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Poisson Distribution


A Poisson distribution is a distribution with the following properties:

1. The number of changes in nonoverlapping intervals are independent for all intervals.

2. The probability of exactly one change in a sufficiently small interval $h \equiv 1/n$ is $P = \nu h \equiv \nu/n$, where $\nu$ is the probability of one change and $n$ is the number of Trials.

3. The probability of two or more changes in a sufficiently small interval $h$ is essentially 0.

The probability of $k$ changes in a given interval is then given by the limit of the Binomial Distribution
P(k) = {n!\over k!(n-k)!} \left({\nu\over n}\right)^k \left({1-{\nu\over n}}\right)^{n-k}
\end{displaymath} (1)

as the number of trials becomes very large,

$\displaystyle \lim_{n\to\infty} P(k)$ $\textstyle =$ $\displaystyle \lim_{n\to\infty}{n(n-1)\cdots(n-k-1)\over n^k}{\nu^k\over k!}\left({1-{\nu\over n}}\right)^n\left({1 - {\nu\over n}}\right)^{-k}$  
  $\textstyle =$ $\displaystyle (1)\left({\nu^k\over k!}\right)(e^{-\nu})(1) = {\nu^ke^{-\nu}\over k!}.$ (2)

This should be normalized so that the sum of probabilities equals 1. Indeed,
\sum_{k=0}^\infty P(k) = e^\nu \sum_{k=0}^\infty {\nu^k\over k!} = e^\nu e^{-\nu} = 1,
\end{displaymath} (3)

as required. The ratio of probabilities is given by
{P(k=i+1)\over P(k=i)} = {{\nu^{i+1}e^{-\nu}\over(i+1)!}\over{i!\over e^{-\nu}\nu^i}}
= {\nu\over i+1}.
\end{displaymath} (4)

The Moment-Generating Function of this distribution is
$\displaystyle M(t)$ $\textstyle =$ $\displaystyle \sum_{k=0}^\infty e^{tk} {\nu^k e^{-\nu}\over k!} = e^{-\nu} \sum_{k=0}^\infty {(\nu e^t)^k\over k!}$  
  $\textstyle =$ $\displaystyle e^{-\nu}e^{\nu e^t} = e^{\nu(e^t-1)}$ (5)
$\displaystyle M'(t)$ $\textstyle =$ $\displaystyle \nu e^te^{\nu (e^t-1)}$ (6)
$\displaystyle M''(t)$ $\textstyle =$ $\displaystyle (\nu e^t)^2 e^{\nu (e^t-1)}+\nu e^t e^{\nu (e^t-1)}$ (7)
$\displaystyle R(t)$ $\textstyle \equiv$ $\displaystyle \ln M(t) = \nu (e^t-1)$ (8)
$\displaystyle R'(t)$ $\textstyle =$ $\displaystyle \nu e^t$ (9)
$\displaystyle R''(t)$ $\textstyle =$ $\displaystyle \nu e^t,$ (10)

$\displaystyle \mu$ $\textstyle =$ $\displaystyle R'(0) = \nu$ (11)
$\displaystyle \sigma^2$ $\textstyle =$ $\displaystyle R''(0) = \nu.$ (12)

The Moments about zero can also be computed directly
$\displaystyle \mu_2'$ $\textstyle =$ $\displaystyle \nu(1+\nu)$ (13)
$\displaystyle \mu_3'$ $\textstyle =$ $\displaystyle \nu(1+3\nu+\nu^2)$ (14)
$\displaystyle \mu_4'$ $\textstyle =$ $\displaystyle \nu(1+7\nu+6\nu^2+\nu^3),$ (15)

as can the Moments about the Mean.
$\displaystyle \mu_1$ $\textstyle =$ $\displaystyle \nu$ (16)
$\displaystyle \mu_2$ $\textstyle =$ $\displaystyle \nu$ (17)
$\displaystyle \mu_3$ $\textstyle =$ $\displaystyle \nu$ (18)
$\displaystyle \mu_4$ $\textstyle =$ $\displaystyle \nu(1+3\nu),$ (19)

so the Mean, Variance, Skewness, and Kurtosis are
$\displaystyle \mu$ $\textstyle =$ $\displaystyle \nu$ (20)
$\displaystyle \sigma^2$ $\textstyle =$ $\displaystyle \nu$ (21)
$\displaystyle \gamma_1$ $\textstyle \equiv$ $\displaystyle {\mu_3\over\sigma^3} = {\nu\over\nu^{3/2}} = \nu^{-1/2}$ (22)
$\displaystyle \gamma_2$ $\textstyle \equiv$ $\displaystyle {\mu_4\over\sigma^4}-3 = {\nu(1+3\nu)\over\nu}-3$  
  $\textstyle =$ $\displaystyle {\nu+3\nu^2-3\nu^2\over\nu^2} = \nu^{-1}.$ (23)

The Characteristic Function is

\end{displaymath} (24)

and the Cumulant-Generating Function is
K(h)=\nu(e^h-1) = \nu(h+{\textstyle{1\over 2!}} h^2+{\textstyle{1\over 3!}} h^3+\ldots),
\end{displaymath} (25)

\kappa_r =\nu.
\end{displaymath} (26)

The Poisson distribution can also be expressed in terms of

\lambda\equiv{\nu\over x},
\end{displaymath} (27)

the rate of changes, so that
P(k) = {(\lambda x)^ke^{-\lambda x}\over k!}.
\end{displaymath} (28)

The Moment-Generating Function of a Poisson distribution in two variables is given by
M(t) = e^{(\nu_1+\nu_2)(e^t-1)}.
\end{displaymath} (29)

If the independent variables $x_1$, $x_2$, ..., $x_N$ have Poisson distributions with parameters $\mu_1$, $\mu_2$, ..., $\mu_N$, then
X=\sum_{j=1}^N x_j
\end{displaymath} (30)

has a Poisson distribution with parameter
\mu=\sum_{j=1}^N \mu_j.
\end{displaymath} (31)

This can be seen since the Cumulant-Generating Function is
\end{displaymath} (32)

K\equiv \sum_j K_j(h)=(e^h-1)\sum_j \mu_j = \mu(e^h-1).
\end{displaymath} (33)


Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 532, 1987.

Press, W. H.; Flannery, B. P.; Teukolsky, S. A.; and Vetterling, W. T. ``Incomplete Gamma Function, Error Function, Chi-Square Probability Function, Cumulative Poisson Function.'' §6.2 in Numerical Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed. Cambridge, England: Cambridge University Press, pp. 209-214, 1992.

Spiegel, M. R. Theory and Problems of Probability and Statistics. New York: McGraw-Hill, p. 111-112, 1992.

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© 1996-9 Eric W. Weisstein