Compound Poisson process

On this page

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 λ > 0 {\displaystyle \lambda >0} and jump size distribution G, is a 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, { N ( t ) : t ≥ 0 } {\displaystyle \{\,N(t):t\geq 0\,\}} is the counting variable of a Poisson process with rate 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\,\}.\,}

When D i {\displaystyle D_{i}} are non-negative integer-valued random variables, then this compound Poisson process is known as a stuttering Poisson process.

We don't have any images related to Compound Poisson process yet.

You can add one yourself here.

We don't have any YouTube videos related to Compound Poisson process yet.

You can add one yourself here.

We don't have any PDF documents related to Compound Poisson process yet.

You can add one yourself here.

We don't have any Books related to Compound Poisson process yet.

You can add one yourself here.

We don't have any archived web articles related to Compound Poisson process yet.

You can submit a link to a page to archive here.

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 ) = λ t 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:

var ⁡ ( Y ( t ) ) = E ⁡ ( var ⁡ ( Y ( t ) ∣ N ( t ) ) ) + var ⁡ ( E ⁡ ( Y ( t ) ∣ N ( t ) ) ) = E ⁡ ( N ( t ) var ⁡ ( D ) ) + var ⁡ ( N ( t ) E ⁡ ( D ) ) = var ⁡ ( D ) E ⁡ ( N ( t ) ) + E ⁡ ( D ) 2 var ⁡ ( N ( t ) ) = var ⁡ ( D ) λ t + E ⁡ ( D ) 2 λ t = λ t ( var ⁡ ( D ) + E ⁡ ( D ) 2 ) = λ t E ⁡ ( D 2 ) . {\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:

Pr ( Y ( t ) = i ) = ∑ n Pr ( Y ( t ) = i ∣ 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)} E ⁡ ( e s Y ) = ∑ i e s i Pr ( Y ( t ) = i ) = ∑ i e s i ∑ n Pr ( Y ( t ) = i ∣ N ( t ) = n ) Pr ( N ( t ) = n ) = ∑ n Pr ( N ( t ) = n ) ∑ i e s i Pr ( Y ( t ) = i ∣ N ( t ) = n ) = ∑ n Pr ( N ( t ) = n ) ∑ i e s i Pr ( D 1 + D 2 + ⋯ + D n = i ) = ∑ n Pr ( N ( t ) = n ) M D ( s ) n = ∑ n Pr ( N ( t ) = n ) e n ln ⁡ ( M D ( s ) ) = M N ( t ) ( ln ⁡ ( M D ( s ) ) ) = e λ t ( M D ( s ) − 1 ) . {\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}}}

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.

μ ( A ) = Pr ( D ∈ A ) . {\displaystyle \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 ⁡ ( λ t ( μ − δ 0 ) ) {\displaystyle \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 ⁡ ( ν ) = ∑ n = 0 ∞ ν ∗ n n ! {\displaystyle \exp(\nu )=\sum _{n=0}^{\infty }{\nu ^{*n} \over n!}}

and

ν ∗ n = ν ∗ ⋯ ∗ ν ⏟ n factors {\displaystyle \nu ^{*n}=\underbrace {\nu *\cdots *\nu } _{n{\text{ factors}}}}

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

Properties of the compound Poisson process

Exponentiation of measures

See also