Gamma Severity Models

Why the Gamma Model Is Useful

Severity models describe the size of a loss, conditional on a loss being observed. The gamma distribution is a flexible continuous model for nonnegative losses: its density can have different shapes, and its moments have compact formulas. The calculations require care because the rate and scale conventions use reciprocal parameters. This tutorial develops the density, probabilities, and moments needed to use the model reliably.

We treat $X$ as a continuous, nonnegative loss. A gamma model assigns probability only to positive values; in particular, it has no probability mass at zero. If the application includes a distinct probability of no loss, a continuous gamma distribution alone does not represent that atom.

Definition: A random variable $X$ has a gamma distribution with shape $\alpha>0$ and rate $\lambda>0$ if its density is $$ f_X(x)=\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x},\qquad x>0, $$ and $f_X(x)=0$ otherwise. Here $\Gamma(\alpha)=\int_0^\infty t^{\alpha-1}e^{-t}\,dt$ is the gamma function. The corresponding scale is $\theta=1/\lambda$; in shape-scale notation the density is $x^{\alpha-1}e^{-x/\theta}/[\Gamma(\alpha)\theta^\alpha]$, for $x>0$.

The shape $\alpha$ controls the form of the density, while the rate $\lambda$ controls the scale of the loss: increasing $\lambda$ concentrates the model at smaller values. The scale $\theta$ acts in the opposite direction: increasing $\theta$ stretches the loss scale. Always identify whether the second parameter is a rate or a scale before substituting it into a formula.

Probabilities and the Cumulative Distribution Function

For any $x\leq 0$, $F_X(x)=0$, since the support is positive. For $x>0$, the cumulative distribution function is the integral of the density from zero to $x$. Setting $t=\lambda y$ in that integral gives a convenient standardized form:

F_X(x)=P(X\leq x) =\frac{1}{\Gamma(\alpha)}\int_0^{\lambda x}t^{\alpha-1}e^{-t}\,dt,\qquad x>0.

This integral is the regularized lower incomplete gamma function. A calculator or statistical software may evaluate it directly. If $\alpha$ is a positive integer, integration by parts also gives a finite expression:

F_X(x)=1-e^{-\lambda x}\sum_{j=0}^{\alpha-1}\frac{(\lambda x)^j}{j!}.

The finite expression is useful when its integer-shape condition holds. It must not be applied by rounding a noninteger shape to an integer. For any shape, an interval probability is found by subtracting cumulative probabilities, and a strict upper-tail probability is $P(X>x)=1-F_X(x)$. Because the distribution is continuous, including or excluding a single endpoint does not change these probabilities.

Worked Example: Probability Between Two Loss Thresholds

A hypothetical portfolio models a positive claim amount $X$, in thousands of dollars, as gamma with shape $\alpha=2$ and rate $\lambda=0.5$ per thousand dollars. Find $P(2<X\leq6)$, $E[X]$, and $f_X(4)$. Give the probability to four decimals and the density to four decimals.

Method. Use the cumulative distribution function for the interval, then use the gamma moment and density formulas. The shape is an integer, so the finite cumulative formula applies. For shape 2 it reduces to $F_X(x)=1-e^{-0.5x}(1+0.5x)$.

At $x=6$, the standardized argument is $\lambda x=0.5(6)=3$, so $F_X(6)=1-e^{-3}(1+3)=1-4e^{-3}=0.8008517$.

At $x=2$, the standardized argument is $\lambda x=0.5(2)=1$, so $F_X(2)=1-e^{-1}(1+1)=1-2e^{-1}=0.2642411$.

Therefore, $P(2<X\leq6)=F_X(6)-F_X(2)=0.8008517-0.2642411=0.5366106$, or 0.5366 to four decimals.

The mean is $\alpha/\lambda=2/0.5=4$ thousand dollars. For the density at $x=4$, $f_X(4)=0.5^2(4)^{2-1}e^{-0.5(4)}/\Gamma(2)$. Since $\Gamma(2)=1$, this is $0.25(4)e^{-2}=e^{-2}=0.1353353$, or 0.1353 per thousand dollars to four decimals.

Check. The interval probability is positive and less than one, and $F_X(2)<F_X(6)$. The density is a height, not a probability; its units are inverse thousands of dollars.

Moments: Derivation and Formulas

The gamma function makes the raw moments particularly direct. For $r$ such that $\alpha+r>0$, integrate $x^r$ against the density and substitute $t=\lambda x$. The condition ensures the integral near zero is finite; the exponential factor ensures convergence at infinity.

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

The second equality follows from $t=\lambda x$, so $x=t/\lambda$ and $dx=dt/\lambda$. The resulting integral is $\Gamma(\alpha+r)$, which proves the raw-moment formula rather than merely identifying it. For positive integer $r$, the recurrence $\Gamma(z+1)=z\Gamma(z)$ simplifies the ratio to a product.

Theorem: For $X$ with shape $\alpha$ and rate $\lambda$, $$ E[X]=\frac{\alpha}{\lambda},\qquad E[X^2]=\frac{\alpha(\alpha+1)}{\lambda^2},\qquad \operatorname{Var}(X)=\frac{\alpha}{\lambda^2}. $$ In shape-scale notation, where $\theta=1/\lambda$, these are $E[X]=\alpha\theta$, $E[X^2]=\alpha(\alpha+1)\theta^2$, and $\operatorname{Var}(X)=\alpha\theta^2$.

