Compound Poisson distribution

Suppose that

${\displaystyle N\sim \operatorname {Poisson} (\lambda ),}$

i.e., N is a random variable whose distribution is a Poisson distribution with expected value λ, and that

${\displaystyle X_{1},X_{2},X_{3},\dots }$

are identically distributed random variables that are mutually independent and also independent of N. Then the probability distribution of the sum of ${\displaystyle N}$ i.i.d. random variables

${\displaystyle Y=\sum _{n=1}^{N}X_{n}}$

is a compound Poisson distribution.

In the case N = 0, then this is a sum of 0 terms, so the value of Y is 0. Hence the conditional distribution of Y given that N = 0 is a degenerate distribution.

The compound Poisson distribution is obtained by marginalising the joint distribution of (Y,N) over N, and this joint distribution can be obtained by combining the conditional distribution Y | N with the marginal distribution of N.

The expected value and the variance of the compound distribution can be derived in a simple way from law of total expectation and the law of total variance. Thus

${\displaystyle \operatorname {E} (Y)=\operatorname {E} \left[\operatorname {E} (Y\mid N)\right]=\operatorname {E} \left[N\operatorname {E} (X)\right]=\operatorname {E} (N)\operatorname {E} (X),}$

${\displaystyle {\begin{aligned}\operatorname {Var} (Y)&=\operatorname {E} \left[\operatorname {Var} (Y\mid N)\right]+\operatorname {Var} \left[\operatorname {E} (Y\mid N)\right]=\operatorname {E} \left[N\operatorname {Var} (X)\right]+\operatorname {Var} \left[N\operatorname {E} (X)\right],\\[6pt]&=\operatorname {E} (N)\operatorname {Var} (X)+\left(\operatorname {E} (X)\right)^{2}\operatorname {Var} (N).\end{aligned}}}$

Then, since E(N) = Var(N) if N is Poisson, these formulae can be reduced to

${\displaystyle \operatorname {E} (Y)=\operatorname {E} (N)\operatorname {E} (X),}$

${\displaystyle \operatorname {Var} (Y)=\operatorname {E} (N)(\operatorname {Var} (X)+{\operatorname {E} (X)}^{2})=\operatorname {E} (N){\operatorname {E} (X^{2})}.}$

The probability distribution of Y can be determined in terms of characteristic functions:

${\displaystyle \varphi _{Y}(t)=\operatorname {E} (e^{itY})=\operatorname {E} _{N}(\left(\operatorname {E} (e^{itX}))^{N}\right)=\operatorname {E} ((\varphi _{X}(t))^{N}),\,}$

and hence, using the probability-generating function of the Poisson distribution, we have

${\displaystyle \varphi _{Y}(t)={\textrm {e}}^{\lambda (\varphi _{X}(t)-1)}.\,}$

An alternative approach is via cumulant generating functions:

${\displaystyle K_{Y}(t)=\ln \operatorname {E} [e^{tY}]=\ln \operatorname {E} [\operatorname {E} [e^{tY}\mid N]]=\ln \operatorname {E} [e^{NK_{X}(t)}]=K_{N}(K_{X}(t)).\,}$

Via the law of total cumulance it can be shown that, if the mean of the Poisson distribution λ = 1, the cumulants of Y are the same as the moments of X₁.

It can be shown that every infinitely divisible probability distribution is a limit of compound Poisson distributions.[1] And compound Poisson distributions is infinitely divisible by the definition.

When ${\displaystyle X_{1},X_{2},X_{3},\dots }$ are non-negative integer-valued i.i.d random variables with ${\displaystyle P(X_{1}=k)=\alpha _{k},\ (k=1,2,\ldots )}$, then this compound Poisson distribution is named discrete compound Poisson distribution[2][3][4] (or stuttering-Poisson distribution[5]) . We say that the discrete random variable ${\displaystyle Y}$ satisfying probability generating function characterization

${\displaystyle P_{Y}(z)=\sum \limits _{i=0}^{\infty }P(Y=i)z^{i}=\exp \left(\sum \limits _{k=1}^{\infty }\alpha _{k}\lambda (z^{k}-1)\right),\quad (|z|\leq 1)}$

has a discrete compound Poisson(DCP) distribution with parameters ${\displaystyle (\alpha _{1}\lambda ,\alpha _{2}\lambda ,\ldots )\in \mathbb {R} ^{\infty }\left(\sum _{i=1}^{\infty }\alpha _{i}=1,\alpha _{i}\geq 0,\lambda >0\right)}$, which is denoted by

${\displaystyle X\sim {\text{DCP}}(\lambda {\alpha _{1}},\lambda {\alpha _{r}},\ldots )}$

Moreover, if ${\displaystyle X\sim {\operatorname {DCP} }(\lambda {\alpha _{1}},\ldots ,\lambda {\alpha _{r}})}$, we say ${\displaystyle X}$ has a discrete compound Poisson distribution of order ${\displaystyle r}$ . When ${\displaystyle r=1,2}$, DCP becomes Poisson distribution and Hermite distribution, respectively. When ${\displaystyle r=3,4}$, DCP becomes triple stuttering-Poisson distribution and quadruple stuttering-Poisson distribution, respectively.[6] Other special cases include: shiftgeometric distribution, negative binomial distribution, Geometric Poisson distribution, Neyman type A distribution, Luria–Delbrück distribution in Luria–Delbrück experiment. For more special case of DCP, see the reviews paper[7] and references therein.

