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{\bf Question}

A man with initial capital $\pounds\!z$ gambles at a casino which
may be assumed infinitely rich.  He plays a series of independent
games and has probabilities $p$ of winning \pounds1 and $q = 1 -p$
of losing his \pounds1 stake.  Write down a difference equation
and boundary condition for the probability $q_z$ of his eventual
ruin. Discuss briefly why these are insufficient for determining
$q_z$.

Consider the following strategy for the gambler.  On the first day
he bets each of his pounds of capital in turn in $z$ games and
puts winnings and retained stake money, totalling $\pounds X_1$,
in a kitty. On the second day he bets with each of the $X_1$
pounds in the kitty and puts winnings and retained stake money,
totalling $\pounds X_2$, in another kitty, and so on  Show that
$\{X_n\}\ \ n = 1,2,\ldots$ is a branching chain with a
probability of ultimate extinction given by $$q_z = \left\{
\begin{array}{cc} \left(\frac{q}{p}\right)^z & \textrm{ if } p  >
q \\ 1 & \textrm{ if } p \leq q
\end{array} \right.$$



\vspace{.25in}

{\bf Answer}
\begin{eqnarray*}q_z & = & pq_{z+1} +  q q_{z-1} \hspace{.2in} z = 1, 2, 3,
\ldots \\ q_0 & = & 1 \end{eqnarray*}  The general solution is
$$q_z = \left\{ \begin{array}{ll} A + \ds B \left( \frac{q}{p}
\right)^z & p \not= q \\ A + Bz & p = q \end{array} \right.$$

Th obtain a particular solution we need two boundary conditions,
whereas we only have one.  Note that because $0 \leq q_z \leq 1$,
when $p=q$ this implies $B=0$ and therefore $A = 1 = q_0$.

Also when $p<q$ we must have $B=0$ and therefore $A = 1 = q_0$. So
it is only when $p>q$ that we have insufficient information.

Consider a particular \pounds 1 and the situation on subsequent
days.

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\bigskip

So we have a branching chain.  Each starting \pounds 1 acts
independently.  The probabilities for the \lq \lq offspring \rq
\rq of any individual \pounds 1 are \begin{eqnarray*} P(Z=0) & = &
q \\ P(Z= 2) & = & p \\ P(Z= k) & = & 0 \, \, {\rm otherwise}
\end{eqnarray*} so the p.g.f. is $$A(s) = q + ps^2.$$

The Fundamental Theorem for branching chains says that the
probability of ultimate extinction is the smallest positive root
of the equation $\ds s = A(s)$

$${\rm So \ \ } s = q + ps^2 \hspace{.2in} ps^2 - s + q = 0$$
$$i.e. \, \, (p s - q)(s-1) = 0 \, \, \, (p+q=1)$$ So the roots
are $$s = 1 \, \, s = \frac{q}{p}$$ So the probability of
extinction with \pounds 1 is $\ds\frac{q}{p}$ if $q<p$ or 1 if
$q\ge p.$



Starting with \pounds $z,$ by independence the probability of
extinction is $$1 {\rm \ if\ \ } q \geq p; \, \, \left(
\frac{q}{p} \right)^z {\rm \ if\ \ } q < p$$




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