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{\bf Question}

Describe what is meant by a compound Poisson process.

${}$

Show that if $A(z)$ is the probability generating function for hte
number of events occurring at each point of the process, then the
random variable $X()t$ - the total number of events occurring in
time $t$ - has probability generating function of the form

$$G_t(z)=\exp(\lambda t A(z) -\lambda t).$$

Suppose that the average number of events occurring at each point
of the process if $\mu$. Show that the average number of events
occurring in time $t$ is $\mu \lambda t$.

${}$

Let the probability that no events occur at a point in the process
be $p$. Find an expression for the probability of no events
occurring in time $t$ in the compound process.



\vspace{.25in}

{\bf Answer}

Suppose that

\begin{description}
\item[(i)]
points occur in a Poisson process $\{N(t): t \geq 0\}$ with rate
$\lambda$.

\item[(ii)]
at the ith point $Y_i$ events occur, where $Y_1,\ Y_2, \cdots$ are
i.i.d random variables.

\item[(iii)]
$Y_i$ and $\{N(t): t \geq 0\}$ are independent.
\end{description}

The total number of events occurring  time interval of length $t$
is

$$X(t)=\ds\sum_{i=1}^{N(t)} y_i$$

$\{X(t): t\geq 0\}$ is said to be a compound Poisson process.

Let the p.g.f of each $Y_i$ be $A(z)$. Then $X(t)$ has p.g.f.

$\begin{array}{rcl} &amp; &amp; \ds\sum_{j=0}^\infty z^j P(X(t)=j)\\ &amp; = &amp;
\ds\sum_{j=0}^\infty \ds\sum_{n=0}^\infty z^j P(X(t)=j\ |\
N(t)=n)P(N(t)=n)\\ &amp; = &amp; \ds\sum_{j=0}^\infty \ds\sum_{n=0}^\infty
z^j P(Y_1+\cdots +Y_n=j)\df{(\lambda t)^ne^{-\lambda t}}{n!}\\ &amp; =
&amp; \ds\sum_{j=0}^\infty \left\{\ds\sum_{n=0}^\infty z^j
P(Y_1+\cdots +Y_n=j)\right\}\df{(\lambda t)^ne^{-\lambda t}}{n!}\\
&amp; = &amp; \ds\sum_{n=0}^\infty [A(z)]^n\df{(\lambda t)^ne^{-\lambda
t}}{n!} \mbox{since the $Y_i$ are independent}\\ &amp; = &amp;
\exp(\lambda t A(z)-\lambda t)\end{array}$

The average number of events in time $t$ is given by $G'(1)$.

$$G'_t(z)=\exp(\lambda t A(z)-\lambda(t))\lambda t A'(z)$$

so $G'_t(1)=\exp(\lambda t A(1)-\lambda t)\cdot \lambda t A'(1)$

Now for any p.g.f. $A(z),\ A(1)=1$ and $A'(1)=\mu$

so $G'_t(1)=\exp(0)\lambda t A'(10)=\mu \lambda t$.

The probability of no events occurring tiem $t$ is given by
$G_t(0)=\exp(\lambda t A(0)-\lambda t)$.

now $A(0)=p$ and so the required probability is $\exp(\lambda t
p-\lambda t)$.




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