the gamma distribution
/ GAM-uh /
If the wait for a single Poisson event is exponential, how long until the kth event? You have to wait for the first, then the second, and so on — you add up k independent exponential waits. The total wait follows the gamma distribution. It is the natural model for the time to accumulate several 'random' arrivals.
Write X ~ Gamma(k, theta), with a shape parameter k > 0 and a scale theta > 0 (some books use the rate beta = 1/theta instead). When k is a whole number, X is the sum of k independent Exponential(rate = 1/theta) waits, with mean E[X] = k theta and variance Var(X) = k theta^2. The density f(x) = x^(k-1) e^(-x/theta) / (theta^k Gamma(k)) for x >= 0 uses the gamma FUNCTION, Gamma(k), purely as the constant that makes the area equal 1. The shape k controls the silhouette: k = 1 gives the pure exponential (a downhill slide), while larger k pushes the bump rightward and makes it more bell-like — a foreshadowing of the normal as many waits pile up.
The gamma is the parent of a small family: integer-k gives the Erlang distribution; k = (degrees of freedom)/2 with theta = 2 gives the chi-squared distribution. It models rainfall totals, insurance claim sizes, and time-to-completion of multi-stage processes, and it serves as a flexible, always-positive, right-skewed shape. A caution: the gamma function Gamma(k) in the formula is a normalizing constant, not the distribution itself — do not confuse the function with the distribution that borrows its name.
Customers arrive as a Poisson process, one every 4 minutes on average (theta = 4). The time until the 3rd customer arrives is Gamma(k = 3, theta = 4), with mean 3 x 4 = 12 minutes and variance 3 x 4^2 = 48.
Sum k exponential waits and you get a gamma; k = 1 is just the exponential again.
Watch the parameterization: shape-scale (k, theta) versus shape-rate (k, beta = 1/theta) are both common and give reciprocal means — mixing them is a frequent source of error.