Common Bühlmann Calculation Errors

Why Small Calculation Errors Matter

A credibility estimate combines a risk's experience with a collective mean. The arithmetic is usually short; the challenge is keeping the model quantities straight. Reversing variance components, averaging periods without their exposure weights, or changing the scale of exposure without changing the variance parameter can materially alter the estimate. A result outside the credibility range from zero to one is also a warning that the inputs or their interpretation need review.

This tutorial focuses on auditing those calculations. The formulas below use the standard Bühlmann and Bühlmann-Straub models. They apply when the stated conditional-mean, conditional-variance, and independence assumptions are appropriate; they are not a substitute for checking those assumptions or the definitions of the supplied inputs.

Definition: In the Bühlmann model, let $\Theta$ denote a risk's latent parameter and $X_j$ its observation in period $j$. Conditional on $\Theta$, observations have common mean $m(\Theta)=E[X_j\mid\Theta]$, common variance $\sigma^2(\Theta)=\operatorname{Var}(X_j\mid\Theta)$, and are independent. Define the collective mean $m=E[m(\Theta)]$, the expected process variance $v=E[\sigma^2(\Theta)]$, and the variance of hypothetical means $a=\operatorname{Var}(m(\Theta))$. For $n$ observations, $k=v/a$, $Z=n/(n+k)$, and the credibility estimate is $Z\bar X+(1-Z)m$, provided $a>0$.

The expected process variance $v$ describes variability within a risk after its latent parameter is fixed. The variance of hypothetical means $a$ describes how much the underlying means differ among risks. They are not interchangeable. Both have squared units of the observations in the ordinary Bühlmann model, but they represent distinct sources of variation.

Definition: In the Bühlmann-Straub model, observation $X_j$ has exposure $w_j>0$, conditional mean $m(\Theta)$, and conditional variance $\sigma^2(\Theta)/w_j$. Conditional independence is assumed. With $W=\sum_j w_j$, the exposure-weighted mean is $\bar X_w=\sum_j w_jX_j/W$. Define $\phi=E[w_j\operatorname{Var}(X_j\mid\Theta)]$, the expected process variance per unit of exposure, and $a=\operatorname{Var}(m(\Theta))$. Then $k=\phi/a$, $Z=W/(W+k)$, and the estimate is $Z\bar X_w+(1-Z)m$. The same $\phi$ applies to each observation when its conditional variance is proportional to $1/w_j$.

Why the Credibility Factor Has This Form

For the ordinary Bühlmann model, consider a linear estimate $\widehat m=\alpha\bar X+(1-\alpha)m$, where $\alpha$ is a weight between zero and one. Under the model, the conditional variance of $\bar X$ is $\sigma^2(\Theta)/n$. Averaging over risks gives $v/n$. Also, $m(\Theta)-m$ has variance $a$. Conditional independence and the fact that $\bar X-m(\Theta)$ has conditional mean zero make the cross-product term vanish. Thus the mean squared error for estimating $m(\Theta)$ is

E[(\widehat m-m(\Theta))^2] =\alpha^2\frac{v}{n}+(1-\alpha)^2a.

Differentiating with respect to $\alpha$, setting the derivative equal to zero, and solving gives

2\alpha\frac{v}{n}-2(1-\alpha)a=0 \quad\Longrightarrow\quad \alpha=\frac{a}{a+v/n} =\frac{n}{n+v/a}.

This explains both the ratio $k=v/a$ and the direction of the credibility effect: more observations increase credibility, while greater process variance relative to variation among risk means decreases it. The Bühlmann-Straub expression replaces the information from $n$ equally exposed observations with total exposure $W$, giving $Z=W/(W+\phi/a)$.

Worked Calculations and Error Checks

Worked Example: Keeping the Variance Components in Their Proper Roles

A hypothetical portfolio model gives expected process variance $v=900$ and variance of hypothetical means $a=300$, both in squared dollars. A risk has $n=6$ independent annual observations with mean $\bar X=\$120$; the collective mean is $m=\$100$. Calculate the credibility estimate and identify the error from reversing the two variance components.

