Gamma

Study guides: Exam P-1, Exam MAS-I

The Gamma Distribution $X \sim \text{Gamma}(\alpha, \theta)$ is a flexible continuous distribution on $(0, \infty)$ generalizing the exponential, used to model skewed loss severities and aggregate claims.

$f(x) = \frac{x^{\alpha-1}\,e^{-x/\theta}}{\theta^{\alpha}\,\Gamma(\alpha)}$

$x > 0$

$\text{where } \alpha > 0 = \text{shape},\; \theta > 0 = \text{scale}$

$E[X] = \alpha\theta$ , $\text{Var}(X) = \alpha\theta^2$ , and coefficient of variation $CV = 1/\sqrt{\alpha}$
Special cases: exponential ( $\alpha = 1$ ) and chi-squared ( $\alpha = n/2$ , $\theta = 2$ )
The sum of $n$ independent $\text{Exp}(\theta)$ variables follows $\text{Gamma}(n, \theta)$
$\Gamma(\alpha) = (\alpha-1)!$ for integer $\alpha$ , and $\Gamma(\alpha+1) = \alpha\,\Gamma(\alpha)$ in general — so the density is elementary whenever $\alpha$ is a whole number
For integer $\alpha$ the CDF has a closed form via the Poisson link (integrating the density by parts $\alpha$ times):

$P(X > x) = \sum_{k=0}^{\alpha-1} \frac{e^{-x/\theta}(x/\theta)^k}{k!}$

Some texts parameterize by rate $\beta = 1/\theta$ , giving $E[X] = \alpha/\beta$ . Read which one a problem uses before substituting.

Mean and Variance of Aggregate Claim Severity Individual losses follow $\text{Gamma}(\alpha = 4, \theta = 250)$ . Find $E[X]$ , $\text{Var}(X)$ , and the coefficient of variation.

Answer $E[X] = \alpha\theta = 4 \times 250 = 1{,}000$ $\text{Var}(X) = \alpha\theta^2 = 4 \times 250^2 = 250{,}000$ $CV = \frac{\sqrt{250{,}000}}{1{,}000} = \frac{500}{1{,}000} = 0.50$

Waiting Time for the Third Claim Claims arrive in a Poisson process at 2 per year, so the waiting time to the 3rd claim is $T \sim \text{Gamma}(\alpha = 3, \theta = 0.5)$ years. Find $P(T > 1)$ .
Answer With integer $\alpha = 3$ and $x/\theta = 1/0.5 = 2$ :
 $\begin{align*} P(T > 1) &= \sum_{k=0}^{2} \frac{e^{-2}\,2^k}{k!} \\ &= e^{-2}\left(1 + 2 + 2\right) \\ &= 5e^{-2} \\ &\approx 0.677 \end{align*}$ 
The identity is just a restatement: "the 3rd claim has not arrived by time 1" is the same event as "at most 2 claims occurred in $[0,1]$ ," and that count is $\text{Poisson}(2)$ . Recognizing this saves integrating $x^2e^{-2x}$ by parts twice.

Finding Gamma Parameters from Mean and Variance A severity distribution is known to be gamma with mean 800 and variance 320,000. Find $\alpha$ and $\theta$ .
Answer Divide the variance by the mean to strip out $\alpha$ :
 $\begin{align*} \frac{\text{Var}(X)}{E[X]} &= \frac{\alpha\theta^2}{\alpha\theta} = \theta \\ \theta &= \frac{320{,}000}{800} = 400 \end{align*}$ 
Then back-substitute: $\alpha = \frac{E[X]}{\theta} = \frac{800}{400} = 2$ So $X \sim \text{Gamma}(2, 400)$ . The ratio trick — variance over mean gives the scale — is a property of the gamma (exponential included), where $\alpha\theta^2/(\alpha\theta) = \theta$ whatever $\alpha$ is. It does not carry over to other scale families: for a uniform on $(0, \theta)$ , variance over mean is $(\theta^2/12)/(\theta/2) = \theta/6$ .