Deriving Gamma Severity Moments

Introduction

A gamma severity model is often specified by its shape and scale, while data summaries are more likely to be given as a mean, variance, or coefficient of variation. Moving between these descriptions is straightforward once the raw moments are derived from the density. The derivation also makes clear which parameter controls relative spread and why a rate parameter must be handled differently from a scale parameter.

This tutorial derives the gamma raw moments from first principles and applies them to parameter conversions. It focuses on moments, not on the distribution’s density profile or its interpretation as a sum of other random variables.

Definition: Let $X$ have a gamma distribution with shape $\alpha>0$ and scale $\theta>0$, written $X\sim\operatorname{Gamma}(\alpha,\theta)$. Its density is $$ f_X(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha},\qquad x>0, $$ and $f_X(x)=0$ for $x\leq 0$. The gamma function is $\Gamma(a)=\int_0^\infty t^{a-1}e^{-t}\,dt$ for $a>0$. In the shape-rate parameterization, the rate is $\beta=1/\theta$, so the same density is $f_X(x)=\frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}$.

The scale $\theta$ has the same units as $X$, such as dollars. The rate $\beta$ has inverse units, such as per dollar. The two parameterizations describe the same family: changing from scale to rate means taking a reciprocal, not leaving the numerical value unchanged.

Deriving the Raw Moments

For a real number $r$, the $r$th raw moment is $E[X^r]$, provided the expectation is finite. Applying the density definition and changing variables from loss $x$ to $t=x/\theta$ gives the general formula.

Start with the expectation integral.
For $r>-\alpha$, the integrand is integrable at zero, and the exponential factor ensures integrability as $x$ tends to infinity: $$ E[X^r]=\int_0^\infty x^r f_X(x)\,dx =\frac{1}{\Gamma(\alpha)\theta^\alpha}\int_0^\infty x^{\alpha+r-1}e^{-x/\theta}\,dx. $$

Rescale the integration variable.
Set $t=x/\theta$, so $x=\theta t$ and $dx=\theta\,dt$. Then $$ E[X^r]=\frac{\theta^r}{\Gamma(\alpha)} \int_0^\infty t^{\alpha+r-1}e^{-t}\,dt. $$

Recognize the gamma function.
The integral is $\Gamma(\alpha+r)$, which is finite when $\alpha+r>0$. Therefore $$ E[X^r]=\theta^r\frac{\Gamma(\alpha+r)}{\Gamma(\alpha)},\qquad r>-\alpha. $$

For nonnegative integer $n$, the gamma-function recurrence $\Gamma(z+1)=z\Gamma(z)$ simplifies this result. The recurrence follows by integration by parts in the defining integral: for $z>0$, take $u=t^z$ and $dv=e^{-t}dt$. The boundary term $-t^ze^{-t}$ is zero at both endpoints, leaving $\Gamma(z+1)=z\Gamma(z)$. Repeating the recurrence $n$ times gives the integer raw moments.

E[X^n] =\theta^n\frac{\Gamma(\alpha+n)}{\Gamma(\alpha)} =\theta^n\alpha(\alpha+1)\cdots(\alpha+n-1),\qquad n=1,2,\ldots

In particular, the first three raw moments are

E[X]=\alpha\theta,\qquad E[X^2]=\alpha(\alpha+1)\theta^2,\qquad E[X^3]=\alpha(\alpha+1)(\alpha+2)\theta^3.

All nonnegative integer raw moments exist for every $\alpha>0$ and $\theta>0$. The condition $r>-\alpha$ matters when using the general formula for a negative real power; it does not restrict ordinary nonnegative integer moments.

Variance and Coefficient of Variation

Variance is a central moment, so it is obtained by subtracting the square of the mean from the second raw moment. Substituting the derived expressions gives

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

Thus the standard deviation is $\sqrt{\alpha}\theta$. The coefficient of variation is the standard deviation divided by the mean, so for a gamma model it depends only on shape:

\operatorname{CV}(X)=\frac{\sqrt{\operatorname{Var}(X)}}{E[X]} =\frac{\sqrt{\alpha}\theta}{\alpha\theta} =\frac{1}{\sqrt{\alpha}}.

Theorem: If $X\sim\operatorname{Gamma}(\alpha,\theta)$, then $E[X]=\alpha\theta$, $\operatorname{Var}(X)=\alpha\theta^2$, and $\operatorname{CV}(X)=1/\sqrt{\alpha}$. Under the shape-rate parameterization $\beta=1/\theta$, these become $E[X]=\alpha/\beta$ and $\operatorname{Var}(X)=\alpha/\beta^2$.

