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\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
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\newcommand{\undb}{\underbrace}
\newcommand{\df}{\ds\frac}
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\begin{document}


{\bf Question}

In a cotton spinning room dust particles are produced according to
a Poisson process of rate $\lambda$. An extractor fan removes the
dust particles at a rate $\mu$ times the number of particles
present.

${}$

Let $p_n(t)$ denote the probability that there are $n$ dust
particles present at time $t$. Obtain the forward differential
equations for $p_n(6)$. The generating function for these
probabilities is defined by

$$G(z,\ t)=\sum_{n=0}^\infty p_n(t)z^n.$$

Show that

$$\df{\pl G}{\pl t}+\mu(z-1)\df{\pl G}{\pl z}=\lambda(z-1)G.$$

Verify that this equation is satisfied by

$$G(z,\ t)=\exp\left(\df{\lambda z}{\mu}\right) \cdot f(e^{-\mu
t}(z-1))$$

where $f$ is an arbitrary differentiable function.

${}$

Find $f$ and hence $G$,\ if the room is free of dust at time
$t=0$.

\vspace{.25in}

{\bf Answer}

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

$n=0$:

$\begin{array}{rcl} P_0(t+\delta t) &amp; = &amp; P(X(t+\delta t)=0\ |\
X(t)=0)p_0(t)\\ &amp; &amp; +P(X(t+\delta t)=0\ |\ X(t)=1)p_1(t)\\ &amp; = &amp;
(1-\lambda\delta t)p_0(t)+(1-\lambda\delta t)\mu \delta t p_1(t)\\
&amp; = &amp; (1-\lambda\delta t)p_0(t)+\mu\delta t p_1(t) \end{array}$

so $p_0'(t)=-\lambda p_0(t)+\mu p_1(t)$.

$n&gt;0$:

$\begin{array}{rcl} P_n(t+\delta t) &amp; = &amp; P(X(t+\delta t)=n\ |\
X(t)=n)p_n(t)\\ &amp; &amp; +P(X(t+\delta t)=n\ |\ X(t)=n+1)p_{n+1}(t)\\ &amp;
&amp; +P(X(t+\delta t)=n\ |\ X(t)=n-1)p_{n-1}(t)\\ &amp; = &amp;
[(1-\lambda\delta t)(1-n\mu\delta t)+\lambda\delta t n\mu\delta
t]p_n(t)\\ &amp; &amp; +[(1-\lambda\delta t)(n+1)\mu \delta t]p_{n+1}(t)\\
&amp; &amp; [\lambda\delta t(1-(n-1)\mu)\delta t]p_{n-1}(t) \end{array}$

$P_n'(t)=0(\lambda+n\mu)p_n(t)+(n+1)\mu p_{n+1}(t)+\lambda
p_{n-1}(t)$

Now let $G(z\, t)=\ds\sum_{n=0}^\infty p_n(t) z^n$

$\begin{array}{rcl} \df{\pl G}{\pl t} &amp; = &amp; \ds\sum_{n=0}^\infty
p_n'(t)z^n\\ &amp; = &amp; \ds\sum_{n=0}^\infty [-(\lambda+\mu
n)p_n(t)+(n+1)\mu p_{n+1}(t)+\lambda p_{n-1}(t)]z^n\\ &amp; &amp;
(\rm{letting}\ p_{-1}(t) \equiv 0)\\ &amp; = &amp; \ds\sum_{n=0}^\infty
-\lambda p_n(t0 z^n-\mu \ds\sum_{n=0}^\infty n p_n(t)z^n\\ &amp; &amp; +
\mu \ds\sum_{n=0}^\infty (n+1)p_{n+1}(t)z^n+\lambda
\ds\sum_{n=0}^\infty p_{n-1}(t)z^n\\ &amp; = &amp; -\lambda
\ds\sum_{n=0}^\infty p_n(t)z^n -\mu z \ds\sum_{n=0}^\infty p_n(t)n
z^{n-1}\\ &amp; &amp; + \mu \ds\sum_{n=1}^\infty p_n(t)nz^{n-1}+\lambda z
\ds\sum_{n=0}^\infty p_n(t) z^n\\ &amp; = &amp; -\mu (z-1) \df{\pl G}{\pl
z}+\lambda(z-1)G\end{array}$

If $G(z,\ t)=e^{\frac{\lambda z}{\mu}} f(e^{-\mu t}(z-1))$

$\df{\pl G}{\pl t}=e^{-\frac{\lambda z}{\mu}}f'(e^{-\mu t}(z-1))
\cdot (z-1)e^{-\mu t}(-\mu)$

$\begin{array}{rcl} \df{\pl G}{\pl z} &amp; = &amp;
\df{\lambda}{\mu}e^{\frac{\lambda z}{\mu}}f(e^{-\mu t}(z-1))\\ &amp; &amp;
+e^{\frac{\lambda z}{\mu}}f'(e^{-\mu t}(z-1))e^{-\mu
t}\end{array}$

$\df{\pl G}{\pl t}+\mu(z-1)\df{\pl G}{\pl
z}=\lambda(z-1)e^{\frac{\lambda z}{\mu}}f(e^{-\lambda
t}(z-1))=\lambda(z-1)G$

${}$

Now if $X(0)=0,\ G(z,\ 0) \equiv 1$

so $1=e^{\frac{\lambda z}{\mu}}f(z-1)$

Thus $f(z-1)=e^{-\frac{\lambda z}{\mu}}$ i.e.
$f(x)=e^{-\frac{\lambda(x+1)}{\mu}}$

${}$

so

$\begin{array}{rcl} G(z,\ t) &amp; = &amp; \exp\left(\df{\lambda
z}{\mu}\right)\exp\left[-\df{\lambda}{\mu}\left(e^{-\mu
t}(z-1)+1\right)\right]\\ &amp; = &amp; \exp\left[\df{\lambda}{\mu}
\left[z-e^{-\mu t}(z-1)-1\right]\right]\\ &amp; = &amp;
\exp\left[\df{\lambda}{\mu}(z-1)(1-e^{-\mu t})\right] \end{array}$

\end{document}
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