Exponential Distribution

$\begin{figure}\begin{center}\BoxedEPSF{ExponentialDistribution.epsf scaled 650}\end{center}\end{figure}$

Given a Poisson Distribution with rate of change $\lambda$ , the distribution of waiting times between successive changes (with ) is

$\displaystyle D(x)$	$\textstyle \equiv$	$\displaystyle P(X \leq x) = 1-P(X > x)$
	$\textstyle =$	$\displaystyle 1 - {(\lambda x)^0e^{-\lambda x}\over 0!} = 1 - e^{-\lambda x}$	(1)
$\displaystyle P(x)$	$\textstyle =$	$\displaystyle D'(x) = \lambda e^{-\lambda x},$	(2)

which is normalized since

$\displaystyle \int_0^\infty P(x)\,dx$	$\textstyle =$	$\displaystyle \lambda \int_0^\infty e^{-\lambda x}\,dx$
	$\textstyle =$	$\displaystyle - [e^{-\lambda x}]^\infty_0 = -(0-1) = 1.$	(3)

This is the only Memoryless Random Distribution. Define the Mean waiting time between successive changes as $\theta\equiv\lambda^{-1}$ . Then

$\begin{displaymath} P(x) = \cases{ {1\over \theta} e^{-x/\theta} & $x \geq 0$\cr 0 & $x < 0$.\cr} \end{displaymath}$

(4)

$\displaystyle M(t)$	$\textstyle =$	$\displaystyle \int_0^\infty e^{tx}\left({1\over \theta}\right)e^{-x/\theta}\, dx = {1\over \theta }\int_0^\infty e^{-(1-\theta t)x/\theta}\, dx$
	$\textstyle =$	$\displaystyle \left[{e^{-(1-\theta t)x/\theta} \over 1-\theta t}\right]_0^\infty = {1\over 1-\theta t}$	(5)
$\displaystyle M'(t)$	$\textstyle =$	$\displaystyle {\theta \over (1-\theta t)^2}$	(6)
$\displaystyle M''(t)$	$\textstyle =$	$\displaystyle {2\theta ^2\over (1-\theta t)^3},$	(7)

$\displaystyle R(t)$	$\textstyle \equiv$	$\displaystyle \ln M(t) = - \ln (1-\theta t)$	(8)
$\displaystyle R'(t)$	$\textstyle =$	$\displaystyle {\theta \over 1-\theta t}$	(9)
$\displaystyle R''(t)$	$\textstyle =$	$\displaystyle {\theta^2\over (1-\theta t)^2}$	(10)
$\displaystyle \mu$	$\textstyle =$	$\displaystyle R'(0) = \theta$	(11)
$\displaystyle \sigma^2$	$\textstyle =$	$\displaystyle R''(0) = \theta^2.$	(12)

The Skewness and Kurtosis are given by

$\displaystyle \gamma_1$	$\textstyle =$	$\displaystyle 2$	(13)
$\displaystyle \gamma_2$	$\textstyle =$	$\displaystyle 6.$	(14)

The Mean and Variance can also be computed directly

$\begin{displaymath} \left\langle{x}\right\rangle{} \equiv \int_0^\infty P(x)\, dx = {1\over s} \int_0^\infty xe^{-x/s}\, dx. \end{displaymath}$

(15)

Use the integral

$\begin{displaymath} \int xe^{ax}\,dx = {e^{ax}\over a^2} (ax-1) \end{displaymath}$

(16)

to obtain

$\displaystyle \left\langle{x}\right\rangle{}$	$\textstyle =$	$\displaystyle {1\over s}\left[{{e^{-x/s}\over\left({-{1\over s}}\right)^2} \left\{{\left({-{1\over s}}\right)x-1}\right\}}\right]^\infty_0$
	$\textstyle =$	$\displaystyle -s\left[{e^{-x/s} \left({1+{x\over s}}\right)}\right]^\infty_0$
	$\textstyle =$	$\displaystyle -s(0-1) = s.$	(17)

Now, to find

$\begin{displaymath} \left\langle{x^2}\right\rangle{} = {1\over s}\int_0^\infty x^2e^{-x/s}\,dx, \end{displaymath}$

(18)

use the integral

$\begin{displaymath} \int x^2e^{-x/s}\,dx = {e^{ax}\over a^3} (2-2ax+a^2x^2) \end{displaymath}$

(19)

$\displaystyle \left\langle{x^2}\right\rangle{}$	$\textstyle =$	$\displaystyle {1\over s}\left[{{e^{-x/s}\over\left({-{1\over s}}\right)^3} \left({2+{2\over s} x+{1\over s^2} x^2}\right)}\right]_0^\infty$
	$\textstyle =$	$\displaystyle -s^2(0-2) = 2s^2,$	(20)

giving

$\displaystyle \sigma^2$	$\textstyle \equiv$	$\displaystyle \left\langle{x^2}\right\rangle{}-{\left\langle{x}\right\rangle{}}^2$
	$\textstyle =$	$\displaystyle 2s^2-s^2 = s^2$	(21)
$\displaystyle \sigma$	$\textstyle \equiv$	$\displaystyle \sqrt{{\rm var}(x)} = s.$	(22)

If a generalized exponential probability function is defined by

$\begin{displaymath} P_{(\alpha,\beta)}(x)={1\over\beta}e^{-(x-\alpha)/\beta}, \end{displaymath}$

(23)

$\begin{displaymath} \phi(t)={e^{i\alpha t}\over 1-i\beta t}, \end{displaymath}$

(24)

References

Balakrishnan, N. and Basu, A. P. The Exponential Distribution: Theory, Methods, and Applications. New York: Gordon and Breach, 1996.

Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, pp. 534-535, 1987.