Calculating Bühlmann Premiums

What a Bühlmann Premium Represents

A class’s observed average loss cost is informative, but it may be based on limited experience. A Bühlmann estimate combines that class’s sample mean with the overall mean: the class experience receives more weight as the number of comparable observations grows, while the overall mean provides a complement when experience is sparse. In this tutorial, “premium” means the estimated expected loss cost, not a final rate inclusive of expenses, profit, or other provisions.

The calculation uses a class’s sample mean, an overall mean, and a Bühlmann credibility factor. These quantities must refer to compatible observations and units. For example, if each observation is dollars of loss per vehicle-year, the result is also dollars per vehicle-year. It is not the aggregate loss amount for a portfolio unless it is multiplied by the corresponding number of exposure units.

Definition: In the Bühlmann model, let $\Theta$ denote a risk’s unknown expected outcome, and let $X_1,\ldots,X_n$ be $n$ observations for that risk. The overall mean is $\mu=E[\Theta]=E[X_j]$. The expected process variance is $EPV=E[\operatorname{Var}(X_j\mid\Theta)]$, and the variance of hypothetical means is $VHM=\operatorname{Var}(\Theta)$. For equal-weight observations, define $K=EPV/VHM$ when $VHM>0$. The Bühlmann credibility factor is $Z=n/(n+K)$, and the estimate is $Z\bar X+(1-Z)\mu$, where $\bar X$ is the risk’s sample mean.

This standard form assumes the observations are conditionally independent and identically distributed given $\Theta$, with conditional mean $E[X_j\mid\Theta]=\Theta$, finite variances, and equal exposure weights. The stated $n$ is therefore a count of comparable observations, not a sum of unequal exposure amounts. If observations have different weights, the equal-weight formula is not the appropriate model; the Bühlmann-Straub model addresses that setting.

If $VHM=0$, there is no variation among the hypothetical means, so individual experience offers no information for distinguishing risks and $Z=0$ is appropriate. If $EPV=0$ and $VHM>0$, observations have no process variation around their risk mean, and any positive amount of experience identifies that mean exactly in the model. Ordinary applications have positive values for both components.

Why the Credibility Blend Has This Form

The Bühlmann factor can be justified by choosing the linear blend that minimizes mean squared error. Consider an estimate written as $\widehat{\Theta}=\mu+Z(\bar X-\mu)$. It equals the overall mean when $Z=0$, and it equals the observed mean when $Z=1$. Conditional independence and equal conditional variances imply $\operatorname{Var}(\bar X\mid\Theta)=EPV/n$. The squared-error expectation separates into the uncertainty between risks and the sampling variation within a risk:

E[(\widehat{\Theta}-\Theta)^2] =(1-Z)^2VHM+Z^2\frac{EPV}{n}.

To minimize this quadratic in $Z$, differentiate with respect to $Z$ and set the derivative equal to zero:

-2(1-Z)VHM+2Z\frac{EPV}{n}=0.

Rearranging gives $VHM=Z(VHM+EPV/n)$, so

Z=\frac{VHM}{VHM+EPV/n} =\frac{n}{n+EPV/VHM} =\frac{n}{n+K}.

This also explains the direction of the weighting. A larger $n$ reduces the sampling variance $EPV/n$, increasing the weight on the class mean. A larger $EPV$ makes the sample mean less stable, increasing $K$ and reducing $Z$. A larger $VHM$ means class means vary more around the overall mean, so experience is more useful for identifying a particular class.

Calculating the Class Estimate

For a given class, first determine the observation units and calculate the sample mean. Then find $K$ and $Z$, or use a supplied $Z$, and apply the credibility blend. The complement is $1-Z$; it is not another credibility factor to calculate independently. The two weights must sum to one.

Confirm the unit and observation count.
Identify what one observation measures and count only comparable, equal-weight observations in $n$.

Calculate the class mean.
Add the class observations and divide by $n$. Keep the result in the same units as the overall mean.

Find the credibility factor.
Use $K=EPV/VHM$ and $Z=n/(n+K)$, or use the $Z$ supplied in the problem.

