# Poisson binomial distribution

Parameters $\mathbf{p}\in [0,1]^n$ — success probabilities for each of the n trials k ∈ { 0, …, n } $\sum\limits_{A\in F_k} \prod\limits_{i\in A} p_i \prod\limits_{j\in A^c} (1-p_j)$ $\sum\limits_{l=0}^k \sum\limits_{A\in F_l} \prod\limits_{i\in A} p_i \prod\limits_{j\in A^c}{(1-p_j)}$ $\sum\limits_{i=1}^n p_i$ $\sigma^2 =\sum\limits_{i=1}^n (1 - {p_i}){p_i}$ $\frac{1}{\sigma^3}\sum\limits_{i=1}^n {\left( 1-2{p_i} \right)\left( 1-{{p}_{i}} \right){{p}_{i}}}$ $\frac{1}{\sigma^4}\sum\limits_{i=1}^n {\left( 1 - 6(1 - p_i){p_i} \right)\left( 1 - p_i \right)p_i}$ $\prod\limits_{j=1}^n (1-{p_j}+{p_j}{e^{t}})$ $\prod\limits_{j=1}^n (1-{p_j}+{p_j}{e^{it}})$

In probability theory and statistics, the Poisson binomial distribution is the discrete probability distribution of a sum of independent Bernoulli trials that are not necessarily identically distributed. The concept is named after Siméon Denis Poisson.

In other words, it is the probability distribution of the number of successes in a sequence of n independent yes/no experiments with success probabilities $p_1, p_2, \dots , p_n$. The ordinary binomial distribution is a special case of the Poisson binomial distribution, when all success probabilities are the same, that is $p_1 = p_2 = \dots = p_n$ .

## Mean and variance

Since a Poisson binomial distributed variable is a sum of n independent Bernoulli distributed variables, its mean and variance will simply be sums of the mean and variance of the n Bernoulli distributions:

$\mu = \sum\limits_{i=1}^n p_i$
$\sigma^2 =\sum\limits_{i=1}^n (1-p_i) p_i$

## Probability mass function

The probability of having k successful trials out of a total of n can be written as the sum [1]

$Pr(K=k) = \sum\limits_{A\in {{F}_{k}}}{\prod\limits_{i\in A}{{{p}_{i}}}\prod\limits_{j\in {{A}^{c}}}{(1-{{p}_{j}})}}$

where $F_k$ is the set of all subsets of k integers that can be selected from {1,2,3,...,n}. For example, if n = 3, then ${{F}_{2}}=\left\{ \{1,2\},\{1,3\},\{2,3\} \right\}$. $A^c$ is the complement of $A$, i.e. $A^c =\{1,2,3,\dots,n\}\backslash A$.

$F_k$ will contain $n!/(n-k)!k!$ elements, the sum over which is infeasible to compute in practice unless the number of trials n is small (e.g. if n=30, $F_{15}$ contains over 1020 elements). There are luckily more efficient ways to calculate $Pr(K=k)$.

As long as none of the success probabilities are equal to one, one can calculate the probability of k successes using the recursive formula [2] [3]

\Pr (K=k)=\left\{ \begin{align} & \prod\limits_{i=1}^{n}{(1-{{p}_{i}})} \qquad k=0 \\ & \frac{1}{k}\sum\limits_{i=1}^{k}{{{(-1)}^{i-1}}\Pr (K=k-i)T(i)} \qquad k>0 \\ \end{align} \right.

where $T(i)=\sum\limits_{j=1}^{n}{{{\left( \frac{{{p}_{j}}}{1-{{p}_{j}}} \right)}^{i}}}$ .

The recursive formula is not numerically stable, and should be avoided if $n$ is greater than approximately 20. Another possibility is using the discrete Fourier transform [4]

$\Pr (K=k)=\frac{1}{n+1}\sum\limits_{l=0}^{n}{{{C}^{-lk}}\prod\limits_{m=1}^{n}{\left( 1+({{C}^{l}}-1){{p}_{m}} \right)}}$

where $C=\exp \left( \frac{2i\pi }{n+1} \right)$

where $i=\sqrt{-1}$

Still other methods are described in .[5]