As a Binomial distribution with infinitesimal time-steps The Poisson distribution can be derived as a limiting case to the
binomial distribution as the number of trials goes to infinity and the
expected number of successes remains fixed — see
law of rare events below. Therefore, it can be used as an approximation of the binomial distribution if is sufficiently large and is sufficiently small. The Poisson distribution is a good approximation of the binomial distribution if is at least 20 and is smaller than or equal to 0.05, and an excellent approximation if and . Letting F_{\mathrm B} and F_{\mathrm P} be the respective
cumulative density functions of the binomial and Poisson distributions, one has: F_\mathrm{B}(k;n, p) \ \approx\ F_\mathrm{P}(k;\lambda=np). One derivation of this uses
probability-generating functions. Consider a
Bernoulli trial (coin-flip) whose probability of one success (or expected number of successes) is \lambda \leq 1 within a given interval. Split the interval into parts, and perform a trial in each subinterval with probability \tfrac{ \lambda }{n}. The probability of successes out of trials over the entire interval is then given by the binomial distribution p_k^{(n)}=\binom nk \left(\frac{\lambda}{n}\right)^{\!k} \left(1{-}\frac{\lambda}{n}\right)^{\! n-k}, whose generating function is: P^{(n)}(x)=\sum_{k=0}^n p_k^{(n)} x^k = \left(1-\frac{\lambda}{n} +\frac{\lambda}{n} x \right)^n. Taking the limit as increases to infinity (with fixed) and applying the product limit definition of the
exponential function, this reduces to the generating function of the Poisson distribution: \lim_{n\to\infty} P^{(n)}(x) = \lim_{n\to\infty} \left(1{+}\tfrac{\lambda(x-1)}{n}\right)^n = e^{\lambda(x-1)} = \sum_{k= 0}^\infty e^{-\lambda}\frac{\lambda^k }{k!} x^k.
General • If X_1 \sim \mathrm{Pois}(\lambda_1)\, and X_2 \sim \mathrm{Pois}(\lambda_2)\, are independent, then the difference Y = X_1 - X_2 follows a
Skellam distribution. • If X_1 \sim \mathrm{Pois}(\lambda_1)\, and X_2 \sim \mathrm{Pois}(\lambda_2)\, are independent, then the distribution of X_1 conditional on X_1+X_2 is a
binomial distribution. Specifically, if X_1+X_2=k, then X_1| X_1+X_2=k\sim \mathrm{Binom}(k, \lambda_1/(\lambda_1+\lambda_2)). More generally, if , , ..., are independent Poisson random variables with parameters , , ..., then given \sum_{j=1}^n X_j=k, it follows that X_i\Big|\sum_{j=1}^n X_j=k \sim \mathrm{Binom}\left(k, \frac{\lambda_i}{\sum_{j=1}^n \lambda_j}\right). In fact, \{X_i\} \sim \mathrm{Multinom}\left(k, \left\{\frac{\lambda_i}{\sum_{j=1}^n\lambda_j}\right\}\right). • If X \sim \mathrm{Pois}(\lambda)\, and the distribution of Y conditional on is a
binomial distribution, Y \mid (X = k) \sim \mathrm{Binom}(k, p), then the distribution of Y follows a Poisson distribution Y \sim \mathrm{Pois}(\lambda \cdot p). In fact, if, conditional on \{X = k\}, \{Y_i\} follows a
multinomial distribution, \{Y_i\} \mid (X = k) \sim \mathrm{Multinom}\left(k, p_i\right), then each Y_i follows an independent Poisson distribution Y_i \sim \mathrm{Pois}(\lambda \cdot p_i), \rho(Y_i, Y_j) = 0. • The Poisson distribution is a
special case of the discrete compound Poisson distribution (or stuttering Poisson distribution) with only a parameter. The discrete compound Poisson distribution can be deduced from the limiting distribution of univariate multinomial distribution. It is also a
special case of a
compound Poisson distribution. • For sufficiently large values of , (say ), the
normal distribution with mean and variance (standard deviation \sqrt{\lambda}) is an excellent approximation to the Poisson. If is greater than about 10, then the normal distribution is a good approximation if an appropriate
continuity correction is performed, i.e., if , where is a non-negative integer, is replaced by . F_\mathrm{Poisson}(x;\lambda) \approx F_\mathrm{normal}(x;\mu=\lambda,\sigma^2=\lambda) •
Variance-stabilizing transformation: If X \sim \mathrm{Pois}(\lambda), then Y = 2 \sqrt{X} \approx \mathcal{N}(2\sqrt{\lambda};1), and Y = \sqrt{X} \approx \mathcal{N}(\sqrt{\lambda};1/4). Under this transformation, the convergence to normality (as \lambda increases) is far faster than the untransformed variable. Other, slightly more complicated, variance stabilizing transformations are available, one of which is
Anscombe transform. See
Data transformation (statistics) for more general uses of transformations. • If for every the number of arrivals in the time interval follows the Poisson distribution with mean , then the sequence of inter-arrival times are independent and identically distributed
exponential random variables having mean . • The
cumulative distribution functions of the Poisson and
chi-squared distributions are related in the following ways: F_\text{Poisson}(k;\lambda) = 1-F_{\chi^2}(2\lambda;2(k+1)) \quad\quad \text{ integer } k, and P(X=k)=F_{\chi^2}(2\lambda;2(k+1)) -F_{\chi^2}(2\lambda;2k).
