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Compound Poisson distribution

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In probability theory, a compound Poisson distribution is the probability distribution of the sum of a number of independent identically-distributed random variables, where the number of terms to be added is itself a Poisson-distributed variable. The result can be either a continuous or a discrete distribution.

Contents

• Definition

• Properties

• Discrete compound Poisson distribution

• Compound Poisson Gamma distribution

• Compound Poisson processes

• Applications

• See also

• References

Definition

Suppose that

N ∼ ∼ Poisson ⁡ ⁡ ( λ λ ) , {\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

X 1 , X 2 , X 3 , … … {\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 N {\displaystyle N} i.i.d. random variables

Y = ∑ ∑ n = 1 N X n {\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.

Properties

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

E ⁡ ⁡ ( Y ) = E ⁡ ⁡ [ E ⁡ ⁡ ( Y ∣ ∣ N ) ] = E ⁡ ⁡ [ N E ⁡ ⁡ ( X ) ] = E ⁡ ⁡ ( N ) E ⁡ ⁡ ( X ) , {\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),}

Var ⁡ ⁡ ( Y ) = E ⁡ ⁡ [ Var ⁡ ⁡ ( Y ∣ ∣ N ) ] + Var ⁡ ⁡ [ E ⁡ ⁡ ( Y ∣ ∣ N ) ] = E ⁡ ⁡ [ N Var ⁡ ⁡ ( X ) ] + Var ⁡ ⁡ [ N E ⁡ ⁡ ( X ) ] , = E ⁡ ⁡ ( N ) Var ⁡ ⁡ ( X ) + ( E ⁡ ⁡ ( X ) ) 2 Var ⁡ ⁡ ( N ) . {\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-distributed, these formulae can be reduced to

E ⁡ ⁡ ( Y ) = E ⁡ ⁡ ( N ) E ⁡ ⁡ ( X ) = λ λ E ⁡ ⁡ ( X ) , {\displaystyle \operatorname {E} (Y)=\operatorname {E} (N)\operatorname {E} (X)=\lambda \operatorname {E} (X),}

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

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

φ φ Y ( t ) = E ⁡ ⁡ ( e i t Y ) = E ⁡ ⁡ ( ( E ⁡ ⁡ ( e i t X ∣ ∣ N ) ) N ) = E ⁡ ⁡ ( ( φ φ X ( t ) ) N ) , {\displaystyle \varphi _{Y}(t)=\operatorname {E} (e^{itY})=\operatorname {E} \left(\left(\operatorname {E} (e^{itX}\mid N)\right)^{N}\right)=\operatorname {E} \left((\varphi _{X}(t))^{N}\right),\,}

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

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

An alternative approach is via cumulant generating functions:

K Y ( t ) = ln ⁡ ⁡ E ⁡ ⁡ [ e t Y ] = ln ⁡ ⁡ E ⁡ ⁡ [ E ⁡ ⁡ [ e t Y ∣ ∣ N ] ] = ln ⁡ ⁡ E ⁡ ⁡ [ e N K X ( t ) ] = K N ( K X ( t ) ) . {\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.

Every infinitely divisible probability distribution is a limit of compound Poisson distributions.cite-ref-1[1] And compound Poisson distributions is infinitely divisible by the definition.

Discrete compound Poisson distribution

When X 1 , X 2 , X 3 , … … {\displaystyle X_{1},X_{2},X_{3},\dots } are positive integer-valued i.i.d random variables with P ( X 1 = k ) = α α k , ( k = 1 , 2 , … … ) {\displaystyle P(X_{1}=k)=\alpha _{k},\ (k=1,2,\ldots )} , then this compound Poisson distribution is named discrete compound Poisson distributioncite-ref-libro-2-0[2]cite-ref-zhang-3-0[3]cite-ref-zhang2-4-0[4] (or stuttering-Poisson distributioncite-ref-kemp-5-0[5]) . We say that the discrete random variable Y {\displaystyle Y} satisfying probability generating function characterization

P Y ( z ) = ∑ ∑ i = 0 ∞ ∞ P ( Y = i ) z i = exp ⁡ ⁡ ( ∑ ∑ k = 1 ∞ ∞ α α k λ λ ( z k − − 1 ) ) , ( | z | ≤ ≤ 1 ) {\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 ( α α 1 λ λ , α α 2 λ λ , … … ) ∈ ∈ R ∞ ∞ {\displaystyle (\alpha _{1}\lambda ,\alpha _{2}\lambda ,\ldots )\in \mathbb {R} ^{\infty }} (where ∑ ∑ i = 1 ∞ ∞ α α i = 1 {\textstyle \sum _{i=1}^{\infty }\alpha _{i}=1} , with α α i ≥ ≥ 0 , λ λ > 0 {\textstyle \alpha _{i}\geq 0,\lambda >0} ), which is denoted by

X ∼ ∼ DCP ( λ λ α α 1 , λ λ α α 2 , … … ) {\displaystyle X\sim {\text{DCP}}(\lambda {\alpha _{1}},\lambda {\alpha _{2}},\ldots )}

Moreover, if X ∼ ∼ DCP ( λ λ α α 1 , … … , λ λ α α r ) {\displaystyle X\sim {\operatorname {DCP} }(\lambda {\alpha _{1}},\ldots ,\lambda {\alpha _{r}})} , we say X {\displaystyle X} has a discrete compound Poisson distribution of order r {\displaystyle r} . When r = 1 , 2 {\displaystyle r=1,2} , DCP becomes Poisson distribution and Hermite distribution, respectively. When r = 3 , 4 {\displaystyle r=3,4} , DCP becomes triple stuttering-Poisson distribution and quadruple stuttering-Poisson distribution, respectively.cite-ref-6[6] Other special cases include: shift geometric 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 papercite-ref-7[7] and references therein.

