Gamma Severity: Mean and Variance by Parameterization

Why Gamma Parameterization Matters

A gamma model can describe a positive claim severity, but its second parameter is not named consistently across references. Some use a scale, which has the same units as the loss; others use a rate, which has reciprocal loss units. The numerical values of these parameters are reciprocals, so inserting a rate where a formula expects a scale can produce a substantially incorrect mean and variance.

This tutorial focuses on calculating the first two moments from either convention and converting between them. As with any severity model, distinguish the units of the loss from the units of its variance: if $X$ is measured in dollars, then $E[X]$ is in dollars and $\operatorname{Var}(X)$ is in dollars squared.

Definition: A gamma severity with shape $\alpha>0$ and scale $\theta>0$, written here as $X\sim\operatorname{Gamma}(\alpha,\theta)$ in the shape-scale convention, has density $$ f_X(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha},\qquad x>0, $$ where $\Gamma(\alpha)=\int_0^\infty t^{\alpha-1}e^{-t}\,dt$. In the shape-rate convention, use $\alpha>0$ and $\beta>0$, with density $$ f_X(x)=\frac{\beta^\alpha x^{\alpha-1}e^{-\beta x}}{\Gamma(\alpha)},\qquad x>0. $$ The same distribution is described by either convention when $\beta=1/\theta$.

The shape parameter $\alpha$ is dimensionless. If severity is measured in dollars, $\theta$ is measured in dollars and $\beta$ is measured in inverse dollars. The symbols $\theta$ and $\beta$ are being used specifically for gamma scale and rate here; they are not the course’s interest-rate symbols.

Mean and Variance Formulas

For a shape-scale gamma severity, calculate the moments directly with $\theta$. For a shape-rate gamma severity, use $\beta$ and its reciprocal relationship with the scale. The formulas apply when the parameters are positive and the stated density is the model for $X$; they should not be applied by treating a rate as though it were a scale.

Theorem: If $X$ has a gamma distribution with shape $\alpha>0$ and scale $\theta>0$, then $$ E[X]=\alpha\theta,\qquad E[X^2]=\alpha(\alpha+1)\theta^2,\qquad \operatorname{Var}(X)=\alpha\theta^2. $$ Equivalently, for shape $\alpha>0$ and rate $\beta>0$, $$ E[X]=\frac{\alpha}{\beta},\qquad E[X^2]=\frac{\alpha(\alpha+1)}{\beta^2},\qquad \operatorname{Var}(X)=\frac{\alpha}{\beta^2}. $$

The second raw moment is included because variance is calculated as $E[X^2]-(E[X])^2$. In either convention, the variance is not the square of the mean: the shape factor in $\operatorname{Var}(X)$ is $\alpha$, not $\alpha^2$.

Why the Moment Formulas Hold

A short derivation makes the role of the scale clear. Start with the shape-scale density and integrate $x f_X(x)$ to obtain the mean. In the integral, substitute $t=x/\theta$, so $x=\theta t$ and $dx=\theta\,dt$. The remaining integral is a gamma-function integral.

\begin{aligned} E[X] &=\int_0^\infty x\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha}\,dx\\ &=\frac{\theta}{\Gamma(\alpha)}\int_0^\infty t^\alpha e^{-t}\,dt\\ &=\theta\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)} =\alpha\theta. \end{aligned}

The last equality uses the gamma-function recurrence $\Gamma(\alpha+1)=\alpha\Gamma(\alpha)$, valid for $\alpha>0$. Applying the same substitution to $E[X^2]$ gives $\theta^2\Gamma(\alpha+2)/\Gamma(\alpha)$. Applying the recurrence twice yields $\alpha(\alpha+1)\theta^2$. Therefore,

\operatorname{Var}(X) =E[X^2]-(E[X])^2 =\alpha(\alpha+1)\theta^2-\alpha^2\theta^2 =\alpha\theta^2.

Since the rate version has $\theta=1/\beta$, substitution gives $E[X]=\alpha/\beta$ and $\operatorname{Var}(X)=\alpha/\beta^2$. This conversion also checks the units: dividing a dimensionless shape by a rate gives loss units for the mean, and dividing by the rate squared gives squared-loss units for the variance.

Worked Calculations

Worked Example: Moments From a Scale Parameter

A hypothetical insurer models an individual covered loss $X$ using a gamma severity with shape $\alpha=3$ and scale $\theta=\$400$. Calculate the mean and variance. No interest or mortality assumption is needed for this severity calculation.

Method. The parameter is explicitly a scale, so use the shape-scale formulas. First record the units: $\alpha=3$ is dimensionless and $\theta=\$400$.

Mean. $$ E[X]=\alpha\theta=3(\$400)=\$1{,}200. $$

Variance. $$ \operatorname{Var}(X)=\alpha\theta^2 =3(\$400)^2 =3(\$160{,}000)=\$480{,}000\text{ dollars}^2. $$

Check. The second raw moment is $E[X^2]=3(4)(\$400)^2=\$1{,}920{,}000\text{ dollars}^2$. Subtracting the squared mean gives $\$1{,}920{,}000-(\$1{,}200)^2=\$1{,}920{,}000-\$1{,}440{,}000=\$480{,}000\text{ dollars}^2$, as required. The common mistake to avoid is squaring the scale in the mean formula; the mean is $\alpha\theta$, not $\alpha\theta^2$.

Worked Example: Moments From a Rate Parameter

A hypothetical equipment-loss model has gamma shape $\alpha=5$ and rate $\beta=0.002$ per dollar. Calculate the mean and variance. No interest or mortality assumption is involved.