Poisson approximation Assume X_1\sim\operatorname{Pois}(\lambda_1), X_2\sim\operatorname{Pois}(\lambda_2), \dots, X_n\sim\operatorname{Pois}(\lambda_n) where \lambda_1 + \lambda_2 + \dots + \lambda_n=1, then (X_1, X_2, \dots, X_n) is
multinomially distributed (X_1, X_2, \dots, X_n) \sim \operatorname{Mult}(N, \lambda_1, \lambda_2, \dots, \lambda_n) conditioned on N = X_1 + X_2 + \dots X_n. This means, among other things, that for any nonnegative function f(x_1, x_2, \dots, x_n), if (Y_1, Y_2, \dots, Y_n)\sim\operatorname{Mult}(m, \mathbf{p}) is multinomially distributed, then \operatorname{E}[f(Y_1, Y_2, \dots, Y_n)] \le e\sqrt{m}\operatorname{E}[f(X_1, X_2, \dots, X_n)] where (X_1, X_2, \dots, X_n)\sim\operatorname{Pois}(\mathbf{p}). The factor of e\sqrt{m} can be replaced by 2 if f is further assumed to be monotonically increasing or decreasing.
Bivariate Poisson distribution This distribution has been extended to the
bivariate case. The
generating function for this distribution is g( u, v ) = \exp[ ( \theta_1 - \theta_{12} )( u - 1 ) + ( \theta_2 - \theta_{12} )(v - 1) + \theta_{12} ( uv - 1 ) ] with \theta_1, \theta_2 > \theta_{ 12 } > 0 The marginal distributions are and and the correlation coefficient is limited to the range 0 \le \rho \le \min\left\{ \sqrt{ \frac{ \theta_1 }{ \theta_2 } }, \sqrt{ \frac{ \theta_2 }{ \theta_1 } } \right\} A simple way to generate a bivariate Poisson distribution X_1,X_2 is to take three independent Poisson distributions Y_1,Y_2,Y_3 with means \lambda_1,\lambda_2,\lambda_3 and then set X_1 = Y_1 + Y_3, X_2 = Y_2 + Y_3. The probability function of the bivariate Poisson distribution is \Pr(X_1=k_1,X_2=k_2) = \exp\left(-\lambda_1-\lambda_2-\lambda_3\right) \frac{\lambda_1^{k_1}}{k_1!} \frac{\lambda_2^{k_2}}{k_2!} \sum_{k=0}^{\min(k_1,k_2)} \binom{k_1}{k} \binom{k_2}{k} k! \left( \frac{\lambda_3}{\lambda_1\lambda_2}\right)^k
Free Poisson distribution The free Poisson distribution with jump size \alpha and rate \lambda arises in
free probability theory as the limit of repeated
free convolution \left( \left(1-\frac{\lambda}{N}\right)\delta_0 + \frac{\lambda}{N}\delta_\alpha\right)^{\boxplus N} as . In other words, let X_N be random variables so that X_N has value \alpha with probability \frac{\lambda}{N} and value 0 with the remaining probability. Assume also that the family X_1, X_2, \ldots are
freely independent. Then the limit as N \to \infty of the law of X_1 + \cdots +X_N is given by the Free Poisson law with parameters \lambda,\alpha. This definition is analogous to one of the ways in which the classical Poisson distribution is obtained from a (classical) Poisson process. The measure associated to the free Poisson law is given by \mu=\begin{cases} (1-\lambda) \delta_0 + \nu,& \text{if } 0\leq \lambda \leq 1 \\ \nu, & \text{if }\lambda >1, \end{cases} where \nu = \frac{1}{2\pi\alpha t}\sqrt{4\lambda \alpha^2 - ( t - \alpha (1+\lambda))^2} \, dt and has support [\alpha (1-\sqrt{\lambda})^2,\alpha (1+\sqrt{\lambda})^2]. This law also arises in
random matrix theory as the
Marchenko–Pastur law. Its
free cumulants are equal to \kappa_n=\lambda\alpha^n.
Some transforms of this law We give values of some important transforms of the free Poisson law; the computation can be found in e.g. in the book
Lectures on the Combinatorics of Free Probability by A. Nica and R. Speicher The R-transform of the free Poisson law is given by R(z) = \frac{\lambda \alpha}{1-\alpha z}. The Cauchy transform (which is the negative of the
Stieltjes transformation) is given by G(z) = \frac{ z + \alpha - \lambda \alpha - \sqrt{ (z-\alpha (1+\lambda))^2 - 4 \lambda \alpha^2}}{2\alpha z} The S-transform is given by S(z) = \frac{1}{z+\lambda} in the case that \alpha = 1. == Statistical inference ==