Deriving the Bühlmann Credibility Factor

Why the Bühlmann Credibility Factor Has This Form

A Bühlmann premium balances two sources of information: a risk’s observed experience and the collective mean across risks. The credibility factor $Z$ measures the weight assigned to the individual experience. Its value depends not only on how many observations are available, but also on how variable observations are within a risk compared with how much underlying risk means differ across the collective.

The derivation below assumes equal exposure in every period. It shows why the experience receives more weight as the number of observations or between-risk variation increases, and less weight as within-risk variation increases. The result is a best linear predictor; it is not necessarily the exact conditional expectation of the next observation.

Definition: Let $\Theta$ denote a risk’s unobserved risk parameter, and let $X_1,\ldots,X_n,X_{n+1}$ be observations for that risk. Assume observations are conditionally independent and identically distributed given $\Theta$, and have finite second moments. Define the collective mean as $\mu=E[X_j]$, the expected process variance as $\mathrm{EPV}=E[\operatorname{Var}(X_j\mid\Theta)]$, and the variance of hypothetical means as $\mathrm{VHM}=\operatorname{Var}(E[X_j\mid\Theta])$. The experience mean is $\bar X=(X_1+\cdots+X_n)/n$.

The EPV describes expected variation among observations for the same risk. The VHM describes variation in the conditional means from one risk to another. Both are nonnegative. For the standard credibility factor $Z=n/(n+\mathrm{EPV}/\mathrm{VHM})$, assume $\mathrm{VHM}>0$. The observations must also be comparable: each one represents the same exposure and period basis.

Deriving the Weight

Consider predicting the next observation with a linear expression $a+b\bar X$, where $a$ and $b$ are constants chosen to minimize mean squared prediction error. The future observation and the experience mean both have expectation $\mu$. The law of total variance gives $\operatorname{Var}(X_j)=\mathrm{EPV}+\mathrm{VHM}$.

Calculate the variance of the experience mean.
Conditional on $\Theta$, the $n$ observations are independent with the same variance. Thus the conditional variance of their mean is $\operatorname{Var}(X_j\mid\Theta)/n$. Applying the law of total variance to $\bar X$ gives $\operatorname{Var}(\bar X)=\mathrm{VHM}+\mathrm{EPV}/n$.

Calculate its covariance with the future observation.
For each past observation, conditional independence makes its conditional covariance with $X_{n+1}$ zero. Their unconditional covariance is therefore the covariance of their conditional means, which is $\mathrm{VHM}$. Averaging the $n$ covariances gives $\operatorname{Cov}(X_{n+1},\bar X)=\mathrm{VHM}$.

Choose the best slope.
For a fixed slope $b$, the prediction error has mean $\mu-a-b\mu$. Its mean squared error is minimized over $a$ when this mean is zero, so $a=(1-b)\mu$. The remaining error variance is minimized when $b=\operatorname{Cov}(X_{n+1},\bar X)/\operatorname{Var}(\bar X)$.

Substituting the covariance and variance from Steps 1 and 2 yields the credibility factor. The intercept makes the predictor a weighted average of the experience mean and the collective mean.

Z=b =\frac{\mathrm{VHM}}{\mathrm{VHM}+\mathrm{EPV}/n} =\frac{n\,\mathrm{VHM}}{n\,\mathrm{VHM}+\mathrm{EPV}} =\frac{n}{n+\mathrm{EPV}/\mathrm{VHM}}, \qquad a=(1-Z)\mu.

Theorem: Under the equal-exposure Bühlmann assumptions above, the best linear predictor of the next observation is the Bühlmann premium $$ Z\bar X+(1-Z)\mu, \qquad Z=\frac{n}{n+K}, \qquad K=\frac{\mathrm{EPV}}{\mathrm{VHM}}, $$ provided $\mathrm{VHM}>0$. Here $K$ is the credibility constant, measured in observation units. The experience weight increases with $n$ and $\mathrm{VHM}$, and decreases with $\mathrm{EPV}$.

