EPV and VHM in Bühlmann Credibility

Why Separate the Two Sources of Variation?

Two insureds with different underlying risk characteristics may have different expected claim counts or average losses. At the same time, even insureds with the same characteristics can have different outcomes from one period to another. The Bühlmann model separates these sources of variability: variation within a risk class and variation between risk classes.

This distinction matters in credibility because repeated observations contain information about a risk, but individual observations also fluctuate randomly. The expected value of process variance (EPV) summarizes the first kind of variation; the variance of hypothetical means (VHM) summarizes the second. This tutorial defines and calculates those components without yet combining them into a credibility estimate. The treatment is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Definition: In the Bühlmann model, $\Theta$ is a latent risk parameter, or risk class, drawn from a population of risks. Given $\Theta$, the observations $X_1,\ldots,X_n$ are assumed to be conditionally independent and identically distributed, with finite conditional mean and variance: $$ E[X_j\mid\Theta]=\mu(\Theta), \qquad \operatorname{Var}(X_j\mid\Theta)=\sigma^2(\Theta). $$ The EPV is $v=E[\sigma^2(\Theta)]$, the population average of the conditional variances. The VHM is $a=\operatorname{Var}(\mu(\Theta))$, the population variance of the conditional means. The symbol $a$ here denotes a variance component, not an annuity.

The word “hypothetical” in VHM refers to the mean that would be associated with a risk if its latent parameter were known. The actual value of $\mu(\Theta)$ varies across risks in the population. EPV, by contrast, averages the random fluctuation that remains after conditioning on the risk parameter.

The moments are taken over the distribution of $\Theta$. Thus, EPV is not the conditional variance for one particular risk, and VHM is not the variance of individual observations within one class. If every risk has the same conditional mean, VHM is zero even when outcomes within each risk are highly variable. If every risk has a fixed outcome given its parameter, EPV is zero even when the risks have different means.

How the Components Relate to Total Variance

The law of total variance gives the connection between these components and the unconditional variance of one observation. Let $m=E[X]$. Since $E[X\mid\Theta]=\mu(\Theta)$, the law of iterated expectations gives $m=E[\mu(\Theta)]$. Applying the law of total variance,

\operatorname{Var}(X) =E[\operatorname{Var}(X\mid\Theta)] +\operatorname{Var}(E[X\mid\Theta]) =v+a.

The first term measures expected variation around each risk’s own conditional mean. The second measures how much those conditional means vary across the population. Their sum accounts for all variation in $X$: an observation can differ from the overall mean because it fluctuates around its own risk mean, because its risk mean differs from the overall mean, or both.

For two different observations $X_j$ and $X_k$ from the same risk, where $j\ne k$, conditional independence gives another useful result:

\operatorname{Cov}(X_j,X_k) =E[\operatorname{Cov}(X_j,X_k\mid\Theta)] +\operatorname{Cov}(E[X_j\mid\Theta],E[X_k\mid\Theta]) =0+\operatorname{Var}(\mu(\Theta)) =a.

The shared latent risk parameter links the observations: a risk with a higher conditional mean tends to produce higher observations in multiple periods. The covariance is VHM under the stated conditional-independence assumption. This result does not say that every pair of observations from different risks has covariance $a$; the shared-$\Theta$ setup is essential.

Derivation: For one observation, write $X-m=(X-\mu(\Theta))+(\mu(\Theta)-m)$. Squaring and taking expectations gives two squared terms and a cross term. The cross term is zero because $$ E[(X-\mu(\Theta))(\mu(\Theta)-m)] =E\big[(\mu(\Theta)-m)E[X-\mu(\Theta)\mid\Theta]\big]=0. $$ The expected first squared term is $E[\operatorname{Var}(X\mid\Theta)]=v$; the expected second squared term is $\operatorname{Var}(\mu(\Theta))=a$. Therefore $\operatorname{Var}(X)=v+a$.

Calculating EPV and VHM

When $\Theta$ has a finite number of possible classes, calculate each class’s conditional mean and variance, then average or take a variance across classes using their population probabilities. A convenient sequence is:

Find conditional moments.
For each possible value of $\Theta$, calculate $\mu(\Theta)=E[X\mid\Theta]$ and $\sigma^2(\Theta)=\operatorname{Var}(X\mid\Theta)$.

