Predictive Variance of a Credibility-Priced Outcome

Why a Credibility Premium Needs a Prediction Variance

A credibility premium is a prediction of a future outcome, not a guarantee that the outcome will equal the premium. Even if the premium is well calculated, a future observation can differ because the risk has its own random process variation and because its underlying mean is not known exactly. Separating these sources of uncertainty makes a prediction more useful: it shows how much variation remains from the future period itself and how much comes from estimating the risk-specific mean.

This tutorial uses the Bühlmann-Straub model with exposure weights. The target is a future per-exposure observation. The variance of that observation depends on its exposure, while uncertainty about the risk-specific mean depends on how much experience is available to estimate it.

Definition: For risk $i$, let $X_{ij}$ be the observed per-exposure outcome in period $j$, with exposure $w_{ij}>0$. Given the risk-specific mean $\mu_i$ and process variance $\sigma_i^2$, assume $$ E[X_{ij}\mid\mu_i,\sigma_i^2]=\mu_i,\qquad \operatorname{Var}(X_{ij}\mid\mu_i,\sigma_i^2)=\frac{\sigma_i^2}{w_{ij}}. $$ Observations from different periods are conditionally independent given the risk parameters. Across risks, define $m=E[\mu_i]$, $v=\operatorname{Var}(\mu_i)$, and $s=E[\sigma_i^2]$. Here $v$ is the variance of hypothetical risk-specific means, and $s$ is the expected process variance for one unit of exposure.

For one risk, write its total historical exposure as $W=\sum_j w_j$, and its exposure-weighted experience mean as $\bar X=\sum_j w_jX_j/W$. When $s>0$ and $v>0$, the Bühlmann-Straub credibility constant is $K=s/v$, the credibility factor is $Z=W/(W+K)$, and the credibility premium for the future per-exposure mean is

\widehat{\mu}=Z\bar X+(1-Z)m.

If the future period has exposure $w_0$, denote its per-exposure observation by $X_0$. The conditional variance of $X_0$ is $\sigma_i^2/w_0$. Thus larger future exposure reduces process variation in the per-exposure outcome, but it does not by itself add historical information to the premium.

Prediction-Error Variance and Its Components

The prediction error is $X_0-\widehat{\mu}$. Its variance, taken over the hypothetical population of risks, historical observations, and the future observation, is the prediction-error variance. Under the model above, it separates into a future process component and an estimation component:

\operatorname{Var}(X_0-\widehat{\mu}) = \underbrace{\frac{s}{w_0}}_{\text{future process risk}} + \underbrace{v(1-Z)}_{\text{uncertainty about the risk-specific mean}}.

The first term is the average conditional variance of the future observation. The second is the mean squared error in using the credibility premium to estimate $\mu_i$. Both terms are needed: a premium can estimate the risk-specific mean accurately while the future observation still has substantial process variation.

Condition on the risk parameters and history.
The future observation has conditional mean $\mu_i$ and variance $\sigma_i^2/w_0$. Its conditional independence from the history means the historical data do not change this conditional process variance.

Average the future process variance.
Taking the expectation over risks gives $E[\sigma_i^2/w_0]=s/w_0$.

Calculate uncertainty in the premium’s estimate of the mean.
The premium combines the portfolio mean and the experience mean. Its mean squared difference from $\mu_i$ is $v(1-Z)$, as derived below.

Add the components.
The future process error has conditional mean zero, so it has zero covariance with the error $\mu_i-\widehat{\mu}$. The two variance contributions therefore sum.

