Gamma Mean And Variance Parameterizations

Why Gamma Parameterizations Need Care

The Gamma distribution is often used to represent positive, continuous insurance quantities such as claim severity. It has several parameterizations in common use. The two most important here use shape and scale or shape and rate. Both describe the same family of distributions, but the second parameter has a reciprocal relationship: rate is the reciprocal of scale. Confusing the two can turn a plausible-looking calculation into a severely incorrect one.

The key discipline is to identify the parameterization before calculating. State what each parameter means, keep its units attached, and use the corresponding formulas for the mean and variance. This tutorial is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: A random variable $X$ has a Gamma distribution with shape $\alpha>0$ and scale $\theta>0$, written $X\sim\operatorname{Gamma}(\alpha,\theta)$ in the shape-scale convention, if its density is $$ f_X(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha},\qquad x>0. $$ The equivalent shape-rate convention uses shape $\alpha>0$ and rate $\lambda>0$: $$ f_X(x)=\frac{\lambda^\alpha x^{\alpha-1}e^{-\lambda x}}{\Gamma(\alpha)},\qquad x>0. $$ Here $\Gamma(\alpha)$ is the Gamma function. Values outside $x>0$ have density zero.

The two densities describe the same distribution when their parameters satisfy $\lambda=1/\theta$. The shape $\alpha$ is unchanged. Scale has the same units as $X$; rate has inverse units. For example, if $X$ is measured in dollars, $\theta$ is in dollars and $\lambda$ is per dollar.

Parameter conversion: In the shape-scale and shape-rate conventions, $$ \lambda=\frac{1}{\theta},\qquad \theta=\frac{1}{\lambda}. $$ Thus $\operatorname{Gamma}(\alpha,\theta)$ in shape-scale notation and $\operatorname{Gamma}(\alpha,\lambda)$ in shape-rate notation refer to the same distribution only when the second parameters are reciprocals. Always state which convention is being used.

Mean and Variance in Each Convention

In shape-scale notation, the mean and variance are $\alpha\theta$ and $\alpha\theta^2$, respectively. In shape-rate notation, they are $\alpha/\lambda$ and $\alpha/\lambda^2$. These formulas agree under $\lambda=1/\theta$, but they are not interchangeable without converting the parameter.

\begin{array}{c|c|c} \text{Convention} & E[X] & \operatorname{Var}(X)\\ \hline \text{Shape-scale }(\alpha,\theta) & \alpha\theta & \alpha\theta^2\\ \text{Shape-rate }(\alpha,\lambda) & \alpha/\lambda & \alpha/\lambda^2 \end{array}

A short derivation shows why these moment formulas hold. For the shape-scale density, use the Gamma-function identity $\Gamma(a+1)=a\Gamma(a)$. Integrating $x f_X(x)$ gives

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

For the second raw moment, the same integral with $x^2$ in place of $x$ gives $E[X^2]=\alpha(\alpha+1)\theta^2$. Therefore,

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

To obtain the shape-rate formulas, substitute $\theta=1/\lambda$. This gives $E[X]=\alpha/\lambda$ and $\operatorname{Var}(X)=\alpha/\lambda^2$. The conversion is also confirmed directly from the densities: replacing $\theta$ by $1/\lambda$ changes $e^{-x/\theta}$ to $e^{-\lambda x}$, and changes $1/\theta^\alpha$ to $\lambda^\alpha$.

Worked Example: Converting a Scale to a Rate

Assume a hypothetical claim severity $X$, measured in dollars, has a Gamma distribution with shape $\alpha=4$ and scale $\theta=225$ dollars. Convert to shape-rate notation and calculate the mean and variance.

Method. Use the reciprocal conversion for the rate, then use the formulas that match the parameterization. This avoids treating a scale as if it were already a rate.

Convert the parameter. The exact rate is $\lambda=1/225$ per dollar. To seven significant digits, this is $0.004444444$ per dollar. The equivalent pairs are therefore shape-scale $(4,225\text{ dollars})$ and shape-rate $(4,1/225\text{ per dollar})$.

