Variance Of An Unequal-Exposure Risk Mean

Why the Variance of a Risk Mean Depends on Exposure

A risk’s experience may be summarized by an exposure-weighted mean of its period-level loss costs. That mean gives more influence to periods with more exposure, but its sampling variance is not found by simply averaging the period variances. The weights affect both the numerator and the denominator, so they must be carried through the variance calculation.

The key result under the standard Bühlmann-Straub assumptions is that the conditional variance of the weighted mean is inversely proportional to total exposure. This result is useful only when the model assumptions hold. With arbitrary period variances, or with dependence between periods, the period-level variances and covariances must be used directly.

Definition: For risk $i$, let $X_{ij}$ be the observed loss cost in period $j$, and let $w_{ij}>0$ be the exposure for that risk-period. The total exposure is $W_i=\sum_j w_{ij}$, and the exposure-weighted risk mean is $$ \bar X_i=\frac{\sum_j w_{ij}X_{ij}}{W_i}. $$ A period-level process variance is $\operatorname{Var}(X_{ij}\mid\Theta_i)$, where $\Theta_i$ represents the risk’s underlying characteristics. It describes period-to-period variation after conditioning on the risk’s underlying level.

For the variance of $\bar X_i$ to be a process variance conditional on $\Theta_i$, assume the exposures are fixed and positive, and condition on the same risk parameter for all periods. In the standard model, the conditional mean is constant across periods, and different periods are conditionally independent. The independence assumption is important: it makes the variance of a weighted sum equal to the sum of the weighted variances.

General Variance Formula

First consider a general set of conditional variances $v_{ij}=\operatorname{Var}(X_{ij}\mid\Theta_i)$, without yet imposing a relationship between variance and exposure. If the periods are conditionally independent, then

\operatorname{Var}(\bar X_i\mid\Theta_i) = \frac{\sum_j w_{ij}^2v_{ij}}{W_i^2}.

The square on each weight follows from the rule $\operatorname{Var}(aX)=a^2\operatorname{Var}(X)$. The denominator is also squared because the entire weighted sum is divided by the fixed total exposure. Thus, large-exposure periods may have large weights in the mean, but their contribution to its variance depends on the square of that weight as well as on their own process variance.

If periods are not conditionally independent, covariance terms must also be included. Writing $c_{ijk}=\operatorname{Cov}(X_{ij},X_{ik}\mid\Theta_i)$ for $j\ne k$, the general expression is

\operatorname{Var}(\bar X_i\mid\Theta_i) = \frac{\sum_j w_{ij}^2v_{ij} +\sum_{j\ne k}w_{ij}w_{ik}c_{ijk}}{W_i^2}.

The second sum is over ordered pairs, so each distinct pair contributes twice when its covariance is symmetric. Equivalently, sum once over unordered pairs and multiply each covariance contribution by $2$. Positive covariance increases the variance of the mean; negative covariance reduces it. The covariance-free formula does not apply when such dependence is present.

Worked Example: Combining Unequal Period Variances

A hypothetical commercial account has three conditionally independent periods with exposures $1$, $2$, and $5$ units. Given the account’s risk parameter, the period-level loss-cost variances are $64$, $32$, and $12.8$ dollars squared, respectively. Find the conditional variance of its exposure-weighted mean.

Method. Use the general independent-period formula. The total exposure is $W=1+2+5=8$.

The weighted-variance contributions in the numerator are $1^2(64)=64$, $2^2(32)=128$, and $5^2(12.8)=320$. Their sum is $64+128+320=512$ dollars squared times exposure squared. The squared denominator is $8^2=64$.

\operatorname{Var}(\bar X\mid\Theta) =\frac{512}{64} =8\text{ dollars}^2.

Check. Each period variance equals $64$ divided by its exposure: $64/1=64$, $64/2=32$, and $64/5=12.8$. Therefore the inverse-exposure model applies with process variance parameter $64$, and that model gives $64/8=8$ dollars squared, matching the direct calculation. The common mistake is to report $4$ dollars squared; squaring the exposure weights and dividing by $W^2$ gives $8$, not $4$.

Why Total Exposure Determines Variance in the Standard Model

The standard Bühlmann-Straub process-variance assumption specifies that, conditional on $\Theta_i$, the period-level observations have a common conditional mean and a variance inversely proportional to exposure:

E[X_{ij}\mid\Theta_i]=\mu(\Theta_i), \qquad \operatorname{Var}(X_{ij}\mid\Theta_i)=\frac{\sigma^2(\Theta_i)}{w_{ij}}.