The derivation explains the ratio: increasing the number of observations reduces the process-variance contribution to the variance of $\bar X$ from $\mathrm{EPV}$ to $\mathrm{EPV}/n$. The between-risk component $\mathrm{VHM}$ remains. Consequently, the experience mean becomes a more reliable indicator of the risk’s underlying mean as $n$ grows.

Applying the Result

For equal-exposure observations, a practical calculation has three parts: determine the collective mean and structural variances, calculate $K$ and $Z$, then blend the observed mean with the collective mean. The examples use hypothetical models; no interest or mortality adjustment is involved.

Worked Example: A Normal Model With Known Structural Variances

A hypothetical portfolio models each risk’s conditional mean $M$ in dollars per period with $E[M]=500$ and $\operatorname{Var}(M)=300$ dollars squared. Given $M$, each period’s loss $X_j$ is Normal with mean $M$ and variance $900$ dollars squared, independently across periods. For one risk, the three observed losses are $500, $560, and $620. Find the Bühlmann premium. There is no discounting or mortality assumption.

Method. The conditional variance is 900 for every risk, so $\mathrm{EPV}=900$. The conditional mean is $M$, so $\mathrm{VHM}=\operatorname{Var}(M)=300$. The collective mean is $\mu=E[M]=500$.

The number of observations is $n=3$. Therefore $K=900/300=3$, and $Z=3/(3+3)=1/2$. The experience mean is $\bar X=(500+560+620)/3=1{,}680/3=560$.

The premium is $Z\bar X+(1-Z)\mu=(1/2)(560)+(1/2)(500)=280+250=\$530.00$.

Check. The premium is halfway between $560 and $500, as required by $Z=1/2$. The common mistake this method avoids is treating the observed mean as fully credible despite having only three observations.

Worked Example: Poisson Claim Counts With Two Risk Types

In a hypothetical book, each risk has a claim count $X_j$ per fixed one-year period. Its unknown claim rate $\Lambda$ is 2 or 6 claims per year, each with probability $1/2$. Given $\Lambda$, annual counts are independent Poisson random variables with mean $\Lambda$. One risk has annual counts 3, 4, 5, 6, 5, and 7 over six years. Calculate its premium for the next year. No interest or mortality assumption applies.

Method. For a Poisson variable with mean $\Lambda$, the conditional mean and variance are both $\Lambda$. Thus $\mu=E[\Lambda]=(2+6)/2=4$, and $\mathrm{EPV}=E[\operatorname{Var}(X_j\mid\Lambda)]=E[\Lambda]=4$.

The VHM is the variance of $\Lambda$. Its deviations from the mean 4 are $-2$ and $2$, so $\mathrm{VHM}=\bigl((-2)^2+2^2\bigr)/2=4$. Consequently $K=4/4=1$, and for $n=6$, $Z=6/(6+1)=6/7$.

The observed total is $3+4+5+6+5+7=30$ claims, so the experience mean is $30/6=5$ claims per year. The premium is $(6/7)(5)+(1/7)(4)=30/7+4/7=34/7=4.8571$ claims per year, to four decimal places.

Check. Since the observed mean exceeds the collective mean, the premium lies between 4 and 5; 4.8571 does. The common mistake this method avoids is using the Poisson variance at an averaged rate to calculate VHM: EPV is 4, while VHM is separately calculated as $\operatorname{Var}(\Lambda)=4$.

Worked Example: Meeting a Credibility Target and Calculating the Premium

A hypothetical model for a period’s net cost has risk-specific mean $M$, where $M$ is $80 or $120 with equal probability. Given $M$, each period’s cost is Normal with mean $M$ and variance 1,200 dollars squared, independently across periods. The Normal model is an approximation for net cost, not a model restricted to nonnegative individual losses. A risk has total cost $1,320 over 12 equal-exposure periods. Determine the smallest observation count needed for $Z$ to reach at least 0.7500, then calculate the Bühlmann premium using the 12 observations. No discounting or mortality assumption applies.

