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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 the
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 a probability generating function of the form

$$G(z) = \exp(\lambda t A(z)-\lambda t)$$

During the working day (8 a.m. to 6 p.m.) the Highfield Patent
Medicine Co. receives telephone calls ordering various numbers of
bottles of Dr. Hirst's Rejuvenating Elixir. The telephone calls
arrive according to a Poisson process with rate $\lambda$ calls
per hour. The number $N$ of bottles ordered by a telephone call
has a geometric distribution, i.e.

$$P(N=n)=p(1-p)^{n-1},\ \ n=1,2,\cdots$$

Find the mean and variance of the number of bottles ordered per
day.


\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 $Y_i$ event occur, where $Y_1, Y_2,
...$ are i.i.d. random variables's
\item[(iii)]$Y_i$ and $\{N(t):t\geq 0 \}$ are independent.
The total number of events occurring in a time interval of length
t i s$$X(t) = \sum_{i=1}^{N(t)}Y_i$$
\end{description}

$\{N(t):t\geq 0 \}$ 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{eqnarray*} \sum_{j=0}^\infty z^jP(X(t)=j) & = &
\sum_{j=0}^\infty \sum_{n=0}^\infty z^j P(X(t)=j| N(t) =
n)P(N(t)=n) \\ & = & \sum_{j=0}^\infty \sum_{n=0}^\infty z^j P(Y_1
+ Y_2 + ... +Y_n=j) \frac{\lambda t^ne^{-\lambda t}}{n!} \\ & = &
\sum_{n=0}^\infty \left\{ \sum_{j=0}^\infty z^j  P(Y_1 + Y_2 + ...
+Y_n=j)\right\} \frac{\lambda t^ne^{-\lambda t}}{n!} \\ & = &
\sum_{n=0}^\infty [A(n)]^n \frac{(\lambda t)^n e^{-\lambda t}}{n!}
\\ & &  {\rm since\ the\ }Y_i{\rm \ are\ independent} \\ & = &
\exp (\lambda t A(z) - \lambda t) \\ & = & G(z)
\end{eqnarray*}


${}$

For the geometric distribution \begin{eqnarray*} A(z) &=&
\sum_{n=1}^ainfty p(1-p)^{n-1}z^n \\ &=& p z \sum_{n=1}^\infty
(1-p)^{n-1}z^{n-1} \\ & = & \frac{p z}{1-(1-p)z} \end{eqnarray*}
\newpage
From 8a.m. to 6p.m. there are 10 hours.  So the p.g.f. for the
number of bottles ordered are day is \begin{eqnarray*} G(z) & = &
\exp \left( \frac{10 \lambda p z }{1 - (1-p)z} - 10\lambda\right)
\\ G'(z) & = & \exp \left( \frac{10 \lambda p z }{1 - (1-p)z} -
10\lambda\right)\\ & & \cdot \frac{(1-(1-p)z) +
z(1-p)}{(1-(1-p)z)^2} \cdot 10 \lambda p \\ & = & \exp \left(
\frac{10 \lambda p z }{1 - (1-p)z} - 10\lambda\right)  \cdot
\frac{10 \lambda p}{(1 - (1-p)z)^2} \\ G''(z) & = & \exp \left(
\frac{10 \lambda p z }{1 - (1-p)z} - 10\lambda\right) \left[
\frac{10 \lambda p}{(1 - (1-p)z)^2} \right] \\ & & + \exp \left(
\frac{10 \lambda p z }{1 - (1-p)z} - 10\lambda\right) \cdot
\frac{2(1-p)\cdot 10 \lambda p}{(1 - (1-p)z)^3} \\ G'(1) & = &
\exp (0) \cdot \frac{10\lambda p}{p^2} \\ {\rm  So\ \ } E(X) & = &
\frac{10\lambda}{p} \\ G''(1) & = &
\left(\frac{10\lambda}{p}\right)^2 + \frac{20\lambda(1-p)}{p^2} \\
Var(X) & = & G''(1) + G'(1) - G'(1)^2 \\ & = &
\frac{20\lambda(1-p)}{p^2} +  \frac{10\lambda}{p}
\\ & = & \frac{20\lambda}{p^2} - \frac{\lambda}{p} \\ & = &
\frac{10\lambda}{p^2}(2-p) \end{eqnarray*}


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