Compound Poisson process

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

$Y(t)=\sum _{i=1}^{N(t)}D_{i}$ where, $\{\,N(t):t\geq 0\,\}$ is a Poisson process with rate $\lambda$ , and $\{\,D_{i}:i\geq 1\,\}$ are independent and identically distributed random variables, with distribution function G, which are also independent of $\{\,N(t):t\geq 0\,\}.\,$ When $D_{i}$ are non-negative integer-valued random variables, then this compound Poisson process is known as a stuttering Poisson process which has the feature that two or more events occur in a very short time .

Properties of the compound Poisson process

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

$\,E(Y(t))=E(D_{1}+...+D_{N(t)})=E(N(t))E(D_{1})=E(N(t))E(D)=\lambda tE(D).$ Making similar use of the law of total variance, the variance can be calculated as:

{\begin{aligned}\operatorname {var} (Y(t))&=E(\operatorname {var} (Y(t)|N(t)))+\operatorname {var} (E(Y(t)|N(t)))\\&=E(N(t)\operatorname {var} (D))+\operatorname {var} (N(t)E(D))\\&=\operatorname {var} (D)E(N(t))+E(D)^{2}\operatorname {var} (N(t))\\&=\operatorname {var} (D)\lambda t+E(D)^{2}\lambda t\\&=\lambda t(\operatorname {var} (D)+E(D)^{2})\\&=\lambda tE(D^{2}).\end{aligned}} Lastly, using the law of total probability, the moment generating function can be given as follows:

$\,\Pr(Y(t)=i)=\sum _{n}\Pr(Y(t)=i|N(t)=n)\Pr(N(t)=n)$ {\begin{aligned}E(e^{sY})&=\sum _{i}e^{si}\Pr(Y(t)=i)\\&=\sum _{i}e^{si}\sum _{n}\Pr(Y(t)=i|N(t)=n)\Pr(N(t)=n)\\&=\sum _{n}\Pr(N(t)=n)\sum _{i}e^{si}\Pr(Y(t)=i|N(t)=n)\\&=\sum _{n}\Pr(N(t)=n)\sum _{i}e^{si}\Pr(D_{1}+D_{2}+\cdots +D_{n}=i)\\&=\sum _{n}\Pr(N(t)=n)M_{D}(s)^{n}\\&=\sum _{n}\Pr(N(t)=n)e^{n\ln(M_{D}(s))}\\&=M_{N(t)}(\ln(M_{D}(s)))\\&=e^{\lambda t\left(M_{D}(s)-1\right)}.\end{aligned}} Exponentiation of measures

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

$\mu (A)=\Pr(D\in A).\,$ Let δ0 be the trivial probability distribution putting all of the mass at zero. Then the probability distribution of Y(t) is the measure

$\exp(\lambda t(\mu -\delta _{0}))\,$ where the exponential exp(ν) of a finite measure ν on Borel subsets of the real line is defined by

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

$\nu ^{*n}=\underbrace {\nu *\cdots *\nu } _{n{\text{ factors}}}$ is a convolution of measures, and the series converges weakly.