Method. The collective mean is $\mu=E[M]=(80+120)/2=\$100$. The conditional variance is 1,200 dollars squared, so $\mathrm{EPV}=1{,}200$. The risk-mean deviations from $100 are $-20$ and $20$, giving $\mathrm{VHM}=((-20)^2+20^2)/2=400$ dollars squared. Thus $K=1{,}200/400=3$.

For a target $Z\geq0.7500$, solve $n/(n+3)\geq3/4$. Since $n+3>0$, this is equivalent to $4n\geq3n+9$, or $n\geq9$. The smallest integer count is 9 periods.

For the premium based on the 12 available observations, $\bar X=1{,}320/12=\$110$ and $Z=12/(12+3)=0.8$. Hence the premium is $(0.8)(\$110)+(0.2)(\$100)=\$88+\$20=\$108.00$.

Check. At 9 periods, $Z=9/(9+3)=0.7500$, while at 8 periods $Z=8/11$, which is less than 0.7500. With 12 periods, the premium is $8 above the collective mean, which is $0.8$ times the experience mean’s $10 excess. This avoids confusing the count that meets the target with the count actually used in the premium.

Common Weighting Errors and Limits

The denominator $n+K$ does not make $K$ the weight on experience. It makes $K$ the amount of credibility-constant exposure needed for the experience to receive half weight: if $n=K$, then $Z=K/(K+K)=1/2$. If $n$ is much larger than $K$, $Z$ approaches 1; if $n$ is small relative to $K$, $Z$ is closer to 0.

A useful check is to verify that $0\leq Z\leq1$, and that the premium lies between $\bar X$ and $\mu$. If $Z$ is close to 1, the premium should be close to the experience mean. If $Z$ is close to 0, it should be close to the collective mean.

Reversing the weights. The factor $n/(n+K)$ weights $\bar X$; its complement weights $\mu$.
Using the wrong variance component. EPV is expected conditional variance; VHM is variance of conditional means. Their ratio determines $K$.
Mixing observation units. The $n$ in this formula counts comparable, equal-exposure observations. It is not automatically a count of years if period exposures differ.
Claiming an exact posterior mean. Bühlmann credibility gives the best linear predictor under the stated model. The exact conditional expectation may be nonlinear in the observations.

Check Your Understanding

For each problem, use the equal-exposure Bühlmann assumptions stated in the question.

A model has $\mathrm{EPV}=200$, $\mathrm{VHM}=50$, and $n=4$. Calculate $K$ and $Z$. (Answer: $K=4$, $Z=0.5000$.)
A collective mean is 80, with $\mathrm{EPV}=600$ and $\mathrm{VHM}=200$. A risk’s six-observation mean is 95. Calculate its premium. (Answer: $Z=2/3$, premium $=\$90.00$.)
For a hypothetical Poisson model, the annual rate $\Lambda$ is 1 or 5 with equal probability. Find EPV and VHM. If a risk has five annual counts totaling 21, calculate its next-year premium. (Answer: $\mathrm{EPV}=3$, $\mathrm{VHM}=4$, $Z=20/23$, premium $=93/23=4.0435$ claims.)
If $\mathrm{EPV}=500$ and $\mathrm{VHM}=100$, what is the smallest integer $n$ for which $Z\geq0.8000$? (Answer: $n=20$.)
Suppose $\mathrm{VHM}=0$, with finite EPV and at least one observation. Explain the implication for a best linear predictor. (Answer: There is no between-risk variation to learn; the best linear weight on experience is 0 and the predictor is the collective mean $\mu$. The standard expression $n/(n+\mathrm{EPV}/\mathrm{VHM})$ is not used when $\mathrm{VHM}=0$.)

Key takeaway: With equal-exposure, conditionally independent observations and $\mathrm{VHM}>0$, the credibility factor is $Z=n/(n+\mathrm{EPV}/\mathrm{VHM})$. It follows from the covariance of future experience with the observed mean relative to that mean’s variance, and it assigns $Z$ to individual experience and $1-Z$ to the collective mean.

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