Limiting Cases of the Credibility Formula

Why Limiting Cases Matter

A credibility premium balances a risk’s experience against the collective mean. The credibility weight indicates how much influence the observed experience receives. Studying the limiting cases makes the direction of that balance clear: very little information should generally leave the premium near the collective mean, while sufficiently informative experience should receive more weight.

The limits also help check calculations. If increasing exposure moves a premium away from the collective mean when the risk’s experience differs from that mean, the weight or formula has likely been reversed. The results below apply to Bühlmann credibility and its exposure-weighted Bühlmann-Straub form, under their standard variance assumptions.

Definition: Let $m=E[E[X\mid\Theta]]$ be the collective mean, let $\text{EPV}=E[\operatorname{Var}(X\mid\Theta)]$ be the expected process variance, and let $\text{VHM}=\operatorname{Var}(E[X\mid\Theta])$ be the variance of hypothetical means. When $\text{VHM}>0$, define $K=\text{EPV}/\text{VHM}$. For Bühlmann credibility with $n$ equal-exposure observations, $Z=n/(n+K)$. For Bühlmann-Straub credibility with total exposure $w$, $Z=w/(w+K)$. In either case the premium is $P=Z\bar X+(1-Z)m$, where $\bar X$ is the relevant experience mean or exposure-weighted mean.

How the Weight Determines the Limit

In Bühlmann credibility, each of the $n$ observations is treated as having equal exposure, so $n$ plays the role of total exposure. In Bühlmann-Straub credibility, observations may have different exposures; $w$ is their total exposure and $\bar X$ is their exposure-weighted mean. Under the Bühlmann-Straub model, an observation with exposure $w_j$ has conditional variance $\text{EPV}/w_j$. The aggregate exposure therefore determines the precision of the weighted mean.

For either formula, the credibility weight can be written as $Z=1/(1+K/w)$, when $w>0$. This form makes the limits transparent. If $K$ stays positive and finite, increasing exposure drives $K/w$ toward zero and $Z$ toward one. If exposure approaches zero, $K/w$ grows without bound and $Z$ approaches zero.

P=Z\bar X+(1-Z)m =m+Z(\bar X-m)

The second expression shows why the weight controls the limiting premium. When $Z$ tends to zero, the experience difference $\bar X-m$ is suppressed and the premium tends to $m$. When $Z$ tends to one, the premium tends to $\bar X$, provided the experience mean is defined and remains fixed in the limit.

Write the weight using the relevant exposure.
Use $n$ for equal-exposure Bühlmann data or $w$ for total Bühlmann-Straub exposure.

Check the structural parameters.
Compute $K=\text{EPV}/\text{VHM}$, if both quantities are positive.

Take the limit of $Z$.
Then use $P=m+Z(\bar X-m)$ to identify the limiting premium.

Exposure Limits

Assume $K$ is positive and finite, and suppose the experience mean $\bar X$ remains fixed as exposure changes. As exposure $w$ approaches zero, $Z=w/(w+K)$ approaches zero, so the premium approaches the collective mean. As $w$ becomes large, $Z$ approaches one, so the premium approaches the risk’s experience mean.

\lim_{w\to 0^+}\frac{w}{w+K}=0, \qquad \lim_{w\to\infty}\frac{w}{w+K}=1

For an exactly zero exposure, there is no exposure-weighted experience mean to calculate from that observation: it contributes neither premium experience nor information. The zero-exposure conclusion describes a limit through positive exposures, or the premium based on the remaining positive-exposure history. It does not license substituting an arbitrary value for an undefined $\bar X$.

Worked Example: Increasing a Risk’s Exposure

A hypothetical insurer uses Bühlmann-Straub credibility to price a group’s annual loss rate. The collective mean is $m=\$80$ per exposure unit, the group’s observed weighted mean is $\bar X=\$68$, $\text{EPV}=900$ dollars squared, and $\text{VHM}=100$ dollars squared. Find the premium at total exposures $w=3$ and $w=90$, and identify the large-exposure limit. Use the standard Bühlmann-Straub assumptions, with the same experience mean at both exposure levels.

Method. Calculate $K$, then $Z$, then substitute in the credibility premium. This method avoids treating exposure as a direct multiplier of the premium.

First, $K=\text{EPV}/\text{VHM}=900/100=9$ exposure units. At $w=3$,

Z=\frac{3}{3+9}=\frac14, \qquad P=\frac14(68)+\frac34(80)=17+60=\$77.00.

At $w=90$,

Z=\frac{90}{90+9}=\frac{10}{11}, \qquad P=\frac{10}{11}(68)+\frac{1}{11}(80) =\frac{760}{11} =\$69.09.

