<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\df}{\ds\frac}
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\begin{document}


{\bf Question}

\begin{description}
\item[(a)]
A gambler with initial capital $£z$ plays a sequence of games
against an opponent with initial capital $£(a-z)$. At each game
the gambler wins $£1$ from his opponent with probability $p$,
loses $£1$ to him with probability $q$ or the game is drawn and
neither player gains any money. The games are independent with
$p+q&lt;1$ and $0&lt;p&lt;q&lt;1$.

Obtain a difference equation for the probability $p_z$, that the
gambler will ruin his opponent eventually, together with two
boundary conditions. Solve this equation and use it to find the
effect of trebling the initial capitals on the gambler's chance of
ruining his opponent.

\item[(b)]
For a Poisson process $(N(t); t \geq 0)$ with rate $\lambda$,
write down the probability that $N(t)$ equals $n\ (n=0,\ 1,\
2,\cdots\ ;\ t&gt;0)$.

Prove that for any $s$ and $t$ such that $ 0 leq s &lt; t,\
N(t)-N(s)$ has a Poisson distribution with parameter
$\lambda(t-s)$.
\end{description}


\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
$p_z=p\cdot p_{z+1}+qp_{z-1}+(1-p-q)p_z$

so $p_z=\df{p}{p+q}p_{z+1}+\df{q}{p+q}p_{z-1}$

with boundary conditions $p_0=0,\ p_a=1$

Substituting $p_z=\lambda^z$ gives

$p_z=\df{p}{p+q}\lambda^{z+1}+\df{q}{p+q}\lambda^{z-1}$

i.e. $\df{p}{p+q}\lambda^2-\lambda-\df{q}{p+q}=0$

$(\lambda-1)\left(\df{p}{p+q}\lambda-\df{q}{p+q}\right)=0$

so $\lambda=1$ or $\lambda=\df{q}{p}$

\newpage
The general solution is

$$p_z=A\left(\df{q}{p}\right)^z+B$$

The boundary conditions give

$0=A+B\ \ \ 1=A\left(\df{q}{p}\right)^a+B$

so $A=\df{1}{(\frac{q}{p})^a-1}$ and $B=-A$

so $p_z=\df{(\frac{q}{p})^z-1}{(\frac{q}{p})^a-1}$

Replacing $z$ and $a$ by $3z$ and $3a$ gives

$p_{3z}=\df{(\frac{q}{p})^{3z}-1}{(\frac{q}{p})^{3a}-1}=p_z \times
\df{(\frac{q}{p})^{2z}+(\frac{q}{p})^z+1}{(\frac{p}{q})^{2a}+(\frac{q}{p})^a+1}&lt;p_z$
for $z&lt;a$ and $q&gt;p$.

so the chances of the opponent being ruined are reduced.

\item[(b)]
$P(N(t)=n)=\df{(\lambda t)^n}{n!}e^{-\lambda t}\ \ n=0\, 1,\ 2,\
\cdots$

$\begin{array}{rcl} P(N(t)-N(s)=n) &amp; = &amp; \ds\sum_{m=0}^\infty
P(N(s)=m\\ &amp; &amp; \rm{and}\ N(t)=m+n)\\ &amp; = &amp; \ds\sum_{m=0}^\infty
P(N(s)=m\\ &amp; &amp; \rm{and}\ N(t-s)=n)\\ &amp; = &amp; \ds\sum_{m=0}^\infty
P(N(s)=m)P(N(t-s)=n)\\ &amp; &amp; \mbox{(Numbers of events in non-}\\ &amp; &amp;
\mbox{overlapping time intervals indep.)}\\ &amp; = &amp;
P(N(t-s)=n)\ds\sum_{m=0}^\infty \undb{P(N(s)=m)}\\ &amp; &amp;
\hspace{2in} =1\\ &amp; = &amp; P(N(t-a)=n)
\end{array}$

so $N(t-s)$ is Poisson with rate $\lambda(t-s)$.
\end{description}




\end{document}
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