Sensitivity of Class Indications to the Structural Parameter

Why Test Sensitivity to the Structural Parameter?

A credibility-weighted class indication depends on both the class data and the assumed relationship between within-class variation and differences among classes. Even when the exposure and observed class loss cost are unchanged, an alternative value of the Bühlmann structural parameter can produce a different credibility factor and indication. Sensitivity analysis makes that dependence visible: it shows whether an indication is stable or materially affected by the parameter choice.

This tutorial holds the class data, exposure, and collective complement fixed while varying the structural parameter. It does not assess how the parameter should be estimated from a portfolio, nor does it change the credibility model. The calculations use the Bühlmann model with a common structural parameter for the classes being compared.

Definition: In the Bühlmann model, the structural parameter is $K=\mathrm{EPV}/\mathrm{VHM}$, where $\mathrm{EPV}$ is the expected process variance and $\mathrm{VHM}$ is the variance of hypothetical means. Under the model's common-parameter assumptions, for class $j$ with $n_j$ comparable exposure units, credibility is $Z_j(K)=n_j/(n_j+K)$. Its indicated loss cost is $I_j(K)=Z_j(K)\bar X_j+[1-Z_j(K)]\mu$, where $\bar X_j$ is the observed class mean loss cost and $\mu$ is the fixed collective mean.

The model assumes that, conditional on a class's underlying risk parameter, its observations are identically distributed and independent, with a conditional mean and variance. Across classes, those hypothetical means have a common distribution, with finite, positive $\mathrm{VHM}$; $\mathrm{EPV}$ is also finite. The simple formula using $n_j$ applies when the observations represent comparable exposure units. The parameter $K$ is measured on the same exposure-unit scale as $n_j$.

A larger $K$ means more process variation relative to variation among hypothetical means. For a fixed amount of class experience, that implies less credibility for the class mean and more weight on the collective mean. A smaller positive $K$ has the opposite effect. These are comparisons under alternative parameter values; they do not mean that the observed class data have changed.

How an Alternative Parameter Changes an Indication

For fixed $\bar X_j$, $\mu$, and $n_j$, the indication can be written in a form that makes the movement especially clear:

I_j(K)=\mu+\frac{n_j}{n_j+K}(\bar X_j-\mu).

The quantity $\bar X_j-\mu$ is the class's observed difference from the collective mean. The credibility factor determines how much of that difference remains in the indication. Raising $K$ reduces the retained proportion; it does not change the direction of the difference.

Result: For fixed class data and exposure, $I_j(K)$ moves toward $\mu$ as $K$ increases. If $\bar X_j>\mu$, the indication decreases toward $\mu$; if $\bar X_j<\mu$, the indication increases toward $\mu$. If $\bar X_j=\mu$, it equals $\mu$ for every $K$.

This result follows by differentiating the indication with respect to $K$:

\frac{dI_j(K)}{dK} = -\frac{n_j}{(n_j+K)^2}(\bar X_j-\mu).

For $n_j>0$ and $K>0$, the coefficient $-n_j/(n_j+K)^2$ is negative. Therefore the derivative is negative when the class mean is above the collective mean and positive when it is below. The derivative is zero when the two means are equal. Also, $n_j/(n_j+K)$ tends to zero as $K$ grows, so $I_j(K)$ tends to $\mu$. This establishes both the direction and the limiting value of the change.

For two parameter values $K_1<K_2$, the absolute change in the indication is

|I_j(K_2)-I_j(K_1)| = |\bar X_j-\mu| \frac{n_j(K_2-K_1)}{(n_j+K_1)(n_j+K_2)}.

This expression follows by subtracting the two forms $\mu+n_j(\bar X_j-\mu)/(n_j+K)$ and taking the absolute value. It shows that sensitivity depends on the size of the class's difference from the collective mean, its exposure, and the range of parameter values being compared. It also provides a direct check on a pair of calculated indications.

Worked Example: Comparing Two Classes at Alternative Values of K

A hypothetical portfolio has a collective mean loss cost of $\$1{,}200$. Class North has $n=6$ exposure units and an observed mean of $\$1{,}500$. Class South has $n=15$ units and an observed mean of $\$1{,}080$. Calculate both indications at $K=3$ and $K=9$, then interpret the changes.

Method. Use $Z=n/(n+K)$, then calculate $I=\mu+Z(\bar X-\mu)$. Keep $\mu$, $n$, and each observed mean fixed across the two calculations.

