Poisson binomial

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

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 collection of n independent yes/no experiments with success probabilities ${\displaystyle p_{1},p_{2},\dots ,p_{n))$. The ordinary binomial distribution is a special case of the Poisson binomial distribution, when all success probabilities are the same, that is ${\displaystyle p_{1}=p_{2}=\cdots =p_{n))$.

Definitions

Probability Mass Function

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

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

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

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

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]}

${\displaystyle \Pr(K=k)={\begin{cases}\prod \limits _{i=1}^{n}(1-p_{i})&k=0\\{\frac {1}{k))\sum \limits _{i=1}^{k}(-1)^{i-1}\Pr(K=k-i)T(i)&k>0\\\end{cases))}$

where

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

The recursive formula is not numerically stable, and should be avoided if ${\displaystyle n}$ is greater than approximately 20. An alternative is to use a divide-and-conquer algorithm: if we assume ${\displaystyle n=2^{b))$ is a power of two, denoting by ${\displaystyle f(p_{i:j})}$ the Poisson binomial of ${\displaystyle p_{i},\dots ,p_{j))$ and ${\displaystyle *}$ the convolution operator, we have ${\displaystyle f(p_{1:2^{b)))=f(p_{1:2^{b-1)))*f(p_{2^{b-1}+1:2^{b)))}$.

Another possibility is using the discrete Fourier transform.^[4]

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

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

Still other methods are described in "Statistical Applications of the Poisson-Binomial and conditional Bernoulli distributions" by Chen and Liu.^[5]

Cumulative distribution function

The cumulative distribution function (CDF) can be expressed as:

${\displaystyle \Pr(K\leq k)=\sum _{l=0}^{k}\sum \limits _{A\in F_{l))\prod \limits _{i\in A}p_{i}\prod \limits _{j\in A^{c))(1-p_{j})}$ ,

where ${\displaystyle F_{l))$ is the set of all subsets of size ${\displaystyle l}$ that can be selected from ${\displaystyle \{1,2,3,...,n\))$.

Properties

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:

${\displaystyle \mu =\sum \limits _{i=1}^{n}p_{i))$

${\displaystyle \sigma ^{2}=\sum \limits _{i=1}^{n}(1-p_{i})p_{i))$

For fixed values of the mean (${\displaystyle \mu }$) and size (n), the variance is maximal when all success probabilities are equal and we have a binomial distribution. When the mean is fixed, the variance is bounded from above by the variance of the Poisson distribution with the same mean which is attained asymptotically ^{[citation needed]} as n tends to infinity.

Entropy

There is no simple formula for the entropy of a Poisson binomial distribution, but the entropy is bounded above by the entropy of a binomial distribution with the same number parameter and the same mean. Therefore, the entropy is also bounded above by the entropy of a Poisson distribution with the same mean.^[6]

The Shepp–Olkin concavity conjecture, due to Lawrence Shepp and Ingram Olkin in 1981, states that the entropy of a Poisson binomial distribution is a concave function of the success probabilities ${\displaystyle p_{1},p_{2},\dots ,p_{n))$.^[7] This conjecture was proved by Erwan Hillion and Oliver Johnson in 2015.^[8] The Shepp–Olkin monotonicity conjecture, also from the same 1981 paper, is that the entropy is monotone increasing in ${\displaystyle p_{i))$, if all ${\displaystyle p_{i}\leq 1/2}$. This conjecture was also proved by Hillion and Johnson, in 2019.^[9]

Chernoff bound

The probability that a Poisson binomial distribution gets large, can be bounded using its moment generating function as follows (valid when ${\displaystyle s\geq \mu }$ and for any ${\displaystyle t>0}$):

${\displaystyle {\begin{aligned}\Pr[S\geq s]&\leq \exp(-st)\operatorname {E} \left[\exp \left[t\sum _{i}X_{i}\right]\right]\\&=\exp(-st)\prod _{i}(1-p_{i}+e^{t}p_{i})\\&=\exp \left(-st+\sum _{i}\log \left(p_{i}(e^{t}-1)+1\right)\right)\\&\leq \exp \left(-st+\sum _{i}\log \left(\exp(p_{i}(e^{t}-1))\right)\right)\\&=\exp \left(-st+\sum _{i}p_{i}(e^{t}-1)\right)\\&=\exp \left(s-\mu -s\log {\frac {s}{\mu ))\right),\end{aligned))}$

where we took ${\textstyle t=\log \left(s/\mu \right)}$. This is similar to the tail bounds of a binomial distribution.

Computational methods

The reference ^[10] discusses techniques of evaluating the probability mass function of the Poisson binomial distribution. The following software implementations are based on it:

• An R package poibin was provided along with the paper,^[10] which is available for the computing of the cdf, pmf, quantile function, and random number generation of the Poisson binomial distribution. For computing the PMF, a DFT algorithm or a recursive algorithm can be specified to compute the exact PMF, and approximation methods using the normal and Poisson distribution can also be specified.
• poibin - Python implementation - can compute the PMF and CDF, uses the DFT method described in the paper for doing so.