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%\setcounter{chapter}{11}

\chapter{Random Walks}\label{chp 12}  

\section{Random Walks in Euclidean Space}\label{sec 12.1}

\par
In the last several chapters, we have studied sums of random variables with the goal being
to describe the distribution and density functions of the sum.  In this chapter, we shall look
at sums of discrete random variables from a different perspective.  We shall be concerned with
properties which can be associated with the sequence of partial sums, such as the number of
sign changes of this sequence, the number of terms in the sequence which equal 0, and the
expected size of the maximum term in the sequence.
\par
We begin with the following definition.
\begin{definition} 
 Let $\{X_k\}_{k = 1}^\infty$ be a sequence of independent, identically distributed discrete
random variables.  For each positive integer $n$, we let $S_n$ denote the sum $X_1 + X_2 +
\cdots + X_n$.  The sequence $\{S_n\}_{n = 1}^\infty$ is called a \emx {random
walk.\index{random walk}}  If the common range of the $X_k$'s is ${\mathbf R}^m$, then we say that
$\{S_n\}$ is a random walk in ${\mathbf R}^m$.
\end{definition} 
\par
We view the sequence of $X_k$'s as being the outcomes of independent experiments.  Since the
$X_k$'s are independent, the probability of any particular (finite) sequence of outcomes can be
obtained by multiplying the probabilities that each $X_k$ takes on the specified value in the
sequence.  Of course, these individual probabilities are given by the common distribution of the
$X_k$'s.  We will typically be interested in finding probabilities for events involving the
related sequence of $S_n$'s.  Such events can be described in terms of the $X_k$'s, so their
probabilities can be calculated using the above idea.
\par
There are several ways to visualize a random walk.  One can imagine that a particle is
placed at the origin in ${\mathbf R}^m$ at time $n = 0$.  The sum $S_n$ represents the position of
the particle at the end of $n$ seconds.  Thus, in the time interval $[n-1, n]$, the particle
moves (or jumps) from position $S_{n-1}$ to $S_{n}$.  The vector representing this motion is just
$S_n - S_{n-1}$, which equals $X_n$.  This means that in a random walk, the jumps are
independent and identically distributed.  If $m = 1$, for example, then one can imagine a 
particle on the real line that starts at the origin, and at the end of each second, jumps one
unit to the right or the left, with probabilities given by the distribution of the $X_k$'s.  If
$m = 2$, one can visualize the process as taking place in a city in which the streets form square
city blocks.  A person starts at one corner (i.e., at an intersection of two streets) and goes in
one of the four possible directions according to the distribution of the $X_k$'s.  If $m = 3$,
one might imagine being in a jungle gym, where one is free to move in any one of six directions
(left, right, forward, backward, up, and down).  Once again, the probabilities of these movements
are given by the distribution of the $X_k$'s.
\par
Another model of a random walk (used mostly in the case where the range is ${\mathbf R}^1$) is a
game, involving two people, which consists of a sequence of independent,
identically distributed moves.  The sum $S_n$ represents the score of the first person, say,
after $n$ moves, with the assumption that the score of the second person is $-S_n$.  For
example, two people might be flipping coins, with a match or non-match representing $+1$ or $-1$,
respectively, for the first player.  Or, perhaps one coin is being flipped, with a head or tail
representing $+1$ or $-1$, respectively, for the first player.
\subsection*{Random Walks on the Real Line}
We shall first consider the simplest non-trivial case
of a random walk in ${\mathbf R}^1$, namely the case where the common distribution function of
the random variables $X_n$ is  given by
$$
f_X(x) = \left \{ \begin{array}{ll}
                        1/2, & \mbox{if $x = \pm 1,$} \\
                          0, & \mbox{otherwise.} 
                  \end{array}
         \right.  
$$
This situation corresponds to a fair coin being flipped, with $S_n$ representing the number of
heads minus the number of tails which occur in the first $n$ flips.  We note that in this
situation, all paths of length $n$ have the same probability, namely $2^{-n}$.
\par
It is sometimes instructive to represent a random walk as a polygonal line, or path, in
the plane, where the horizontal axis represents time and the vertical axis represents the value
of $S_n$.  Given a sequence $\{S_n\}$ of partial sums, we first plot the points $(n, S_n)$, and
then for each $k < n$, we connect $(k, S_k)$ and $(k+1, S_{k+1})$ with a straight line segment.
The \emx {length} of a path is just the difference in the time values of the beginning