As $w$ tends to infinity, $Z$ tends to one and the premium tends to $\bar X=\$68$. The computed premium at $w=90$ is between $\$68$ and $\$80$, as it should be; increasing exposure has moved it closer to the group’s experience mean.

EPV and VHM Limits

At fixed positive exposure, $K$ summarizes the balance between process variation and variation among risk means. If $\text{EPV}$ approaches zero while $\text{VHM}$ stays positive, then $K$ approaches zero and $Z$ approaches one: observations become highly informative about the risk’s mean. If $\text{EPV}$ becomes very large while $\text{VHM}$ stays positive, then $K$ becomes very large and $Z$ approaches zero: experience receives little weight.

The same logic applies from the other side of the ratio. If $\text{VHM}$ approaches zero while $\text{EPV}$ stays positive, $K$ becomes very large and $Z$ approaches zero. With little variation in underlying risk means, an individual’s experience gives little reason to depart from the collective mean. If $\text{VHM}$ becomes very large while $\text{EPV}$ stays positive, $K$ approaches zero and $Z$ approaches one for fixed positive exposure.

Change, holding the other quantities fixed	Effect on $K$	Limiting weight	Limiting premium
Exposure $w\to 0^+$, with $K>0$	Unchanged	$Z\to 0$	Collective mean $m$
Exposure $w\to\infty$, with $K$ finite	Unchanged	$Z\to 1$	Experience mean $\bar X$
EPV $\to 0$, with VHM positive	$K\to 0$	$Z\to 1$	Experience mean $\bar X$
EPV $\to\infty$, with VHM positive	$K\to\infty$	$Z\to 0$	Collective mean $m$
VHM $\to 0$, with EPV positive	$K\to\infty$	$Z\to 0$	Collective mean $m$
VHM $\to\infty$, with EPV positive	$K\to 0$	$Z\to 1$	Experience mean $\bar X$

These rows assume the quantities not being varied remain fixed and the exposure is positive where parameter limits are considered. If an endpoint makes a structural parameter exactly zero, the formula involving $K$ may not be defined as written. Interpret such an endpoint through the stated limit rather than inserting zero into a ratio that may be undefined.

Worked Example: Changing EPV and VHM Separately

For a hypothetical risk, let the collective mean be $m=\$100$, the experience mean be $\bar X=\$70$, and the Bühlmann-Straub exposure be $w=12$. Compare the premium for (a) $\text{EPV}=400$, $\text{VHM}=100$; (b) $\text{EPV}=25$, $\text{VHM}=100$; and (c) $\text{EPV}=400$, $\text{VHM}=25$. All variances are in dollars squared.

Method. Recalculate $K$ and $Z$ for each pair, keeping $w$, $m$, and $\bar X$ fixed. This isolates the effect of each structural parameter.

(a) $K=400/100=4$. Therefore $Z=12/(12+4)=3/4$, and

P=\frac34(70)+\frac14(100)=52.50+25.00=\$77.50.

(b) $K=25/100=0.25$. Thus $Z=12/(12+0.25)=12/12.25=48/49$. The premium is

P=\frac{48}{49}(70)+\frac{1}{49}(100) =\frac{3,460}{49} =\$70.61.

Reducing EPV has raised the weight on experience and moved the premium closer to $\$70$.

P=\frac37(70)+\frac47(100) =30+57.142857\ldots =\$87.14.

Reducing VHM has lowered the experience weight and moved the premium closer to $\$100$. Each result lies between the two means, confirming the direction of the credibility adjustment.

Justifying the Bühlmann-Straub Weight

The weight also follows from a linear prediction argument. Let $\Theta$ denote the risk parameter, let $M=E[X\mid\Theta]$ be its hypothetical mean, and consider predicting $M$ by $P=Z\bar X+(1-Z)m$. Assume conditional unbiasedness $E[\bar X\mid\Theta]=M$ and conditional variance $\operatorname{Var}(\bar X\mid\Theta)=\text{EPV}/w$, as in the Bühlmann-Straub model. The conditional error is

P-M=(1-Z)(m-M)+Z(\bar X-M).

Conditional on $\Theta$, the second term has mean zero, so its cross-product with the first term has expectation zero. Taking the unconditional mean squared error gives

E[(P-M)^2] =(1-Z)^2\operatorname{Var}(M) +Z^2\frac{\text{EPV}}{w} =(1-Z)^2\text{VHM}+Z^2\frac{\text{EPV}}{w}.

Differentiate this quadratic with respect to $Z$, set the derivative to zero, and solve:

-2(1-Z)\text{VHM} +2Z\frac{\text{EPV}}{w}=0, \qquad Z=\frac{\text{VHM}}{\text{VHM}+\text{EPV}/w} =\frac{w}{w+\text{EPV}/\text{VHM}}.

