Compound Poisson process: Difference between revisions

Revision as of 16:00, 2 February 2020

Revision as of 14:39, 25 June 2020

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

:<math>\,operatorname E(Y(t)) = \operatorname E(D_{1}D_1 + ...\cdots + D_{N(t)}) = \operatorname E(N(t))\operatorname E(D_{1}D_1) = \operatorname E(N(t)) \operatorname E(D) = \lambda t \operatorname E(D).</math>

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

:<math>

\begin{align}

\operatorname{var}(Y(t)) &= \operatorname E(\operatorname{var}(Y(t)|\mid N(t))) + \operatorname{var}(\operatorname E(Y(t)|\mid N(t))) \\

&= \operatorname E(N(t)\operatorname{var}(D)) + \operatorname{var}(N(t) \operatorname E(D)) \\

&= \operatorname{var}(D) \operatorname E(N(t)) + \operatorname E(D)^2 \operatorname{var}(N(t)) \\

&= \operatorname{var}(D)\lambda t + \operatorname E(D)^2\lambda t \\

&= \lambda t(\operatorname{var}(D) + \operatorname E(D)^2) \\

&= \lambda t \operatorname E(D^2).

\end{align}

</math>

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

:<math>\,\Pr(Y(t)=i) = \sum_{n}sum_n \Pr(Y(t)=i|\mid N(t)=n)\Pr(N(t)=n) </math>

:<math>

\begin{align}

\operatorname E(e^{sY}) & = \sum_i e^{si} \Pr(Y(t)=i) \\

& = \sum_i e^{si} \sum_{n} \Pr(Y(t)=i|\mid N(t)=n)\Pr(N(t)=n) \\

& = \sum_n \Pr(N(t)=n) \sum_i e^{si} \Pr(Y(t)=i|\mid 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 \\