and  ending points on the path.  The reader is referred to Figure \ref{fig 12.1}.  This
figure, and the process it illustrates, are identical with the example, given in Chapter
1, of two people playing heads or tails.
\putfig{3.5truein}{PSfig1-3}{A random walk of length 40.}{fig 12.1}
\subsection*{Returns and First Returns}
We say that an \emx {equalization}\index{equalization} has occurred, or there is a \emx {return
to the origin}\index{return to the origin} at time $n$, if $S_n = 0$.  We note that this can only
occur if $n$ is an even integer.  To  calculate the probability of an equalization at time $2m$,
we  need only count the number of paths of length
$2m$ which begin and end at the origin.  The number of such paths is clearly
$${2m \choose m}\ .$$
Since each path has probability $2^{-2m}$, we have the following theorem.
\begin{theorem}\label{thm 12.1.1} 
The probability of a return to
the origin at time $2m$ is given by 
$$u_{2m} = {2m \choose m}2^{-2m}\ .$$
The probability of a return to the origin at an odd time is 0.
\end{theorem}%\endthm
A random walk is said to have a \emx {first return} to the origin\index{first return to the
origin} \index{return to the origin!first} at time
$2m$ if
$m > 0$, and
$S_{2k} \ne 0$ for all $k < m$.  In Figure~\ref{fig 12.1}, the first return occurs at time 2. 
We define $f_{2m}$ to be the probability of this event.  (We also define $f_0 = 0$.)  One can think of
the expression
$f_{2m}2^{2m}$ as the number of paths of length
$2m$ between the points $(0, 0)$ and $(2m, 0)$ that do not touch the horizontal axis except at
the endpoints.  Using this idea, it is easy to prove the following theorem.
\begin{theorem}\label{thm 12.1.2} 
For $n \ge 1$, the probabilities $\{u_{2k}\}$ and $\{f_{2k}\}$ are related by the equation
$$u_{2n} = f_0 u_{2n} + f_2 u_{2n-2} + \cdots + f_{2n}u_0\ .$$
\proof
There are $u_{2n}2^{2n}$ paths of length $2n$ which have endpoints $(0, 0)$ and $(2n, 0)$.  The
collection of such paths can be partitioned into $n$ sets, depending upon the time of the
first return to the origin.  A path in this collection which has a first return to the origin
at time $2k$ consists of an initial segment from $(0, 0)$ to $(2k, 0)$, in which no interior
points are on the horizontal axis, and a terminal segment from $(2k, 0)$ to $(2n, 0)$, with no
further restrictions on this segment.  Thus, the number of paths in the collection which have a
first return to the origin at time $2k$ is given by
$$f_{2k}2^{2k}u_{2n-2k}2^{2n-2k} = f_{2k}u_{2n-2k}2^{2n}\ .$$
If we sum over $k$, we obtain the equation
$$u_{2n}2^{2n} = f_0u_{2n} 2^{2n} + f_2u_{2n-2}2^{2n} + \cdots + f_{2n}u_0 2^{2n}\ .$$
Dividing both sides of this equation by $2^{2n}$ completes the proof.
\end{theorem}
The expression in the right-hand side of the above theorem should remind the reader of a sum
that appeared in Definition~\ref{defn 7.1} of the convolution of two distributions.  The
convolution of two sequences is defined in a similar manner.  The above theorem says that the
sequence $\{u_{2n}\}$ is the convolution of itself and the sequence
$\{f_{2n}\}$.  Thus, if we represent each of these sequences by an ordinary generating
function, then we can use the above relationship to determine the value $f_{2n}$.
\begin{theorem}\label{thm 12.1.3}
For $m \ge 1$, the probability of a first return to the origin at time $2m$ is given by
$$f_{2m} = {{u_{2m}}\over{2m-1}} = {{2m \choose m}\over{(2m-1)2^{2m}}}\ .$$
\proof
We begin by defining the generating functions
$$U(x) = \sum_{m = 0}^\infty u_{2m}x^m$$
and
$$F(x) = \sum_{m = 0}^\infty f_{2m}x^m\ .$$
Theorem~\ref{thm 12.1.2} says that
\begin{equation} 
U(x) = 1 + U(x)F(x)\ .
\label{eq 12.1.1}  
\end{equation} 
(The presence of the 1 on the right-hand side is due to the fact that $u_0$ is defined to be 1,
but Theorem~\ref{thm 12.1.2} only holds for $m \ge 1$.)  We note that both generating functions
certainly converge on the interval $(-1, 1)$, since all of the coefficients are at most 1 in
absolute value.  Thus, we can solve the above equation for $F(x)$, obtaining
$$F(x) = {{U(x) - 1}\over{U(x)}}\ .$$
Now, if we can find a closed-form expression for the function $U(x)$, we will also have a
closed-form expression for $F(x)$.  From Theorem~\ref{thm 12.1.1}, we have
$$U(x) = \sum_{m = 0}^\infty {2m \choose m}2^{-2m}x^m\ .$$
In Wilf,\footnote{H.~S.~Wilf, \emx {Generatingfunctionology,}\index{WILF, H. S.}
(Boston: Academic Press, 1990), p. 50.} we find that
$${1\over{\sqrt {1 - 4x}}} = \sum_{m = 0}^\infty {2m \choose m} x^m\ .$$
The reader is asked to prove this statement in Exercise~\ref{exer 12.1.1}.  If we replace $x$ by $x/4$
in the last equation, we see that
$$U(x) = {1\over{\sqrt {1-x}}}\ .$$
Therefore, we have
\begin{eqnarray*}
F(x) &=& {{U(x) - 1}\over{U(x)}}\\
     &=& {{(1-x)^{-1/2} - 1}\over{(1-x)^{-1/2}}}\\
     &=& 1 - (1-x)^{1/2}\ . 
\end{eqnarray*}

