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.

Definition

Suppose that

N ∼ Poisson ⁡ ( λ ) , {\displaystyle N\sim \operatorname {Poisson} (\lambda ),} {\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 } {\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} {\displaystyle N} i.i.d. random variables

Y = ∑ n = 1 N X n {\displaystyle Y=\sum _{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),} {\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}}} {\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 E ⁡ ( X ) , {\displaystyle \operatorname {E} (Y)=\operatorname {E} (N)\operatorname {E} (X)=\lambda \operatorname {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 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})}.} {\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 Y ( t ) = E ⁡ ( e i t Y ) = E ⁡ ( ( E ⁡ ( e i t X ∣ N ) ) N ) = E ⁡ ( ( 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),\,} {\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 Y ( t ) = e λ ( φ X ( t ) − 1 ) . {\displaystyle \varphi _{Y}(t)={\textrm {e}}^{\lambda (\varphi _{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)).\,} {\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.[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 } {\displaystyle X_{1},X_{2},X_{3},\dots } are positive integer-valued i.i.d random variables with P ( X 1 = k ) = α k ,   ( k = 1 , 2 , … ) {\displaystyle P(X_{1}=k)=\alpha _{k},\ (k=1,2,\ldots )} {\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 Y {\displaystyle Y} {\displaystyle Y} satisfying probability generating function characterization

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

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

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

Feller's characterization of the compound Poisson distribution states that a non-negative integer valued r.v. X {\displaystyle X} {\displaystyle X} is infinitely divisible if and only if its distribution is a discrete compound Poisson distribution.[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 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 α k {\displaystyle \alpha _{k}} {\displaystyle \alpha _{k}} are negative, it is the discrete pseudo compound Poisson distribution.[3] We define that any discrete random variable Y {\displaystyle Y} {\displaystyle Y} satisfying probability generating function characterization

G Y ( z ) = ∑ i = 0 ∞ P ( Y = i ) z i = exp ⁡ ( ∑ k = 1 ∞ α k 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)} {\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 1 , λ 2 , … ) =: ( α 1 λ , α 2 λ , … ) ∈ R ∞ {\displaystyle (\lambda _{1},\lambda _{2},\ldots )=:(\alpha _{1}\lambda ,\alpha _{2}\lambda ,\ldots )\in \mathbb {R} ^{\infty }} {\displaystyle (\lambda _{1},\lambda _{2},\ldots )=:(\alpha _{1}\lambda ,\alpha _{2}\lambda ,\ldots )\in \mathbb {R} ^{\infty }} where ∑ i = 1 ∞ α i = 1 {\textstyle \sum _{i=1}^{\infty }{\alpha _{i}}=1} {\textstyle \sum _{i=1}^{\infty }{\alpha _{i}}=1} and ∑ i = 1 ∞ | α i | < ∞ {\textstyle \sum _{i=1}^{\infty }{\left|{\alpha _{i}}\right|}<\infty } {\textstyle \sum _{i=1}^{\infty }{\left|{\alpha _{i}}\right|}<\infty }, with α i ∈ R , λ > 0 {\displaystyle {\alpha _{i}}\in \mathbb {R} ,\lambda >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).[10] To be more explicit, if

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

and

X i ∼ Γ ⁡ ( α , β ) {\displaystyle X_{i}\sim \operatorname {\Gamma } (\alpha ,\beta )} {\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}} {\displaystyle Y=\sum _{i=1}^{N}X_{i}}

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

E ⁡ [ Y ] = λ α β =: μ , Var ⁡ [ Y ] = λ α ( 1 + α ) β 2 =: σ 2 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}}} {\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 , σ 2 , p {\displaystyle \mu ,\sigma ^{2},p} {\displaystyle \mu ,\sigma ^{2},p} to the Poisson and Gamma parameters λ , α , β {\displaystyle \lambda ,\alpha ,\beta } {\displaystyle \lambda ,\alpha ,\beta } is the following:

λ = μ 2 − p ( 2 − p ) σ 2 , α = 2 − p p − 1 , β = μ 1 − p ( p − 1 ) 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}}} {\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 > 0 {\displaystyle \lambda >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\,\}} {\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},} {\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\,\}} {\displaystyle \{\,N(t):t\geq 0\,\}} is a Poisson process with rate λ {\displaystyle \lambda } {\displaystyle \lambda }, and { D i : i ≥ 1 } {\displaystyle \{\,D_{i}:i\geq 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\,\}.\,} {\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]

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.[13] Thompson applied the same model to monthly total rainfalls.[14]

