Poisson binomial distribution

Poisson binomial

Parameters	$\mathbf {p} \in {{[0,1]}^{n}}$ — success probabilities for each of the n trials
Support	k ∈ { 0, …, n }
PMF	$\sum \limits _{A\in {{F}_{k}}}{\prod \limits _{i\in A}{{p}_{i}}\prod \limits _{j\in {{A}^{c}}}{(1-{{p}_{j}})}}$
CDF	$\sum \limits _{l=0}^{k}{\sum \limits _{A\in {{F}_{l}}}{\prod \limits _{i\in A}{{p}_{i}}\prod \limits _{j\in {{A}^{c}}}{(1-{{p}_{j}})}}}$
Mean	$\sum \limits _{i=1}^{n}{{p}_{i}}$
Variance	${{\sigma }^{2}}=\sum \limits _{i=1}^{n}{(1-{{p}_{i}}){{p}_{i}}}$
Skewness	${\frac {1}{{\sigma }^{3}}}\sum \limits _{i=1}^{n}{\left(1-2{{p}_{i}}\right)\left(1-{{p}_{i}}\right){{p}_{i}}}$
Excess kurtosis	${\frac {1}{{\sigma }^{4}}}\sum \limits _{i=1}^{n}{\left(1-6(1-{{p}_{i}}){{p}_{i}}\right)\left(1-{{p}_{i}}\right){{p}_{i}}}$
MGF	$\prod \limits _{i=1}^{n}{(1-{{p}_{i}}+{{p}_{i}}{{e}^{t}})}$
CF	$\prod \limits _{i=1}^{n}{(1-{{p}_{i}}+{{p}_{i}}{{e}^{it}})}$

In probability theory and statistics, the Poisson binomial distribution is the discrete probability distribution of the number of successes in a sequence of n independent yes/no experiments (also called Bernoulli trials), each of which has a probability p_i of success, where i is an integer from 1 to n. The ordinary Binomial distribution is a special case of the Poisson binomial distribution, when all success probabilities p_i are the same.

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

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

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. $A^{c}=\{1,2,3,...,n\}\backslash A$.

$F_{k}$ will contain $n!/(n-k)!$ 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 n successes using the recursive formula

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

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

Another possibility is using the discrete Fourier transform

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

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

Still other methods are described in .

See also

• Le Cam's theorem
• Binomial distribution
• Poisson distribution