Gamma Coefficient Of Variation

Why Relative Variability Matters

For positive claim severities, the standard deviation measures variability in the same units as the loss. But it does not, by itself, tell us how large that variability is compared with the average severity. A standard deviation of $1,000 has a different interpretation when the mean is $2,000 than when the mean is $20,000. The coefficient of variation (CV) puts spread on a relative scale by dividing the standard deviation by the mean.

For a Gamma severity distribution, the CV has a particularly useful property: it depends on the shape parameter, not on the scale parameter. Thus, two Gamma distributions can have different average claim amounts but the same relative variability. This tutorial derives that result and applies it to hypothetical severity distributions. The material is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: If $X$ has finite mean $\mu=E[X]>0$ and standard deviation $\sigma=\sqrt{\operatorname{Var}(X)}$, its coefficient of variation is $$ \operatorname{CV}(X)=\frac{\sigma}{\mu} =\frac{\sqrt{\operatorname{Var}(X)}}{E[X]}. $$ The CV is dimensionless. For a positive Gamma severity, it measures standard deviation relative to mean severity.

Gamma Parameters and the CV

Use the shape-scale convention in which $X$ has a Gamma distribution with shape $\alpha>0$ and scale $\theta>0$. Its support is $x>0$, and its density is

f_X(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha}, \qquad x>0.

Under this convention, the mean is $\alpha\theta$ and the variance is $\alpha\theta^2$. These moments can also be written in terms of the mean $\mu$ and the dispersion parameter $\phi$, where $\mu=\alpha\theta$ and $\phi=1/\alpha$. Then $\operatorname{Var}(X)=\phi\mu^2$. The two parameterizations describe the same distribution; the scale $\theta$ is not the rate $1/\theta$.

Theorem: For a Gamma distribution with shape $\alpha$ and scale $\theta$, $$ \operatorname{CV}(X)=\frac{1}{\sqrt{\alpha}}. $$ Equivalently, in the mean-dispersion form $\operatorname{Var}(X)=\phi\mu^2$, its coefficient of variation is $\sqrt{\phi}$, with $\phi=1/\alpha$.

The result follows directly from the Gamma moments. Since $\alpha>0$ and $\theta>0$, the standard deviation is $\sqrt{\alpha}\theta$, and the mean is $\alpha\theta$. Dividing cancels the scale:

\operatorname{CV}(X) =\frac{\sqrt{\alpha\theta^2}}{\alpha\theta} =\frac{\sqrt{\alpha}\theta}{\alpha\theta} =\frac{1}{\sqrt{\alpha}}.

The cancellation is the key justification: changing $\theta$ stretches or shrinks the severity amounts and their standard deviation in the same proportion, so it does not change their relative spread. Increasing $\alpha$, in contrast, lowers the CV. The equivalent derivation from the mean-dispersion form is $\operatorname{CV}(X)=\sqrt{\phi\mu^2}/\mu=\sqrt{\phi}$, because $\mu>0$.

Using the Shape to Measure Relative Spread

A smaller Gamma shape means a larger CV and therefore more variability relative to the mean. A larger shape means less relative variability. This conclusion concerns relative spread, not necessarily the dollar amount of the standard deviation. When means differ, always calculate or compare both quantities if the question asks about both absolute and relative variability.

The following sequence keeps the parameterization and interpretation clear:

Identify the parameterization.
Check whether the Gamma distribution is given by shape and scale, or by mean and dispersion. Do not confuse scale with rate.

Find the CV.
Use $1/\sqrt{\alpha}$ when shape is given, or $\sqrt{\phi}$ when dispersion is given. If only moments are supplied, use $\sqrt{\operatorname{Var}(X)}/E[X]$.

Interpret the comparison.
A larger CV means greater variability relative to the mean. A larger standard deviation means greater variability in the original severity units; the two comparisons need not agree.

Worked Example: Finding the CV From Shape and Scale

A hypothetical property severity model assumes $X$ has a Gamma distribution with shape $\alpha=4$ and scale $\theta=\$750$. Calculate its mean, variance, standard deviation, and CV. State what the CV says about relative spread.

