Gamma Severity Components

Why Gamma Severity Components Depend on the Risk Mean

In a Bühlmann model, a risk has an unknown characteristic that determines its expected claim severity. Even when two risks have different expected severities, their claim amounts may also fluctuate differently from their own expected values. For Gamma severities with a common shape parameter, that process variation depends directly on the risk-specific mean. Consequently, the expected process variance requires the second moment of the risk mean, not merely its average.

This tutorial derives the two variance components for this model and shows how they fit into the total variance decomposition. The calculations are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Model definition: Let $\Theta$ denote a risk characteristic and let $M=m(\Theta)>0$ be the conditional mean severity for that risk. Assume that, conditional on $\Theta$, a claim severity $X$ has a Gamma distribution with a common shape parameter $k>0$ and scale parameter $M/k$. Equivalently, its rate parameter is $k/M$. Thus $E[X\mid\Theta]=M$ and $\operatorname{Var}(X\mid\Theta)=M^2/k$. The model requires $M$ to have a finite second moment for the variance components below to be finite.

The shape-scale parameterization matters. A Gamma random variable with shape $\alpha$ and scale $\beta$ has mean $\alpha\beta$ and variance $\alpha\beta^2$. Taking $\alpha=k$ and $\beta=M/k$ gives conditional mean $k(M/k)=M$ and conditional variance $k(M/k)^2=M^2/k$. The scale changes with the risk mean so that the conditional mean is $M$, while the shape remains fixed.

The Two Variance Components

Write $m_0=E[M]$ for the mean of the hypothetical risk means. The expected process variance is the average conditional variance across risks. The variance of hypothetical means measures how much the conditional means differ across risks.

Definitions: For the conditional Gamma model with common shape $k$, $$ \text{EPV}=E[\operatorname{Var}(X\mid\Theta)] =\frac{E[M^2]}{k} =\frac{\operatorname{Var}(M)+(E[M])^2}{k}, \qquad \text{VHM}=\operatorname{Var}(E[X\mid\Theta]) =\operatorname{Var}(M). $$ The unconditional mean severity is $E[X]=E[M]$, and the unconditional variance is $\operatorname{Var}(X)=\text{EPV}+\text{VHM}$.

The expression for EPV follows because the conditional variance for a particular risk is $M^2/k$. Averaging over the distribution of risks gives $E[M^2]/k$. The identity $E[M^2]=\operatorname{Var}(M)+(E[M])^2$ then gives the alternative form. This distinction is important: the squared average risk mean, $(E[M])^2$, is not the average squared risk mean, $E[M^2]$, unless $M$ is constant.

Why the Total Variance Splits This Way

The total variance identity separates variation within a risk from variation between risk means. In this setting the within-risk contribution is EPV and the between-risk contribution is VHM.

Center around the unconditional mean.
Write $X-E[X]=(X-M)+(M-E[M])$, since $E[X]=E[M]$.

Square and take expectations.
The squared expression has two squared terms and a cross-product. The cross-product has expectation zero because $E[X-M\mid\Theta]=0$.

Evaluate the remaining terms.
The expected value of $(X-M)^2$ is $E[\operatorname{Var}(X\mid\Theta)]$; the expected value of $(M-E[M])^2$ is $\operatorname{Var}(M)$.

In equation form, the cross-product vanishes by conditioning on $\Theta$:

E[(X-M)(M-E[M])] =E\!\left[(M-E[M])E[X-M\mid\Theta]\right]=0.

Therefore $\operatorname{Var}(X)=E[\operatorname{Var}(X\mid\Theta)]+\operatorname{Var}(M)$. Substituting the Gamma conditional variance gives $\operatorname{Var}(X)=E[M^2]/k+\operatorname{Var}(M)$. This is also a useful arithmetic check: the unconditional variance must equal the sum of the two nonnegative components.

Worked Applications

Worked Example: Two Risk Means and a Common Shape

A hypothetical liability portfolio has two equally likely risk types. Their conditional mean severities are $M=\$400$ and $M=\$1{,}000$, and each type has Gamma severity with common shape $k=2$ and scale $M/2$. Find EPV, VHM, and the unconditional variance. A candidate instead uses $(E[M])^2/k$ for EPV; quantify the resulting error. No interest discounting or mortality model is involved.

Method and conditional parameters. For the $\$400$ type, the Gamma shape is 2 and scale is $\$400/2=\$200$. For the $\$1{,}000$ type, the shape is 2 and scale is $\$1{,}000/2=\$500$. Use $E[M^2]/k$ for EPV.

Calculate the mean and second moment of $M$.

$$ E[M]=0.5(400)+0.5(1{,}000)=700, $$

E[M^2]=0.5(400^2)+0.5(1{,}000^2) =0.5(160{,}000)+0.5(1{,}000{,}000) =580{,}000.

Calculate the components. EPV is $580{,}000/2=\$290{,}000$. The variance of $M$ is $E[M^2]-(E[M])^2=580{,}000-700^2=580{,}000-490{,}000=\$90{,}000$, so VHM is $\$90{,}000$. Hence the unconditional variance is $\$290{,}000+\$90{,}000=\$380{,}000$.

Check the candidate’s calculation. Using $(E[M])^2/k$ gives $700^2/2=\$245{,}000$. This understates EPV by $\$290{,}000-\$245{,}000=\$45{,}000$. The error is $ \operatorname{Var}(M)/k=90{,}000/2=\$45{,}000$, exactly the part lost by replacing $E[M^2]$ with $(E[M])^2$. The component sum checks: $\$290{,}000+\$90{,}000=\$380{,}000$.

