Sensitivity of Credibility Estimates to Model Inputs

Why Sensitivity Matters

A credibility estimate is not determined by experience alone. Its weight depends on the balance between process variance, which measures variation within a risk class, and between-class variance, which measures variation among class means. In Bühlmann-Straub credibility, the total exposure weight also matters. A reasonable estimate can therefore move when an analyst revises a variance component, corrects an exposure measure, or updates the complement.

Sensitivity analysis asks two related questions: which way does the estimate move when an input changes, and by how much? It is important to distinguish a change to the credibility factor from a change to the experience mean. In particular, changing period exposure weights can affect both.

Definition: Let $\mu$ be the collective mean, $v$ the expected process variance per unit of exposure, and $a$ the variance of the risk-class means. Define $K=v/a$, when $a>0$. For total exposure weight $w$, the Bühlmann-Straub credibility factor and estimate are $$ Z=\frac{w}{w+K}=\frac{wa}{wa+v}, \qquad \widehat{\theta}=Z\bar{X}_w+(1-Z)\mu. $$ Here $\bar{X}_w$ is the exposure-weighted mean for the class. In the ordinary Bühlmann model with $n$ equally weighted observations, use $w=n$.

The standard Bühlmann-Straub interpretation assumes that, conditional on a class mean $\Theta$, observations have mean $\Theta$, process variance $v/w_j$ for period $j$, and are conditionally uncorrelated. The class means have common mean $\mu$ and variance $a$, and the collective parameters are treated as fixed for the calculation. The results below describe this model; if the variance structure, exposure definition, or dependence assumptions do not hold, the resulting sensitivity conclusions may not describe the data well.

Direction of Change

The ratio $K=v/a$ summarizes how much process variation there is relative to variation among class means. A larger $K$ means more exposure is needed before a class’s experience receives a given credibility weight. For fixed $w$, increasing $v$ increases $K$ and reduces $Z$; increasing $a$ reduces $K$ and increases $Z$. More exposure increases $Z$, provided the experience and model assumptions are otherwise held fixed.

The estimate is a weighted average. If $\bar{X}_w>\mu$, raising $Z$ pulls the estimate upward toward experience; if $\bar{X}_w<\mu$, raising $Z$ pulls it downward toward experience. If $\bar{X}_w=\mu$, changing $Z$ does not change the estimate. Thus, the direction of the credibility-factor change alone does not tell the direction of the estimate without knowing where experience lies relative to the complement.

\frac{\partial Z}{\partial v}=-\frac{wa}{(wa+v)^2}<0,\qquad \frac{\partial Z}{\partial a}=\frac{wv}{(wa+v)^2}>0,\qquad \frac{\partial Z}{\partial w}=\frac{av}{(wa+v)^2}>0.

These derivatives hold with the other inputs fixed and with positive $w$, $a$, and $v$. They show local, or small-change, sensitivity. They do not imply that equal percentage changes in different inputs cause equal changes in $Z$. For finite changes, recalculate $Z$ from the revised inputs rather than relying on a derivative approximation.

Why the Credibility Formula Has This Sensitivity

The formula can be justified by finding the best linear estimator of the class mean under the stated variance assumptions. Consider an estimator of the form $\mu+Z(\bar{X}_w-\mu)$. The class mean has variance $a$, while the exposure-weighted experience mean has conditional process variance $v/w$. The mean squared error of this estimator is therefore

\operatorname{MSE}(Z)=(1-Z)^2a+Z^2\frac{v}{w}.

The first term is the contribution from not fully following the class mean; the second is the contribution from process variation in the experience. To minimize the mean squared error, differentiate with respect to $Z$, set the derivative equal to zero, and solve:

\begin{aligned} \frac{d}{dZ}\operatorname{MSE}(Z) &=-2(1-Z)a+2Z\frac{v}{w}=0,\\ a&=Z\left(a+\frac{v}{w}\right),\\ Z&=\frac{a}{a+v/w} =\frac{wa}{wa+v} =\frac{w}{w+v/a}. \end{aligned}

The second derivative is $2a+2v/w>0$, so this stationary point minimizes the mean squared error. This derivation explains the balance behind the factor: greater exposure reduces the process-variance contribution $v/w$, while greater between-class variance makes it more valuable to distinguish this class from the collective mean. The result is a best linear estimator under these assumptions; it is not, without additional distributional assumptions, a claim that the estimator is the exact conditional expectation.