The quadratic has positive second derivative $2\text{VHM}+2\text{EPV}/w$ when both variance components are positive, so this stationary value minimizes mean squared error. The derivation explains both the ratio $K$ and why exposure increases credibility: it reduces the process variance of $\bar X$.

Simultaneous Limits and a Common Trap

Separate one-quantity limits have clear answers, but simultaneous limits need not. If both exposure and $K$ approach zero, the weight $w/(w+K)$ depends on their relative rates. For example, if $w=K$, the weight remains $1/2$; if $K$ is much larger than $w$, the weight approaches zero; if $K$ is much smaller than $w$, it approaches one. Thus “exposure goes to zero” alone does not settle a joint limit in which $K$ also changes.

A frequent exam error is reversing the ratio and using $K=\text{VHM}/\text{EPV}$. For $\text{EPV}=900$, $\text{VHM}=100$, and $w=3$, that error gives $K=1/9$ and $Z=3/(3+1/9)=27/28$, incorrectly assigning almost all weight to only three exposure units. The correct ratio is $K=900/100=9$, giving $Z=3/(3+9)=1/4$. The direction check catches the error: large process variation relative to between-risk variation should reduce, not increase, credibility.

Worked Example: A Joint Limit Depends on the Path

A hypothetical Bühlmann-Straub portfolio has collective mean $m=\$50$ and an experience mean held at $\bar X=\$110$ as exposure $w$ and $K$ shrink together. Compare the limiting premium along the path $w=K$, and along paths where $K=10w$ and $K=w/10$, as $w\to0^+$.

Method. Substitute each path into $Z=w/(w+K)$, then calculate the premium. This avoids assuming that small exposure automatically implies zero credibility when $K$ is also shrinking.

Along $w=K$, $Z=w/(w+w)=1/2$ for every positive $w$. Hence

P=\frac12(110)+\frac12(50)=\$80.00.

Along $K=10w$, $Z=w/(w+10w)=1/11$, so

P=\frac{1}{11}(110)+\frac{10}{11}(50) =\frac{610}{11} =\$55.45.

Along $K=w/10$, $Z=w/(w+w/10)=10/11$, so

P=\frac{10}{11}(110)+\frac{1}{11}(50) =\frac{1,150}{11} =\$104.55.

Each path has $w\to0^+$ and $K\to0^+$, but the limiting premiums differ because the ratio $w/K$ differs. The results remain between $\$50$ and $\$110$, which checks the calculations.

Check Your Understanding

For each problem, identify the relevant limit or calculate the requested premium. Round dollar amounts to the nearest cent.

A Bühlmann-Straub risk has $m=\$120$, $\bar X=\$100$, $K=8$, and $w=2$. Find its premium. (Answer: $Z=0.2$, $P=\$116.00$.)
With $m=\$75$, fixed positive $K$, and a fixed experience mean of $\$90$, what is the limiting premium as Bühlmann-Straub exposure grows without bound? (Answer: $\$90$.)
At fixed exposure $w=10$, $m=\$60$, and $\bar X=\$90$, let EPV approach zero while VHM remains positive. What is the limiting premium? (Answer: $\$90$.)
A risk has exposure $w=6$, $m=\$40$, $\bar X=\$70$, EPV $=180$, and VHM $=60$. Calculate $K$, $Z$, and the premium. (Answer: $K=3$, $Z=2/3$, $P=\$60.00$.)
Suppose $w\to0^+$, $m=\$30$, and $\bar X=\$90$, while $K=2w$. Find the limiting weight and premium. (Answer: $Z\to1/3$, $P\to\$50.00$.)

Change, holding the other quantities fixed	Effect on \(K\)	Limiting weight	Limiting premium
Exposure \(w\to 0^+\), with \(K>0\)	Unchanged	\(Z\to 0\)	Collective mean \(m\)
Exposure \(w\to\infty\), with \(K\) finite	Unchanged	\(Z\to 1\)	Experience mean \(\bar X\)
EPV \(\to 0\), with VHM positive	\(K\to 0\)	\(Z\to 1\)	Experience mean \(\bar X\)
EPV \(\to\infty\), with VHM positive	\(K\to\infty\)	\(Z\to 0\)	Collective mean \(m\)
VHM \(\to 0\), with EPV positive	\(K\to\infty\)	\(Z\to 0\)	Collective mean \(m\)
VHM \(\to\infty\), with EPV positive	\(K\to 0\)	\(Z\to 1\)	Experience mean \(\bar X\)

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