Calculate the mean. In shape-scale notation, $E[X]=\alpha\theta=4(225)=900$ dollars. In shape-rate notation, $E[X]=\alpha/\lambda=4/(1/225)=4(225)=900$ dollars.

Calculate the variance. In shape-scale notation, $\operatorname{Var}(X)=\alpha\theta^2=4(225^2)=4(50{,}625)=202{,}500$ dollars squared. In shape-rate notation, $\operatorname{Var}(X)=\alpha/\lambda^2=4/(1/225)^2=4(50{,}625)=202{,}500$ dollars squared.

Check. Both parameterizations give the same mean and variance, and the rate has inverse-dollar units. The standard deviation, if needed, is $\sqrt{202{,}500}=450$ dollars.

Units and a Reliable Calculation Sequence

Units provide a useful check on both parameter conversion and moment calculations. If $X$ is measured in dollars, then $E[X]$ must be in dollars and $\operatorname{Var}(X)$ must be in dollars squared. Consequently, $\theta$ is in dollars and $\lambda$ is per dollar. In the rate formulas, dividing a dimensionless shape by a rate produces dollars for the mean; dividing by the squared rate produces dollars squared for the variance.

Identify the convention.
Check whether the density uses $e^{-x/\theta}$ (scale) or $e^{-\lambda x}$ (rate). A parameter name alone may not identify the convention.

Record units.
Write the units of $X$, then give scale those same units and rate their reciprocal units.

Convert if necessary.
Use $\lambda=1/\theta$ or $\theta=1/\lambda$, keeping an exact reciprocal until the final rounding.

Use matching moment formulas.
For scale, use $\alpha\theta$ and $\alpha\theta^2$; for rate, use $\alpha/\lambda$ and $\alpha/\lambda^2$.

Check the result.
Verify units and, when possible, calculate a moment in both conventions to confirm that the answers agree.

Worked Example: Calculating Moments From a Rate

Suppose a hypothetical portfolio models individual claim severities $X$, in thousands of dollars, by a Gamma distribution with shape $\alpha=5$ and rate $\lambda=0.25$ per thousand dollars. Find the equivalent scale, mean, and variance.

Method. Convert the rate to its reciprocal scale, or use the shape-rate moment formulas directly. The common mistake to avoid is using $0.25$ as a scale even though the problem specifies it as a rate.

Convert to scale. $\theta=1/\lambda=1/0.25=4$ thousand dollars. The equivalent parameter pairs are $(5,4)$ in shape-scale notation and $(5,0.25)$ in shape-rate notation, with units as stated.

Find the mean. Using rate notation, $E[X]=\alpha/\lambda=5/0.25=20$ thousand dollars. The scale calculation agrees: $5(4)=20$ thousand dollars.

Find the variance. Using rate notation, $\operatorname{Var}(X)=\alpha/\lambda^2=5/(0.25^2)=5/0.0625=80$ thousand-dollars squared. The scale calculation agrees: $5(4^2)=5(16)=80$ thousand-dollars squared.

Check. The standard deviation is $\sqrt{80}\approx8.9443$ thousand dollars. The variance has squared units, as it must, and the two conventions produce identical results.

Recovering Parameters From Moments

Sometimes a problem gives a target mean $m>0$ and variance $s^2>0$, then asks for the Gamma parameters that produce them. Equating the target moments to the shape-scale formulas gives $m=\alpha\theta$ and $s^2=\alpha\theta^2$. Dividing the variance equation by the mean equation yields $\theta=s^2/m$; substituting this into the mean equation yields $\alpha=m^2/s^2$. In rate notation, the equivalent result is $\lambda=m/s^2$.