Worked Sensitivity Calculations

Worked Example: Changing Process and Between-Class Variance

A hypothetical class has exposure weight $w=18$, experience mean $\bar{X}_w=\$120$, and complement $\mu=\$100$. The process variance is $v=900$ dollars squared per exposure unit, and the between-class variance is $a=100$ dollars squared. Calculate the estimate, then assess separate 25% increases in $v$ and $a$. Use the Bühlmann-Straub model and round final estimates to the nearest cent.

Method. Calculate $K=v/a$, then $Z=w/(w+K)$, and finally $\widehat{\theta}=\mu+Z(\bar{X}_w-\mu)$. Change one input at a time so the effects can be compared.

Initially, $K=900/100=9$, so $Z=18/(18+9)=18/27=2/3$. The experience differs from the complement by $\$120-\$100=\$20$. Thus,

\widehat{\theta}=100+\frac{2}{3}(20)=\$113.33.

If $v$ increases by 25%, its revised value is $900(1.25)=1{,}125$, while $a$ remains 100. Then $K=1{,}125/100=11.25$, and $Z=18/(18+11.25)=18/29.25=8/13$. The revised estimate is

100+\frac{8}{13}(20)=\$112.31.

The higher process variance reduces the weight on the class’s above-complement experience, so the estimate moves down toward $\$100$.

Instead, keep $v=900$ and increase $a$ by 25%, to $100(1.25)=125$. Now $K=900/125=7.2$, and $Z=18/(18+7.2)=18/25.2=5/7$. The estimate is

100+\frac{5}{7}(20)=\$114.29.

The increased between-class variance raises credibility, moving the estimate farther toward $\$120$. Common mistake avoided: increasing $a$ increases $Z$; it does not increase $K=v/a$.

Worked Example: Exposure Changes Both Weight and Experience

A hypothetical risk class has two periods of experience. Period 1 has exposure $w_1=4$ and mean loss cost $X_1=\$90$; period 2 has $w_2=12$ and mean loss cost $X_2=\$110$. Let $v=800$ dollars squared per exposure unit, $a=100$ dollars squared, and $\mu=\$100$. Calculate the estimate, then recalculate if period 2 exposure is revised to $w_2=20$, with both period means unchanged. Round final estimates to the nearest cent.

Method. In each case, first calculate the total weight and weighted mean. Then calculate $K$, $Z$, and the credibility estimate. This is necessary because revising one period’s weight changes not only total credibility but also the experience mean being weighted.

Initially, total exposure is $w=4+12=16$. The weighted mean is

\bar{X}_w=\frac{4(90)+12(110)}{16} =\frac{360+1{,}320}{16} =\$105.

Here $K=800/100=8$, so $Z=16/(16+8)=2/3$. The estimate is $100+(2/3)(105-100)=\$103.33$.

With revised period 2 exposure, total exposure is $w=4+20=24$. The new weighted mean is

\bar{X}_w=\frac{4(90)+20(110)}{24} =\frac{360+2{,}200}{24} =\$106.6667.

The revised credibility factor is $Z=24/(24+8)=3/4$. The estimate, retaining the unrounded weighted mean in the calculation, is

100+\frac{3}{4}(106.6667-100)=\$105.00.

The estimate rises from $\$103.33$ to $\$105.00$. Both effects are upward here: the higher exposure gives the class more credibility, and it gives greater weight to the period with the higher mean. Common mistake avoided: changing $w_2$ while holding the old weighted mean of $\$105$ fixed would omit the change in the experience being weighted.

Worked Example: Competing Changes and Local Sensitivity

A hypothetical class has $w=12$, $v=600$ dollars squared per exposure unit, $a=100$ dollars squared, $\bar{X}_w=\$130$, and $\mu=\$100$. Calculate the original estimate. Then calculate it if process variance rises by 15% and between-class variance rises by 10% at the same time. Also use derivatives to interpret the separate small-change effects. Round final estimates to the nearest cent and derivative values to six decimal places.

Method. Use $Z=wa/(wa+v)$. For the derivative interpretation, hold $w$, experience, and complement fixed. Since $wa=12(100)=1{,}200$ and $wa+v=1{,}800$, the original factor is $1{,}200/1{,}800=2/3$, and the estimate is $100+(2/3)(30)=\$120.00$.

The factor sensitivities at the original inputs are

\frac{\partial Z}{\partial v} =-\frac{1{,}200}{1{,}800^2} =-\frac{1}{2{,}700} \approx-0.000370, \qquad \frac{\partial Z}{\partial a} =\frac{12(600)}{1{,}800^2} =\frac{1}{450} \approx0.002222.