Although it is possible to compute the value of $f_{2m}$ using the Binomial Theorem, it is
easier to note that $F'(x) = U(x)/2$, so that the coefficients $f_{2m}$ can be found by
integrating the series for $U(x)$.  We obtain, for $m \ge 1$,

\begin{eqnarray*}
f_{2m} &=& {u_{2m-2}\over{2m}}\\
       &=& {2m-2 \choose m-1}\over{m2^{2m-1}}\\
       &=& {2m \choose m}\over{(2m-1)2^{2m}}\\
       &=& {u_{2m}\over{2m-1}}\ ,
\end{eqnarray*}

since
$${2m-2 \choose m-1} = {m\over{2(2m-1)}}{2m\choose m}\ .$$
This completes the proof of the theorem.
\end{theorem}
\subsection*{Probability of Eventual Return}\index{return to the origin!probability of eventual}
In the symmetric random walk process in ${\mathbf R}^m$, what is the probability that the particle
eventually returns to the origin?  We first examine this question in the case that $m = 1$, and
then we consider the general case.  The results in the next two examples are due to
P\'olya.\footnote{G.~P\'olya, ``\"Uber eine Aufgabe der Wahrscheinlichkeitsrechnung betreffend
die Irrfahrt im Strassennetz,'' Math. Ann., vol.~84 (1921), pp. 149-160.}\index{P\'OLYA, G.}
\begin{example}\label{exam 12.1.0.5}(Eventual Return in ${\mathbf R}^1$)
One has to approach the idea of eventual return with some care, since the
sample space seems to be the set of all walks of infinite length, and this set is
non-denumerable.  To avoid difficulties, we will define $w_n$ to be the probability that a first
return has occurred no later than time $n$.  Thus, $w_n$ concerns the sample space of all walks
of length $n$, which is a finite set.  In terms of the $w_n$'s, it is reasonable to define the
probability that the particle eventually returns to the origin to be 
$$w_* = \lim_{n \rightarrow \infty} w_n\ .$$
This limit clearly exists and is at most one, since the sequence $\{w_n\}_{n = 1}^\infty$ is an
increasing sequence, and all of its terms are at most one.  
\par
In terms of the $f_n$ probabilities, we see that
$$w_{2n} = \sum_{i = 1}^n f_{2i}\ .$$
Thus, 
$$w_* = \sum_{i = 1}^\infty f_{2i}\ .$$
In the proof of Theorem~\ref{thm 12.1.3}, the generating function 
$$F(x) = \sum_{m = 0}^\infty f_{2m}x^m$$ 
was introduced.  There it was noted that this series converges for $x \in (-1, 1)$.  In fact, it
is possible to show that this series also converges for $x = \pm 1$ by using Exercise~\ref{exer
12.1.4}, together with the fact that
$$f_{2m} = {{u_{2m}}\over{2m-1}}\ .$$
(This fact was proved in the proof of Theorem~\ref{thm 12.1.3}.)  Since we also know that 
$$F(x) = 1 - (1-x)^{1/2}\ ,$$
we see that 
$$w_* = F(1) = 1\ .$$
Thus, with probability one, the particle returns to the origin.
\par
An alternative proof of the fact that $w_* = 1$ can be obtained by using the
results in Exercise~\ref{exer 12.1.2}.
\end{example}
\par
\begin{example}\label{exam 12.1.0.6}(Eventual Return in ${\mathbf R}^m$)
We now turn our attention to the case that the random walk takes place in more than one dimension.
We define $f^{(m)}_{2n}$ to be the probability that the first return to the origin in ${\mathbf
R}^m$ occurs at time $2n$.  The quantity $u^{(m)}_{2n}$ is defined in a similar manner.  Thus,
$f^{(1)}_{2n}$ and $u^{(1)}_{2n}$ equal $f_{2n}$ and $u_{2n}$, which were defined earlier.  If,
in addition, we define $u^{(m)}_0 = 1$ and $f^{(m)}_0 = 0$, then one can mimic the proof of
Theorem~\ref{thm 12.1.2}, and show that for all $m \ge 1$,
\begin{equation}
u^{(m)}_{2n} = f^{(m)}_0 u^{(m)}_{2n} + f^{(m)}_2 u^{(m)}_{2n-2} + \cdots +
f^{(m)}_{2n}u^{(m)}_0\ .