Method. Because the parameter is a rate, use the shape-rate formulas. As a conversion check, the scale is $$ \theta=\frac{1}{\beta}=\frac{1}{0.002\text{ per dollar}}=\$500. $$

Mean using the rate. $$ E[X]=\frac{\alpha}{\beta} =\frac{5}{0.002\text{ per dollar}} =\$2{,}500. $$

Variance using the rate. $$ \operatorname{Var}(X)=\frac{\alpha}{\beta^2} =\frac{5}{(0.002\text{ per dollar})^2} =\frac{5}{0.000004\text{ per dollar}^2} =\$1{,}250{,}000\text{ dollars}^2. $$

Check by scale. The scale formulas give $E[X]=5(\$500)=\$2{,}500$ and $\operatorname{Var}(X)=5(\$500)^2=5(\$250{,}000)=\$1{,}250{,}000\text{ dollars}^2$. The results agree. The common mistake is to use $5(0.002)$ for the mean; that treats a rate as a scale and also fails the units check.

Worked Example: Converting a Rate Given Per Thousand Dollars

A hypothetical portfolio model specifies $X$, measured in thousands of dollars, as gamma with shape $\alpha=4$ and rate $\beta=0.5$ per thousand dollars. Find the mean and variance in thousands of dollars and then express them in dollars. No interest or mortality assumption is needed.

Step 1: Convert the rate to a scale. Since the loss unit is one thousand dollars, $$ \theta=\frac{1}{0.5\text{ per thousand dollars}} =2\text{ thousand dollars}. $$

Step 2: Calculate the mean. $$ E[X]=\frac{4}{0.5}=8\text{ thousand dollars}=\$8{,}000. $$

Step 3: Calculate the variance. $$ \operatorname{Var}(X)=\frac{4}{(0.5)^2} =\frac{4}{0.25} =16\text{ thousand-dollars}^2. $$ Because one thousand dollars is $\$1{,}000$, one thousand-dollars squared is $(\$1{,}000)^2=\$1{,}000{,}000\text{ dollars}^2$. Thus the variance is $16(\$1{,}000{,}000)=\$16{,}000{,}000\text{ dollars}^2$.

Check. Using the scale, the mean is $4(2)=8$ thousand dollars and the variance is $4(2^2)=16$ thousand-dollars squared. Both match the rate calculation. The common mistake is to convert the mean to dollars but leave the variance as though it were in dollars squared; the variance conversion requires squaring the factor of $1{,}000$.

Parameter and Unit Checks

Before calculating, identify which convention the problem states. The labels “scale” and “rate” are more informative than a parameter’s letter, because sources can use different letters. Then carry the units through each formula. These checks often reveal an error before any further calculation is needed.

Identify the convention.
A scale multiplies the shape in the mean; a rate is used in the denominator. Do not select a formula based only on a parameter symbol.

Convert if useful.
Use $\theta=1/\beta$ or $\beta=1/\theta$. The two parameters must be reciprocals expressed in compatible units.

Calculate and check dimensions.
The mean must have loss units, and the variance must have squared-loss units. When practical, recalculate with the equivalent parameterization.

For example, if a loss is measured in thousands of dollars, a rate of $0.5$ per thousand dollars corresponds to a scale of $2$ thousand dollars, not $\$2$. The numerical formulas work with the stated loss unit, but any conversion to dollars must be made consistently. In particular, converting a variance requires squaring the unit-conversion factor.

Common Exam Trap

Suppose a gamma model has shape $2$ and rate $0.004$ per dollar. The incorrect calculation $E[X]=2(0.004)=\$0.008$ multiplies by the rate as though it were a scale. The units expose the problem: shape times a per-dollar rate has units of inverse dollars, not dollars.

The correction is to divide by the rate, or first convert to the scale: $$ \theta=\frac{1}{0.004\text{ per dollar}}=\$250,\qquad E[X]=\frac{2}{0.004}=\$500. $$ The variance is $$ \operatorname{Var}(X)=\frac{2}{(0.004)^2} =\$125{,}000\text{ dollars}^2. $$ A second check uses the scale: $2(\$250)=\$500$ and $2(\$250)^2=\$125{,}000\text{ dollars}^2$. This distinction between reciprocal parameterizations is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Key takeaway: For shape $\alpha$, scale $\theta$, and rate $\beta=1/\theta$, the mean is $\alpha\theta=\alpha/\beta$ and the variance is $\alpha\theta^2=\alpha/\beta^2$. Confirm the parameter convention and units before substituting.

Check Your Understanding

For each hypothetical severity model, identify the parameter convention, calculate the requested moments, and keep the stated units consistent.

A gamma loss has shape $2.5$ and scale $\$600$. Find its mean and variance in dollars and dollars squared. (Answer: mean $\$1{,}500$; variance $\$900{,}000\text{ dollars}^2$)
A gamma severity has shape $6$ and rate $0.003$ per dollar. Find its equivalent scale, mean, and variance. (Answer: scale $\$333.333\ldots$; mean $\$2{,}000$; variance $\$666{,}666.666\ldots\text{ dollars}^2$)
A claim amount measured in thousands of dollars has gamma shape $3$ and scale $1.5$ thousand dollars. Find its mean in dollars and its variance in dollars squared. (Answer: mean $\$4{,}500$; variance $\$6{,}750{,}000\text{ dollars}^2$)
A severity model has shape $4$ and rate $2$ per thousand dollars. Convert to scale, then find the mean and variance in dollars and dollars squared. (Answer: scale $0.5$ thousand dollars; mean $\$2{,}000$; variance $\$1{,}000{,}000\text{ dollars}^2$)
A gamma model has shape $2$ and a stated parameter of $0.005$ per dollar. Determine whether it is a scale or rate from its units, then calculate the mean and variance. (Answer: rate; mean $\$400$; variance $\$80{,}000\text{ dollars}^2$)

What You'll Learn