Blend and interpret.
Calculate $Z\bar X+(1-Z)\mu$. If a total expected loss amount is requested, multiply the per-exposure estimate by the matching number of exposure units.

Worked Example: Annual Loss Cost for a Small Class

A hypothetical insurer studies one class with four equal-exposure observations of annual loss cost: $820, $940, $760, and $1,080 per vehicle-year. The overall mean is $750 per vehicle-year. Suppose the model is $\Theta\sim N(750,60000)$ and $X_j\mid\Theta\sim N(\Theta,360000)$, where the second parameter is the variance. Assume conditional independence, no discounting, and no mortality adjustment. Calculate the Bühlmann estimate. Identify the method and the common mistake it avoids.

Method: equal-weight Bühlmann blending. There are $n=4$ comparable observations, each in dollars per vehicle-year. Their sum is $820+940+760+1{,}080=3{,}600$, so the class sample mean is $3{,}600/4=\$900$ per vehicle-year. Under the stated model, $EPV=360{,}000$ and $VHM=60{,}000$, both in squared dollars per vehicle-year. Thus $K=360{,}000/60{,}000=6$.

The credibility factor is $Z=4/(4+6)=0.4$, and the complement is $1-Z=0.6$. Therefore,

\widehat{\Theta}=0.4(900)+0.6(750)=360+450=\$810 \quad\text{per vehicle-year}.

Check: the weights sum to $0.4+0.6=1$, and the estimate, $810, lies between the class mean of $900 and the overall mean of $750. It is closer to the overall mean because the class receives less than half the weight. A common mistake is to reverse the weights and calculate $0.6(900)+0.4(750)=\$840$. That gives too much weight to the class experience: $Z=0.4$, not $0.6$, is the weight on its sample mean.

From a Loss Cost to a Portfolio Amount

A Bühlmann estimate is in the units of the observations used to calculate the sample mean. If the observations are per-exposure loss costs, the estimate remains a per-exposure loss cost. To obtain the expected loss amount for a stated portfolio, multiply by the matching exposure count. This conversion does not change the estimated loss cost itself, and it does not add expenses or other rate provisions.

For instance, a result in dollars per policy-year can be multiplied by the number of policy-years represented by the portfolio. It should not be multiplied by a count of policies if the relevant exposure unit is something else, such as vehicle-years. The quantity being counted must match the denominator in the estimated loss cost.

Worked Example: Unequal Class Experience Versus the Overall Mean

A hypothetical equipment insurer has three equal-weight annual loss-cost observations for a class: $1,400, $1,600, and $1,800 per insured unit-year. The overall mean is $1,200 per insured unit-year. Suppose $\Theta\sim N(1200,50000)$ and $X_j\mid\Theta\sim N(\Theta,250000)$, with the second parameter in each distribution denoting variance. Assume conditional independence and no discounting or mortality adjustment. Calculate the estimate. State the method and the error the calculation guards against.

Method: calculate the class mean, then apply the Bühlmann factor. The sum is $1{,}400+1{,}600+1{,}800=4{,}800$, so $\bar X=4{,}800/3=\$1{,}600$ per insured unit-year. The model gives $EPV=250{,}000$ and $VHM=50{,}000$, so $K=250{,}000/50{,}000=5$. With $n=3$, $Z=3/(3+5)=3/8=0.375$, leaving $1-Z=0.625$ for the overall mean.

\widehat{\Theta} =0.375(1{,}600)+0.625(1{,}200) =600+750 =\$1{,}350\text{ per insured unit-year}.

Check: $1,350 lies between $1,200 and $1,600, and it is closer to $1,200 because the complement weight, 0.625, is larger. The method avoids the error of treating $1-Z$ as the class-experience weight. Doing so would give $0.625(1{,}600)+0.375(1{,}200)=\$1{,}450$, which incorrectly gives the sample mean the larger weight.

Worked Example: Converting an Estimated Loss Cost to a Total

A hypothetical property class has two equal-exposure observations of $500 and $580 per location-year. The overall mean is $600 per location-year. Assume $\Theta\sim N(600,40000)$ and $X_j\mid\Theta\sim N(\Theta,240000)$, with conditional independence, no discounting, and no mortality adjustment. The portfolio has 2,400 location-years of exposure. Calculate both the Bühlmann estimate per location-year and the corresponding total expected loss amount. Name the method and the unit trap it avoids.