North at $K=3$. The factor is $Z=6/(6+3)=6/9=2/3$. The class difference is $\$1{,}500-\$1{,}200=\$300$. Thus $I=\$1{,}200+(2/3)(\$300)=\$1{,}200+\$200=\$1{,}400.00$.

North at $K=9$. The factor is $Z=6/(6+9)=6/15=0.4$. Thus $I=\$1{,}200+0.4(\$300)=\$1{,}320.00$. The indication falls by $\$80.00$, toward the collective mean.

South at $K=3$. The factor is $Z=15/(15+3)=15/18=5/6$. The class difference is $\$1{,}080-\$1{,}200=-\$120$. Thus $I=\$1{,}200+(5/6)(-\$120)=\$1{,}200-\$100=\$1{,}100.00$.

South at $K=9$. The factor is $Z=15/(15+9)=15/24=5/8$. Thus $I=\$1{,}200+(5/8)(-\$120)=\$1{,}200-\$75=\$1{,}125.00$. The indication rises by $\$25.00$, again toward the collective mean.

Check and interpretation. At $K=9$, each credibility factor is lower than at $K=3$: North's falls from $2/3$ to $0.4$, while South's falls from $5/6$ to $5/8$. The indications move toward $\$1{,}200$ from opposite sides, as required. The larger change for North is consistent with its larger distance from the collective mean and smaller exposure. The trap avoided is treating a larger $K$ as more class credibility.

Reading Sensitivity Across Different Class Sizes

A class's sensitivity is not determined by exposure alone. A small class generally has a lower credibility factor for a given $K$, but its indication may change little if its observed mean is close to the collective mean. Conversely, a class with a substantial difference from the complement can show a noticeable change even when its credibility factor changes moderately. The amount and direction of the class difference must be considered alongside the credibility factor.

The next example isolates this comparison. It uses the same two values of $K$ for both classes, and it reports the changes in dollars so that the scale of the indications remains clear.

Worked Example: Identifying the More Sensitive Indication

A hypothetical program has a fixed collective mean loss cost of $\$980$. Class Cedar has $n=3$ exposure units and an observed mean of $\$1{,}580$. Class Birch has $n=18$ units and an observed mean of $\$930$. Compare the indicated costs at $K=2$ and $K=8$, and identify which class is more sensitive in dollars.

Cedar at $K=2$. Its credibility factor is $3/(3+2)=3/5=0.6$. Its difference from the collective mean is $\$600$, so the indication is $\$980+0.6(\$600)=\$1{,}340.00$.

Cedar at $K=8$. Its factor is $3/(3+8)=3/11$. The indication is $\$980+(3/11)(\$600)=\$980+\$1800/11=\$1{,}143.64$, rounded to the nearest cent. The decrease is $\$1{,}340.00-\$1{,}143.64=\$196.36$.

Birch at $K=2$. Its factor is $18/(18+2)=18/20=0.9$. Its difference from the collective mean is $-\$50$, so the indication is $\$980+0.9(-\$50)=\$935.00$.

Birch at $K=8$. Its factor is $18/(18+8)=18/26=9/13$. The indication is $\$980+(9/13)(-\$50)=\$980-\$450/13=\$945.38$, rounded to the nearest cent. It rises by $\$10.38$.

Check and interpretation. Cedar is more sensitive in dollars: its indication changes by $\$196.36$, compared with Birch's $\$10.38$. Cedar's class mean is $\$600$ above the complement, while Birch is only $\$50$ below it. The directions are also correct: Cedar moves down and Birch moves up as $K$ increases. The trap avoided is assuming that the class with more exposure must always have the less sensitive indication, without considering the distance from the collective mean.

What a Sensitivity Comparison Does—and Does Not—Show

A sensitivity table is useful when the selected $K$ is uncertain or when an analyst wants to understand the consequences of plausible alternatives. For each class, it can display $Z_j(K)$ and $I_j(K)$ at the same set of parameter values. Reading across a row shows how that class's indication responds; reading down a column compares classes under one common assumption.

The comparison is conditional on the inputs being held fixed. In particular, the collective mean $\mu$ is not recalculated at each $K$, and neither the class exposures nor observed class means change. If an analysis changes the complement, exposure basis, or observed data as well as $K$, the resulting difference cannot be attributed to $K$ alone.

The parameter should also be applied on the correct exposure scale. If $n_j$ counts annual exposure units, $K$ must be expressed on that same basis. Substituting a parameter defined for a different exposure unit without conversion changes the credibility calculation. The formula is not a general substitute for a model with unequal exposure weights or class-specific structural parameters.

