Gamma distribution

The story

One Exponential wait is the time until the next event; add up $k$ independent Exponential waits, each with mean $\theta$ , and you get the time until the $k$ -th event. That sum is $\text{Gamma}(k, \theta)$ . The definition extends smoothly to non-integer $k$ , which makes the Gamma a general-purpose model for positive, skewed quantities.

The mean is $k\theta$ and the variance is $k\theta^2$ . The shape $k$ controls the form: below 1 the density shoots up towards infinity at zero, at exactly 1 it is the Exponential, and above 1 it rises to a peak at $(k - 1)\theta$ before tailing off. As $k$ grows, the skew fades and the curve looks more and more Normal. The scale $\theta$ simply stretches the horizontal axis.

Formulas

Support

x > 0

PDF

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

CDF

F(x) = \frac{\gamma(k,\, x/\theta)}{\Gamma(k)}

Expectation

\mathbb{E}[X] = k\theta

Variance

\operatorname{Var}(X) = k\theta^2

Mode

(k - 1)\theta \quad \text{for } k \ge 1

Parameters

Parameter	What it controls
$k$	The shape: roughly, how many Exponential waits are being added up. Larger $k$ means a less skewed, more bell-like curve.
$\theta$	The scale: the mean of each underlying Exponential wait. Doubling $\theta$ doubles the mean and quadruples the variance.

Where it shows up

The total time to serve the next three customers in a queue.
Monthly rainfall totals, which are positive and right-skewed.
Insurance claim sizes.
A prior for a Poisson rate or a Normal precision in Bayesian statistics.

Try this

Fix $\theta = 1$ and sweep $k$ from 0.5 to 10. Watch the curve go from a spike at zero, through the Exponential at $k = 1$ , to an almost symmetric hump with its mode at $k - 1$ .
Set $k = 1.5$ and $\theta = 2$ , then open the Chi-squared page with $k = 3$ . The two curves are the same, because Chi-squared(3) is $\text{Gamma}(3/2, 2)$ .
Choose $k = 2$ , $\theta = 1.5$ , press Draw 1000 and check the sample mean against $k\theta = 3$ and the sample variance against $k\theta^2 = 4.5$ .

Related distributions

The Exponential with rate $\lambda$ is Gamma with $k = 1$ and $\theta = 1/\lambda$ .
Chi-squared with $k$ degrees of freedom is $\text{Gamma}(k/2, 2)$ .
Independent Gammas with the same $\theta$ add: $\text{Gamma}(k_1, \theta) + \text{Gamma}(k_2, \theta)$ is $\text{Gamma}(k_1 + k_2, \theta)$ .
For large $k$ it is approximately Normal with mean $k\theta$ and variance $k\theta^2$ .