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

See also

References

  1. Lukacs, E. (1970). Characteristic functions. London: Griffin. ISBN 0-85264-170-2.
  2. Johnson, N.L., Kemp, A.W., and Kotz, S. (2005) Univariate Discrete Distributions, 3rd Edition, Wiley, ISBN 978-0-471-27246-5.
  3. Huiming, Zhang; Yunxiao Liu; Bo Li (2014). "Notes on discrete compound Poisson model with applications to risk theory". Insurance: Mathematics and Economics. 59: 325–336. doi:10.1016/j.insmatheco.2014.09.012.
  4. Huiming, Zhang; Bo Li (2016). "Characterizations of discrete compound Poisson distributions". Communications in Statistics - Theory and Methods. 45 (22): 6789–6802. doi:10.1080/03610926.2014.901375. S2CID 125475756.
  5. Kemp, C. D. (1967). ""Stuttering – Poisson" distributions". Journal of the Statistical and Social Enquiry of Ireland. 21 (5): 151–157. hdl:2262/6987.
  6. Patel, Y. C. (1976). Estimation of the parameters of the triple and quadruple stuttering-Poisson distributions. Technometrics, 18(1), 67-73.
  7. Wimmer, G., Altmann, G. (1996). The multiple Poisson distribution, its characteristics and a variety of forms. Biometrical journal, 38(8), 995-1011.
  8. Feller, W. (1968). An Introduction to Probability Theory and its Applications. Vol. I (3rd ed.). New York: Wiley.
  9. Adelson, R. M. (1966). "Compound Poisson Distributions". Journal of the Operational Research Society. 17 (1): 73–75. doi:10.1057/jors.1966.8.
  10. Jørgensen, Bent (1997). The theory of dispersion models. Chapman & Hall. ISBN 978-0412997112.
  11. S. M. Ross (2007). Introduction to Probability Models (ninth ed.). Boston: Academic Press. ISBN 978-0-12-598062-3.
  12. Ata, N.; Özel, G. (2013). "Survival functions for the frailty models based on the discrete compound Poisson process". Journal of Statistical Computation and Simulation. 83 (11): 2105–2116. doi:10.1080/00949655.2012.679943. S2CID 119851120.
  13. Revfeim, K. J. A. (1984). "An initial model of the relationship between rainfall events and daily rainfalls". Journal of Hydrology. 75 (1–4): 357–364. Bibcode:1984JHyd...75..357R. doi:10.1016/0022-1694(84)90059-3.
  14. Thompson, C. S. (1984). "Homogeneity analysis of a rainfall series: an application of the use of a realistic rainfall model". Journal of Climatology. 4 (6): 609–619. Bibcode:1984IJCli...4..609T. doi:10.1002/joc.3370040605.
  15. Jørgensen, Bent; Paes De Souza, Marta C. (January 1994). "Fitting Tweedie's compound poisson model to insurance claims data". Scandinavian Actuarial Journal. 1994 (1): 69–93. doi:10.1080/03461238.1994.10413930.
  16. Smyth, Gordon K.; Jørgensen, Bent (29 August 2014). "Fitting Tweedie's Compound Poisson Model to Insurance Claims Data: Dispersion Modelling". ASTIN Bulletin. 32 (1): 143–157. doi:10.2143/AST.32.1.1020.
  17. Whiting, Bruce R. (3 May 2002). Antonuk, Larry E.; Yaffe, Martin J. (eds.). "Signal statistics in x-ray computed tomography". Medical Imaging 2002: Physics of Medical Imaging. 4682. International Society for Optics and Photonics: 53–60. Bibcode:2002SPIE.4682...53W. doi:10.1117/12.465601. S2CID 116487704.
  18. Elbakri, Idris A.; Fessler, Jeffrey A. (16 May 2003). Sonka, Milan; Fitzpatrick, J. Michael (eds.). "Efficient and accurate likelihood for iterative image reconstruction in x-ray computed tomography". Medical Imaging 2003: Image Processing. 5032. SPIE: 1839–1850. Bibcode:2003SPIE.5032.1839E. CiteSeerX 10.1.1.419.3752. doi:10.1117/12.480302. S2CID 12215253.
  19. Whiting, Bruce R.; Massoumzadeh, Parinaz; Earl, Orville A.; O'Sullivan, Joseph A.; Snyder, Donald L.; Williamson, Jeffrey F. (24 August 2006). "Properties of preprocessed sinogram data in x-ray computed tomography". Medical Physics. 33 (9): 3290–3303. Bibcode:2006MedPh..33.3290W. doi:10.1118/1.2230762. PMID 17022224.