The coefficient of variation is dimensionless, which is useful when severity amounts are expressed in different currency units. Multiplying every loss by a positive constant changes the mean and standard deviation by that constant but leaves their ratio unchanged.

Recovering Parameters from Moment Summaries

Suppose a gamma model is matched to a positive mean $m$ and variance $s^2$. Dividing the squared mean by the variance eliminates the scale:

m=\alpha\theta,\qquad s^2=\alpha\theta^2 \quad\Longrightarrow\quad \alpha=\frac{m^2}{s^2},\qquad \theta=\frac{s^2}{m}.

The corresponding rate is $\beta=m/s^2$. If instead the mean $m$ and coefficient of variation $c$ are given, then $\alpha=1/c^2$ and $\theta=mc^2$. These relationships require $m>0$, $s^2>0$, and $c>0$; they describe a nondegenerate gamma model.

Given information	Shape $\alpha$	Scale $\theta$	Rate $\beta$
Mean $m$, variance $s^2$	$m^2/s^2$	$s^2/m$	$m/s^2$
Mean $m$, coefficient of variation $c$	$1/c^2$	$mc^2$	$1/(mc^2)$

Worked Applications

Worked Example: Deriving and Evaluating Three Raw Moments

A hypothetical individual claim amount has $X\sim\operatorname{Gamma}(\alpha=2,\theta=\$1{,}500)$, with scale parameterization. Derive the raw-moment formula from the density and calculate the mean, variance, and first three raw moments. Give monetary moments to the nearest dollar, and the coefficient of variation to four decimal places.

Method. Use the density integral and gamma-function recurrence, then use the second raw moment to check the variance.

Step 1: Establish the raw-moment expression. Inserting $\alpha=2$ into the general density calculation gives $$ E[X^r]=\frac{1}{\Gamma(2)(1500)^2} \int_0^\infty x^{r+1}e^{-x/1500}\,dx =(1500)^r\frac{\Gamma(r+2)}{\Gamma(2)}. $$ Here $\Gamma(2)=1$, and the expression is finite for $r>-2$.

Step 2: Calculate the mean and variance. $$ E[X]=2(1500)=\$3{,}000. $$ $$ \operatorname{Var}(X)=2(1500)^2=2(2{,}250{,}000)=4{,}500{,}000\text{ dollars}^2. $$ The standard deviation is $\sqrt{4{,}500{,}000}=\$2{,}121.32$, and the coefficient of variation is $2121.32/3000=0.7071$.

Step 3: Calculate the second and third raw moments. $$ E[X^2]=2(3)(1500)^2=6(2{,}250{,}000)=13{,}500{,}000\text{ dollars}^2. $$ $$ E[X^3]=2(3)(4)(1500)^3=24(3{,}375{,}000{,}000) =81{,}000{,}000{,}000\text{ dollars}^3. $$

Check and common mistake avoided. The variance check is $E[X^2]-(E[X])^2=13{,}500{,}000-9{,}000{,}000=4{,}500{,}000$, as required. A common error is to report $E[X^2]$ as the variance; the subtraction of the squared mean is essential.

Worked Example: Recovering Shape and Rate from Mean and Variance

A hypothetical portfolio is modeled by a gamma claim severity with mean $\$8{,}000$ and variance $32{,}000{,}000$ dollars squared. Find its shape, scale, and rate, then calculate $E[X^3]$ and the coefficient of variation. Report $E[X^3]$ in dollars cubed.

Method. Match the model mean and variance to the gamma formulas, then apply the raw third-moment formula.

Step 1: Recover the shape. $$ \alpha=\frac{m^2}{s^2} =\frac{(8000)^2}{32{,}000{,}000} =\frac{64{,}000{,}000}{32{,}000{,}000}=2. $$

Step 2: Recover scale and rate. $$ \theta=\frac{s^2}{m} =\frac{32{,}000{,}000}{8{,}000} =\$4{,}000,\qquad \beta=\frac{1}{\theta}=\frac{1}{4000}=0.00025\text{ per dollar}. $$ The scale and rate are reciprocals, with the stated units.

Step 3: Calculate the third raw moment and coefficient of variation. $$ E[X^3]=2(3)(4)(4000)^3 =24(64{,}000{,}000{,}000) =1{,}536{,}000{,}000{,}000\text{ dollars}^3. $$ Also, $\operatorname{CV}(X)=1/\sqrt{2}=0.7071$.

Check and common mistake avoided. The implied mean is $2(4000)=8000$, and the implied variance is $2(4000)^2=32{,}000{,}000$, so both match the data. A common error is to use $s^2/m$ as the rate; it is the scale. The rate is its reciprocal.

