Estimating EPV From Within-Risk Variation

Why Estimate Within-Risk Variation?

In the Bühlmann-Straub model, observed experience varies for two distinct reasons: risks can have different underlying expected outcomes, and observations for a given risk can fluctuate around that risk’s expectation. The EPV describes the second source: the expected conditional variance after accounting for the amount of exposure. Estimating it from repeated observations means comparing observations within the same risk, not comparing different risks’ averages.

This tutorial develops the pooled within-risk estimator when observation exposures are unequal. The distinction matters because a rate based on more exposure is assumed to be less variable. The estimator below is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: For risk $i$, let $\Theta_i$ denote its risk-specific parameter. Let $X_{ij}$ be its observed outcome in period $j$, and let $w_{ij}>0$ be the exposure for that observation. Under the Bühlmann-Straub model, observations are conditionally independent given $\Theta_i$, with $$ E[X_{ij}\mid\Theta_i]=\mu(\Theta_i), \qquad \operatorname{Var}(X_{ij}\mid\Theta_i)=\frac{\phi(\Theta_i)}{w_{ij}}. $$ The EPV is $E[\phi(\Theta)]$, the expected conditional variance per unit of exposure. The model requires the stated inverse-exposure variance relationship; if the observations do not satisfy it, this estimator need not estimate the intended EPV.

For each risk with $n_i$ observed periods, its exposure-weighted mean is

\bar X_i=\frac{\sum_{j=1}^{n_i}w_{ij}X_{ij}}{\sum_{j=1}^{n_i}w_{ij}}.

The within-risk weighted sum of squares is

Q_i=\sum_{j=1}^{n_i}w_{ij}(X_{ij}-\bar X_i)^2.

Each risk with $n_i$ observations contributes $n_i-1$ degrees of freedom. Pooling across risks gives the estimator

\widehat{\operatorname{EPV}} = \frac{\displaystyle\sum_i Q_i} {\displaystyle\sum_i(n_i-1)} = \frac{\displaystyle\sum_i\sum_{j=1}^{n_i}w_{ij}(X_{ij}-\bar X_i)^2} {\displaystyle\sum_i(n_i-1)}.

A risk with one observation has no within-risk comparison: it contributes zero to the numerator and zero to the denominator. The estimator cannot be calculated if every risk has at most one observation, because there are no within-risk degrees of freedom.

Why the Estimator Uses This Denominator

The weighted mean removes the risk’s unknown conditional mean. The remaining weighted deviations estimate its conditional variance factor. The following justification shows why the expected within-risk sum of squares is $(n_i-1)\phi(\Theta_i)$, even when exposures differ.

Center on the conditional mean.
For a fixed risk $i$, write $\mu_i=\mu(\Theta_i)$, $\phi_i=\phi(\Theta_i)$, and $W_i=\sum_jw_{ij}$. Conditional independence and the model give $E[X_{ij}-\mu_i\mid\Theta_i]=0$ and $\operatorname{Var}(X_{ij}\mid\Theta_i)=\phi_i/w_{ij}$.

Use the weighted residual identity.
Expanding around the weighted mean gives $Q_i=\sum_j w_{ij}(X_{ij}-\mu_i)^2-W_i(\bar X_i-\mu_i)^2$. This follows by expanding each square and using $\sum_jw_{ij}(X_{ij}-\bar X_i)=0$.

Take the conditional expectation.
The first term has expectation $\sum_j w_{ij}(\phi_i/w_{ij})=n_i\phi_i$. Also, $\operatorname{Var}(\bar X_i\mid\Theta_i)=\sum_j(w_{ij}/W_i)^2(\phi_i/w_{ij})=\phi_i/W_i$, so the second term has expectation $W_i(\phi_i/W_i)=\phi_i$.

Count the remaining degrees of freedom.
Subtracting the two expectations gives $E[Q_i\mid\Theta_i]=(n_i-1)\phi_i$. Across representative sampled risks, dividing the pooled total by $\sum_i(n_i-1)$ therefore estimates the average conditional variance factor, $E[\phi(\Theta)]$.

This argument also explains the exposure weights: multiplying a squared deviation by $w_{ij}$ offsets its modeled variance reduction by exposure. Simply using unweighted squared deviations would not have the same expectation when exposures differ.

Worked Calculations

Worked Example: Pooling Two Risks With Unequal Exposures

A hypothetical portfolio has two risks. For risk A, the observed rates are 10, 14, and 12, with exposures 1, 2, and 1. For risk B, the rates are 8 and 12, with exposures 1 and 3. Under the Bühlmann-Straub inverse-exposure variance model, estimate the EPV. Give the result to four decimal places.

Method. Calculate each risk’s exposure-weighted mean and weighted within-risk sum of squares, then divide their total by the pooled within-risk degrees of freedom.

For risk A, total exposure is $1+2+1=4$, and $$ \bar X_A=\frac{1(10)+2(14)+1(12)}{4}=\frac{50}{4}=12.5. $$ Thus $$ Q_A=1(10-12.5)^2+2(14-12.5)^2+1(12-12.5)^2 =6.25+4.50+0.25=11. $$ Risk A contributes $3-1=2$ degrees of freedom.

For risk B, total exposure is $1+3=4$, and $$ \bar X_B=\frac{1(8)+3(12)}{4}=\frac{44}{4}=11. $$ Therefore $$ Q_B=1(8-11)^2+3(12-11)^2=9+3=12. $$ Risk B contributes $2-1=1$ degree of freedom. The pooled estimate is $$ \widehat{\operatorname{EPV}}=\frac{11+12}{2+1}=\frac{23}{3}=7.6667. $$ Check. The pooled numerator is 23 and the pooled degrees of freedom are 3; their ratio is 7.6667. The estimate is in squared rate units per exposure unit. The common mistake avoided here is replacing either exposure-weighted mean with an ordinary arithmetic mean.

