Compound Poisson process

A compound Poisson process is a continuous-time stochastic process with jumps. The jumps arrive randomly according to a Poisson process and the size of the jumps is also random, with a specified probability distribution. To be precise, a compound Poisson process, parameterised by a rate ${\displaystyle \lambda >0}${\displaystyle \lambda >0} and jump size distribution G, is a process ${\displaystyle \{,円Y(t):t\geq 0,円\}}${\displaystyle \{,円Y(t):t\geq 0,円\}} given by

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

where, ${\displaystyle \{,円N(t):t\geq 0,円\}}${\displaystyle \{,円N(t):t\geq 0,円\}} is the counting variable of a Poisson process with rate ${\displaystyle \lambda }${\displaystyle \lambda }, and ${\displaystyle \{,円D_{i}:i\geq 1,円\}}${\displaystyle \{,円D_{i}:i\geq 1,円\}} are independent and identically distributed random variables, with distribution function G, which are also independent of ${\displaystyle \{,円N(t):t\geq 0,円\}.,円}${\displaystyle \{,円N(t):t\geq 0,円\}.,円}

When ${\displaystyle D_{i}}${\displaystyle D_{i}} are non-negative integer-valued random variables, then this compound Poisson process is known as a stuttering Poisson process. ^{[citation needed ]}

The expected value of a compound Poisson process can be calculated using a result known as Wald's equation as:

${\displaystyle \operatorname {E} (Y(t))=\operatorname {E} (D_{1}+\cdots +D_{N(t)})=\operatorname {E} (N(t))\operatorname {E} (D_{1})=\operatorname {E} (N(t))\operatorname {E} (D)=\lambda t\operatorname {E} (D).}${\displaystyle \operatorname {E} (Y(t))=\operatorname {E} (D_{1}+\cdots +D_{N(t)})=\operatorname {E} (N(t))\operatorname {E} (D_{1})=\operatorname {E} (N(t))\operatorname {E} (D)=\lambda t\operatorname {E} (D).}

Making similar use of the law of total variance, the variance can be calculated as:

${\displaystyle {\begin{aligned}\operatorname {var} (Y(t))&=\operatorname {E} (\operatorname {var} (Y(t)\mid N(t)))+\operatorname {var} (\operatorname {E} (Y(t)\mid N(t)))\\[5pt]&=\operatorname {E} (N(t)\operatorname {var} (D))+\operatorname {var} (N(t)\operatorname {E} (D))\\[5pt]&=\operatorname {var} (D)\operatorname {E} (N(t))+\operatorname {E} (D)^{2}\operatorname {var} (N(t))\\[5pt]&=\operatorname {var} (D)\lambda t+\operatorname {E} (D)^{2}\lambda t\\[5pt]&=\lambda t(\operatorname {var} (D)+\operatorname {E} (D)^{2})\\[5pt]&=\lambda t\operatorname {E} (D^{2}).\end{aligned}}}${\displaystyle {\begin{aligned}\operatorname {var} (Y(t))&=\operatorname {E} (\operatorname {var} (Y(t)\mid N(t)))+\operatorname {var} (\operatorname {E} (Y(t)\mid N(t)))\\[5pt]&=\operatorname {E} (N(t)\operatorname {var} (D))+\operatorname {var} (N(t)\operatorname {E} (D))\\[5pt]&=\operatorname {var} (D)\operatorname {E} (N(t))+\operatorname {E} (D)^{2}\operatorname {var} (N(t))\\[5pt]&=\operatorname {var} (D)\lambda t+\operatorname {E} (D)^{2}\lambda t\\[5pt]&=\lambda t(\operatorname {var} (D)+\operatorname {E} (D)^{2})\\[5pt]&=\lambda t\operatorname {E} (D^{2}).\end{aligned}}}

Lastly, using the law of total probability, the moment generating function can be given as follows:

${\displaystyle \Pr(Y(t)=i)=\sum _{n}\Pr(Y(t)=i\mid N(t)=n)\Pr(N(t)=n)}${\displaystyle \Pr(Y(t)=i)=\sum _{n}\Pr(Y(t)=i\mid N(t)=n)\Pr(N(t)=n)}

${\displaystyle {\begin{aligned}\operatorname {E} (e^{sY})&=\sum _{i}e^{si}\Pr(Y(t)=i)\\[5pt]&=\sum _{i}e^{si}\sum _{n}\Pr(Y(t)=i\mid N(t)=n)\Pr(N(t)=n)\\[5pt]&=\sum _{n}\Pr(N(t)=n)\sum _{i}e^{si}\Pr(Y(t)=i\mid N(t)=n)\\[5pt]&=\sum _{n}\Pr(N(t)=n)\sum _{i}e^{si}\Pr(D_{1}+D_{2}+\cdots +D_{n}=i)\\[5pt]&=\sum _{n}\Pr(N(t)=n)M_{D}(s)^{n}\\[5pt]&=\sum _{n}\Pr(N(t)=n)e^{n\ln(M_{D}(s))}\\[5pt]&=M_{N(t)}(\ln(M_{D}(s)))\\[5pt]&=e^{\lambda t\left(M_{D}(s)-1\right)}.\end{aligned}}}${\displaystyle {\begin{aligned}\operatorname {E} (e^{sY})&=\sum _{i}e^{si}\Pr(Y(t)=i)\\[5pt]&=\sum _{i}e^{si}\sum _{n}\Pr(Y(t)=i\mid N(t)=n)\Pr(N(t)=n)\\[5pt]&=\sum _{n}\Pr(N(t)=n)\sum _{i}e^{si}\Pr(Y(t)=i\mid N(t)=n)\\[5pt]&=\sum _{n}\Pr(N(t)=n)\sum _{i}e^{si}\Pr(D_{1}+D_{2}+\cdots +D_{n}=i)\\[5pt]&=\sum _{n}\Pr(N(t)=n)M_{D}(s)^{n}\\[5pt]&=\sum _{n}\Pr(N(t)=n)e^{n\ln(M_{D}(s))}\\[5pt]&=M_{N(t)}(\ln(M_{D}(s)))\\[5pt]&=e^{\lambda t\left(M_{D}(s)-1\right)}.\end{aligned}}}

Let N, Y, and D be as above. Let μ be the probability measure according to which D is distributed, i.e.

${\displaystyle \mu (A)=\Pr(D\in A).,円}${\displaystyle \mu (A)=\Pr(D\in A).,円}

Let δ₀ be the trivial probability distribution putting all of the mass at zero. Then the probability distribution of Y(t) is the measure

${\displaystyle \exp(\lambda t(\mu -\delta _{0})),円}${\displaystyle \exp(\lambda t(\mu -\delta _{0})),円}

where the exponential exp(ν) of a finite measure ν on Borel subsets of the real line is defined by

${\displaystyle \exp(\nu )=\sum _{n=0}^{\infty }{\nu ^{*n} \over n!}}${\displaystyle \exp(\nu )=\sum _{n=0}^{\infty }{\nu ^{*n} \over n!}}

and

${\displaystyle \nu ^{*n}=\underbrace {\nu *\cdots *\nu } _{n{\text{ factors}}}}${\displaystyle \nu ^{*n}=\underbrace {\nu *\cdots *\nu } _{n{\text{ factors}}}}

is a convolution of measures, and the series converges weakly.

• Poisson process
• Poisson distribution
• Compound Poisson distribution
• Non-homogeneous Poisson process
• Campbell's formula for the moment generating function of a compound Poisson process

• Poisson point processes
• Lévy processes

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