\label{eq 12.1.1.5}
\end{equation}
\par
We continue to generalize previous work by defining
$$U^{(m)}(x) = \sum_{n = 0}^\infty u^{(m)}_{2n} x^n$$
and
$$F^{(m)}(x) = \sum_{n = 0}^\infty f^{(m)}_{2n} x^n\ .$$
Then, by using Equation~\ref{eq 12.1.1.5}, we see that
$$U^{(m)}(x) = 1 + U^{(m)}(x) F^{(m)}(x)\ ,$$
as before.  These functions will always converge in the interval $(-1, 1)$, since all of their
coefficients are at most one in magnitude.  In fact, since 
$$w^{(m)}_* = \sum_{n = 0}^\infty f^{(m)}_{2n} \le 1$$
for all $m$, the series for $F^{(m)}(x)$ converges at $x = 1$ as well, and $F^{(m)}(x)$ is
left-continuous at $x = 1$, i.e., 
$$\lim_{x \uparrow 1} F^{(m)}(x) = F^{(m)}(1)\ .$$
Thus, we have
\begin{equation}
w^{(m)}_* = \lim_{x \uparrow 1} F^{(m)}(x) = \lim_{x \uparrow 1} 
\frac{U^{(m)}(x) - 1}{U^{(m)}(x)}\ ,
\label{eq 12.1.1.6}
\end{equation}
so to determine $w^{(m)}_*$, it suffices to determine 
$$\lim_{x \uparrow 1} U^{(m)}(x)\ .$$
We let $u^{(m)}$ denote this limit.
\par
We claim that 
$$u^{(m)} = \sum_{n = 0}^\infty u^{(m)}_{2n}\ .$$
(This claim is reasonable; it says that to find out what happens to the function $U^{(m)}(x)$ at
$x = 1$, just let $x = 1$ in the power series for $U^{(m)}(x)$.)  To prove the claim, we note
that the coefficients $u^{(m)}_{2n}$ are non-negative, so $U^{(m)}(x)$ increases monotonically on
the interval $[0, 1)$.  Thus, for each $K$, we have
$$\sum_{n = 0}^K u^{(m)}_{2n} \le \lim_{x \uparrow 1} U^{(m)}(x) = u^{(m)} \le \sum_{n = 0}^\infty
u^{(m)}_{2n}\ .$$
By letting $K \rightarrow \infty$, we see that
$$u^{(m)} = \sum_{2n}^\infty u^{(m)}_{2n}\ .$$
This establishes the claim.
\par
From Equation~\ref{eq 12.1.1.6}, we see that if $u^{(m)} < \infty$, then the probability of an
eventual return is
$$\frac {u^{(m)} - 1}{u^{(m)}}\ ,$$
while if $u^{(m)} = \infty$, then the probability of eventual return is 1.
\par
To complete the example, we must estimate the sum
$$\sum_{n = 0}^\infty u^{(m)}_{2n}\ .$$  
In Exercise~\ref{exer 12.1.12}, the reader is asked to show that
$$u^{(2)}_{2n} = \frac 1 {4^{2n}} {{2n}\choose n}^2\ .$$
Using Stirling's Formula, it is easy to show that (see Exercise~\ref{exer 12.1.13})
$${{2n}\choose n} \sim \frac {2^{2n}}{\sqrt {\pi n}}\ ,$$
so
$$u^{(2)}_{2n} \sim \frac 1{\pi n}\ .$$
From this it follows easily that
$$\sum_{n = 0}^\infty u^{(2)}_{2n}$$
diverges, so $w^{(2)}_* = 1$, i.e., in ${\mathbf R}^2$, the probability of an eventual return is
1.
\par
When $m = 3$, Exercise~\ref{exer 12.1.12} shows that
$$u^{(3)}_{2n} = \frac 1{2^{2n}}{{2n}\choose n} \sum_{j,k} 
\biggl(\frac 1{3^n}\frac{n!}{j!k!(n-j-k)!}\biggr)^2\ .$$
Let $M$ denote the largest value of 
$$\frac 1{3^n}\frac {n!}{j!k!(n - j - k)!}\ ,$$
over all non-negative values of $j$ and $k$ with $j + k \le n$.  It is easy, using Stirling's
Formula, to show that 
$$M \sim \frac cn\ ,$$
for some constant $c$.  Thus, we have
$$u^{(3)}_{2n} \le \frac 1{2^{2n}}{{2n}\choose n} \sum_{j,k} 
\biggl(\frac M{3^n}\frac{n!}{j!k!(n-j-k)!}\biggr)\ .$$
Using Exercise~\ref{exer 12.1.14}, one can show that the right-hand expression is at most
$$\frac {c'}{n^{3/2}}\ ,$$
where $c'$ is a constant.  Thus,
$$\sum_{n = 0}^\infty u^{(3)}_{2n}$$
converges, so $w^{(3)}_*$ is strictly less than one.  This means that in ${\mathbf R}^3$, the
probability of an eventual return to the origin is strictly less than one (in fact, it is
approximately .34).
\par
One may summarize these results by stating that one should not get drunk in more than two
dimensions.
\end{example}