Common mistake avoided: Do not square the average risk mean when calculating EPV. That would omit the contribution from variation in risk means, scaled by $1/k$.

Worked Example: A Three-Point Distribution of Risk Means

For a hypothetical equipment coverage, $M$ equals $\$300$, $\$700$, or $\$1{,}100$, with probabilities $0.25$, $0.50$, and $0.25$, respectively. Conditional on the risk type, severity is Gamma with common shape $k=5$ and scale $M/5$. Calculate EPV, VHM, and the unconditional variance. No interest discounting or mortality model applies.

Method and parameters. The three conditional Gamma distributions all have shape 5. Their respective scales are $\$60$, $\$140$, and $\$220$. First find $E[M]$ and $E[M^2]$; then use the component formulas.

Mean of the risk means:

$$ E[M]=0.25(300)+0.50(700)+0.25(1{,}100) =75+350+275=700. $$

Second moment:

E[M^2]=0.25(90{,}000)+0.50(490{,}000)+0.25(1{,}210{,}000) =22{,}500+245{,}000+302{,}500=570{,}000.

Components and check. EPV is $570{,}000/5=\$114{,}000$. VHM is $570{,}000-700^2=570{,}000-490{,}000=\$80{,}000$. The unconditional variance is therefore $\$194{,}000$. As a check, the component sum is $114{,}000+80{,}000=194{,}000$. The common mistake here is to use the mean $700$ alone to stand in for the three risk means; the second moment calculation retains the variation among them.

Worked Example: A Continuous Distribution of Risk Means

In a hypothetical property portfolio, the risk mean $M$ is uniformly distributed from $\$200$ to $\$800$. Conditional severity is Gamma with common shape $k=4$ and scale $M/4$. Find the two components and the unconditional variance. No interest discounting or mortality assumption is needed.

Method and distribution parameters. Here $M\sim\operatorname{Uniform}(200,800)$. Its mean is $(200+800)/2=\$500$, and its variance is $(800-200)^2/12=\$30{,}000$. Conditional on each $M$, severity has shape 4 and scale $M/4$, which ranges from $\$50$ to $\$200$.

Find the second moment of $M$. Since $E[M^2]=\operatorname{Var}(M)+(E[M])^2$,

E[M^2]=30{,}000+500^2=30{,}000+250{,}000=280{,}000.

Calculate and check. EPV is $280{,}000/4=\$70{,}000$. VHM is $\$30{,}000$. Thus the unconditional variance is $\$100{,}000$, equal to $70{,}000+30{,}000$. Using only $500^2/4=\$62{,}500$ for EPV would understate it by $\$7{,}500$, which is $\operatorname{Var}(M)/4$.

Conditions, Interpretation, and Common Traps

The compact formula $\text{EPV}=E[M^2]/k$ depends on the model’s assumptions: the conditional severity is Gamma, its conditional mean is $M$, and the shape $k$ is common across risks. The conditional variance is then $M^2/k$. If the shape varies with risk, say it is $k(\Theta)$, the correct expression is $E[M^2/k(\Theta)]$, not $E[M^2]/k$ for some arbitrarily selected shape. The VHM remains $\operatorname{Var}(M)$ whenever $M=E[X\mid\Theta]$.

The components have units of squared dollars when severity is measured in dollars. The shape $k$ is dimensionless, so dividing $E[M^2]$ by $k$ preserves those units. This provides a useful dimensional check. Both components are nonnegative, and EPV is positive when conditional severity has positive variance.

A second frequent trap is confusing conditional and unconditional variation. For a particular risk mean $M$, the process variance is $M^2/k$; the expected process variance averages this quantity over the risk distribution. VHM, by contrast, is the variance of the conditional means themselves. Neither is the unconditional variance alone: both are needed to reconstruct it.

Key takeaway: With common Gamma shape $k$ and risk-specific mean $M$, calculate EPV from the second moment $E[M^2]$, calculate VHM from $\operatorname{Var}(M)$, and verify that their sum is the unconditional severity variance.

Check Your Understanding

For each problem, identify the relevant model quantities before applying the component formulas.

A hypothetical class has equally likely risk means of $\$200$ and $\$600$, with common Gamma shape $k=2$. Find EPV, VHM, and the unconditional variance. (Answer: EPV $\$100{,}000$; VHM $\$40{,}000$; unconditional variance $\$140{,}000$)
A hypothetical coverage has $M=\$400$ with probability $0.20$ and $M=\$900$ with probability $0.80$; conditional severity has common shape $k=3$. Find EPV and VHM. (Answer: EPV $\$226{,}666.67$; VHM $\$40{,}000$)
For a hypothetical portfolio, $M\sim\operatorname{Uniform}(100,500)$, and conditional severity has common Gamma shape $k=4$. Find EPV, VHM, and the unconditional variance. (Answer: EPV $\$25{,}833.33$; VHM $\$13{,}333.33$; unconditional variance $\$39{,}166.67$)
A common-shape Gamma model has $E[M]=\$600$, $k=3$, and EPV $\$150{,}000$. Infer VHM and the unconditional variance. (Answer: VHM $\$90{,}000$; unconditional variance $\$240{,}000$)
A hypothetical portfolio has two equally likely risk types: $(M,k)=(\$300,2)$ and $(\$900,6)$. The conditional Gamma shape differs by type. Find EPV and VHM, and explain why the common-shape formula cannot be used. (Answer: EPV $\$90{,}000$; VHM $\$90{,}000$; use $E[M^2/k(\Theta)]$ because shape varies)

What You'll Learn