Feller's characterization of the compound Poisson distribution states that a non-negative integer valued r.v. X {\displaystyle X} is infinitely divisible if and only if its distribution is a discrete compound Poisson distribution.cite-ref-8[8] 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, ..., Xn 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 queuecite-ref-kemp-5-1[5]cite-ref-9[9]). The discrete compound Poisson distribution is also widely used in actuarial science for modelling the distribution of the total claim amount.cite-ref-zhang-3-1[3]

When some α α k {\displaystyle \alpha _{k}} are negative, it is the discrete pseudo compound Poisson distribution.cite-ref-zhang-3-2[3] We define that any discrete random variable Y {\displaystyle Y} satisfying probability generating function characterization

G Y ( z ) = ∑ ∑ i = 0 ∞ ∞ P ( Y = i ) z i = exp ⁡ ⁡ ( ∑ ∑ k = 1 ∞ ∞ α α k λ λ ( z k − − 1 ) ) , ( | z | ≤ ≤ 1 ) {\displaystyle G_{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 pseudo compound Poisson distribution with parameters ( λ λ 1 , λ λ 2 , … … ) =: ( α α 1 λ λ , α α 2 λ λ , … … ) ∈ ∈ R ∞ ∞ {\displaystyle (\lambda _{1},\lambda _{2},\ldots )=:(\alpha _{1}\lambda ,\alpha _{2}\lambda ,\ldots )\in \mathbb {R} ^{\infty }} where ∑ ∑ i = 1 ∞ ∞ α α i = 1 {\textstyle \sum _{i=1}^{\infty }{\alpha _{i}}=1} and ∑ ∑ i = 1 ∞ ∞ | α α i | < ∞ ∞ {\textstyle \sum _{i=1}^{\infty }{\left|{\alpha _{i}}\right|}<\infty } , with α α i ∈ ∈ R , λ λ > 0 {\displaystyle {\alpha _{i}}\in \mathbb {R} ,\lambda >0} .

Compound Poisson Gamma distribution

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 is a Tweedie distribution with variance power 1 < p < 2 (proof via comparison of characteristic function).cite-ref-j-rgensen-1997-10-0[10] To be more explicit, if

N ∼ ∼ Poisson ⁡ ⁡ ( λ λ ) , {\displaystyle N\sim \operatorname {Poisson} (\lambda ),}

and

X i ∼ ∼ Γ Γ ⁡ ⁡ ( α α , β β ) {\displaystyle X_{i}\sim \operatorname {\Gamma } (\alpha ,\beta )}

i.i.d., then the distribution of

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

is a reproductive exponential dispersion model E D ( μ μ , σ σ 2 ) {\displaystyle ED(\mu ,\sigma ^{2})} with

E ⁡ ⁡ [ Y ] = λ λ α α β β =: μ μ , Var ⁡ ⁡ [ Y ] = λ λ α α ( 1 + α α ) β β 2 =: σ σ 2 μ μ p . {\displaystyle {\begin{aligned}\operatorname {E} [Y]&=\lambda {\frac {\alpha }{\beta }}=:\mu ,\\[4pt]\operatorname {Var} [Y]&=\lambda {\frac {\alpha (1+\alpha )}{\beta ^{2}}}=:\sigma ^{2}\mu ^{p}.\end{aligned}}}

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

λ λ = μ μ 2 − − p ( 2 − − p ) σ σ 2 , α α = 2 − − p p − − 1 , β β = μ μ 1 − − p ( p − − 1 ) σ σ 2 . {\displaystyle {\begin{aligned}\lambda &={\frac {\mu ^{2-p}}{(2-p)\sigma ^{2}}},\\[4pt]\alpha &={\frac {2-p}{p-1}},\\[4pt]\beta &={\frac {\mu ^{1-p}}{(p-1)\sigma ^{2}}}.\end{aligned}}}

Compound Poisson processes

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

Y ( t ) = ∑ ∑ i = 1 N ( t ) D i , {\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, { N ( t ) : t ≥ ≥ 0 } {\displaystyle \{\,N(t):t\geq 0\,\}} is a Poisson process with rate λ λ {\displaystyle \lambda } , and { D i : i ≥ ≥ 1 } {\displaystyle \{\,D_{i}:i\geq 1\,\}} are independent and identically distributed random variables, with distribution function G, which are also independent of { N ( t ) : t ≥ ≥ 0 } . {\displaystyle \{\,N(t):t\geq 0\,\}.\,} cite-ref-11[11]

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

Applications

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.cite-ref-revf-13-0[13] Thompson applied the same model to monthly total rainfalls.cite-ref-thom-14-0[14]

There have been applications to insurance claimscite-ref-15[15]cite-ref-16[16] and x-ray computed tomography.cite-ref-17[17]cite-ref-18[18]cite-ref-19[19]