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\begin{document}


{\bf Question}

The oven in the baking room at the Glen Eyre Baked Bean factory
produces molecules of a noxious gas according to a Poisson process
of rate $\lambda$. Because of complaint by the workers a catalytic
deodorizer has been installed which neutralizes the gas in the
room at a rate of $\mu$ times the number of molecules present.

${}$

Let $p_n(t)$ denote the probability that $n$ such molecules are
present at time $t$. Obtain the forward differential equations for
$p_n(t)$.

${}$

Assuming that a long-term equilibrium distribution exists, use the
forward equations to obtain difference equations for the
equilibrium probabilities. Solve these equations and deduce that
the equilibrium distribution is a Poisson distribution.

${}$

The local Health and Safety Executive insists that the baking room
should be totally free of this noxious gas for at least $90\%$ of
the time. What does this imply about the relationship between
$\lambda$ and $\mu$?

\vspace{.25in}

{\bf Answer}

Note: all $o(\delta t)$ terms omitted in this \lq solution'.

Let $X(t)$ denote the number of molecules present at time $t$.

$\begin{array}{rcl}n=0: p_0(t+\delta t) & = & P(X(t+\delta t)=0\
|\ X(t)=0)p_0(t)\\ & & +P(X(t+\delta t)=0\ |\ X(t)=1)p_1(t)\\ & =
& (1-\lambda\delta t)p_0(t)+(1-\lambda \delta t)\mu\delta t
p_1(t)\\ & = & p_0(t)-\lambda p_0(t)\delta t+\mu\delta t p_1(t)
\end{array}$

We deduce $p_0'(t)=-\lambda p_0(t)+\mu p_1(t)$

$\begin{array}{rcl}n>0: p_n(t+\delta t) & = & P(X(t+\delta t)=n\
|\ X(t)=n)p_n(t)\\ & & +P(X(t+\delta t)=n\ |\
X(t)=n+1)p_{n+1}(t)\\ & & +P(X(t+\delta t)=n\ |\
X(t)=n-1)p_{n-1}(t)\\ & = & [(1-\lambda\delta t)(1-n\mu\delta
t)+\lambda\delta t\cdot n \mu\delta t] p_n(t)\\ & & +[(1-\lambda
\delta t)(n+1)\mu\delta t] p_{n+1}(t)\\ & & +[\lambda\delta
t(1-(n-1)\mu\delta t]p_{n-1}(t)\\ & = & (1-(\lambda+n\mu)\delta
t)p_n(t)+(n+1)\mu \delta tp_{n+1}t\\ & & +\lambda \delta
tp_{n-1}(t)\end{array}$

We deduce $P_n'(t)=-(\lambda+_n\mu)p_n(t)+(n+1)\mu
p_{n+1}(t)+\lambda p_{n-1}(t)$.

Assuming that $p_n(t)to\pi_n$ as $t\to\infty$ then $p_n'(t)\to 0$
as $t\to\infty$ for all $n$.

We obtain

$0=-\lambda \pi_0+\mu\pi_1$\ \ i.e. $\mu\pi_1=\lambda\pi_0$

$0=-(\lambda+n\,mu)\pi_n+(n+1)\mu\pi_{n+1}+\lambda\pi_{n-1}$

$\begin{array}{rcl} (n+1)\mu\pi_{n+1} & = &
(\lambda+n\mu)\pi_n-\lambda\pi_{n-1}\\ n=1: 2\mu\pi_2 & = &
(\lambda+\mu)\pi_1-\lambda\pi_0=\lambda\pi_1\\ n=2: 3\mu\pi_3 & =
& (\lambda+2\mu)\pi_2-\lambda\pi_1=\lambda\pi_2 \end{array}$

By induction we can then establish

$$n\mu\pi_n=\lambda\pi_{n-1}\ \rm{i.e.}\ \pi_n=\ds
\frac{1}{n}\cdot\ds \frac{\lambda}{\mu}\pi_{n-1}$$

This then gives $\pi_n=\ds \frac{1}{n!}\left(\ds
\frac{\lambda}{\mu}\right)^n\pi_0$.

For this to be a probability distribution $\ds\sum_{n=0}^\infty
\pi_n=1$.

so $\pi_0\ds\sum_{n=0}^\infty \ds \frac{1}{n!}\left(\ds
\frac{\lambda}{\mu}\right)^n=1$ i.e.
$\pi_0=e^{-\frac{\lambda}{\mu}}$

and $\pi_n=\ds \frac{1}{n!}\left(\ds
\frac{\lambda}{\mu}\right)^ne^{-\frac{\lambda}{\mu}}$

so the equilibrium distribution is Poisson with parameter $\ds
\frac{\lambda}{\mu}$.

The proportion of time for which gases are not emitted is
$\pi_0=e^{-\frac{\lambda}{\mu}}$.

So we must have $e^{-\frac{\lambda}{\mu}} \geq \ds \frac{9}{10}$.

$$\rm{i.e.}\ \lambda \leq \mu \ln\ds \frac{10}{9} (\mu \leq
9.49\cdots \times \lambda)$$

\end{document}