\subsection*{Expected Number of Equalizations}
We now give another example of the use of generating functions to find a general formula for
terms in a sequence, where the sequence is related by recursion relations to other sequences.
Exercise~\ref{exer 12.1.9} gives still another example.
\begin{example}(Expected Number of Equalizations)\label{exam
12.1.1}\index{equalizations!expected number of} 
In this example, we will derive a formula for the expected number of equalizations in a random
walk of length $2m$.  As in the proof of Theorem~\ref{thm 12.1.3}, the method has four main parts. 
First, a recursion is found which relates the $m$th term in the unknown sequence to earlier
terms in the same sequence and to terms in other (known) sequences.  An example of such
a recursion is given in Theorem~\ref{thm 12.1.2}.  Second, the recursion is used to derive a
functional equation involving the generating functions of the unknown sequence and one or more
known sequences.  Equation~\ref{eq 12.1.1} is an example of such a functional equation.  Third, the
functional equation is solved for the unknown generating function.  Last, using a device such
as the Binomial Theorem, integration, or differentiation, a formula for the $m$th coefficient
of the unknown generating function is found.
\par
We begin by defining $g_{2m}$ to be the number of equalizations among all of the random walks
of length $2m$.  (For each random walk, we disregard the equalization at time 0.)  We
define $g_0 = 0$.  Since  the number of walks of length $2m$ equals $2^{2m}$, the expected number
of equalizations among all such random walks is $g_{2m}/2^{2m}$.  Next, we define
the generating function
$G(x)$:
$$G(x) = \sum_{k = 0}^\infty g_{2k}x^k\ .$$
Now we need to find a recursion which relates the sequence $\{g_{2k}\}$ to one or both of the
known sequences $\{f_{2k}\}$ and $\{u_{2k}\}$.  We consider $m$ to be a fixed positive integer,
and consider the set of all paths of length $2m$ as the disjoint union
$$ E_2 \cup E_4 \cup \cdots \cup E_{2m} \cup H\ ,$$
where $E_{2k}$ is the set of all paths of length $2m$ with first equalization at time $2k$,
and $H$ is the set of all paths of length $2m$ with no equalization.  It is easy to show (see
Exercise~\ref{exer 12.1.3}) that
$$|E_{2k}| = f_{2k} 2^{2m}\ .$$
We claim that the number of equalizations among all paths belonging to the set $E_{2k}$ is equal
to 
\begin{equation}
|E_{2k}| + 2^{2k} f_{2k} g_{2m - 2k}\ .
\label{eq 12.1.2}  
\end{equation}
Each path in $E_{2k}$ has one equalization at time $2k$, so the total number of such
equalizations is just $|E_{2k}|$.  This is the first summand in expression Equation~\ref{eq 12.1.2}. 
There are $2^{2k} f_{2k}$ different initial segments of length $2k$ among the paths in
$E_{2k}$.  Each of these initial segments can be augmented to a path of length $2m$ in
$2^{2m-2k}$ ways, by adjoining all possible paths of length $2m - 2k$.  The
number of equalizations obtained by adjoining all of these paths to any one initial segment is
$g_{2m - 2k}$, by definition.  This gives the second summand in Equation~\ref{eq 12.1.2}.  Since $k$
can range from 1 to $m$, we obtain the recursion
\begin{equation}  
g_{2m} = \sum_{k = 1}^m \Bigl(|E_{2k}| + 2^{2k}f_{2k}g_{2m - 2k}\Bigr)\ .
\label{eq 12.1.3}  
\end{equation}                                                                        
\par
The second summand in the typical term above should remind the reader of a convolution.  In
fact, if we multiply the generating function $G(x)$ by the generating function 
$$F(4x) = \sum_{k = 0}^\infty 2^{2k}f_{2k} x^k\ ,$$
the coefficient of $x^m$ equals
$$\sum_{k = 0}^m 2^{2k}f_{2k}g_{2m-2k}\ .$$
Thus, the product $G(x)F(4x)$ is part of the functional equation that we are seeking.  The
first summand in the typical term in Equation~\ref{eq 12.1.3} gives rise to the sum
$$2^{2m}\sum_{k = 1}^m f_{2k}\ .$$
From Exercise~\ref{exer 12.1.2}, we see that this sum is just $(1 - u_{2m})2^{2m}$.    Thus, we need to
create a generating function whose $m$th coefficient is this term; this generating function is
$$\sum_{m = 0}^\infty (1- u_{2m})2^{2m} x^m\ ,$$
or
$$\sum_{m = 0}^\infty 2^{2m} x^m - \sum_{m = 0}^\infty u_{2m}2^{2m} x^m\ .$$
The first sum is just $(1-4x)^{-1}$, and the second sum is $U(4x)$.  So, the functional
equation which we have been seeking is
$$G(x) = F(4x)G(x) + {1\over{1-4x}} - U(4x)\ .$$
If we solve this recursion for $G(x)$, and simplify, we obtain
\begin{equation}  
G(x) = {1\over{(1-4x)^{3/2}}} - {1\over{(1-4x)}}\ .
\label{eq 12.1.4}  
\end{equation}                                                                        
\par
We now need to find a formula for the coefficient of $x^m$.  The first summand in 
Equation~\ref{eq 12.1.4} is $(1/2)U'(4x)$, so the coefficient of $x^m$ in this function is
$$u_{2m+2} 2^{2m+1}(m+1)\ .$$
The second summand in Equation~\ref{eq 12.1.4} is the sum of a geometric series with common 
ratio $4x$, so the coefficient of $x^m$ is $2^{2m}$.  Thus, we obtain

\begin{eqnarray*}
g_{2m} &=& u_{2m+2}2^{2m+1}(m+1) - 2^{2m}\\
       &=& {1\over 2}{{2m+2} \choose {m+1}} (m+1)  - 2^{2m}\ .
\end{eqnarray*}