Method and parameter check. This is the shape-scale convention, so $E[X]=\alpha\theta$ and $\operatorname{Var}(X)=\alpha\theta^2$. The scale is measured in dollars.

Mean. $E[X]=4(\$750)=\$3{,}000.$

Variance. $\operatorname{Var}(X)=4(\$750)^2=4(\$562{,}500)=\$2{,}250{,}000.$

Standard deviation. $\sqrt{\$2{,}250{,}000}=\$1{,}500,$ since $1{,}500^2=2{,}250{,}000.$

Coefficient of variation. $\operatorname{CV}(X)=\$1{,}500/\$3{,}000=0.5.$ Equivalently, $1/\sqrt{4}=0.5$. The standard deviation is half the mean, so the modeled severity has a CV of 50%.

Check and mistake avoided. The moment calculation gives the same CV as the shape formula. A common error is to use $\theta$ as the mean; here the mean is $\alpha\theta=\$3{,}000$, not $\$750$.

From Mean and Variance to Shape and CV

Sometimes the severity information is provided as a mean and variance rather than as a Gamma shape. For a Gamma distribution, the dispersion form gives a direct route: $\phi=\operatorname{Var}(X)/\mu^2$, so $\operatorname{CV}=\sqrt{\phi}$ and $\alpha=1/\phi$. This use of the Gamma relationship requires a positive mean and variance and the assumption that the distribution is Gamma; the same moment ratio can be calculated for other distributions, but it need not equal the reciprocal square root of a Gamma shape.

Worked Example: Inferring Gamma Shape From Moments

A hypothetical equipment severity model is Gamma with mean $\$2{,}400$ and variance $\$1{,}440{,}000$. Find its CV, shape, scale, and standard deviation. Check the result by recovering the stated moments.

Method. First calculate the standard deviation and CV from the supplied moments. Then use $\operatorname{Var}(X)=\phi\mu^2$, $\alpha=1/\phi$, and $\theta=\mu/\alpha$.

Standard deviation. Since $1{,}200^2=1{,}440{,}000$, the standard deviation is $\$1{,}200$.

CV and dispersion. $\operatorname{CV}(X)=\$1{,}200/\$2{,}400=0.5.$ Thus, $\phi=\operatorname{CV}(X)^2=0.25$.

Shape and scale. $\alpha=1/\phi=1/0.25=4.$ Then $\theta=\mu/\alpha=\$2{,}400/4=\$600.$

Moment check. The implied mean is $4(\$600)=\$2{,}400$. The implied variance is $4(\$600)^2=4(\$360{,}000)=\$1{,}440{,}000$, matching the supplied values. The implied CV is $1/\sqrt{4}=0.5$.

Mistake avoided. Do not use variance divided by mean as the dispersion. The correct calculation is $\phi=\operatorname{Var}(X)/\mu^2=\$1{,}440{,}000/(\$2{,}400)^2=0.25$; the mean must be squared.

Comparing Relative and Absolute Variability

When two Gamma severities have different means, their CVs answer a different question from their standard deviations. The CV comparison describes spread relative to the respective means. The standard deviation comparison describes spread in dollars. A distribution can have a lower CV and still have a larger standard deviation if its mean is sufficiently larger.

Worked Example: Ranking Two Hypothetical Severity Models

Two hypothetical commercial coverage segments are modeled with Gamma severities. Segment A has mean $\$1{,}200$ and shape $\alpha_A=6.25$. Segment B has mean $\$3{,}000$ and shape $\alpha_B=16$. For each segment, calculate the CV, standard deviation, and scale. Identify which has greater relative variability and which has greater absolute variability.

Method. For each Gamma model, calculate $\operatorname{CV}=1/\sqrt{\alpha}$, then $\sigma=\mu\operatorname{CV}$, and $\theta=\mu/\alpha$.