Method. Use the ordinary Bühlmann formula. First calculate $k=v/a=900/300=3$ years. Then

Z=\frac{n}{n+k}=\frac{6}{6+3}=\frac{2}{3}.

The estimate is

\widehat m =\frac{2}{3}(120)+\frac{1}{3}(100) =80+33.3333 =\$113.3333.

Rounded to the nearest cent, the estimate is \$113.33. As a check, it lies between the observed mean of \$120 and collective mean of \$100, and it gives the observed mean greater weight.

Common mistake. Reversing the components gives $k=300/900=1/3$ and $Z=6/(6+1/3)=18/19\approx0.9474$. That would produce $0.9474(120)+0.0526(100)\approx\$118.95$, an unjustifiably high weight on the experience. The numerator of $k$ is the expected process variance; its denominator is the variance of hypothetical means.

Worked Example: Weighting Unequal Periods and Using the Collective Mean

A hypothetical class has observations of \$80, \$100, and \$140, with exposures 1, 2, and 5 units, respectively. Its model parameters are $\phi=400$ squared dollars per exposure unit and $a=100$ squared dollars. The collective mean is \$110. Calculate the Bühlmann-Straub estimate. Explain why neither a simple period average nor the class's own experience average should replace the collective mean.

Method. Calculate total exposure and the exposure-weighted class mean:

W=1+2+5=8,\qquad \bar X_w=\frac{1(80)+2(100)+5(140)}{8} =\frac{980}{8}=\$122.50.

The risk parameter is $k=\phi/a=400/100=4$ exposure units, so

Z=\frac{8}{8+4}=\frac{2}{3}, \qquad \widehat m=\frac{2}{3}(122.50)+\frac{1}{3}(110) =81.6667+36.6667=\$118.3333.

Rounded to the nearest cent, the estimate is \$118.33. It lies between the weighted experience mean and the collective mean. The unweighted period average, $(80+100+140)/3=\$106.67$, gives the low-exposure first period the same influence as the five-unit third period, so it is not the correct experience mean here.

Common mistake. Using \$122.50 as both the class mean and the collective mean eliminates the intended pull toward the portfolio mean and returns \$122.50 regardless of $Z$. The collective mean is the model's mean across risks; it is not the experience mean of the particular class being rated.

Worked Example: Changing Exposure Units Consistently

A hypothetical risk has total exposure $W=12$ units, weighted mean $\bar X_w=\$130$, and collective mean $m=\$100$. The model parameters are $\phi=600$ squared dollars per exposure unit and $a=150$ squared dollars. Re-express exposure in units that are 100 times smaller, and verify that the estimate is unchanged.

Method. In the original units, $k=600/150=4$ and

Z=\frac{12}{12+4}=0.75,\qquad \widehat m=0.75(130)+0.25(100)=\$122.50.

When each new exposure unit is one-hundredth as large, the numerical total exposure becomes $W'=100(12)=1200$. To preserve the conditional variance $\phi/w$, the numerical process parameter must also become $\phi'=100(600)=60{,}000$: then $\phi'/w'=\phi/w$. Thus $k'=60{,}000/150=400$, and

Z'=\frac{1200}{1200+400}=0.75.

The experience mean and estimate are unchanged, so the result remains \$122.50. The check is that both total exposure and $\phi$ scaled by 100, leaving $W/(W+\phi/a)$ unchanged.

Common mistake. Changing $W$ to 1200 but leaving $\phi=600$ gives $Z=1200/(1200+4)=0.9967$. This mixes units: the converted exposure and the unconverted process parameter no longer describe the same conditional variance model.

Worked Example: Separating Within-Risk and Between-Risk Variation

A hypothetical model has two equally likely risk types. Their conditional means are \$90 and \$130, and their conditional variances are 400 and 800 squared dollars, respectively. A risk has four independent observations with mean \$125, and the collective mean is to be determined from the model. Calculate the two variance components, the credibility estimate, and a check against using total variance as the process variance.