Average the conditional variances.
Weight each $\sigma^2(\Theta)$ by the population probability of its class. The result is EPV, $v$.

Calculate the variance of conditional means.
First find $m=E[\mu(\Theta)]$; then calculate the probability-weighted squared deviations of $\mu(\Theta)$ from $m$. The result is VHM, $a$.

Check against total variance.
Use $\operatorname{Var}(X)=v+a$ as a consistency check, or calculate the unconditional variance independently if the distribution makes that practical.

Worked Example: Bernoulli Claim Counts in Two Risk Classes

In a hypothetical portfolio, $\Theta$ is either class A or class B, each with probability $0.5$. In one year, the claim count $X$ is Bernoulli with probability $0.10$ in class A and $0.30$ in class B. Find EPV, VHM, and the unconditional variance. The method is to calculate conditional Bernoulli moments and then apply the population weights.

For a Bernoulli variable with claim probability $p$, the conditional mean is $p$ and the conditional variance is $p(1-p)$. Therefore, for class A:

\mu(A)=0.10,\qquad \sigma^2(A)=0.10(1-0.10)=0.09.

For class B:

\mu(B)=0.30,\qquad \sigma^2(B)=0.30(1-0.30)=0.21.

Weighting the conditional variances by the class probabilities gives:

$$ v=0.5(0.09)+0.5(0.21)=0.045+0.105=0.15. $$

The overall mean is:

$$ m=0.5(0.10)+0.5(0.30)=0.05+0.15=0.20. $$

The VHM is the weighted variance of the two conditional means:

$$ a=0.5(0.10-0.20)^2+0.5(0.30-0.20)^2 =0.5(0.01)+0.5(0.01)=0.01. $$

Thus, $\operatorname{Var}(X)=v+a=0.15+0.01=0.16$. As an independent check, the unconditional count is Bernoulli with probability $0.20$, so its variance is $0.20(1-0.20)=0.16$. Both calculations agree. A common mistake would be to report only $0.15$, the average within-class variance, as the total variance; that omits the $0.01$ of between-class variation.

Worked Example: Poisson Counts with a Gamma-Distributed Risk Parameter

Suppose a hypothetical insured’s annual claim count $X$, conditional on its latent rate $\Theta=\theta$, has a Poisson distribution with mean $\theta$. Across risks, let $\Theta\sim\operatorname{Gamma}(2,0.1)$, using the shape-rate parameterization: shape $2$, rate $0.1$. Calculate EPV, VHM, and the unconditional variance.

For a Poisson distribution, the conditional mean and variance both equal its parameter:

E[X\mid\Theta=\theta]=\theta,\qquad \operatorname{Var}(X\mid\Theta=\theta)=\theta.

A Gamma distribution with shape $\alpha$ and rate $\beta$ has mean $\alpha/\beta$ and variance $\alpha/\beta^2$. Here:

E[\Theta]=\frac{2}{0.1}=20,\qquad \operatorname{Var}(\Theta)=\frac{2}{(0.1)^2}=200.

Because the conditional variance equals $\Theta$, the EPV is:

v=E[\operatorname{Var}(X\mid\Theta)] =E[\Theta]=20.

Because the conditional mean also equals $\Theta$, the VHM is:

a=\operatorname{Var}(E[X\mid\Theta]) =\operatorname{Var}(\Theta)=200.

The unconditional variance is $20+200=220$. This also agrees with the variance of a Poisson mixture: $E[X]=20$, and the law of total variance gives $\operatorname{Var}(X)=E[\Theta]+\operatorname{Var}(\Theta)=220$. The key distinction is that $20$ is average Poisson fluctuation within a risk, while $200$ arises from rate differences across risks.

Worked Example: Exponential Claim Severity by Risk Class

Consider a hypothetical severity $X$ for an individual claim. The latent risk class is equally likely to be A or B. Conditional on class A, $X$ is exponential with rate $0.001$ per dollar; conditional on class B, it is exponential with rate $0.0005$ per dollar. Find EPV and VHM in squared dollars, then check the total variance.