Feller's characterization of the compound Poisson distribution states that a non-negative integer valued r.v. ${\displaystyle X}$ is infinitely divisible if and only if its distribution is a discrete compound Poisson distribution.[8] It can be shown that the negative binomial distribution is discrete infinitely divisible, i.e., if X has a negative binomial distribution, then for any positive integer n, there exist discrete i.i.d. random variables X₁, ..., X_n whose sum has the same distribution that X has. The shift geometric distribution is discrete compound Poisson distribution since it is a trivial case of negative binomial distribution.

This distribution can model batch arrivals (such as in a bulk queue[5][9]). The discrete compound Poisson distribution is also widely used in actuarial science for modelling the distribution of the total claim amount.[3]

When some ${\displaystyle \alpha _{k}}$ are non-negative, it is the discrete pseudo compound Poisson distribution.[3] We define that any discrete random variable ${\displaystyle Y}$ satisfying probability generating function characterization

${\displaystyle G_{Y}(z)=\sum \limits _{n=0}^{\infty }P(Y=n)z^{n}=\exp \left(\sum \limits _{k=1}^{\infty }\alpha _{k}\lambda (z^{k}-1)\right),\quad (|z|\leq 1)}$

has a discrete pseudo compound Poisson distribution with parameters ${\displaystyle (\lambda _{1},\lambda _{2},\ldots )=:(\alpha _{1}\lambda ,\alpha _{2}\lambda ,\ldots )\in \mathbb {R} ^{\infty }\left({\sum \limits _{k=1}^{\infty }{\alpha _{k}}=1,\sum \limits _{k=1}^{\infty }{\left|{\alpha _{k}}\right|}<\infty ,{\alpha _{k}}\in {\mathbb {R} },\lambda >0}\right)}$.

If X has a gamma distribution, of which the exponential distribution is a special case, then the conditional distribution of Y | N is again a gamma distribution. The marginal distribution of Y can be shown to be a Tweedie distribution[10] with variance power 1<p<2 (proof via comparison of characteristic function (probability theory)). To be more explicit, if

${\displaystyle N\sim \operatorname {Poisson} (\lambda ),}$

and

${\displaystyle X_{i}\sim \operatorname {\Gamma } (\alpha ,\beta )}$

i.i.d., then the distribution of

${\displaystyle Y=\sum _{i=1}^{N}X_{i}}$

is a reproductive exponential dispersion model ${\displaystyle ED(\mu ,\sigma ^{2})}$ with

${\displaystyle {\begin{aligned}\operatorname {E} [Y]&=\lambda {\frac {\alpha }{\beta }}=:\mu ,\\\operatorname {Var} [Y]&=\lambda {\frac {\alpha (1+\alpha )}{\beta ^{2}}}=:\sigma ^{2}\mu ^{p}.\end{aligned}}}$

The mapping of parameters Tweedie parameter ${\displaystyle \mu ,\sigma ^{2},p}$ to the Poisson and Gamma parameters ${\displaystyle \lambda ,\alpha ,\beta }$ is the following:

${\displaystyle {\begin{aligned}\lambda &={\frac {\mu ^{2-p}}{(2-p)\sigma ^{2}}},\\\alpha &={\frac {2-p}{p-1}},\\\beta &={\frac {\mu ^{1-p}}{(p-1)\sigma ^{2}}}.\end{aligned}}}$

A compound Poisson process with rate ${\displaystyle \lambda >0}$ and jump size distribution G is a continuous-time stochastic process ${\displaystyle \{\,Y(t):t\geq 0\,\}}$ given by

${\displaystyle Y(t)=\sum _{i=1}^{N(t)}D_{i},}$

where the sum is by convention equal to zero as long as N(t)=0. Here, ${\displaystyle \{\,N(t):t\geq 0\,\}}$ is a Poisson process with rate ${\displaystyle \lambda }$, and ${\displaystyle \{\,D_{i}:i\geq 1\,\}}$ are independent and identically distributed random variables, with distribution function G, which are also independent of ${\displaystyle \{\,N(t):t\geq 0\,\}.\,}$[11]

For the discrete version of compound Poisson process, it can be used in survival analysis for the frailty models.[12]

A compound Poisson distribution, in which the summands have an exponential distribution, was used by Revfeim to model the distribution of the total rainfall in a day, where each day contains a Poisson-distributed number of events each of which provides an amount of rainfall which has an exponential distribution.[13] Thompson applied the same model to monthly total rainfalls.[14]

There has been applications to insurance claims[15][16] and x-ray computed tomography.[17][18][19]