Worked Example: Unequal Numbers of Observations

For a second hypothetical portfolio, risk A has rates 4 and 8, each with exposure 1. Risk B has rates 10, 12, and 8, with exposures 1, 3, and 2. Risk C has just one observed rate, 20, with exposure 4. Estimate the EPV to four decimal places.

Method. Pool all within-risk sums of squares and all within-risk degrees of freedom. Do not give a single-observation risk a within-risk variance contribution.

For A, $\bar X_A=(4+8)/2=6$, so $$ Q_A=1(4-6)^2+1(8-6)^2=4+4=8, \qquad n_A-1=1. $$ For B, total exposure is $1+3+2=6$, and $$ \bar X_B=\frac{1(10)+3(12)+2(8)}{6}=\frac{62}{6}=\frac{31}{3}. $$ The weighted sum of squares is $$ Q_B=1\left(10-\frac{31}{3}\right)^2 +3\left(12-\frac{31}{3}\right)^2 +2\left(8-\frac{31}{3}\right)^2 =\frac{1+75+98}{9}=\frac{58}{3}. $$ Risk B contributes $3-1=2$ degrees of freedom. Risk C has $n_C-1=0$, so it contributes neither numerator nor denominator. Hence $$ \widehat{\operatorname{EPV}} =\frac{8+\frac{58}{3}}{1+2} =\frac{\frac{82}{3}}{3} =\frac{82}{9}=9.1111. $$ Check. The estimator pools $3$ degrees of freedom, not $4$ observations minus one overall. Risk C supplies no within-risk information, despite having a recorded rate. The common mistake avoided is treating one observation as if it could estimate variation around its own risk mean.

Worked Example: Separating Exposure Weighting From Ordinary Averaging

A hypothetical insurer records rates for two risks. Risk A has rates 6 and 10, with exposures 2 and 2. Risk B has rates 14, 16, and 18, each with exposure 2. Estimate the EPV to four decimal places and explain what would go wrong if the numerator were left unweighted.

Method. Apply the specified exposure weights to both the means and the squared deviations. Here the weights happen to be equal within each risk, so the weighted mean equals the ordinary mean, but the weights still scale the sums of squares.

For risk A, $\bar X_A=(2(6)+2(10))/4=8$. Thus $$ Q_A=2(6-8)^2+2(10-8)^2=8+8=16, \qquad n_A-1=1. $$ For risk B, $\bar X_B=(14+16+18)/3=16$. Its weighted sum of squares is $$ Q_B=2(14-16)^2+2(16-16)^2+2(18-16)^2 =8+0+8=16, \qquad n_B-1=2. $$ Therefore $$ \widehat{\operatorname{EPV}}=\frac{16+16}{1+2}=\frac{32}{3}=10.6667. $$ Check. If the exposures were incorrectly omitted, the unweighted sums of squares would be $8$ for A and $8$ for B, producing $16/3=5.3333$. That is not the model-based estimate: the data specify exposure 2 for every observation, so the weighted numerator is twice as large. The common mistake avoided is silently changing the exposure scale while retaining the same interpretation of EPV.

Common Traps and Interpretation

The estimator is a pooled estimate of within-risk conditional variance per exposure unit. It is not the variance of all observed outcomes around one portfolio-wide mean. A portfolio-wide calculation mixes variation around each risk’s own mean with differences among the risks’ means, so it does not isolate EPV.

A frequent denominator error is to subtract one only once from the total number of observations. Each risk has its own estimated mean, so each risk uses one degree of freedom. The correct denominator is $\sum_i(n_i-1)=N-R$ when all $R$ risks have at least one observation, not $N-1$. For the first worked example, using $23/(5-1)=5.75$ is wrong; the two separate risk means leave $2+1=3$ degrees of freedom, giving $23/3=7.6667$.

Before using the result, check the model’s exposure convention. If exposures are measured in policy-years, the EPV is per policy-year under the assumed variance relationship. Rescaling all exposure weights changes the numerical variance-per-unit scale correspondingly. A formula that ignores the declared exposure units can yield a numerically plausible but incorrectly interpreted answer.

Key takeaway: For each risk, center observations at their exposure-weighted mean, multiply squared deviations by their exposures, and pool the resulting sums of squares over $\sum_i(n_i-1)$. The estimator uses repeated observations within risks; a risk with only one observation supplies no within-risk degrees of freedom.

Check Your Understanding

For each problem, use the Bühlmann-Straub inverse-exposure variance model and the pooled within-risk estimator. Round estimates to four decimal places where needed.

A risk has two rates, 5 and 9, each with exposure 1. Find its contribution to the pooled EPV numerator and its degrees of freedom. (Answer: numerator 8; 1 degree of freedom.)
Risk A has rates 2 and 6 with exposures 1 and 3. Risk B has rates 10 and 14 with exposure 2 each. Find the pooled EPV estimate. (Answer: 14.0000.)
Risk A has rates 3, 7, and 5 with exposures 1, 2, and 1. A second risk has one observation, rate 9, with exposure 5. Find the pooled estimate and state the degrees of freedom. (Answer: 5.5000; 2 degrees of freedom.)
Risk A has rates 4 and 8, with exposures 2 and 2. Risk B has rates 10 and 14, with exposures 1 and 3. Find the pooled EPV estimate, using each risk’s weighted mean. (Answer: 12.0000.)

What You'll Learn