Gamma Distributions For Claim Severity

Why Use a Gamma Model for Claim Severity?

Claim severities are generally positive amounts, and their distributions are often right-skewed: many claims are moderate, while a smaller number are much larger. The Gamma distribution is one continuous model for this pattern. It assigns probability only to positive values and can represent a range of right-skewed shapes, making it a useful candidate for modeling a positive claim amount.

A Gamma model is a distributional assumption, not a statement that actual losses must follow that exact pattern. In a ratemaking analysis, an actuary should check whether the model is appropriate for the particular claim definition, coverage, and data. The formulas below describe calculations conditional on the stated model.

Definition: In this tutorial, $X$ is a positive claim severity with a Gamma distribution having shape parameter $\alpha>0$ and scale parameter $\theta>0$, 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 $x\leq 0$. Here $\Gamma(\alpha)$ is the Gamma function, which for a positive integer $n$ satisfies $\Gamma(n)=(n-1)!$. The scale $\theta$ has the same units as $X$, such as dollars; the shape $\alpha$ is dimensionless.

The parameters are not interchangeable. Some sources specify a Gamma distribution using a rate $\lambda$ rather than a scale. The equivalent rate form uses $\lambda=1/\theta$, so $\lambda$ has reciprocal units, such as per dollar. This tutorial uses the shape–scale form throughout. If a problem gives a rate, convert it to a scale before substituting into the formulas here.

For this parameterization, the expected severity is $E[X]=\alpha\theta$. For positive integer shape $k$, the survival probability has a particularly convenient finite-sum expression. That expression makes many exam-style probability calculations possible without evaluating a general incomplete Gamma function.

Probabilities When the Shape Is an Integer

Let the shape be a positive integer $k$, and let $z=x/\theta$. The cumulative distribution function is $F_X(x)=P(X\leq x)$; the survival function is $S_X(x)=P(X>x)=1-F_X(x)$. For $x\geq 0$,

S_X(x)=e^{-x/\theta}\sum_{j=0}^{k-1}\frac{(x/\theta)^j}{j!}.

This finite sum is especially useful for calculating the probability of exceeding a threshold or falling in an interval. For example, when $k=1$, the sum contains only 1, so $S_X(x)=e^{-x/\theta}$. When $k=2$, it is $e^{-x/\theta}(1+x/\theta)$. For a noninteger shape, the Gamma CDF or survival function is generally evaluated with a numerical tool or table rather than this finite sum.

Interpretation: The ratio $x/\theta$ is dimensionless. The threshold and the scale must be expressed in the same units before forming this ratio. For a continuous Gamma severity, $P(X=x)=0$, so $P(a<X\leq b)=F_X(b)-F_X(a)=S_X(a)-S_X(b)$.

Worked Example: A Mean and a Threshold Probability

A hypothetical equipment insurer models the positive repair severity $X$, in dollars, as $\operatorname{Gamma}(2,600)$ in the shape–scale parameterization. Calculate the expected severity, the probability a claim exceeds $600, and the probability it does not exceed $600. Give probabilities to four decimal places.

Method and parameter check. The shape is $k=2$, and the scale is $\theta=\$600$. The threshold is also in dollars, so $z=600/600=1$.

Expected severity. For this parameterization, $E[X]=\alpha\theta$, hence $$ E[X]=2(600)=\$1{,}200. $$

Exceedance probability. The integer-shape survival formula gives $$ P(X>600)=S_X(600)=e^{-1}\left(1+\frac{600}{600}\right)=2e^{-1}=0.7357588823\ldots. $$ Thus, to four decimal places, $P(X>600)=0.7358$.

Non-exceedance probability. By the complement rule, $$ P(X\leq 600)=1-0.7357588823\ldots=0.2642411177\ldots, $$ which is $0.2642$ to four decimal places.

Check and common mistake. The two probabilities sum to $0.7357588823\ldots+0.2642411177\ldots=1$. A common error is to report $S_X(600)$ as the probability below $600. It is the probability above the threshold; the probability at or below it is the complement.

