Comparing Bühlmann and Bühlmann-Straub Estimates

Why Compare the Two Credibility Models?

Bühlmann and Bühlmann-Straub both blend a risk class’s observed experience with a collective mean. The key distinction is how they represent the amount of experience. In the Bühlmann model, each observation contributes one comparable unit of experience. In the Bühlmann-Straub model, observations can have different exposure weights, so their contributions need not be equal.

This difference matters whenever a class’s observations cover unequal numbers of policies, vehicles, or other exposure units. Counting periods alone can then give the wrong balance between the class’s own experience and the collective mean. The comparison is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Definition: In the Bühlmann model, let $X_j$ be the observation for risk class $i$ in period $j$, conditional on the class’s latent risk parameter $\Theta_i$. Assume observations are conditionally independent and identically distributed, with conditional mean $m(\Theta_i)$ and conditional variance $s^2(\Theta_i)$. Define the collective mean $\mu=E[m(\Theta_i)]$, the expected process variance $\text{EPV}=E[s^2(\Theta_i)]$, and the variance of hypothetical means $\text{VHM}=\operatorname{Var}(m(\Theta_i))$. For $n$ observations, the Bühlmann estimate is $Z\bar X+(1-Z)\mu$, where $\bar X$ is their arithmetic mean and $Z=n/(n+K)$, with $K=\text{EPV}/\text{VHM}$.

For Bühlmann-Straub, an observation is a rate or loss cost $X_{ij}$ based on exposure $w_{ij}$. Conditional on $\Theta_i$, assume the observation has mean $m(\Theta_i)$ and variance $s^2(\Theta_i)/w_{ij}$; observations are conditionally independent. Greater exposure therefore reduces the conditional variance of the observed rate. Define total exposure $W_i=\sum_j w_{ij}$ and the exposure-weighted mean $\bar X_{iw}=\sum_j w_{ij}X_{ij}/W_i$.

Definition: Under the Bühlmann-Straub assumptions, the credibility estimate is $Z_i\bar X_{iw}+(1-Z_i)\mu$, where $Z_i=W_i/(W_i+K)$ and $K=\text{EPV}/\text{VHM}$. Here, $K$ is measured in the same exposure units as $W_i$. The formula presumes positive VHM and the stated variance relationship; if VHM is zero, there is no between-class variation to support credibility in this form.

Why Total Exposure Determines Credibility

The Bühlmann-Straub factor follows from the uncertainty in the weighted mean. Write $m=m(\Theta_i)$ and $W=W_i$. Conditional independence and the variance assumption give

\operatorname{Var}(\bar X_{iw}\mid \Theta_i) =\sum_j\left(\frac{w_{ij}}{W}\right)^2\frac{s^2(\Theta_i)}{w_{ij}} =\frac{s^2(\Theta_i)}{W}.

Averaging over risks gives an expected process variance of $\text{EPV}/W$ for the weighted mean. Its between-risk variance is VHM. In a linear credibility estimate, the coefficient assigned to the experience mean is the between-risk variance divided by the total variance of that mean:

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

Thus, exposure enters twice: it determines the weights in the experience average and the credibility factor. This is why replacing total exposure with the number of periods is generally incorrect.

When the Estimates Coincide

If every observation has exposure 1, then total exposure equals the number of observations and the weighted mean equals the ordinary arithmetic mean. Under matching model assumptions and variance components, the Bühlmann and Bühlmann-Straub formulas are identical.

If every observation instead has the same exposure $c$, the Bühlmann-Straub weighted mean is still the ordinary mean, but total exposure is $nc$, not $n$. The models agree if the Bühlmann model is parameterized for observations of exposure $c$. In that model, the process variance per observation is $\text{EPV}/c$, so its $K$ value is $K/c$. Its credibility factor is $n/(n+K/c)=nc/(nc+K)$, which matches Bühlmann-Straub. Using the unit-exposure $K$ with only $n$ observations does not make that adjustment.