To verify the estimation component, condition on the risk parameters. The weighted experience mean has conditional mean $\mu_i$ and conditional variance $\sigma_i^2/W$, because its observations are conditionally independent and have variances $\sigma_i^2/w_j$. Hence $E[(\bar X-\mu_i)^2]=s/W$. Also, $\bar X-\mu_i$ has conditional mean zero, so its covariance with $\mu_i-m$ is zero. Expanding the squared premium error gives

\begin{aligned} E[(\widehat{\mu}-\mu_i)^2] &=E\left[\left\{Z(\bar X-\mu_i)-(1-Z)(\mu_i-m)\right\}^2\right]\\ &=Z^2\frac{s}{W}+(1-Z)^2v. \end{aligned}

Since $Z=W/(W+K)$ and $K=s/v$, we have $s/W=v(1-Z)/Z$. Substituting this identity gives

Z^2\frac{s}{W}+(1-Z)^2v =Zv(1-Z)+(1-Z)^2v =v(1-Z).

Finally, write $X_0-\widehat{\mu}=(X_0-\mu_i)+(\mu_i-\widehat{\mu})$. The first term has conditional mean zero given the risk parameters and history. Its covariance with the second term is therefore zero, and its expected conditional variance is $s/w_0$. This proves the stated decomposition.

Worked Applications

Worked Example: A Future Period With More Than One Exposure Unit

A hypothetical portfolio has $m=\$500$, $s=40{,}000$ dollars squared per exposure unit, and $v=10{,}000$ dollars squared. One risk has total historical exposure $W=4$ and exposure-weighted experience mean $\bar X=\$560$. Find its credibility premium and the prediction-error variance for a future period with exposure $w_0=2.5$.

Method and credibility premium. First calculate $K=s/v=40{,}000/10{,}000=4$. Thus $Z=W/(W+K)=4/(4+4)=0.5$, and

\widehat{\mu}=0.5(560)+0.5(500)=280+250=\$530.

Variance components. Future process risk is $s/w_0=40{,}000/2.5=16{,}000$ dollars squared. Uncertainty about the risk-specific mean is $v(1-Z)=10{,}000(0.5)=5{,}000$ dollars squared. Therefore,

\operatorname{Var}(X_0-\widehat{\mu})=16{,}000+5{,}000=21{,}000\text{ dollars squared}.

The prediction-error standard deviation is $\sqrt{21{,}000}=\$144.91$, rounded to the nearest cent. Check: the future process term is lower than $s$ because the future exposure exceeds one unit. The method avoids using $s$ without adjusting for $w_0$.

Worked Example: Unequal Historical Exposures

A hypothetical risk has three observed per-exposure outcomes: $X_1=12$ at exposure $w_1=1$, $X_2=15$ at $w_2=2$, and $X_3=16$ at $w_3=3$. Let $m=12$, $s=900$, and $v=300$. Find the credibility premium and prediction-error variance for a future period with exposure $w_0=3$.

Weighted experience mean. Total exposure is $W=1+2+3=6$. The exposure-weighted total is $1(12)+2(15)+3(16)=12+30+48=90$, so $\bar X=90/6=15$.

Credibility premium. The credibility constant is $K=900/300=3$, and $Z=6/(6+3)=2/3$. Therefore,

\widehat{\mu}=\frac{2}{3}(15)+\frac{1}{3}(12)=10+4=14.

Prediction-error variance. Future process risk is $s/w_0=900/3=300$. Mean uncertainty is $v(1-Z)=300(1/3)=100$. The total is $300+100=400$ squared units of outcome per exposure. Check: the history contributes total exposure $6$, not three periods; the unequal weights must be used in both $W$ and $\bar X$. This avoids treating each period as if it had the same exposure.

Worked Example: Prediction Error Is Not the Marginal Variance

Suppose $m=100$, $s=16$, and $v=9$. A risk has four equal-exposure observations $90,100,110,100$, each with exposure 1. Find the credibility premium, the prediction-error variance for a future exposure of 1, and the marginal variance of the future observation.

The total historical exposure is $W=4$, and the historical mean is $(90+100+110+100)/4=100$. The credibility constant is $K=16/9$, so

Z=\frac{4}{4+16/9}=\frac{4}{52/9}=\frac{9}{13}, \qquad \widehat{\mu}=\frac{9}{13}(100)+\frac{4}{13}(100)=100.

