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

$$P(k) = {n!\over k!(n-k)!} \left({\nu\over n}\right)^k \left({1-{\nu\over n}}\right)^{n-k}$$ (1)

as the number of trials becomes very large,

$$\lim_{n\to\infty} P(k) = \lim_{n\to\infty}{n(n-1)\cdots(n-k-1)\over n^k}{\nu^k\over k!}\left({1-{\nu\over n}}\right)^n\left({1 - {\nu\over n}}\right)^{-k}$$

$$= (1)\left({\nu^k\over k!}\right)(e^{-\nu})(1) = {\nu^ke^{-\nu}\over k!}.$$ (2)

This should be normalized so that the sum of probabilities equals 1. Indeed,

$$\sum_{k=0}^\infty P(k) = e^\nu \sum_{k=0}^\infty {\nu^k\over k!} = e^\nu e^{-\nu} = 1,$$ (3)

as required. The ratio of probabilities is given by

$${P(k=i+1)\over P(k=i)} = {{\nu^{i+1}e^{-\nu}\over(i+1)!}\over{i!\over e^{-\nu}\nu^i}} = {\nu\over i+1}.$$ (4)

The Moment-Generating Function of this distribution is

$$M(t) = \sum_{k=0}^\infty e^{tk} {\nu^k e^{-\nu}\over k!} = e^{-\nu} \sum_{k=0}^\infty {(\nu e^t)^k\over k!}$$

$$= e^{-\nu}e^{\nu e^t} = e^{\nu(e^t-1)}$$ (5)

$$M'(t) = \nu e^te^{\nu (e^t-1)}$$ (6)

$$M''(t) = (\nu e^t)^2 e^{\nu (e^t-1)}+\nu e^t e^{\nu (e^t-1)}$$ (7)

$$R(t) \equiv \ln M(t) = \nu (e^t-1)$$ (8)

$$R'(t) = \nu e^t$$ (9)

$$R''(t) = \nu e^t,$$ (10)

so

$$\mu = R'(0) = \nu$$ (11)

$$\sigma^2 = R''(0) = \nu.$$ (12)

The Moments about zero can also be computed directly

$$\mu_2' = \nu(1+\nu)$$ (13)

$$\mu_3' = \nu(1+3\nu+\nu^2)$$ (14)

$$\mu_4' = \nu(1+7\nu+6\nu^2+\nu^3),$$ (15)

as can the Moments about the Mean.

$$\mu_1 = \nu$$ (16)

$$\mu_2 = \nu$$ (17)

$$\mu_3 = \nu$$ (18)

$$\mu_4 = \nu(1+3\nu),$$ (19)

so the Mean, Variance, Skewness, and Kurtosis are

$$\mu = \nu$$ (20)

$$\sigma^2 = \nu$$ (21)

$$\gamma_1 \equiv {\mu_3\over\sigma^3} = {\nu\over\nu^{3/2}} = \nu^{-1/2}$$ (22)

$$\gamma_2 \equiv {\mu_4\over\sigma^4}-3 = {\nu(1+3\nu)\over\nu}-3$$

$$= {\nu+3\nu^2-3\nu^2\over\nu^2} = \nu^{-1}.$$ (23)

The Characteristic Function is

$$\phi(t)=e^{m(e^{it}-1)}$$ (24)

and the Cumulant-Generating Function is

$$K(h)=\nu(e^h-1) = \nu(h+{\textstyle{1\over 2!}} h^2+{\textstyle{1\over 3!}} h^3+\ldots),$$ (25)

so

$$\kappa_r =\nu.$$ (26)

The Poisson distribution can also be expressed in terms of

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

the rate of changes, so that

$$P(k) = {(\lambda x)^ke^{-\lambda x}\over k!}.$$ (28)

The Moment-Generating Function of a Poisson distribution in two variables is given by

$$M(t) = e^{(\nu_1+\nu_2)(e^t-1)}.$$ (29)

If the independent variables $x_1$, $x_2$, ..., $x_N$ have Poisson distributions with parameters $\mu_1$, $\mu_2$, ..., $\mu_N$, then

$$X=\sum_{j=1}^N x_j$$ (30)

has a Poisson distribution with parameter

$$\mu=\sum_{j=1}^N \mu_j.$$ (31)

This can be seen since the Cumulant-Generating Function is

$$K_j(h)=\mu_j(e^h-1),$$ (32)

$$K\equiv \sum_j K_j(h)=(e^h-1)\sum_j \mu_j = \mu(e^h-1).$$ (33)

References

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.