# Hyperexponential distribution

(Redirected from Hyper-exponential distribution)

In probability theory, a hyper-exponential distribution is a continuous probability distribution whose probability density function of the random variable X is given by

$f_X(x) = \sum_{i=1}^n f_{Y_i}(x)\;p_i,$

where each Yi is an exponentially distributed random variable with rate parameter λi, and pi is the probability that X will take on the form of the exponential distribution with rate λi.[1] It is named the hyper-exponential distribution since its coefficient of variation is greater than that of the exponential distribution, whose coefficient of variation is 1, and the hypoexponential distribution, which has a coefficient of variation less than one. While the exponential distribution is the continuous analogue of the geometric distribution, the hyper-exponential distribution is not analogous to the hypergeometric distribution. The hyper-exponential distribution is an example of a mixture density.

An example of a hyper-exponential random variable can be seen in the context of telephony, where, if someone has a modem and a phone, their phone line usage could be modeled as a hyper-exponential distribution where there is probability p of them talking on the phone with rate λ1 and probability q of them using their internet connection with rate λ2.

## Properties of the hyper-exponential distribution

Since the expected value of a sum is the sum of the expected values, the expected value of a hyper-exponential random variable can be shown as

$E[X] = \int_{-\infty}^\infty x f(x) \, dx= \sum_{i=1}^n p_i\int_0^\infty x\lambda_i e^{-\lambda_ix} \, dx = \sum_{i=1}^n \frac{p_i}{\lambda_i}$

and

$E\!\left[X^2\right] = \int_{-\infty}^\infty x^2 f(x) \, dx = \sum_{i=1}^n p_i\int_0^\infty x^2\lambda_i e^{-\lambda_ix} \, dx = \sum_{i=1}^n \frac{2}{\lambda_i^2}p_i,$

from which we can derive the variance:[2]

$\operatorname{Var}[X] = E\!\left[X^2\right] - E\!\left[X\right]^2 = \sum_{i=1}^n \frac{2}{\lambda_i^2}p_i - \left[\sum_{i=1}^n \frac{p_i}{\lambda_i}\right]^2 = \left[\sum_{i=1}^n \frac{p_i}{\lambda_i}\right]^2 + \sum_{i=1}^n \sum_{j=1}^n p_i p_j \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_j} \right)^2.$

The standard deviation exceeds the mean in general (except for the degenerate case of all the λs being equal), so the coefficient of variation is greater than 1.

The moment-generating function is given by

$E\!\left[e^{tx}\right] = \int_{-\infty}^\infty e^{tx} f(x) \, dx= \sum_{i=1}^n p_i \int_0^\infty e^{tx}\lambda_i e^{-\lambda_i x} \, dx = \sum_{i=1}^n \frac{\lambda_i}{\lambda_i - t}p_i.$