\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

If a car driver has had $k$ accidents in the time interval $(0,\
t]$, then in the time interval $(t,t+\delta t)$ she has a
probability of

\begin{description}
\item[(i)]
$(\beta+\gamma k)\delta t+o(\delta t)$ of having one accident,

\item[(ii)]
$1-(\beta+\gamma k)\delta t+o(\delta t)$ of having no accidents,

\item[(iii)]
$o(\delta t)$ of having more than one accident.
\end{description}

Let $P_k(t)$ denote the probability that the driver has $k$
accidents in the time interval $(0,t]$. Show that the generating
function

$$G(z,t)=\ds\sum_{k=0}^\infty p_k(t)z^k$$

satisfies the differential equation

$$\ds \frac{\pl G}{\pl t}+\gamma z(1-z)\ds \frac{\pl G}{\pl
z}+\beta(1-z)G(z,t)=0,$$

and explain why $G(z,0) \equiv 1$.

${}$

Verify that

$$G(z,t)=e^{-\beta t}\left[1-\left(1-e^{-\gamma
t}\right)z\right]^{-\frac{\beta}{\gamma}}$$

satisfies these equations.

\vspace{.25in}

{\bf Answer}

The unusual arguments concerning conditional probabilities leads
to (omitting $\sigma(\delta t)$ terms) $$p_0(t + \delta t) =
p_0(t)(1-\beta \delta t)$$ we deduce $$p_0'(t) = -\beta p_0(t)$$
For $k>0$ $$p_k(t+\delta t) = p_k(t)(1-(\beta + \gamma k)\delta t)
+ p_{k-1}(t)(\beta + \gamma(k+1)\delta t)$$ We deduce that
$$p_k'(t) = -(\beta + \gamma k)p_k(t) + (\beta +
\gamma(k-1))p_{k-1}(t)$$Now \begin{eqnarray*} \frac{\pl G}{\pl t}
& = & \sum_{k=0}^\infty p_k'(t)z^k \\ & = & -\beta
\sum_{k=0}^\infty p_k(t)z^k - \gamma z \sum_{k=0}^\infty p_k(t) k
z^{k-1} \\ & & + \beta z \sum_{k=1}^\infty p_{k-1}(t)z^{k-1} +
\gamma z^2 \sum_{k-1}^\infty p_{k-1}(t)(k-1)z^{k-2} \\ & = &
-\beta \sum_{k=0}^\infty p_k(t)z^k - \gamma z^2 \sum_{k=0}^\infty
p_k(t)kz^{k-1} \\ & & + \beta z \sum_{k=0}^\infty p_k(t)z^{k} +
\gamma z^2 \sum_{k=0}^\infty p_k(t)kz^{k-1} \\ & = & -\beta G -
\gamma z \frac{\pl G}{\pl z} + \beta z G + \gamma z^2 \frac{\pl
G}{\pl z} \end{eqnarray*} Hence $$\frac{\pl G}{\pl t} + \gamma
z(1-z) \frac{\pl G}{\pl z} + \beta(1-z) = 0$$ Now $p_0(0) = 1$ and
$p_k(0) = 0$ for $k \geq 1$ Thus $$G(z,0) \equiv 1$$


Let $\ds G(z,t) = e^{-\beta t}[1 - (1 - e^{-\gamma
t})z]^{-\frac{\beta}{\gamma}}$ Then $G(z,0) \equiv 1,$ and,
\begin{eqnarray*} \frac{\pl G}{\pl z} & = & e^{-\beta t} [1 - (1 - e^{-\gamma
t})z]^{-\frac{\beta}{\gamma}-1} \frac{\beta}{\gamma}(1 -
e^{-\gamma t}) \\ \frac{\pl G}{\pl t} & = & -\beta e^{-\beta t} [1
- (1 - e^{-\gamma t})z]^{-\frac{\beta}{\gamma}} \\ & & + e^{-\beta
t} [1 - (1 - e^{-\gamma t})z]^{-\frac{\beta}{\gamma}-1}
\frac{\beta}{\gamma} \cdot \gamma e^{-\gamma t} z \end{eqnarray*}
Thus

$\ds e^{\beta t}(\beta G - \beta z G + \frac{\pl G}{\pl t} +
\gamma z \frac{\pl G}{\pl z} - \gamma z^2\frac{\pl G}{\pl z})$
\begin{eqnarray*} & = & \beta [1 - (1 - e^{-\gamma
t})z]^{-\frac{\beta}{\gamma}} - \beta z [1 - (1 - e^{-\gamma
t})z]^{-\frac{\beta}{\gamma}}  \\ & & -\beta [1 - (1 - e^{-\gamma
t})z]^{-\frac{\beta}{\gamma}}  + [1 - (1 - e^{-\gamma
t})z]^{-\frac{\beta}{\gamma}-1} \beta z e^{-\gamma t} \\ & & +
\beta z [1 - (1 - e^{-\gamma t})z]^{-\frac{\beta}{\gamma}-1}(1 -
e^{-\gamma t}) \\ & & -\beta z^2 [1 - (1 - e^{-\gamma
t})z]^{-\frac{\beta}{\gamma}-1} (1 - e^{-\gamma t}) \\ & = &
-\beta z [1 - (1 - e^{-\gamma t})z]^{-\frac{\beta}{\gamma}-1}[1 -
(1 - e^{-\gamma t})z] \\ & & + \beta z [1 - (1 - e^{-\gamma
t})z]^{-\frac{\beta}{\gamma}-1} -  \beta z^2 [1 - (1 - e^{-\gamma
t})z]^{-\frac{\beta}{\gamma}-1}(1 - e^{-\gamma t})\\ & =&0
\end{eqnarray*} Hence the given equation satisfies the differential
equation.


\end{document}