worked Example: Unequal Process Variances Across Risk Types

For a hypothetical annual claim-cost measure, again with no interest or mortality assumptions, a risk is low or high with probability \(0.5\) each. Conditional on being low, each independent annual observation equals 40 or 60 with equal probability. Conditional on being high, it equals 60 or 120 with equal probability. One risk has observed values \(60,60,120\). Calculate its credibility estimate.

1. Calculate conditional means and variances. The low mean is \((40+60)/2=50\); its variance is \(0.5(10^2)+0.5(10^2)=100\). The high mean is \((60+120)/2=90\); its variance is \(0.5(30^2)+0.5(30^2)=900\).

2. Calculate EPV, the collective mean, and VHM. The probability-weighted EPV is \(0.5(100)+0.5(900)=500\). The collective mean is \(0.5(50)+0.5(90)=70\). The type means are 20 below and 20 above 70, so \(\text{VHM}=0.5(20^2)+0.5(20^2)=400\).

3. Calculate \(K\), \(Z\), and the estimate. \(K=500/400=1.25\). The sample average is \(\bar X=(60+60+120)/3=80\), with \(n=3\). Therefore \(Z=3/(3+1.25)=3/4.25=12/17\), or approximately \(0.7059\). The complement is \(1-Z=5/17\). Then

\widehat m=\frac{12}{17}(80)+\frac{5}{17}(70) =\frac{960+350}{17} =\frac{1310}{17} \approx 77.06.

The estimate is 7.06 above the collective mean and 2.94 below the sample average; it lies between them, as a weighted average should. The result is rounded only at the final step to two decimal places. Averaging the low and high conditional variances without their probabilities, or substituting VHM for EPV, would produce a different and incorrect \(K\).

Worked Example: Comparing Estimates at Different Experience Volumes

Suppose hypothetical annual loss-cost observations follow this collective model, with no interest or mortality assumptions. A risk's type is low or high, each with probability \(0.5\), and the corresponding conditional means are 90 and 110. Given its type, each observation equals its conditional mean minus or plus \(10\sqrt{3}\), each with probability \(0.5\), independently over years. Three separate risks follow this same model. Risk A has one observation, \(110+10\sqrt{3}\). Risk B has four observations, two of each possible value for the low type. Risk C is high type and has nine observations, five equal to \(110+10\sqrt{3}\) and four equal to \(110-10\sqrt{3}\). Calculate all three credibility estimates.

1. Establish common model quantities. Each conditional variance is \((10\sqrt{3})^2=300\), so \(\text{EPV}=300\). The collective mean is \(0.5(90)+0.5(110)=100\), and \(\text{VHM}=0.5(10^2)+0.5(10^2)=100\). Thus \(K=300/100=3\).

2. Calculate Risk A. With \(n=1\), \(Z=1/(1+3)=1/4\). Its sample average is \(110+10\sqrt{3}\). Therefore

\widehat m_A=100+\frac14(10+10\sqrt{3}) =102.5+2.5\sqrt{3} \approx 106.83.

3. Calculate Risk B. Its two observations of each value average to 90, and \(n=4\). Thus \(Z=4/(4+3)=4/7\), giving \(\widehat m_B=(4/7)(90)+(3/7)(100)=94.29\) to two decimal places.

4. Calculate Risk C. Its sample average is

\bar X_C=\frac{5(110+10\sqrt{3})+4(110-10\sqrt{3})}{9} =110+\frac{10\sqrt{3}}{9}.

With \(n=9\), \(Z=9/(9+3)=3/4\). Hence \(\widehat m_C=100+(3/4)(10+10\sqrt{3}/9)=107.5+(5\sqrt{3}/6)\approx108.94\). Each estimate is rounded only at the final step to two decimal places. Risk B's experience is below the collective mean, while A's and C's are above it; all estimates move toward 100 rather than necessarily reaching the type-specific mean.

Interpreting the Estimate and Avoiding Traps

The credibility factor is between zero and one when EPV and VHM are positive. Holding the variance components fixed, \(Z=n/(n+K)\) increases as \(n\) increases. A larger sample average therefore receives more weight, while a smaller sample receives more weight on the collective mean. In the limit as \(n\) grows, \(Z\) approaches one; for \(n=0\), there is no individual experience to contribute and the estimate is the collective mean.

The model separates two sources of variation for a reason. Large EPV means observations on the same risk can fluctuate substantially, making a short history less persuasive. Large VHM means risks' expected values differ substantially, making observed experience more useful for distinguishing a risk from the collective. The credibility constant \(K=\text{EPV}/\text{VHM}\) expresses their relative scale in units of observations.

A common exam trap is to calculate the sample average correctly but apply the weights in the wrong direction. For the second example, an incorrect calculation might use \(Z=12/17\) on the collective mean and \(5/17\) on experience:

\text{Wrong: }\frac{12}{17}(70)+\frac{5}{17}(80)=72.94.

This reverses the meaning of \(Z\). The correction is to put \(Z\) on the experience average and \(1-Z\) on the collective mean: \((12/17)(80)+(5/17)(70)=77.06\). A quick check is that, for \(0\le Z\le1\), the result must lie between \(\bar X\) and \(\mu\). The wrong answer 72.94 does lie between 70 and 80, so that check alone does not catch every reversal; writing the definition of \(Z\) and naming which quantity it weights is essential.

Key takeaway: First compute the collective mean, EPV, and VHM from the risk model. Then calculate \(K\), use the number of observations to obtain \(Z\), and apply \(Z\) to the experience average. This tutorial's calculations are aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Check Your Understanding

Work each problem using the stated conditional model. Round final monetary estimates to cents where needed.

A hypothetical risk type is low or high with equal probabilities. Conditional means are 30 and 50, and each conditional variance is 100. Find the collective mean, EPV, VHM, and \(K\). (Answer: \(\mu=40\), EPV \(=100\), VHM \(=100\), \(K=1\).)
Using the model in Question 1, a risk has two observations with average 45. Calculate \(Z\) and the Bühlmann estimate. (Answer: \(Z=2/3\); estimate \(=43.33\).)
A hypothetical annual measure has two equally likely types with conditional means 60 and 100 and conditional variances 200 and 600. A risk has four observations averaging 90. Calculate EPV, VHM, \(K\), \(Z\), and its estimate. (Answer: EPV \(=400\), VHM \(=400\), \(K=1\), \(Z=0.8000\), estimate \(=88.00\).)
For a model with \(\mu=120\), EPV \(=300\), and VHM \(=100\), a risk has six observations averaging 100. Calculate its estimate and state whether it is closer to the sample average or collective mean. (Answer: \(K=3\), \(Z=2/3\), estimate \(=106.67\); it is closer to the sample average.)
Two separate risks share \(\mu=200\), EPV \(=500\), and VHM \(=250\). Risk A has two observations averaging 230; Risk B has eight observations averaging 190. Calculate both estimates and identify which receives greater experience weight. (Answer: \(K=2\); Risk A has \(Z=0.5000\) and estimate \(=215.00\); Risk B has \(Z=0.8000\) and estimate \(=192.00\); Risk B receives greater experience weight.)

What You'll Learn