Given a Poisson Distribution with rate of change
, the distribution of waiting times between successive
changes (with
) is
which is normalized since
This is the only Memoryless Random Distribution. Define the Mean
waiting time between successive changes as
. Then
![\begin{displaymath}
P(x) = \cases{
{1\over \theta} e^{-x/\theta} & $x \geq 0$\cr
0 & $x < 0$.\cr}
\end{displaymath}](e_2679.gif) |
(4) |
The Moment-Generating Function is
so
The Skewness and Kurtosis are given by
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}](e_2697.gif) |
(15) |
Use the integral
![\begin{displaymath}
\int xe^{ax}\,dx = {e^{ax}\over a^2} (ax-1)
\end{displaymath}](e_2698.gif) |
(16) |
to obtain
Now, to find
![\begin{displaymath}
\left\langle{x^2}\right\rangle{} = {1\over s}\int_0^\infty x^2e^{-x/s}\,dx,
\end{displaymath}](e_2703.gif) |
(18) |
use the integral
![\begin{displaymath}
\int x^2e^{-x/s}\,dx = {e^{ax}\over a^3} (2-2ax+a^2x^2)
\end{displaymath}](e_2704.gif) |
(19) |
giving
If a generalized exponential probability function is defined by
![\begin{displaymath}
P_{(\alpha,\beta)}(x)={1\over\beta}e^{-(x-\alpha)/\beta},
\end{displaymath}](e_2712.gif) |
(23) |
then the Characteristic Function is
![\begin{displaymath}
\phi(t)={e^{i\alpha t}\over 1-i\beta t},
\end{displaymath}](e_2713.gif) |
(24) |
and the Mean, Variance, Skewness, and Kurtosis are
See also Double Exponential Distribution
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.
Spiegel, M. R. Theory and Problems of Probability and Statistics. New York: McGraw-Hill, p. 119, 1992.
© 1996-9 Eric W. Weisstein
1999-05-25