Worked Example: Checking a Reported Change Between K Values

A hypothetical class has $n=8$, an observed mean loss cost of $\$1{,}450$, and a fixed collective mean of $\$1{,}250$. An analyst compares $K=2$ with $K=5$. Calculate the two indications and verify their difference using the direct sensitivity formula.

At $K=2$. The factor is $8/(8+2)=0.8$. The class difference is $\$200$, so the indication is $\$1{,}250+0.8(\$200)=\$1{,}410.00$.

At $K=5$. The factor is $8/(8+5)=8/13$. The indication is $\$1{,}250+(8/13)(\$200)=\$1{,}250+\$1{,}600/13=\$1{,}373.08$, rounded to the nearest cent. The decrease is $\$1{,}410.00-\$1{,}373.08=\$36.92$.

Direct check. The formula gives $|\$1{,}450-\$1{,}250|[8(5-2)/((8+2)(8+5))]=\$200(24/130)=\$36.923\ldots$, which rounds to $\$36.92$. The computed change is therefore consistent. The indication decreases because the observed class mean exceeds the collective mean.

Common Calculation Traps

The most important direction check is straightforward: increasing $K$ lowers $Z$. It must therefore pull the indication closer to the collective mean, not farther away. For example, for $n=4$, $K=2$, $\mu=\$1{,}000$, and $\bar X=\$1{,}300$, the correct calculation is $Z=4/(4+2)=2/3$, then $I=\$1{,}000+(2/3)(\$300)=\$1{,}200$. A result of $\$1{,}100$ is not consistent with that factor and difference.

A common incorrect calculation uses $K/(n+K)$ as the class credibility factor. In the same example, that mistake gives $Z=2/6=1/3$ and the incorrect indication $\$1{,}100$. The $K$ term is the amount of collective weight relative to exposure in the equivalent expression $I=[n/(n+K)]\bar X+[K/(n+K)]\mu$; it is not the class weight.

Reversing the direction. A larger $K$ means a smaller class credibility factor and a stronger pull toward $\mu$.
Changing the complement by accident. To isolate parameter sensitivity, keep the collective mean fixed across the comparison.
Comparing factors without comparing indications. Two classes can have different factor changes but very different loss-cost changes because their observed deviations differ.
Rounding too early. Retain exact fractions or unrounded factors until calculating the final dollar indication; then round money to cents.

Key takeaway: With class data and the collective mean fixed, $K$ controls how much of the observed class-to-collective difference survives in the indication. Raising $K$ reduces credibility and moves every class indication toward the collective mean; the size of the movement depends on both exposure and the class's distance from that mean.

Check Your Understanding

For each problem, hold the stated class data and collective mean fixed across the parameter values. Money answers are rounded to the nearest cent.

A hypothetical class has $n=4$, $\bar X=\$1{,}200$, and $\mu=\$900$. Calculate its indication at $K=2$ and $K=6$. (Answer: $\$1{,}100.00$ at $K=2$; $\$1{,}020.00$ at $K=6$.)
A class has $n=10$, $\bar X=\$1{,}350$, and $\mu=\$1{,}500$. Calculate its indication at $K=2$ and $K=8$, and state the direction of change. (Answer: $\$1{,}375.00$ at $K=2$; $\$1{,}416.67$ at $K=8$; it rises toward the collective mean.)
A class has $n=5$, $\bar X=\$2{,}400$, and $\mu=\$2{,}000$. Compare the indications at $K=1$ and $K=4$. (Answer: $\$2{,}333.33$ at $K=1$; $\$2{,}222.22$ at $K=4$; a decrease of $\$111.11$.)
A class has $n=6$, $\bar X=\$1{,}400$, and $\mu=\$1{,}000$. Its indicated loss cost is $\$1{,}160$. Find the implied $K$, assuming the Bühlmann formula applies. (Answer: $Z=(1{,}160-1{,}000)/(1{,}400-1{,}000)=0.4$, so $K=6(1-Z)/Z=9$.)
A hypothetical portfolio has $\mu=\$1{,}100$. Class A has $n=3$, $\bar X=\$1{,}700$; Class B has $n=12$, $\bar X=\$1{,}000$. Compare each indication at $K=2$ and $K=10$, and identify the larger dollar change. (Answer: Class A is $\$1{,}460.00$ at $K=2$ and $\$1{,}238.46$ at $K=10$, a decrease of $\$221.54$. Class B is $\$1{,}014.29$ at $K=2$ and $\$1{,}045.45$ at $K=10$, an increase of $\$31.17$. Class A has the larger change.)

What You'll Learn