For an exponential distribution with rate $\lambda$, its mean is $1/\lambda$ and its variance is $1/\lambda^2$. Thus, the conditional moments for class A are:

\mu(A)=\frac{1}{0.001}=1{,}000,\qquad \sigma^2(A)=\frac{1}{(0.001)^2}=1{,}000{,}000.

For class B, they are:

\mu(B)=\frac{1}{0.0005}=2{,}000,\qquad \sigma^2(B)=\frac{1}{(0.0005)^2}=4{,}000{,}000.

The EPV is the equally weighted average of the conditional variances:

$$ v=0.5(1{,}000{,}000)+0.5(4{,}000{,}000)=2{,}500{,}000. $$

The overall mean is $m=0.5(1{,}000)+0.5(2{,}000)=1{,}500$. Therefore:

a=0.5(1{,}000-1{,}500)^2+0.5(2{,}000-1{,}500)^2 =0.5(250{,}000)+0.5(250{,}000)=250{,}000.

The total variance is $2{,}500{,}000+250{,}000=2{,}750{,}000$ dollars squared. Each component has variance units; taking a square root would produce dollars, but the components themselves should not be reported in dollars.

Interpretation and Common Traps

EPV and VHM describe different features of the same population model. A large EPV means individual outcomes are noisy even after conditioning on risk class. A large VHM means the conditional means differ substantially among risks. Neither component alone is the unconditional variance unless the other component is zero.

Do not interchange the operations. EPV is the expectation of conditional variances, $E[\operatorname{Var}(X\mid\Theta)]$. VHM is the variance of conditional means, $\operatorname{Var}(E[X\mid\Theta])$.
Use population weights. If risk classes have unequal probabilities, both calculations must reflect those probabilities. An unweighted average is only correct when the classes are equally likely.
Do not use a variance of class labels. VHM uses the numerical conditional means, not arbitrary labels such as A and B.
Keep the conditioning clear. $\operatorname{Var}(X\mid\Theta)$ describes variation within a fixed risk class; $\operatorname{Var}(X)$ includes both sources.
Check assumptions before using the covariance result. The result $\operatorname{Cov}(X_j,X_k)=a$ for distinct observations from one risk relies on conditional independence and a common conditional mean given $\Theta$.

These quantities are structural summaries of a model, not automatically estimates from observed data. Their values depend on the assumed distribution of the latent risk parameter and on the conditional model for observations. The same observed portfolio can therefore have different EPV and VHM values under different modeling assumptions.

Check Your Understanding

For each problem, identify the conditional moments first, then separate within-risk and between-risk variation.

A hypothetical account is in class A with probability $0.25$ and class B with probability $0.75$. Its annual claim count is conditionally Bernoulli, with probabilities $0.05$ and $0.25$, respectively. Calculate EPV, VHM, and the unconditional variance. (Answer: EPV $=0.1525$, VHM $=0.0075$, total variance $=0.1600$.)
A hypothetical claim count is conditionally Poisson with rate $\Theta$. The rate is $4$ with probability $0.4$ and $10$ with probability $0.6$. Calculate the overall mean, EPV, VHM, and total variance. (Answer: mean $=7.6$, EPV $=7.6$, VHM $=8.64$, total variance $=16.24$.)
A hypothetical severity is exponential conditional on class. The two equally likely classes have means $\$500$ and $\$1{,}500$. Calculate EPV, VHM, and total variance in squared dollars. (Answer: EPV $=\$1{,}250{,}000$ squared, VHM $=\$250{,}000$ squared, total variance $=\$1{,}500{,}000$ squared.)
A hypothetical risk is in class A with probability $0.2$ and class B with probability $0.8$. The conditional means of $X$ are $50$ and $100$, and the conditional variances are $400$ and $900$, respectively. Calculate the overall mean, EPV, VHM, and total variance. (Answer: mean $=90$, EPV $=800$, VHM $=400$, total variance $=1{,}200$.)
For one hypothetical risk, $\Theta$ is $2$ or $6$, each with probability $0.5$, and observations $X_1,X_2$ are conditionally independent Poisson variables with rate $\Theta$. Find the covariance of $X_1$ and $X_2$, then find the variance of $X_1+X_2$. (Answer: $\operatorname{Cov}(X_1,X_2)=4$; $\operatorname{Var}(X_1+X_2)=24$.)

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