Method: credibility-weight the class mean, then apply exposure. The sample mean is $(500+580)/2=\$540$ per location-year. Here $EPV=240{,}000$, $VHM=40{,}000$, and $K=240{,}000/40{,}000=6$. With $n=2$, $Z=2/(2+6)=0.25$, so $1-Z=0.75$.

\widehat{\Theta} =0.25(540)+0.75(600) =135+450 =\$585\text{ per location-year}.

The total expected loss amount for 2,400 location-years is $585\times2{,}400=\$1{,}404{,}000$. Check: the per-location-year estimate lies between the sample mean and the overall mean, and the total equals that rate of loss cost multiplied by the stated exposure. This avoids the unit error of reporting $585 as the total portfolio loss amount; it is a per-location-year estimate.

Common Calculation Checks

Before accepting a result, check both the arithmetic and the interpretation. A credibility estimate is a weighted average, so for $0\leq Z\leq1$ it must lie between $\bar X$ and $\mu$. If it does not, the factor, complement, or arithmetic has likely been mishandled. The result should also retain the sample mean’s units unless the calculation explicitly converts it to a total.

Use the stated observation count. In the equal-weight model, $n$ counts comparable observations. Do not substitute the number of exposures within each observation unless the model specifies that those are separate observations.
Keep the weights attached to the right means. $Z$ multiplies the class sample mean; $1-Z$ multiplies the overall mean.
Do not confuse loss cost with total premium. The blend estimates the expected loss measure being modeled. Expense, profit, or other provisions are not included unless separately specified.
Match the portfolio conversion to the unit. Multiply by the number of exposure units that correspond to one observation’s denominator.
Do not force this formula onto unequal weights. A collection of observations with different exposure weights requires a model that accounts for those weights rather than the equal-weight $n/(n+K)$ calculation.

Key takeaway: For equal-weight experience, calculate $Z=n/(n+K)$, then estimate the class mean as $Z\bar X+(1-Z)\mu$. Preserve the observations’ units, and multiply by matching exposure only when a portfolio total is requested.

Check Your Understanding

For each problem, calculate the requested Bühlmann estimate and keep track of the observation units. Each scenario is hypothetical.

A class has five equal-exposure annual loss-cost observations: $700, $800, $900, $750, and $850 per unit-year. The overall mean is $700 per unit-year and $K=5$. Find $Z$ and the credibility estimate. (Answer: $Z=0.5$; $750 per unit-year.)
A class has four equal-weight observations with sample mean $910 per policy-year. The overall mean is $700 per policy-year, $EPV=180{,}000$, and $VHM=60{,}000$. Find $K$, $Z$, and the estimate. (Answer: $K=3$, $Z=4/7$, and $820 per policy-year.)
A class’s sample mean is $1,050 per vehicle-year, the overall mean is $900 per vehicle-year, and $n=3$. Given $EPV=240{,}000$ and $VHM=60{,}000$, calculate the estimate and explain which mean receives more weight. (Answer: $K=4$, $Z=3/7$, estimate $\$964.29$ per vehicle-year; the overall mean receives more weight.)
A class has two equal-weight observations of $400 and $600 per location-year. The overall mean is $500 per location-year, $EPV=120{,}000$, and $VHM=40{,}000$. If the portfolio has 1,500 location-years, calculate the estimated loss cost and the total expected loss amount. (Answer: $K=3$, $Z=0.4$, estimate $500 per location-year, and total expected loss amount $750,000.)
A class has three observations with sample mean $1,300 per unit-year; the overall mean is $1,000 per unit-year. The model has $EPV=200{,}000$ and $VHM=50{,}000$. A candidate uses $1-Z$ on the class mean. Calculate the correct estimate and the candidate’s incorrect estimate. (Answer: $K=4$, $Z=3/7$; correct estimate $\$1{,}128.57$ per unit-year; incorrect reversed-weight estimate $\$1{,}171.43$ per unit-year.)

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