Here $\sigma^2(\Theta_i)$ is the risk-specific process-variance parameter on the chosen exposure scale. Periods are assumed conditionally independent. This model is appropriate when each additional unit of exposure contributes information in the same way, so a period with twice the exposure has half the conditional variance. It does not say that every unequal-exposure problem automatically has variance $\sigma^2/W_i$; the inverse-exposure condition must be checked.

Substitute the model variance for $v_{ij}$ in the general formula. Each term in the numerator simplifies because one factor of exposure cancels:

\begin{aligned} \operatorname{Var}(\bar X_i\mid\Theta_i) &=\frac{\sum_j w_{ij}^2\{\sigma^2(\Theta_i)/w_{ij}\}}{W_i^2}\\ &=\frac{\sigma^2(\Theta_i)\sum_j w_{ij}}{W_i^2}\\ &=\frac{\sigma^2(\Theta_i)}{W_i}. \end{aligned}

This derivation explains why total exposure, rather than the number of periods, controls the conditional sampling variance in the standard model. Splitting a fixed total exposure into more periods does not change the variance if the inverse-exposure assumption and conditional independence remain valid. Doubling total exposure halves the conditional variance.

The conditional mean of the weighted mean is also the risk-specific mean:

E[\bar X_i\mid\Theta_i] =\frac{\sum_jw_{ij}\mu(\Theta_i)}{W_i} =\mu(\Theta_i).

The equality uses the same conditional mean in every period. If the underlying mean changes systematically over time, or if the observations have different conditional means, the weighted average is still calculable, but it is not an unbiased summary of one common risk mean under this model.

Worked Example: Verifying the Inverse-Exposure Model

A hypothetical equipment insurer observes one risk over periods with exposures $2$, $3$, and $5$ units. Conditional on its risk parameter, the loss-cost variances are $90$, $60$, and $36$ dollars squared, and the periods are conditionally independent. Find the variance of the weighted mean, and determine whether total exposure alone can be used.

Check the model. Multiply each variance by its exposure: $2(90)=180$, $3(60)=180$, and $5(36)=180$. Thus each period variance is $180/w_j$, so the periods satisfy the inverse-exposure model with $\sigma^2(\Theta)=180$ dollars squared.

The total exposure is $W=2+3+5=10$. The model formula gives $\operatorname{Var}(\bar X\mid\Theta)=180/10=18$ dollars squared.

Direct check. The numerator of the general formula is $2^2(90)+3^2(60)+5^2(36)=360+540+900=1{,}800$. The denominator is $10^2=100$, so $1{,}800/100=18$ dollars squared. The common mistake is to treat the period variances as though they were averaged with weights $w_j/W$; the variance formula uses squared weights in the numerator.

Conditional and Unconditional Variance

The preceding formulas condition on the risk parameter. Across a population of risks, $\Theta_i$ varies, so the unconditional variance of the risk mean has a second source of variation: differences in the underlying risk means. The law of total variance states that

\operatorname{Var}(\bar X_i) = E[\operatorname{Var}(\bar X_i\mid\Theta_i)] +\operatorname{Var}(E[\bar X_i\mid\Theta_i]).

Under the standard Bühlmann-Straub assumptions, $E[\bar X_i\mid\Theta_i]=\mu(\Theta_i)$ and $\operatorname{Var}(\bar X_i\mid\Theta_i)=\sigma^2(\Theta_i)/W_i$. Therefore, for fixed total exposure $W_i$,

\operatorname{Var}(\bar X_i) = \operatorname{Var}(\mu(\Theta_i)) +\frac{E[\sigma^2(\Theta_i)]}{W_i}.

The first term is variation between risks in their underlying means. The second is expected within-risk process variance, reduced by total exposure. This distinction matters: more exposure reduces sampling noise around a risk’s own mean, but it does not remove genuine differences between risks. The variance decomposition is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Worked Example: Adding Between-Risk Variation

Suppose a hypothetical risk portfolio follows the standard model. The variance of the underlying risk means is $25$ dollars squared, and the expected process-variance parameter is $100$ dollars squared. A particular risk has total exposure $4$. Find the unconditional variance of its weighted mean.