The variance follows from $E[X^2]-(E[X])^2$: $\alpha(\alpha+1)/\lambda^2-\alpha^2/\lambda^2=\alpha/\lambda^2$. The standard deviation is $\sqrt{\alpha}/\lambda$, or $\sqrt{\alpha}\theta$. These quantities inherit the units of $X$ for the mean and standard deviation; variance has squared units.

Worked Example: Upper-Tail Probability and a Third Moment

A hypothetical repair-loss amount $X$, in thousands of dollars, follows a gamma model with shape $\alpha=3$ and rate $\lambda=2$ per thousand dollars. Find $P(X>2)$, $E[X]$, $\operatorname{Var}(X)$, and $E[X^3]$. Report the probability to four decimals.

Method. Standardize the threshold using $\lambda x$ and use the integer-shape cumulative formula. Use the raw-moment result for the third moment; do not confuse it with the cube of the mean.

At the threshold $x=2$, $\lambda x=2(2)=4$. For shape 3, $F_X(2)=1-e^{-4}(1+4+4^2/2)=1-13e^{-4}$. Thus $P(X>2)=13e^{-4}=0.2381033$, or 0.2381 to four decimals.

The mean is $3/2=1.5$ thousand dollars, and the variance is $3/2^2=0.75$ squared thousand dollars.

For the third raw moment, $E[X^3]=\Gamma(6)/[\Gamma(3)2^3]$. Using $\Gamma(6)=5!=120$ and $\Gamma(3)=2!=2$, this equals $120/(2\cdot8)=7.5$ cubic thousand dollars.

Check. The tail probability is below one and is positive. Also $E[X^3]$ is a raw moment, not $1.5^3=3.375$; the two quantities need not be equal.

Working With the Scale Parameter

Some descriptions provide shape and scale rather than shape and rate. The density then contains $\theta^\alpha$ in the denominator and $e^{-x/\theta}$ in the exponential. To use a rate-based formula, set $\lambda=1/\theta$. An equally reliable approach is to use the scale formulas directly. The two methods must agree because they describe the same distribution.

Worked Example: Converting Scale to Rate

A hypothetical claim amount $X$, in thousands of dollars, has shape $\alpha=2$ and scale $\theta=1.5$. Find its rate, mean, variance, density at $x=3$, and $P(X\leq3)$. Give the density and probability to four decimals.

Method. Convert the scale to its reciprocal rate, then use the formulas for shape 2. The standardized threshold is $x/\theta$, equivalently $\lambda x$.

The rate is $\lambda=1/\theta=1/1.5=2/3$ per thousand dollars. The mean is $\alpha\theta=2(1.5)=3$ thousand dollars. The variance is $\alpha\theta^2=2(1.5^2)=2(2.25)=4.5$ squared thousand dollars.

At $x=3$, the density is $f_X(3)=3^{2-1}e^{-3/1.5}/[\Gamma(2)(1.5^2)]$. Substituting $\Gamma(2)=1$ gives $3e^{-2}/2.25=(4/3)e^{-2}=0.1804470$, or 0.1804 per thousand dollars to four decimals.

For the probability, $x/\theta=3/1.5=2$. Hence $F_X(3)=1-e^{-2}(1+2)=1-3e^{-2}=0.5939942$, or 0.5940 to four decimals.

Check. The probability below the mean is plausible but need not be one-half; a gamma distribution need not be symmetric. The rate $2/3$ is the reciprocal of the scale, not the scale itself.

Common Exam Traps

Swapping rate and scale. If the model gives scale $\theta$, the exponential term is $e^{-x/\theta}$. If it gives rate $\lambda$, it is $e^{-\lambda x}$. Substituting a scale directly as a rate changes the distribution.
Forgetting the support. The density is zero for $x\leq0$, and a cumulative probability at a nonpositive threshold is zero. The gamma model does not assign negative losses.
Using a density as a probability. A density value such as $f_X(3)$ is not $P(X=3)$. For a continuous model, $P(X=3)=0$; probabilities come from integrating the density over an interval or evaluating a CDF difference.
Applying the integer formula to every shape. The finite sum for $F_X(x)$ above requires a positive integer shape. For a noninteger shape, use the incomplete-gamma CDF or an appropriate calculator function.
Mixing raw moments and central moments. The variance is $E[X^2]-(E[X])^2$; a raw third moment is not the cube of the mean.

A reliable calculation starts by writing the support and parameter convention, then checks whether a probability requires a CDF difference, an upper tail, or a density integral. For applied severity work, also retain units: a rate has reciprocal-loss units, while a scale and mean have loss units. This material is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each model, identify the appropriate formula before calculating. Probabilities are rounded to four decimals when needed.

A loss $X$ has shape $\alpha=2$ and rate $\lambda=1$. Find its mean, variance, and second raw moment. (Answer: $E[X]=2$, $\operatorname{Var}(X)=2$, $E[X^2]=6$.)
A claim amount has shape $\alpha=3$ and rate $\lambda=0.5$. Find the probability that it exceeds 4. (Answer: $5e^{-2}=0.6767$.)
A loss has shape $\alpha=2$ and scale $\theta=2$. Find the rate and density at $x=2$. (Answer: $\lambda=0.5$, $f_X(2)=e^{-1}/2=0.1839$.)
A severity $X$ has shape $\alpha=1.5$ and rate $\lambda=3$. Find its mean, variance, and second raw moment. (Answer: $E[X]=0.5$, $\operatorname{Var}(X)=1/6$, $E[X^2]=5/12$.)
A loss has shape $\alpha=2$ and rate $\lambda=0.25$. Find $P(2<X\leq6)$. (Answer: $e^{-0.5}(1+0.5)-e^{-1.5}(1+1.5)=0.3520$.)

What You'll Learn