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 a sum of independent Bernoulli trials.

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

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

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 $n!/(n-k)!$ 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

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

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