Method. Apply the law of total variance. The expected conditional process variance is $100/4=25$ dollars squared. Add the between-risk variance of $25$ dollars squared: $25+25=50$ dollars squared.

\operatorname{Var}(\bar X_i)=50\text{ dollars}^2.

Check. The conditional variance component is $25$, while the variation in underlying risk means contributes another $25$. The result is $50$, not $25$: using only the sampling variance would omit between-risk variation. Conversely, reporting only the between-risk component would omit process uncertainty.

What Changes When Periods Are Dependent?

Conditional independence is a substantive assumption, not a consequence of unequal exposures. For example, a shared weather event, common operational disruption, or claims-handling change could cause loss costs in different periods to move together. When that happens, use the covariance formula rather than $\sigma^2/W$, unless the dependence has already been incorporated in a suitable model.

Worked Example: Including Conditional Covariance

A hypothetical risk has two periods, each with exposure $2$. Conditional on its risk parameter, each period has variance $300$ dollars squared, and the conditional covariance between the period observations is $120$ dollars squared. Find the variance of the exposure-weighted mean.

Method. Because the periods are dependent, include the two covariance contributions. The total exposure is $W=2+2=4$, so the denominator is $16$. The two variance terms sum to $2^2(300)+2^2(300)=1{,}200+1{,}200=2{,}400$. The covariance contribution is $2(2)(2)(120)=960$.

\operatorname{Var}(\bar X\mid\Theta) =\frac{2{,}400+960}{16} =\frac{3{,}360}{16} =210\text{ dollars}^2.

Check. The weighted sum has variance $3{,}360$ before division by the squared total exposure, and $3{,}360/16=210$. Omitting covariance would give $2{,}400/16=150$ dollars squared and understate the variance. The standard independent-period formula does not apply to these data.

Common Exam Traps and Practical Checks

Using the wrong power of a weight. In the variance numerator, each period contributes $w_{ij}^2v_{ij}$, not $w_{ij}v_{ij}$. The denominator is $W_i^2$.
Assuming all period variances are inversely proportional to exposure. Check whether $w_{ij}v_{ij}$ is constant across periods before using $\sigma^2/W_i$. Otherwise use the general formula.
Ignoring dependence. Conditional covariance terms matter whenever periods are not conditionally independent. Positive covariance raises the variance of the weighted mean.
Confusing conditional and unconditional variance. Conditional variance measures process fluctuation for a fixed risk parameter. Unconditional variance also includes variation between risks.
Mixing exposure units. Exposures must use a consistent unit, such as policy-years throughout. If the exposure scale changes, the corresponding process-variance parameter changes as well.

Key takeaway: With arbitrary conditionally independent period variances, calculate $\sum_j w_{ij}^2v_{ij}/W_i^2$. Only under the inverse-exposure process-variance model does this simplify to $\sigma^2(\Theta_i)/W_i$. Add covariance terms for dependent periods, and use the law of total variance when the target is unconditional variance across risks.

Check Your Understanding

For each problem, identify whether the inverse-exposure model is available before using a total-exposure shortcut.

A hypothetical risk has independent periods with exposures $1$ and $3$, and conditional variances $48$ and $16$ dollars squared. Find the conditional variance of its weighted mean and verify whether the inverse-exposure model applies. (Answer: $12$ dollars squared; yes, since $1(48)=3(16)=48$.)
A risk has independent periods with exposures $2$, $4$, and $6$, and conditional variances $75$, $37.5$, and $25$ dollars squared. Find the variance of its weighted mean using total exposure, after checking the model. (Answer: $12.5$ dollars squared; $\sigma^2=150$, $W=12$, and $150/12=12.5$.)
Three independent periods have exposures $1$, $2$, and $3$, with conditional variances $20$, $10$, and $8$ dollars squared. Calculate the variance of the weighted mean directly; do not assume inverse proportionality. (Answer: $132/36=11/3$, or $3.6667$ dollars squared.)
Two dependent periods have exposures $1$ and $2$, conditional variances $40$ and $10$ dollars squared, and conditional covariance $5$ dollars squared. Find the conditional variance of the weighted mean. (Answer: $(40+40+20)/9=100/9$, or $11.1111$ dollars squared.)
Under the standard model, a risk has total exposure $4$, expected process-variance parameter $100$ dollars squared, and between-risk variance of the underlying means $25$ dollars squared. Find the unconditional variance of its weighted mean. (Answer: $25+100/4=50$ dollars squared.)

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