Worked Example: Inferring Parameters from Mean and Relative Spread

For a hypothetical gamma claim severity, the mean is $\$6{,}000$ and the coefficient of variation is $0.5$. Determine the shape, scale, rate, variance, and third raw moment. Give the variance in dollars squared and the third moment in dollars cubed.

Method. Use the shape-only coefficient-of-variation relationship, then use the mean to find scale.

Step 1: Find the shape. $$ \alpha=\frac{1}{c^2}=\frac{1}{(0.5)^2} =\frac{1}{0.25}=4. $$

Step 2: Find scale and rate. $$ \theta=\frac{m}{\alpha}=\frac{6000}{4}=\$1{,}500,\qquad \beta=\frac{1}{1500}=0.0006667\text{ per dollar}. $$

Step 3: Calculate variance and the third raw moment. $$ \operatorname{Var}(X)=4(1500)^2 =4(2{,}250{,}000)=9{,}000{,}000\text{ dollars}^2. $$ $$ E[X^3]=4(5)(6)(1500)^3 =120(3{,}375{,}000{,}000) =405{,}000{,}000{,}000\text{ dollars}^3. $$

Check and common mistake avoided. The standard deviation is $\sqrt{9{,}000{,}000}=\$3{,}000$, and $3000/6000=0.5$, matching the stated coefficient of variation. Using $c^2$ instead of $1/c^2$ would give the wrong shape; the correct relationship is $\alpha=1/c^2$.

Common Errors and Practical Checks

The formulas are compact, but several errors can produce plausible-looking answers. Check both the parameter convention and the type of moment before substituting.

Confusing scale and rate. For a rate $\beta$, the mean is $\alpha/\beta$ and the variance is $\alpha/\beta^2$. For a scale $\theta$, they are $\alpha\theta$ and $\alpha\theta^2$.
Confusing raw and central moments. $E[X^2]$ is a raw moment, whereas $\operatorname{Var}(X)=E[X^2]-(E[X])^2$.
Using the wrong power of the scale. The $n$th raw moment contains $\theta^n$; the units must correspond to $X^n$.
Using the wrong coefficient-of-variation relationship. For gamma shape $\alpha$, $\operatorname{CV}=1/\sqrt{\alpha}$, so $\alpha=1/\operatorname{CV}^2$.
Ignoring the applicability condition. The general real-order moment formula requires $r>-\alpha$. For nonnegative integer moments, this condition is automatically satisfied.

Key takeaway: Deriving $E[X^r]=\theta^r\Gamma(\alpha+r)/\Gamma(\alpha)$ from the density gives the integer raw moments by recurrence. From those moments, $\operatorname{Var}(X)=\alpha\theta^2$, $\operatorname{CV}(X)=1/\sqrt{\alpha}$, and the moment-based parameter conversions follow directly.

Check Your Understanding

For each hypothetical gamma severity problem, identify the parameterization before calculating. Answers use exact values where convenient.

A severity has shape $5$ and scale $\$400$. Find the mean, variance, and coefficient of variation. (Answer: mean $\$2{,}000$, variance $800{,}000$ dollars squared, coefficient of variation $0.4472$.)
A gamma severity has mean $\$15{,}000$ and variance $225{,}000{,}000$ dollars squared. Find shape, scale, rate, and $E[X^2]$. (Answer: shape $1$, scale $\$15{,}000$, rate $1/15{,}000$ per dollar, and $E[X^2]=450{,}000{,}000$ dollars squared.)
A severity has shape $4$ and rate $0.0005$ per dollar. Find its scale, mean, variance, and third raw moment. (Answer: scale $\$2{,}000$, mean $\$8{,}000$, variance $16{,}000{,}000$ dollars squared, and $E[X^3]=960{,}000{,}000{,}000$ dollars cubed.)
A gamma severity has mean $\$9{,}000$ and coefficient of variation $0.6$. Find shape, scale, and variance. (Answer: shape $25/9$, scale $\$3{,}240$, variance $29{,}160{,}000$ dollars squared.)
A gamma severity has mean $\$8{,}000$ and second raw moment $100{,}000{,}000$ dollars squared. Find its variance, shape, scale, and coefficient of variation. (Answer: variance $36{,}000{,}000$ dollars squared, shape $16/9$, scale $\$4{,}500$, coefficient of variation $0.75$.)

Given information	Shape \(\alpha\)	Scale \(\theta\)	Rate \(\beta\)
Mean \(m\), variance \(s^2\)	\(m^2/s^2\)	\(s^2/m\)	\(m/s^2\)
Mean \(m\), coefficient of variation \(c\)	\(1/c^2\)	\(mc^2\)	\(1/(mc^2)\)

What You'll Learn