Segment A. Since $\sqrt{6.25}=2.5$, its CV is $1/2.5=0.4$. Its standard deviation is $0.4(\$1{,}200)=\$480$. Its scale is $\$1{,}200/6.25=\$192$, because $6.25(\$192)=\$1{,}200$.

Segment B. Since $\sqrt{16}=4$, its CV is $1/4=0.25$. Its standard deviation is $0.25(\$3{,}000)=\$750$. Its scale is $\$3{,}000/16=\$187.50$, because $16(\$187.50)=\$3{,}000$.

Comparison and check. Segment A has the greater relative variability because $0.4>0.25$. Segment B has the greater standard deviation because $\$750>\$480$. Checking with the variance formula, A has variance $6.25(\$192)^2=\$230{,}400$, whose square root is $\$480$; B has variance $16(\$187.50)^2=\$562{,}500$, whose square root is $\$750$.

Mistake avoided. Ranking by standard deviation alone would incorrectly identify B as having greater relative variability. The CV is the appropriate comparison for relative spread, while standard deviation ranks absolute spread.

Common Exam Traps and Practical Interpretation

The Gamma CV is simple to calculate, but several small errors can reverse or distort an interpretation:

Using the wrong parameterization. For shape $\alpha$ and scale $\theta$, the CV is $1/\sqrt{\alpha}$. The scale cancels; it is not itself the mean or the CV. If the model uses rate $\beta=1/\theta$, convert carefully rather than substituting $\beta$ for $\theta$.
Confusing variance with standard deviation. The CV uses the square root of the variance divided by the mean. Dividing variance directly by mean does not produce the CV.
Comparing only dollar standard deviations. A larger standard deviation identifies larger absolute spread, not necessarily larger spread relative to the mean.
Applying the Gamma formula without the Gamma assumption. The definition of CV applies broadly, but the relationship $\operatorname{CV}=1/\sqrt{\alpha}$ is specific to the Gamma shape parameterization.
Forgetting units. The mean and standard deviation use the same units, so their ratio is dimensionless. A CV such as 0.4 may be reported as 40%.

In a severity analysis, a CV comparison can help distinguish distributions with different relative spread even when their monetary scales differ. It does not establish which model better fits observed data, nor does it determine tail behavior by itself. It is a moment-based summary that should be interpreted alongside the stated distributional assumption and the purpose of the comparison.

Key takeaway: For a Gamma severity with shape $\alpha$ and scale $\theta$, $\operatorname{CV}=1/\sqrt{\alpha}$. The scale changes the mean and standard deviation proportionally, while the shape determines relative variability. Compare CVs for relative spread and standard deviations for absolute spread.

Check Your Understanding

For each problem, use the Gamma shape-scale convention unless the question gives moments directly. Keep relative variability separate from absolute variability.

A hypothetical Gamma severity has shape $\alpha=9$ and scale $\theta=\$400$. Find its mean, standard deviation, and CV. (Answer: mean $\$3{,}600$, standard deviation $\$1{,}200$, CV $1/3$ or about $0.3333$.)
A Gamma severity has mean $\$500$ and variance $\$62{,}500$. Find its CV and shape. (Answer: CV $0.5$, shape $\alpha=4$.)
Two Gamma models have shapes 4 and 16, respectively. Which has greater relative variability, and what are their CVs? (Answer: the shape-4 model; its CV is $0.5$, compared with $0.25$ for the shape-16 model.)
A hypothetical Gamma model has mean $\$1{,}800$ and CV $0.6$. Find its dispersion $\phi$, shape, standard deviation, and scale. (Answer: $\phi=0.36$, $\alpha=25/9\approx2.7778$, standard deviation $\$1{,}080$, scale $\$648$.)
Model X is Gamma with mean $\$2{,}400$ and shape 9; Model Y is Gamma with mean $\$3{,}600$ and shape 4. Compare their CVs and standard deviations. (Answer: X has CV $1/3$ and standard deviation $\$800$; Y has CV $0.5$ and standard deviation $\$1{,}800$. Y has greater relative and absolute variability.)

What You'll Learn