Why the Integer-Shape Survival Formula Holds

The finite-sum expression follows directly from integrating the density, not from an additional approximation. For integer $k$, substitute $t=u/\theta$ in the tail integral. Since $\Gamma(k)=(k-1)!$, this gives

S_X(x)=\frac{1}{(k-1)!}\int_{x/\theta}^{\infty} t^{k-1}e^{-t}\,dt.

Write $z=x/\theta$. Integration by parts, taking $t^{k-1}$ as the polynomial factor and $e^{-t}\,dt$ as the factor to integrate, yields

\int_z^\infty t^{k-1}e^{-t}\,dt =z^{k-1}e^{-z}+(k-1)\int_z^\infty t^{k-2}e^{-t}\,dt.

The boundary term at infinity is zero because $t^{k-1}e^{-t}$ tends to zero as $t$ tends to infinity. Repeating this recurrence reduces the integral to $\int_z^\infty e^{-t}\,dt=e^{-z}$. The result is

\int_z^\infty t^{k-1}e^{-t}\,dt =(k-1)!e^{-z}\sum_{j=0}^{k-1}\frac{z^j}{j!}.

Dividing by $(k-1)!$ proves $S_X(x)=e^{-z}\sum_{j=0}^{k-1}z^j/j!$. The derivation also explains why the formula is finite for integer shape: repeated integration by parts terminates after finitely many steps.

Worked Example: Probability in a Severity Interval

A hypothetical specialty insurer models a positive claim severity $X$, in dollars, as $\operatorname{Gamma}(3,400)$. Calculate the probability that a claim is more than $400 and no more than $1,200. Give the final probability to four decimal places.

Method. Since the integer shape is $k=3$, calculate the survival probability at each endpoint using $$ S_X(x)=e^{-x/400}\left(1+\frac{x}{400}+\frac{(x/400)^2}{2}\right). $$ The requested interval probability is $S_X(400)-S_X(1{,}200)$.

At $400, $z=400/400=1$, so, retaining full precision in the calculation, $$ S_X(400)=e^{-1}\left(1+1+\frac{1^2}{2}\right) =0.9196986029\ldots. $$

At $1,200, $z=1{,}200/400=3$, and $$ S_X(1{,}200)=e^{-3}\left(1+3+\frac{3^2}{2}\right) =0.4231900811\ldots. $$

Therefore, $$ P(400<X\leq 1{,}200)=0.9196986029\ldots-0.4231900811\ldots =0.4965085218\ldots. $$ To four decimal places, the probability is $0.4965$.

Check and common mistake. The probability is positive and less than 1, as required. Also, the survival probability at $400 is larger than the one at $1,200, as a survival function must be nonincreasing. Rounding each exponential too early can alter the final displayed digits; keep full precision through subtraction and round only the final probability.

Expected Amount Above a Deductible

A severity probability is not itself an expected payment. For a per-loss deductible $d$, the payment above the deductible is $(X-d)_+=\max(X-d,0)$. Its expected value accounts both for the probability of exceeding the deductible and for how far the severity exceeds it. For a Gamma model with integer shape $k$, the expectation can be calculated using the same tail integral.

First, write the positive part as $(X-d)_+=X\,\mathbf{1}_{\{X>d\}}-d\,\mathbf{1}_{\{X>d\}}$, where $\mathbf{1}_{\{X>d\}}$ equals 1 when $X>d$ and 0 otherwise. Taking expectations gives

E[(X-d)_+]=E[X\mathbf{1}_{\{X>d\}}]-dS_X(d).

For $X\sim\operatorname{Gamma}(k,\theta)$, substituting $t=x/\theta$ in the first term gives

E[X\mathbf{1}_{\{X>d\}}] =\frac{\theta}{(k-1)!}\int_{d/\theta}^{\infty}t^ke^{-t}\,dt =k\theta e^{-z}\sum_{j=0}^{k}\frac{z^j}{j!}, \qquad z=\frac{d}{\theta}.

The last equality is the same integration-by-parts result, now applied with power $k$. Combining it with $S_X(d)=e^{-z}\sum_{j=0}^{k-1}z^j/j!$ gives an expected payment per claim in the same currency units as $\theta$ and $d$. This formula assumes a payment of $(X-d)_+$ with no upper limit; it does not calculate a payment capped by a policy limit.

