Hyperexponential distribution

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

${\displaystyle E[X]=\int _{-\infty }^{\infty }xf(x)\,dx=\sum _{i=1}^{n}p_{i}\int _{0}^{\infty }x\lambda _{i}e^{-\lambda _{i}x}\,dx=\sum _{i=1}^{n}{\frac {p_{i}}{\lambda _{i}}}}$

and

${\displaystyle E\!\left[X^{2}\right]=\int _{-\infty }^{\infty }x^{2}f(x)\,dx=\sum _{i=1}^{n}p_{i}\int _{0}^{\infty }x^{2}\lambda _{i}e^{-\lambda _{i}x}\,dx=\sum _{i=1}^{n}{\frac {2}{\lambda _{i}^{2}}}p_{i},}$

from which we can derive the variance:[2]

${\displaystyle \operatorname {Var} [X]=E\!\left[X^{2}\right]-E\!\left[X\right]^{2}=\sum _{i=1}^{n}{\frac {2}{\lambda _{i}^{2}}}p_{i}-\left[\sum _{i=1}^{n}{\frac {p_{i}}{\lambda _{i}}}\right]^{2}=\left[\sum _{i=1}^{n}{\frac {p_{i}}{\lambda _{i}}}\right]^{2}+\sum _{i=1}^{n}\sum _{j=1}^{n}p_{i}p_{j}\left({\frac {1}{\lambda _{i}}}-{\frac {1}{\lambda _{j}}}\right)^{2}.}$

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

${\displaystyle E\!\left[e^{tx}\right]=\int _{-\infty }^{\infty }e^{tx}f(x)\,dx=\sum _{i=1}^{n}p_{i}\int _{0}^{\infty }e^{tx}\lambda _{i}e^{-\lambda _{i}x}\,dx=\sum _{i=1}^{n}{\frac {\lambda _{i}}{\lambda _{i}-t}}p_{i}.}$

A given probability distribution, including a heavy-tailed distribution, can be approximated by a hyperexponential distribution by fitting recursively to different time scales using Prony's method.[3]

• Phase-type distribution
• Hyper-Erlang distribution
• Lomax distribution (continuous mixture of exponentials)