Gamma functions are useful in proving the following result.

Exponential mean and variance

If \(X\) has an \(\ExponDistn(\lambda) \) distribution with pdf

\[ f(x) \;\;=\;\; \lambda e^{-\lambda x} \]

then its mean and variance are

\[ E[X] \;\;=\;\; \frac 1 {\lambda} \spaced{and} \Var(X) \;\;=\;\; \frac 1 {\lambda^2} \]

(Proved in full version)