Moment-based parameter conversion: For a Gamma distribution with mean $m$ and variance $s^2$, $$ \alpha=\frac{m^2}{s^2},\qquad \theta=\frac{s^2}{m},\qquad \lambda=\frac{m}{s^2}. $$ These formulas require $m>0$ and $s^2>0$. They give the same distribution because $\theta\lambda=1$.

Worked Example: Finding Both Parameterizations From Moments

A hypothetical severity model has mean $1{,}200$ dollars and variance $720{,}000$ dollars squared. Assuming a Gamma distribution, determine its shape, scale, and rate.

Method. Solve for the parameters from the specified moments, then verify by substituting them back. This avoids choosing a parameterization based on an unlabeled second parameter.

Find the shape. $\alpha=m^2/s^2=1{,}200^2/720{,}000=1{,}440{,}000/720{,}000=2$. The shape is dimensionless.

Find the scale. $\theta=s^2/m=720{,}000/1{,}200=600$ dollars.

Find the rate. $\lambda=m/s^2=1{,}200/720{,}000=1/600$ per dollar, approximately $0.001666667$ per dollar to seven significant digits. This is the reciprocal of the $600$-dollar scale.

Check the moments. The mean is $2(600)=1{,}200$ dollars. The variance is $2(600^2)=2(360{,}000)=720{,}000$ dollars squared. The rate formulas give $2/(1/600)=1{,}200$ dollars and $2/(1/600)^2=720{,}000$ dollars squared, confirming both parameterizations.

Common Parameterization Trap

Suppose a problem states that $X$ is Gamma with shape $4$ and scale $225$ dollars. A wrong calculation treats $225$ as a rate and uses $E[X]=\alpha/\lambda=4/225$. This gives approximately $0.0178$, and its units would be dollars only if $225$ had units per dollar. But the given $225$ is a scale measured in dollars, not a rate.

The correction is to use the shape-scale formula directly: $E[X]=4(225)=900$ dollars. If rate notation is preferred, first convert the scale to $\lambda=1/225$ per dollar; then $\alpha/\lambda=4/(1/225)=900$ dollars. The same discipline applies to the variance: use $\alpha\theta^2$ for scale or $\alpha/\lambda^2$ for rate, not a mixture of the two.

Key takeaway: Shape is common to both Gamma conventions; scale and rate are reciprocals. Keep the convention and units explicit, use the matching mean and variance formulas, and verify that conversions preserve the distribution’s moments.

Check Your Understanding

For each problem, identify the parameterization before calculating. Answers are given for checking.

A hypothetical claim severity $X$, measured in dollars, has shape $\alpha=2$ and scale $\theta=300$ dollars. Find the equivalent rate, mean, and variance. (Answer: $\lambda=1/300$ per dollar; $E[X]=600$ dollars; $\operatorname{Var}(X)=180{,}000$ dollars squared.)
A severity $X$, in dollars, has shape $\alpha=6$ and rate $\lambda=0.03$ per dollar. Find the scale, mean, and variance. (Answer: $\theta=100/3$ dollars; $E[X]=200$ dollars; $\operatorname{Var}(X)=6{,}666.67$ dollars squared.)
A Gamma model for a hypothetical loss measured in dollars has shape $\alpha=2.5$ and scale $\theta=800$ dollars. Give the equivalent rate and calculate both moments. (Answer: $\lambda=0.00125$ per dollar; $E[X]=2{,}000$ dollars; $\operatorname{Var}(X)=1{,}600{,}000$ dollars squared.)
A Gamma severity has mean $900$ dollars and variance $405{,}000$ dollars squared. Find the shape-scale and shape-rate parameter pairs. (Answer: shape $2$, scale $450$ dollars, rate $1/450$ per dollar.)
A hypothetical model specifies a Gamma severity with shape $3$ and mean $2{,}400$ dollars. Find its scale, rate, and variance, and state which convention uses each second parameter. (Answer: shape-scale $(3,800\text{ dollars})$; shape-rate $(3,1/800\text{ per dollar})$; variance $1{,}920{,}000$ dollars squared.)

What You'll Learn