With unequal exposures, the weighted mean generally differs from the arithmetic mean, and the appropriate credibility factor depends on total exposure. The ordinary Bühlmann assumptions of equal conditional variances no longer describe those rate observations unless they are transformed or the model is otherwise re-specified. The estimates can still coincide by coincidence—for example, if the two experience means equal the collective mean—but that does not make the unweighted calculation valid under unequal exposure.

Worked Example: Equal Unit Exposures

A hypothetical class has four annual loss-cost observations, each based on one exposure unit: $90$, $110$, $150$, and $170$ dollars per exposure. The collective mean is $\mu=\$120$, $\text{EPV}=400$, and $\text{VHM}=100$, with both variance components expressed in squared dollars per exposure. Calculate both estimates.

Method: compare the formulas with matching unit exposures. First, $K=400/100=4$. The arithmetic mean is $(90+110+150+170)/4=520/4=\$130$.

For Bühlmann, $n=4$, so $Z=4/(4+4)=0.5$. The estimate is $0.5(130)+0.5(120)=65+60=\$125$.

For Bühlmann-Straub, $W=1+1+1+1=4$, and the weighted mean is also $\$130$. Thus $Z=4/(4+4)=0.5$, giving $0.5(130)+0.5(120)=\$125$. The results agree, as expected. Common mistake avoided: treating the two factors as different just because the formulas are written with different symbols; with unit exposures, $W=n$ and the weighted mean is the ordinary mean.

Worked Example: Equal Non-Unit Exposures

A hypothetical class has two rate observations, $\$110$ and $\$150$ per exposure, each based on two exposure units. Use $\mu=\$120$, $\text{EPV}=400$, and $\text{VHM}=100$, with $K=4$ exposure units. Compare Bühlmann-Straub with a Bühlmann calculation that mistakenly treats each observation as one unit, then align the Bühlmann parameterization.

Bühlmann-Straub. Total exposure is $W=2+2=4$. The weighted mean is $(2(110)+2(150))/4=(220+300)/4=\$130$. Hence $Z=4/(4+4)=0.5$, and the estimate is $0.5(130)+0.5(120)=\$125$.

Unadjusted Bühlmann calculation. Using $n=2$ and the unit-exposure $K=4$ gives $Z=2/(2+4)=1/3$. The estimate is $(1/3)(130)+(2/3)(120)=43.3333+80=\$123.33$, rounded to cents. This does not match because it treats each two-unit observation as if it provided only one unit of experience.

Aligned Bühlmann calculation. Each observation has process variance $\text{EPV}/2=200$, so the corresponding $K$ is $200/100=2$. Then $Z=2/(2+2)=0.5$, and the estimate is $0.5(130)+0.5(120)=\$125$. This matches Bühlmann-Straub. Common mistake avoided: using $n$ with a $K$ expressed in unit-exposure terms when the observations each represent two units.

Worked Example: Unequal Exposures

A hypothetical class has rate observations of $\$160$, $\$110$, and $\$130$, based on exposures $1$, $3$, and $2$, respectively. Use $\mu=\$120$, $\text{EPV}=400$, and $\text{VHM}=100$, so $K=4$. Calculate the Bühlmann-Straub estimate and compare it with an unweighted Bühlmann calculation.

Bühlmann-Straub. Total exposure is $W=1+3+2=6$. The weighted total is $1(160)+3(110)+2(130)=160+330+260=750$, so $\bar X_{iw}=750/6=\$125$. The factor is $Z=6/(6+4)=0.6$. The estimate is $0.6(125)+0.4(120)=75+48=\$123$.

Unweighted calculation for comparison. The arithmetic mean is $(160+110+130)/3=400/3=\$133.3333$. If treated as three equal-variance observations, the Bühlmann factor would be $3/(3+4)=3/7$, producing $(3/7)(400/3)+(4/7)(120)=400/7+480/7=880/7=\$125.71$, rounded to cents.

