Variance of Hypothetical Means

Why Variation Across Risk Means Matters

A portfolio can contain risks with different underlying expected outcomes. For example, two groups of insureds may have different expected annual claim counts, even though individual observations within either group remain random. To describe this between-risk heterogeneity, we consider the expected outcome for each risk and then measure how those expected outcomes vary across the risk population.

The important distinction is between the expected outcome for a particular risk and the outcomes that risk may actually produce. A risk’s expected claim count is fixed once its risk characteristics are specified; the count observed in a year is still random. The variance of hypothetical means measures variation in those underlying expected counts across risks, not the year-to-year randomness within one risk.

Definition: Let $\Theta$ represent a risk’s characteristics, with a specified distribution over the risk population. For an outcome $X$, the hypothetical mean for a risk with characteristics $\Theta$ is $m(\Theta)=E[X\mid\Theta]$. The variance of hypothetical means, often abbreviated VHM, is $$ \operatorname{VHM}=\operatorname{Var}(m(\Theta)) =E\left[(m(\Theta)-E[m(\Theta)])^2\right], $$ provided the relevant second moments are finite. It measures how much the underlying expected outcomes differ across risks.

The mean used to center the hypothetical means is the risk-population average $E[m(\Theta)]$, with each risk type weighted according to its share of the population. VHM is expressed in squared units of $X$. If $X$ is an annual claim count, VHM is in squared claims; if $X$ is a dollar loss, VHM is in dollars squared. It is nonnegative, and it is zero exactly when the hypothetical mean is constant across the risk population, apart from risk types with zero probability.

VHM does not by itself describe all uncertainty in observed outcomes. It isolates differences in conditional expectations. For a claim-count model, for instance, it concerns differences in expected counts across risks, not the randomness of a count around its own risk-specific expectation. This distinction is central to the Bühlmann model components and is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Calculating VHM for Discrete Risk Types

Suppose the population has $k$ risk types. Type $j$ occurs with probability $p_j$, where $\sum_{j=1}^{k}p_j=1$, and has hypothetical mean $m_j$. The population mean of the hypothetical means is the probability-weighted mean $\bar m=\sum_j p_jm_j$. VHM is the probability-weighted average of squared deviations from $\bar m$.

\bar m=\sum_{j=1}^{k}p_jm_j, \qquad \operatorname{VHM}=\sum_{j=1}^{k}p_j(m_j-\bar m)^2.

The weights matter: a type that represents a larger share of the population contributes more to the population variance. This is a population variance, so the probabilities sum to one and there is no division by $k-1$. The calculation is not a sample variance from a set of equally weighted observed records.

Worked Example: Three Claim-Count Risk Types

A hypothetical portfolio has three risk types. Their population proportions are 0.20, 0.50, and 0.30, and their expected annual claim counts are 2, 5, and 9, respectively. Assume annual claim counts for a risk of type $j$ follow a Poisson distribution with parameter equal to that type’s stated mean. There is no discounting or mortality assumption because this is an annual claim-count calculation. Find the population mean of the hypothetical means and VHM.

Step 1: Identify the hypothetical means and weights. The type-specific means are $m_1=2$, $m_2=5$, and $m_3=9$. The weights are $p_1=0.20$, $p_2=0.50$, and $p_3=0.30$, which sum to $1.00$.

Step 2: Calculate the population mean.

\bar m=0.20(2)+0.50(5)+0.30(9) =0.40+2.50+2.70=5.60\text{ claims}.

Step 3: Calculate each weighted squared deviation. The deviations from 5.60 are $-3.60$, $-0.60$, and $3.40$ claims. Their squared deviations are $12.96$, $0.36$, and $11.56$ squared claims. Applying the population weights gives

\begin{aligned} \operatorname{VHM} &=0.20(12.96)+0.50(0.36)+0.30(11.56)\\ &=2.592+0.180+3.468\\ &=6.240\text{ squared claims}. \end{aligned}

Check. The weights sum to one, the weighted mean is 5.60 claims, and the weighted squared deviations sum to 6.24 squared claims. The conditional Poisson variances are not needed to calculate VHM: this calculation uses only the risk-type probabilities and their conditional means. The method avoids treating the three types as equally common.

Why the Two-Type Formula Has This Form

For two types, the weighted-deviation formula simplifies to a useful expression. Let type 1 have probability $p$ and mean $m_1$, and type 2 have probability $1-p$ and mean $m_2$. Their population mean is $\bar m=pm_1+(1-p)m_2$. Subtracting this mean from each type mean gives

m_1-\bar m=(1-p)(m_1-m_2), \qquad m_2-\bar m=-p(m_1-m_2).

Substitute these deviations into the definition and factor out the squared difference between the two type means:

\begin{aligned} \operatorname{VHM} &=p(1-p)^2(m_1-m_2)^2 +(1-p)p^2(m_1-m_2)^2\\ &=p(1-p)\big((1-p)+p\big)(m_1-m_2)^2\\ &=p(1-p)(m_1-m_2)^2. \end{aligned}

This expression makes the sources of between-type variation visible. VHM is zero if the type means are equal, or if one type has population probability zero. For a fixed difference between the means, VHM is largest when the two types are equally common because $p(1-p)$ is largest at $p=0.5$.

Worked Example: Finding Both Possible Type Proportions

A hypothetical annual claim-count portfolio has two risk types. Type A has probability $p$ and a Poisson claim count with mean 3; type B has probability $1-p$ and a Poisson claim count with mean 7. There is no discounting or mortality assumption. The VHM is 3 squared claims. Find every possible value of $p$ in the interval from 0 to 1.

Step 1: Apply the two-type formula. Here $m_1=3$, $m_2=7$, so the squared difference is $(7-3)^2=16$. Therefore,

3=p(1-p)(16), \qquad p(1-p)=\frac{3}{16}.