Method. The collective mean is $m=(90+130)/2=\$110$. The variance of hypothetical means is

a=\frac{(90-110)^2+(130-110)^2}{2} =\frac{400+400}{2}=400.

The expected process variance is the average of the conditional variances:

v=\frac{400+800}{2}=600.

Consequently, $k=v/a=600/400=1.5$, $Z=4/(4+1.5)=8/11$, and

\widehat m=\frac{8}{11}(125)+\frac{3}{11}(110) =\frac{1000+330}{11} =\$120.9091.

Rounded to the nearest cent, the estimate is \$120.91. As a decomposition check, the unconditional variance of one observation is $v+a=600+400=1000$. Using that total as $v$ would wrongly count variation among risk means as within-risk noise; it would give $Z=4/(4+1000/400)=8/13$ and an estimate of approximately \$119.23 instead.

Common mistake. The variance of all observations pooled together is not the expected conditional variance. Separate the average conditional variance from the variance of conditional means before calculating credibility.

Audit Checks and Frequent Traps

A calculated credibility factor should be interpreted in the context of its model inputs. With $n>0$, $k\geq0$, $W>0$, $\phi\geq0$, and $a>0$, both standard formulas give $0<Z\leq1$. If $a=0$, all risks have the same model mean, so the credibility of risk-specific experience is zero and the estimate is the collective mean. If a fitted variance-component estimate is negative, it is not a valid theoretical variance; review the estimation method and follow its prescribed nonnegative constraint rather than using the negative value to generate a credibility above one or below zero.

Before accepting an answer, check the following:

Definitions: Is the process variance conditional on the risk parameter and averaged across risks? Is $a$ the variance of conditional means?
Mean: Is $m$ the collective mean, rather than the observed mean for the risk being estimated?
Weights: In Bühlmann-Straub, were observations averaged using their exposures, and was $W$ calculated in the same units?
Parameters: Does $\phi$ use the exposure unit specified by the model? Does $k$ therefore have units of exposure, while $Z$ is dimensionless?
Bounds: With admissible nonnegative variance components, is $Z$ between zero and one? Does the estimate lie between the experience mean and the collective mean?

These checks are useful because a plausible-looking weighted average can still come from an invalid $Z$, a mismatched exposure scale, or the wrong mean. The formulas are only as reliable as the definitions and units of their inputs.

Check Your Understanding

For each problem, identify the relevant model quantities before calculating. Monetary answers are rounded to the nearest cent where needed.

A Bühlmann model has $v=720$, $a=180$, $n=5$, $\bar X=\$72$, and $m=\$60$. Find $k$, $Z$, and the credibility estimate. (Answer: $k=4$, $Z=5/9\approx0.5556$, estimate $\$66.67$.)
A Bühlmann-Straub class has total exposure $W=9$, $\phi=270$, $a=90$, weighted mean $\bar X_w=\$44$, and collective mean $m=\$40$. Find the estimate. (Answer: $k=3$, $Z=0.75$, estimate $\$43.00$.)
Two periods have exposures 2 and 3, with observations \$30 and \$50. A Bühlmann-Straub model has $\phi=500$, $a=100$, and collective mean \$38. Calculate the exposure-weighted mean and estimate. (Answer: $\bar X_w=\$42$, $W=5$, $k=5$, $Z=0.5$, estimate $\$40.00$.)
A model uses $W=7$ exposure units, $\phi=350$, $a=70$, $\bar X_w=\$96$, and $m=\$84$. The exposure unit is changed so the numerical exposure is multiplied by 100. What must happen to $\phi$, and what is the estimate before and after conversion? (Answer: $\phi$ must become $35{,}000$; $Z=7/12$ in either unit system and the estimate is $\$91.00$.)
Two equally likely risk types have conditional means \$20 and \$40 and conditional variances 100 and 300. For $n=3$, $\bar X=\$36$, calculate $m$, $v$, $a$, $Z$, and the estimate. (Answer: $m=\$30$, $v=200$, $a=100$, $k=2$, $Z=0.6$, estimate $\$33.60$.)

What You'll Learn