We recall that the quotient $g_{2m}/2^{2m}$ is the expected number of equalizations among all
paths of length $2m$.  Using Exercise~\ref{exer 12.1.4}, it is easy to show that
$${{g_{2m}}\over{2^{2m}}} \sim \sqrt{2\over \pi}\sqrt {2m}\ .$$
In particular, this means that the average number of equalizations among all paths of length
$4m$ is not twice the average number of equalizations among all paths of length $2m$.  In order
for the average number of equalizations to double, one must quadruple the lengths of the random
walks.
\end{example}
It is interesting to note that if we define
$$M_n = \max_{0 \le k \le n} S_k\ ,$$
then we have
$$E(M_n) \sim \sqrt{2\over \pi}\sqrt n\ .$$
This means that the expected number of equalizations and the expected maximum value for random
walks of length $n$ are asymptotically equal as $n \rightarrow \infty$.  (In fact, it can be
shown that the two expected values differ by at most $1/2$ for all positive integers $n$.  See
Exercise~\ref{exer 12.1.9}.)

\exercises
\begin{LJSItem}
 
\i\label{exer 12.1.1} Using the Binomial Theorem, show that
$${1\over{\sqrt {1 - 4x}}} = \sum_{m = 0}^\infty {2m \choose m} x^m\ .$$
What is the interval of convergence of this power series?

\i\label{exer 12.1.2}
\begin{enumerate}

\item Show that for $m \ge 1$, 
$$f_{2m} = u_{2m-2} - u_{2m}\ .$$

\item Using part (a), find a closed-form expression for the sum
$$f_2 + f_4 + \cdots + f_{2m}\ .$$

\item Using part (b), show that
$$\sum_{m = 1}^\infty f_{2m} = 1\ .$$
(One can also obtain this statement from the fact that 
$$F(x) = 1 - (1-x)^{1/2}\ .)$$

\item Using parts (a) and (b), show that the probability of no equalization in the 
first $2m$ outcomes equals the probability of an equalization at time $2m$.
\end{enumerate}

\i\label{exer 12.1.3} Using the notation of Example~\ref{exam 12.1.1}, show that
$$|E_{2k}| = f_{2k} 2^{2m}\ .$$

\i\label{exer 12.1.4} Using Stirling's Formula, show that 
$$u_{2m} \sim {1\over{\sqrt {\pi m}}}\ .$$

\i\label{exer 12.1.5} A \emx {lead change}\index{lead change} in a random walk occurs at time
$2k$ if
$S_{2k-1}$ and $S_{2k+1}$ are of opposite sign.  
\begin{enumerate}
\item Give a rigorous argument which proves that
among all walks of length $2m$ that have an equalization at time $2k$, exactly half have a lead
change at time $2k$.

\item  Deduce that the total number of lead changes among all walks of length $2m$ equals
$${1\over 2}(g_{2m} - u_{2m})\ .$$

\item Find an asymptotic expression for the average number of lead changes in a random walk
of length $2m$. 
\end{enumerate}

\i\label{exer 12.1.6} 
\begin{enumerate}
\item 
Show that the probability that a random walk of length $2m$ has a last return to the origin
\index{last return to the origin}\index{return to the origin!last}
at time $2k$, where $0 \le k \le m$, equals
$$
{{{2k}\choose k}{{2m-2k}\choose {m-k}}\over{2^{2m}}} = u_{2k}u_{2m - 2k}\ .
$$  
(The case $k = 0$ consists of all paths that do not
return to the origin at any positive time.) \emx {Hint}:  A path whose last return to the origin
occurs at time $2k$ consists of two paths glued together, one path of which is of length $2k$ and
which begins and ends at the origin, and the other path of which is of length $2m - 2k$ and which
begins at the origin but never returns to the origin.  Both types of paths can be counted using
quantities which appear in this section.  
\item
Using part (a), show that if $m$ is odd, the probability that a walk of length $2m$ has no
equalization in the last $m$ outcomes is equal to $1/2$, regardless of the value of $m$.  \emx
{Hint}:  The answer to part a) is symmetric in $k$ and $m-k$.
\end{enumerate}

\i\label{exer 12.1.7} Show that the probability of no equalization in a walk of length
$2m$ equals $u_{2m}$.

\istar\label{exer 12.1.8} Show that
$$P(S_1 \ge 0,\ S_2 \ge 0,\ \ldots,\ S_{2m} \ge 0) = u_{2m}\ .$$
\emx {Hint}:  First explain why
\begin{eqnarray*}
&&P(S_1 > 0,\ S_2 > 0,\ \ldots,\ S_{2m} > 0) \\
&& \;\;\;\;\;\;\;\;\;\;\;\;\; = {1\over 2}P(S_1 \ne 0,\ S_2 \ne 0,\ \ldots,\ S_{2m} \ne 0)  \ .
\end{eqnarray*}
Then use Exercise~\ref{exer 12.1.7}, together with the observation that if no equalization occurs in the
first $2m$ outcomes, then the path goes through the point $(1,1)$ and remains on or above the
horizontal line $x = 1$.