Future process risk is $s/w_0=16/1=16$. Mean uncertainty is $v(1-Z)=9(4/13)=36/13$. Thus the prediction-error variance is $16+36/13=244/13\approx18.77$, rounded to two decimal places.

The marginal variance of the future observation itself is different: by the law of total variance, it is $E[\operatorname{Var}(X_0\mid\mu_i,\sigma_i^2)]+\operatorname{Var}(E[X_0\mid\mu_i,\sigma_i^2])=s+v=16+9=25$. The value $18.77$ is the variance of the future outcome’s error around a credibility premium that varies with historical experience; $25$ is the marginal variance of the future outcome around the portfolio mean. Check: these quantities answer different questions, so they need not agree.

Interpretation and Exam Traps

The term $s/w_0$ is process risk for the future exposure period. The term $v(1-Z)$ reflects uncertainty about the risk-specific mean after using experience with credibility $Z$. More historical exposure increases $Z$ and reduces this estimation component; more future exposure reduces the process component. Neither effect should be substituted for the other.

The result is an unconditional prediction-error variance under the Bühlmann-Straub assumptions and known structural parameters. It is not automatically the exact conditional variance of $X_0$ given a particular observed history. Obtaining that conditional distribution generally requires a fully specified model for the risk parameters and observations; first and second moments alone do not determine it.

Using the wrong future exposure. For a per-exposure outcome, divide $s$ by the future exposure $w_0$. Do not multiply by $w_0$ or use a historical weight in its place.
Calling $s+v$ the prediction-error variance. That is the marginal variance of a one-unit future observation around $m$. Prediction error around a credibility premium also depends on the history through $v(1-Z)$.
Confusing the mean and the observation. The premium predicts the risk-specific mean. A future observation has additional process variation, represented by $s/w_0$.
Dropping the exposure weights. In the Bühlmann-Straub model, both $W$ and the historical mean are exposure-weighted. A simple average is valid only when the historical exposures are equal.

Key takeaway: Under the Bühlmann-Straub model, prediction-error variance for a future per-exposure outcome is $s/w_0+v(1-Z)$: average future process variance plus uncertainty remaining in the credibility estimate of the risk-specific mean.

Check Your Understanding

Use the Bühlmann-Straub assumptions stated above. For each problem, find the requested credibility premium and/or prediction-error variance.

A risk has $s=64$, $v=16$, historical total exposure $W=4$, and future exposure $w_0=2$. Its experience mean is 30 and $m=26$. Find the credibility premium and prediction-error variance. (Answer: $Z=0.5$, premium $=28$, variance $=32+8=40$.)
Let $s=100$, $v=100$, $W=5$, $w_0=1$, $m=20$, and $\bar X=24$. Find the premium and prediction-error variance. (Answer: $Z=5/6$, premium $=70/3\approx23.33$, variance $=100+100/6\approx116.67$.)
Historical per-exposure outcomes are 8 at exposure 2 and 14 at exposure 4. Let $m=10$, $s=48$, $v=12$, and $w_0=3$. Find the weighted mean and prediction-error variance. (Answer: $\bar X=12$, $Z=6/(6+4)=0.6$, variance $=16+4.8=20.8$.)
For a model with $s=25$, $v=9$, $W=3$, and future exposure $w_0=1$, compare the marginal variance of the future observation with its prediction-error variance. (Answer: $K=25/9$, $Z=27/52$; marginal variance $=34$, prediction-error variance $=25+9(25/52)=29.3269$, approximately.)
For $s=80$, $v=20$, $W=8$, and future exposure $w_0=4$, identify each component of prediction-error variance and their total. (Answer: $K=4$, $Z=2/3$; process risk $=20$, mean uncertainty $=20/3$, total $=80/3\approx26.67$.)

What You'll Learn