Worked Example: Expected Payment Above a Deductible

A hypothetical insurer models a positive claim severity $X$ in dollars as $\operatorname{Gamma}(2,500)$. The policy pays $(X-500)_+$, with no upper limit. Calculate the expected payment per claim to the nearest cent.

Set up the parameters. Here $k=2$, $\theta=\$500$, $d=\$500$, and $z=d/\theta=1$. The survival probability at the deductible is $$ S_X(500)=e^{-1}(1+1)=\frac{2}{e}=0.7357588823\ldots. $$

Calculate the truncated severity amount. Using the integer-shape tail integral, $$ E[X\mathbf{1}_{\{X>500\}}] =2(500)e^{-1}\left(1+1+\frac{1}{2}\right) =\frac{2{,}500}{e} =\$919.6986029\ldots. $$

Subtract the deductible on claims that exceed it. The expected deductible amount subtracted is $$ 500S_X(500)=500\left(\frac{2}{e}\right) =\frac{1{,}000}{e} =\$367.8794412\ldots. $$ Therefore, $$ E[(X-500)_+]=\frac{2{,}500}{e}-\frac{1{,}000}{e} =\frac{1{,}500}{e} =\$551.8191618\ldots. $$ Rounded only at the final step to the nearest cent, the expected payment is $\$551.82$ per claim.

Check and common mistake. The payment is nonnegative and its expected value is below the mean severity $E[X]=2(500)=\$1{,}000$, which is sensible. A common error is to calculate only $P(X>500)$ and treat that probability as the expected payment. The probability is dimensionless; the expected payment must have dollar units and depends on the sizes of claims above the deductible.

Model Checks and Common Traps

Gamma severity calculations are straightforward when the parameterization and units are explicit. Before calculating, identify whether the problem gives a scale or a rate, check that dollar thresholds and scales use the same monetary units, and determine whether the requested quantity is a probability, a mean severity, or an expected payment.

Scale versus rate: in the shape–scale form, $x/\theta$ is the dimensionless threshold. If a rate $\lambda$ is supplied, use $\theta=1/\lambda$; do not substitute $\lambda$ as though it were a scale.
Tail versus cumulative probability: $S_X(x)$ is $P(X>x)$, while $F_X(x)$ is $P(X\leq x)=1-S_X(x)$.
Interval order: for $a<b$, $P(a<X\leq b)=S_X(a)-S_X(b)$, not the reverse.
Support: a standard Gamma severity is positive and continuous. It cannot represent a probability mass at zero or negative outcomes. If the data include zero outcomes, a model that handles those outcomes is needed rather than silently applying a positive-severity Gamma model to them.
Parameter scope: this tutorial uses the shape–scale form and its basic mean property. A supplied distribution must be interpreted under the parameterization actually stated in the problem.

The Gamma distribution is therefore a candidate for positive claim amounts, not a universal model for every insurance response. A correct calculation depends first on confirming the support, parameterization, and target quantity; only then should the density or survival formula be applied.

Check Your Understanding

For each problem, use the shape–scale convention unless a rate is explicitly given. Probabilities are rounded to four decimal places and dollar amounts to the nearest cent.

A hypothetical repair severity has distribution $\operatorname{Gamma}(2,300)$, with amounts in dollars. Find $E[X]$ and $P(X>600)$. (Answer: $E[X]=\$600$; $P(X>600)=0.4060$)
A hypothetical claim severity has distribution $\operatorname{Gamma}(3,200)$. Find $P(200<X\leq600)$. (Answer: $0.4965$)
A hypothetical loss severity is exponential with rate $0.004$ per dollar, equivalently Gamma with shape 1. Find the probability that the severity is at most $500. (Answer: $0.8647$)
A hypothetical positive equipment-loss severity has distribution $\operatorname{Gamma}(4,100)$, with amounts in dollars. Calculate its expected severity. (Answer: $\$400$)
A hypothetical claim severity has distribution $\operatorname{Gamma}(2,400)$, and a policy pays $(X-400)_+$ with no upper limit. Calculate the expected payment per claim to the nearest cent. (Answer: $\$441.46$)

What You'll Learn