\istar\label{exer 12.1.9} In Feller,\footnote{W.~Feller, \emx {Introduction to Probability
Theory and its Applications,} vol.~I, 3rd~ed. (New York: John Wiley \& Sons, 1968).} one finds the
following theorem:  Let $M_n$ be the random variable which gives the maximum value of $S_k$, for
$1 \le k \le n$.  Define
$$p_{n, r} = {n\choose {{n+r}\over{2}}}2^{-n}\ .$$
If $r \ge 0$, then
$$P(M_n = r) = \left \{ \begin{array}{ll}
                    p_{n, r}\,,&\mbox{if $r \equiv n\,    (\mbox{mod}\ 2)$}, \\   
                  p_{n, r+1}\,,&\mbox{if $r \not\equiv n\,(\mbox{mod}\ 2)$}.
                      \end{array}
             \right.   
$$

\begin{enumerate}

\item Using this theorem, show that
$$E(M_{2m}) = {1\over{2^{2m}}}\sum_{k = 1}^m (4k-1){2m \choose m+k}\ ,$$
and if $n = 2m+1$, then
$$E(M_{2m+1}) = {1\over {2^{2m+1}}} \sum_{k = 0}^m (4k+1){2m+1\choose m+k+1}\ .$$

\item
For $m \ge 1$, define 
$$r_m = \sum_{k = 1}^m k {2m\choose m+k}$$
and
$$s_m = \sum_{k = 1}^m k {2m+1\choose m+k+1}\ .$$
By using the identity
$${n\choose k} = {n-1\choose k-1} + {n-1\choose k}\ ,$$
show that
$$s_m = 2r_m - {1\over 2}\biggl(2^{2m} - {2m \choose m}\biggr)$$
and
$$r_m = 2s_{m-1} + {1\over 2}2^{2m-1}\ ,$$
if $m \ge 2$.

\item
Define the generating functions
$$R(x) = \sum_{k = 1}^\infty r_k x^k$$
and
$$S(x) = \sum_{k = 1}^\infty s_k x^k\ .$$
Show that
$$S(x) = 2 R(x) - {1\over 2}\biggl({1\over{1- 4x}}\biggr) + {1\over 2}\biggl(\sqrt{1-4x}\biggr)$$
and
$$R(x) = 2xS(x) + x\biggl({1\over{1-4x}}\biggr)\ .$$

\item
Show that
$$R(x) = {x\over{(1-4x)^{3/2}}}\ ,$$
and
$$S(x) = {1\over 2}\biggl({1\over{(1- 4x)^{3/2}}}\biggr) - 
{1\over 2}\biggl({1\over{1- 4x}}\biggr)\ .$$

\item
Show that
$$r_m = m{2m-1\choose m-1}\ ,$$
and
$$s_m = {1\over 2}(m+1){2m+1\choose m} - {1\over 2}(2^{2m})\ .$$

\item
Show that
$$E(M_{2m}) = {m\over{2^{2m-1}}}{2m\choose m} + {1\over{2^{2m+1}}}{2m\choose m} - {1\over 2}\ ,$$
and
$$E(M_{2m+1}) = {{m+1}\over{2^{2m+1}}}{2m+2\choose m+1} - {1\over 2}\ .$$
The reader should compare these formulas with the expression for \linebreak $g_{2m}/2^{(2m)}$ in
Example~\ref{exam 12.1.1}.
\end{enumerate}

\istar\label{exer 12.1.10}
(from K. Levasseur\footnote{K. Levasseur, ``How to Beat Your Kids at Their
Own Game," \emx {Mathematics Magazine} vol.~61, no. 5 (December, 1988),
pp.~301-305.})\index{LEVASSEUR, K.} A parent and his child play the following game.  A deck of $2n$
cards, $n$ red and $n$ black, is shuffled.  The cards are turned up one at a time.  Before each card
is turned up, the parent and the child guess whether it will be red or black.  Whoever makes more
correct guesses wins the game.  The child is assumed to guess each color with the same probability,
so she will have a score of
$n$, on average.  The parent keeps track of how many cards of each color have already been
turned up.  If more black cards, say, than red cards remain in the deck, then the parent will
guess black, while if an equal number of each color remain, then the parent guesses each
color with probability 1/2.  What is the expected number of correct guesses that will be made
by the parent?  \emx {Hint}:  Each of the ${{2n}\choose n}$ possible orderings of red and
black cards corresponds to a random walk of length $2n$ that returns to the origin at time
$2n$.  Show that between each pair of successive equalizations, the parent will be right
exactly once more than he will be wrong.  Explain why this means that the average number of
correct guesses by the parent is greater than $n$ by exactly one-half the average number of
equalizations.  Now define the random variable $X_i$ to be 1 if there is an equalization at
time $2i$, and 0 otherwise.  Then, among all relevant paths, we have
$$E(X_i) = P(X_i = 1) = \frac{{{2n-2i}\choose{n-i}}{{2i}\choose{i}}}{{{2n}\choose{n}}}\ .$$
Thus, the expected number of equalizations equals
$$E\biggl(\sum_{i = 1}^n X_i\biggr) = \frac 1{{{2n}\choose{n}}}\sum_{i = 1}^n 
{{2n-2i}\choose{n-i}}{{2i}\choose{i}}\ .$$
One can now use generating functions to find the value of the sum.
\par
It should be noted that in a game such as this, a more interesting question than the one
asked above is what is the probability that the parent wins the game?  For this game, this
question was answered by D. Zagier.\footnote{D. Zagier, ``How Often Should You Beat
Your Kids?" \emx{Mathematics Magazine} vol.~63, no.\ 2 (April 1990), pp. 89-92.}\index{ZAGIER, D.} 
He showed that the probability of winning is asymptotic (for large
$n$) to the quantity
$$\frac 12 + \frac 1{2\sqrt 2}\ .$$

