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{\bf Question}

\begin{description}
\item[(a)]
Suppose that events occur randomly in time, and that the time
intervals between successive events are independent and
identically distrusted, having a negative exponential
distribution.  Prove that such a sequence of events forms a
Poisson process.

${}$

You may assume without proof that a sum of i.i.d. negative
exponential random variables has a gamma distribution,
$\Gamma(n,\lambda)$ with p.d.f.

$$\frac{(\lambda x)^{n-1}e^{-\lambda x}}{(n-1)!}$$

Light bulbs have an average lifetime of 200 days, and are replaced
as soon as they fail.  The time intervals between replacements are
independent and identically distrusted, having a negative
exponential distribution.  What is the probability that at least
1000 days have elapsed since
\begin{description}
\item[(i)] the last new light bulb was fitted,
\item[(ii)] the next to last light bulb was fitted?
\end{description}


\item[(b)]
A machine needs two transistors to function, one of type A and one
of type B.  Both types fail independently according to a Posson
processes, type B twice as often as type A on average.  Gives that
the machine has failed 10 times, what is that probability that 5
failures are due to type A and 5 are due to type B.  Justify your
conclusion.


\end{description}



\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
Let $N(t)$ denote the number of events which occur in time t.  Let
$W_n$ denote the waiting time till the n-th event.  $W_n
\sim\Gamma(n,\lambda)$ for some $\lambda$.

So \begin{eqnarray*} P(N(t)=n)& = & P(W_n \leq t) - P(W_{n+1} \leq
t) \\ & = & \int_0^t\frac{\lambda\cdot(\lambda x)^{n-1}e^{-\lambda
x}}{(n-1)!} \, dx - \int_0^t \frac{\lambda(\lambda x)^n
e^{-\lambda x}}{n!} \, dx \\ & = & \frac{(\lambda t)^n e^{-\lambda
t}}{n!} \\ & & \hspace{.2in}{\rm (Integrating\ the\ 1st\ by\
parts)} \end{eqnarray*}

This is the Poisson probability.

If light bulbs have an average lifetime of 200 days the they fail
according to a poisson process of the rate $\lambda =
\frac{1}{200}$.  The number failing in 100 days has a poisson
distribution with parameter 5.
\begin{description}
\item[(i)] The required probability is $$P(N=0) = e^{-5} \approx
0.0067$$
\item[(ii)] The required probability is $$P(N=0) + P(N=1) = e^{-5} + 5e^{-5} \approx
0.0404$$

\end{description}
\item[(b)] Suppose $\lambda$ is the Poisson Parameter for type A
failure.  Then $2\lambda$ is the poisson parameter for type B
failure.  Let $N_A$ and $N_B$ denote the number of failures of
each type in time t.  Then \begin{eqnarray*} P(N_A = 5| N_A + N_B
+10) & = & \frac{P(N_A = 5 {\rm \ and \ }N_B = 10)}{P(N_A + N_B =
10)} \\ & = & \frac{\ds \frac{e^{-\lambda t}(\lambda
t)^5}{5!}\frac{e^{-2\lambda t}(2\lambda t)^5}{5!}}{\ds
\frac{e^{-3\lambda t}(3\lambda t)^{10}}{10!}} \\ & = &
\frac{10!}{5!5!} \cdot \frac{2^5}{3^{10}} \\ & = & \frac{252
\times 32}{59049} \\ & \approx & 0.137 \end{eqnarray*}

\end{description}


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