Because the experience exceeds the complement by $\$30$, a small factor change $\Delta Z$ changes the estimate by approximately $30\Delta Z$. A process-variance increase of 90 gives the local approximation $\Delta Z\approx(-1/2{,}700)(90)=-0.033333$, or an estimate change of about $-\$1.00$. A between-class variance increase of 10 gives $\Delta Z\approx(1/450)(10)=0.022222$, or an estimate change of about $+\$0.67$. These are local approximations, not exact changes for finite revisions.

For the exact simultaneous revision, $v=600(1.15)=690$ and $a=100(1.10)=110$. Then $wa=12(110)=1{,}320$, and

Z_{\text{new}}=\frac{1{,}320}{1{,}320+690} =\frac{44}{67}\approx0.656716, \qquad \widehat{\theta}_{\text{new}} =100+\frac{44}{67}(30) \approx\$119.70.

The exact revised estimate is $\$0.30$ below the original. The two changes work in opposite directions, and their combined finite effect is not exactly the sum of the separate derivative approximations. Common mistake avoided: applying a first-order derivative approximation as though it were an exact answer for simultaneous, finite changes.

Changing the Complement

For fixed $Z$ and $\bar{X}_w$, the estimate can be written as $\widehat{\theta}=Z\bar{X}_w+(1-Z)\mu$. A change $\Delta\mu$ in the complement therefore changes the estimate by exactly $(1-Z)\Delta\mu$, provided the other inputs remain fixed. The complement receives the remaining weight, so the estimate does not generally move dollar-for-dollar with $\mu$.

\Delta\widehat{\theta}=(1-Z)\Delta\mu \quad\text{when }Z\text{ and }\bar{X}_w\text{ are fixed.}

This calculation isolates the mechanical effect of changing the complement. In practice, an updated complement may also indicate revised collective data or model parameters, in which case $v$, $a$, or other inputs may need to be reviewed rather than held fixed.

Common Exam Traps and Practical Checks

Reversing the variance ratio. Use $K=v/a$, not $a/v$. More process variance lowers credibility; more between-class variance raises it.
Changing exposure units inconsistently. The exposure weights and $v$ must use compatible units. Rescaling weights without making the corresponding process-variance interpretation consistent can produce a misleading $K$ or factor.
Changing a period weight but keeping the old weighted mean. Recalculate $\bar{X}_w$ whenever a period’s exposure changes, unless the problem explicitly says the aggregate experience mean is held fixed.
Misinterpreting the direction of the final estimate. A lower $Z$ moves the estimate toward $\mu$, not necessarily downward. If experience is below the complement, moving toward the complement raises the estimate.
Confusing local and exact sensitivity. Derivatives approximate small changes near the original inputs. For stated percentage revisions, recompute the credibility factor and estimate from the revised values.

Key takeaway: For fixed exposure, $Z=wa/(wa+v)$: it decreases with process variance and increases with between-class variance. More exposure increases credibility, but a changed exposure weight may also change the weighted experience mean. The final estimate always moves according to the balance between that experience and its complement.

Check Your Understanding

For each problem, state the relevant changed inputs and calculate the requested credibility estimate. Round dollar estimates to the nearest cent.

A class has $w=9$, $v=720$, $a=120$, $\bar{X}_w=\$150$, and $\mu=\$130$. Find $K$, $Z$, and the class estimate. (Answer: $K=6$, $Z=0.6000$, estimate $\$142.00$)
A class has $w=12$, $v=800$, $a=100$, $\bar{X}_w=\$115$, and $\mu=\$100$. If $v$ rises by 25% and all other inputs remain fixed, find the revised $Z$ and estimate. (Answer: $Z=6/11\approx0.5455$, estimate $\$108.18$)
A class has $w=10$, $v=600$, $a=100$, $\bar{X}_w=\$124$, and $\mu=\$100$. If $a$ rises by 50%, find the revised credibility factor and estimate. (Answer: $Z=5/7\approx0.7143$, estimate $\$117.14$)
Two periods have $(w_1,X_1)=(5,\$80)$ and $(w_2,X_2)=(15,\$120)$. Let $v=800$, $a=100$, and $\mu=\$100$. Find the estimate, then find it again if $w_1$ rises to 10 and both period means stay fixed. (Answer: original estimate $\$107.14$; revised weighted mean $\$104.00$, revised $Z=25/33$, revised estimate $\$103.03$)
A class has $w=16$, $v=800$, $a=100$, and $\bar{X}_w=\$130$. Its complement rises from $\$105$ to $\$110$, with the other inputs fixed. Find the new estimate and its change from the original. (Answer: new estimate $\$123.33$, an increase of $\$1.67$)

What You'll Learn