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\begin{document}


{\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 a time
interval of length $t$ - has probability generating function

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

Points in time occur in a Poisson process with rate $\lambda$. At
each point two fair coins are tossed. Find the probability
generating function for the total number of heads occurring in a
time interval of length $t$. Find the mean number of heads
occurring in a time interval of length $t$.

Let $W$ denote the waiting time before any heads occur.

Show that

$$P(W>t)=\exp\left(\df{-3\lambda t}{4}\right)$$



\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 in a time interval of length
$t$ is

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

$\{X(t): t \geq 0\}$ 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{eqnarray*} \sum_{j=0}^\infty z^j P(X(t)=j) & = &
\sum_{j=0}^\infty\sum_{n=0}^\infty x^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+\cdots+Y_n=j)\df{(\lambda t)^ne^{-\lambda t}}{n!}\\ & = &
\sum_{j=0}^\infty \left\{\sum_{n=0}^\infty z^j
P(Y_1+\cdots+Y_n=j)\right\}\df{(\lambda t)^ne^{-\lambda t}}{n!}\\
& = & \sum_{n=0}^\infty [A(z)]^n \df{(\lambda t)^ne^{-\lambda
t}}{n!}\\ & & \mbox{since the $Y_i$ are independent}\\ & = &
\exp(\lambda t A(z)-\lambda t)\end{eqnarray*}

${}$

Now $A(x)=\df{1}{4}+\df{1}{2}x+\df{1}{4}z^2$ in this case.

So the number of heads in a time interval of length $t$ has p.g.f.

$\begin{array}{rcl} G_t(z) & = & \exp\left(\lambda
t\left(\df{1}{4}+\df{1}{2}z+\df{1}{4}z^2\right)-\lambda t\right)\\
& = & \exp\left(\lambda
t\left(\df{1}{4}z^2+\df{1}{2}z-\df{3}{4}\right)\right)
\end{array}$

The mean number of heads is $G'(1)$.

$$G'(z)=\exp\left(\lambda
t\left(\df{1}{4}z^2+\df{1}{2}z-\df{3}{4}\right)\right) \cdot
\lambda t \left(\df{1}{2} z+\df{1}{2}\right)$$

so $G'(1)=e^0 \lambda t=\lambda t$

Let $W$ be the waiting time before a head is recorded.

${}$

\un{EITHER}

$\begin{array}{rcl} P(W>t) & = & \mbox{P(no events in compound
process in $(0,\ t]$)}\\ & = & G_t(0)=\exp\left(-\df{3}{4}\lambda
t\right) \end{array}$

\un{OR}

$\begin{array}{rcl} P(W>t) & = & \mbox{P(no events in Poisson
process)}\\ & & + \ds\sum_{n=1}^\infty\ \mbox{($n$ events and 2
tails at each event)}\\ & = & e^{-\lambda t}+\ds\sum_{n=1}^\infty
\left(\df{1}{4}\right)^n \df{(\lambda t)^n e^{-\lambda t}}{n!}\\ &
= & e^{-\lambda t}+e^{-\lambda t}\left[e^{\lambda
\frac{t}{4}}-1\right]\\ & = & e^{-\frac{3\lambda t}{4}}
\end{array}$



\end{document}