Step 2: Solve the quadratic equation. Rearranging gives

p^2-p+\frac{3}{16}=0.

The discriminant is $1-4(1)(3/16)=1-3/4=1/4$, so the two roots are

p=\frac{1\pm\sqrt{1/4}}{2} =\frac{1\pm 1/2}{2},

which gives $p=0.25$ or $p=0.75$. Both are valid probabilities and both satisfy $p(1-p)=3/16$. The method avoids reporting only one root: the variance is unchanged when the two population proportions are exchanged while the two type means remain fixed.

Check. At either root, $p(1-p)=0.25(0.75)=0.1875=3/16$, and $16(3/16)=3$ squared claims.

VHM When Risk Characteristics Are Continuous

Risk types need not be a finite list. If $\Theta$ has a continuous distribution and the hypothetical mean is a function $m(\Theta)$, VHM is the variance of that function under the distribution of $\Theta$. A useful calculation is to write the mean as a constant plus a multiple of the risk characteristic: adding a constant does not change variance, while multiplying by $b$ multiplies variance by $b^2$.

\operatorname{Var}(a+b\Theta)=b^2\operatorname{Var}(\Theta).

Worked Example: A Continuously Distributed Risk Characteristic

For a hypothetical portfolio, a risk characteristic $\Theta$ is uniformly distributed from 0 to 4. Conditional on $\Theta=\theta$, the annual claim count follows a Poisson distribution with mean $m(\theta)=2+1.5\theta$. Find the mean of the hypothetical means and VHM. No discounting or mortality assumption applies.

Step 1: Find the mean of $\Theta$. For a uniform distribution on $[0,4]$, $E[\Theta]=(0+4)/2=2$. Thus,

E[m(\Theta)]=2+1.5E[\Theta]=2+1.5(2)=5\text{ claims}.

Step 2: Find the variance of $\Theta$. The variance of a uniform distribution on $[a,b]$ is $(b-a)^2/12$. Therefore,

\operatorname{Var}(\Theta)=\frac{(4-0)^2}{12} =\frac{16}{12}=\frac{4}{3}.

Step 3: Transform to the variance of the hypothetical means. Since $m(\Theta)=2+1.5\Theta$, the constant 2 does not affect the variance, and the coefficient 1.5 is squared:

\operatorname{VHM} =\operatorname{Var}(2+1.5\Theta) =(1.5)^2\left(\frac{4}{3}\right) =2.25\left(\frac{4}{3}\right) =3\text{ squared claims}.

Check. The conditional mean ranges from 2 to 8 claims as $\Theta$ ranges from 0 to 4, and its population average is 5 claims. The calculated VHM is the variance of these conditional means under the uniform risk distribution; it is not the variance of $\Theta$ itself.

Interpretation and Common Errors

VHM describes the degree of heterogeneity in underlying expected outcomes. A small VHM means the risk-specific expected outcomes tend to be close to their population average. A larger VHM means they are more spread out. Its numerical size must be interpreted in the units of the outcome squared, so comparisons are most meaningful when the outcome definition and scale are the same.

VHM depends both on how different the hypothetical means are and on how the population is distributed across those means. For example, two risk types with very different means can still have a small VHM if one type is rare. Conversely, a moderate difference between means can matter substantially if both types make up sizeable portions of the population. The distribution of risks is part of the quantity being calculated, not merely background information.

Using equal weights without justification. Risk types should be weighted by their population probabilities, not by the number of rows in a displayed list.
Centering at the wrong value. Use the weighted population mean of the hypothetical means, not one type’s mean or an unweighted average.
Confusing variance and standard deviation. VHM is a variance in squared units. Taking its square root gives the standard deviation of hypothetical means, a different quantity.
Confusing between-risk and within-risk variation. VHM concerns variation in conditional expected outcomes across risks. It is not the conditional variance of outcomes for an individual risk.
Discarding a valid quadratic root. When solving for an unknown two-type proportion, check every root against the probability constraints and the original equation.

Key takeaway: Define the risk population, identify each risk’s hypothetical mean, and calculate the probability-weighted variance of those means. The resulting VHM measures between-risk heterogeneity in expected outcomes, with units equal to the square of the outcome’s units.

Check Your Understanding

For each problem, treat the stated probabilities as the distribution of risks in the hypothetical population. Report a variance in squared units of the stated outcome.

A two-type hypothetical portfolio has expected annual claim counts of 4 and 8, with probabilities 0.25 and 0.75. Find the population mean of the hypothetical means and VHM. (Answer: mean 7 claims; VHM 3 squared claims.)
Three hypothetical risk types have expected annual claim counts 1, 4, and 10, with probabilities 0.20, 0.50, and 0.30. Find the mean and VHM. (Answer: mean 5.2 claims; VHM 11.16 squared claims.)
Two hypothetical types have conditional Poisson means 2 and 8. Type 1 has probability $p$, and VHM is 9 squared claims. Find all possible values of $p$ between 0 and 1. (Answer: $p=0.50$.)
A hypothetical risk characteristic $\Theta$ is uniform on $[0,2]$, and the conditional expected annual loss is $m(\Theta)=5+2\Theta$, measured in thousands of dollars. Find the mean of the hypothetical means and VHM. (Answer: mean 7 thousand dollars; VHM $4/3$ squared thousand dollars.)
A hypothetical portfolio has two equally likely risk types. Their conditional expected annual claim counts are 2 and 6. For each type, the conditional claim count is Poisson. Find VHM and the probability-weighted average of the conditional variances. State which is between-risk variation. (Answer: VHM $4$ squared claims; average conditional variance $4$ squared claims; VHM is the between-risk variation.)

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