The estimates differ: the higher-exposure observations receive more weight under Bühlmann-Straub, and the credibility factor uses six exposure units rather than three periods. The ordinary calculation is not justified by the stated model: the conditional variances of these rate observations are $400/1$, $400/3$, and $400/2$, not a common variance. Common mistake avoided: averaging the three rates equally and treating the resulting estimate as the appropriate unequal-exposure credibility estimate.

Practical Comparison and Exam Traps

Before comparing two answers, check that their units and assumptions align. In particular, identify whether observations are totals or rates, whether exposures are equal, and whether $K$ is expressed per period or per exposure unit. A numerical difference can reflect a genuine model difference—or merely inconsistent scaling.

Data situation	Experience mean	Credibility amount	Comparison
All exposures equal to 1	Weighted and ordinary means agree	$W=n$	Models agree with matching components
All exposures equal to $c$	Weighted and ordinary means agree	$W=nc$	Agree when Bühlmann components are scaled to exposure $c$
Exposures unequal	Weighted and ordinary means can differ	Use $W=\sum_j w_{ij}$	Bühlmann-Straub generally gives the suitable estimate under its assumptions

A useful check is to set all weights to 1 in the Bühlmann-Straub formula. It must reduce to the Bühlmann formula. Another is to confirm that $Z$ lies between 0 and 1: positive $W$ and $K$ give $0<Z<1$, so the estimate is a blend of the experience mean and collective mean. If a calculation gives a factor outside this range, revisit the exposure units, variance components, and arithmetic.

Key takeaway: Bühlmann uses the number of comparable observations; Bühlmann-Straub uses total exposure and an exposure-weighted mean. They coincide for unit exposures, and also for common non-unit exposures when the Bühlmann model is scaled consistently. Unequal exposures generally require the Bühlmann-Straub weighting and variance structure.

Check Your Understanding

For each problem, calculate the requested estimate and check whether the model assumptions support an ordinary or exposure-weighted mean.

A class has three unit-exposure observations $90$, $120$, and $150$. Let $\mu=\$100$, $\text{EPV}=300$, and $\text{VHM}=100$. Calculate both estimates. (Answer: Both are $\$110$.)
Two observations of $\$90$ and $\$130$ each have exposure 4. Let $\mu=\$100$, $\text{EPV}=400$, and $\text{VHM}=100$. Calculate the Bühlmann-Straub estimate. (Answer: $\$106.67$.)
Observations $\$150$, $\$100$, and $\$110$ have exposures 1, 2, and 5. Let $\mu=\$100$, $\text{EPV}=300$, and $\text{VHM}=100$. Calculate the Bühlmann-Straub estimate. (Answer: $\$109.09$.)
Observations $\$100$, $\$120$, and $\$140$ have exposures 1, 2, and 3. Let $\mu=\$120$, $\text{EPV}=400$, and $\text{VHM}=100$. Calculate the weighted mean, total exposure, credibility factor, and estimate. (Answer: Weighted mean $\$126.67$, total exposure $6$, credibility factor $0.6$, estimate $\$124.00$.)
A class has total exposure $W=15$, with $\mu=\$200$, $\text{EPV}=500$, and $\text{VHM}=100$. Its weighted mean is $\$220$. Calculate $K$, $Z$, and the Bühlmann-Straub estimate. (Answer: $K=5$, $Z=0.75$, estimate $\$215.00$.)

Data situation	Experience mean	Credibility amount	Comparison
All exposures equal to 1	Weighted and ordinary means agree	\(W=n\)	Models agree with matching components
All exposures equal to \(c\)	Weighted and ordinary means agree	\(W=nc\)	Agree when Bühlmann components are scaled to exposure \(c\)
Exposures unequal	Weighted and ordinary means can differ	Use \(W=\sum_j w_{ij}\)	Bühlmann-Straub generally gives the suitable estimate under its assumptions

What You'll Learn