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{\bf Question}

A gambler with initial capital $£z$ plays against an infinitely
rich opponent. At each play the gambler wins $£2$ with probability
$p$, and loses $£1$ with probability $q=1-p$. Letting $q_z$ denote
the probability that the gambler will eventually be ruined, show
that, for $z \geq 1$,

$$q_z=pq_{z+2}+qq_{z-1}$$

giving a careful explanation of your reasoning. What is the value
of $q_0$?

${}$

Find the general solution of the above difference equation, and
include a discussion of repeated roots when they occur. (The
auxiliary equation has 1 as a root.)

${}$

Use the assumption that $q_z \to 0$ as $z \to \infty$, together
with the value of $q_0$, to show that

$$q_z=1\ \rm{if}\ q \geq 2p,$$

$$q_z=\left[-\ds \frac{1}{2}+\left(\ds \frac{1}{4}+\ds
\frac{q}{p}\right)^\frac{1}{2}\right]\ \rm{if}\ q<2p.$$


\vspace{.25in}

{\bf Answer}

we argue conditionally on the result of the first play, as
follows: \begin{eqnarray*} q_z & = & P({\rm gambler\ wins\ 1st\
bet\ and\ subsequently\ ruined}) \\ & + & P({\rm gambler\ loses\
1st\ bet\ and\ subsequently\ ruined}) \\ \\ & = & P({\rm ruin\ |
wins\ 1st\ bet}) \cdot P({\rm wins\ 1st\ bet}) \\ & + & P({\rm
ruin\ | loses\ 1st\ bet}) \cdot P({\rm loses\ 1st\ bet}) \\ \\ & =
& q_{z+2} \cdot p + q_{z-1} \cdot q \end{eqnarray*} Now $\ds q_0 =
1$.

To solve the difference equation.  The auxiliary equation is
$$p\lambda^3 - \lambda + q = 0$$i.e. $$(\lambda-1)(p\lambda^2
+p\lambda-q)\hspace{.2in}({\rm using\ }p+q=1)$$ So the roots are
$$\lambda_1 = -\frac{1}{2} + \sqrt{\frac{1}{4} + \frac{q}{p}}
\hspace{.2in} \lambda_2 = -\frac{1}{2} - \sqrt{\frac{1}{4} +
\frac{q}{p}} \hspace{.2in} \lambda_3 = 1$$  Now $\lambda_2<-1$,
and so the only possibilities for a repeated root is if $\lambda_1
= 1$ in that case $$\sqrt{\frac{1}{4} + \frac{q}{p}}= \frac{3}{2}
\, \, \, \, {\rm \ which\ gives\ \ } q = 2p.$$ The general
solution is $$q_z = A + Bz + C\lambda_2^z$$ Since$\lambda_2<-1$
this oscillates unboundedly if $c\not= 0$ and $\to \infty$ if $c =
0 \& B \not= 0$

Thus $q_z = A$ and so since $q_0 = 1$, $$q_z = 1 {\rm \ \ \ for\
all\ }z$$

Now if $p\not= q \hspace{.2in} q_z = A + B\lambda_1^z +
C\lambda_2^z$

Now $|\lambda_2| >|\lambda_1| $ and so if $C\not=0$ $q_z$
oscillates unboubdely as $z \to \infty.$  So $C+0$.

If $q>2p$ then $\lambda_1 >1$ and so $q_z \to \pm \infty $ as $ z
\to \infty$  if $B \not= 0$.  Thus A=0 by the given assumption and
$q_0 = 1$ gives $B=1$.

${}$

To summaries:

If $$q \geq 2p {\rm \ then\ \ } q_z = 1$$  If $$q<2p \, \, q_z =
\lambda_1^z = \left[ -\frac{1}{2} + \sqrt{\frac{1}{4} +
\frac{q}{p}}\right]^z$$



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