\item\label{exer 12.1.11}
Prove that
$$u^{(2)}_{2n} = \frac 1{4^{2n}} \sum_{k = 0}^n \frac {(2n)!}{k!k!(n-k)!(n-k)!}\ ,$$
and
$$u^{(3)}_{2n} = \frac 1{6^{2n}} \sum_{j,k} \frac {(2n)!}{j!j!k!k!(n-j-k)!(n-j-k)!}\ ,$$
where the last sum extends over all non-negative $j$ and $k$ with $j+k \le n$.  Also show
that this last expression may be rewritten as
$$\frac 1{2^{2n}}{{2n}\choose n} \sum_{j,k} 
\biggl(\frac 1{3^n}\frac{n!}{j!k!(n-j-k)!}\biggr)^2\ .$$

\item\label{exer 12.1.12}
Prove that if $n \ge 0$, then
$$\sum_{k = 0}^n {n \choose k}^2 = {{2n} \choose n}\ .$$
\emx {Hint}: Write the sum as 
$$\sum_{k = 0}^n {n \choose k}{n \choose {n-k}}$$
and explain why this is a coefficient in the product
$$(1 + x)^n (1 + x)^n\ .$$
Use this, together with Exercise~\ref{exer 12.1.11}, to show that
$$u^{(2)}_{2n} = \frac 1{4^{2n}}{{2n}\choose n}\sum_{k = 0}^n {n \choose k}^2 = 
\frac 1 {4^{2n}} {{2n}\choose n}^2\ .$$

\item\label{exer 12.1.13}
Using Stirling's Formula, prove that
$${{2n}\choose n} \sim \frac {2^{2n}}{\sqrt {\pi n}}\ .$$

\item\label{exer 12.1.14}
Prove that
$$\sum_{j,k} 
\biggl(\frac 1{3^n}\frac{n!}{j!k!(n-j-k)!}\biggr) = 1\ ,$$
where the sum extends over all non-negative $j$ and $k$ such that $j + k \le n$.  \emx {Hint}: 
Count how many ways one can place $n$ labelled balls in 3 labelled urns.

\item\label{exer 12.1.15}
Using the result proved for the random walk in ${\mathbf R}^3$ in Example~\ref{exam 12.1.0.6}, 
explain why the probability of an eventual return in ${\mathbf R}^n$ is strictly less than one,
for all $n \ge 3$.  \emx {Hint}: Consider a random walk in ${\mathbf R}^n$ and disregard all but
the first three coordinates of the particle's position.

\end{LJSItem}

\section{Gambler's Ruin\label{sec 12.2}}\index{Gambler's Ruin}
\par
In the last section, the simplest kind of symmetric random walk in ${\mathbf R}^1$ was studied. 
In this section, we remove the assumption that the random walk is symmetric.  Instead, we assume
that
$p$ and $q$ are non-negative real numbers with $p+q = 1$, and that the common distribution function
of the jumps of the random walk is
$$f_X(x) = \left \{ \begin{array}{ll}
               p, & \mbox{if $x = 1$},\\ 
               q, & \mbox{if $x = -1$}.
                      \end{array}
             \right.   
$$ 
One can imagine the random walk as representing a sequence of tosses of a weighted coin, with a
head appearing with probability $p$ and a tail appearing with probability $q$.  An alternative
formulation of this situation is that of a gambler playing a sequence of games against an
adversary (sometimes thought of as another person, sometimes called ``the house") where, in
each game, the gambler has probability $p$ of winning.

\subsection*{The Gambler's Ruin Problem}

The above formulation of this type of random walk leads to a problem known as the Gambler's Ruin
problem.\index{Gambler's Ruin}  This problem was introduced in Exercise~\ref{exer 11.2.22}, but
we will give the description of the problem again.  A gambler starts with a ``stake" of size~$s$. 
She plays until her capital reaches the value $M$ or the value 0.  In the language of Markov 
chains, these two values correspond to absorbing states.  We
are interested in studying the probability of occurrence of each of these two outcomes.  
\par
One can also assume that the gambler is playing against an ``infinitely rich" adversary.  In this
case, we  would say that there is only one absorbing state, namely when the gambler's stake is 0. 
Under this assumption, one can ask for the probability that the gambler is eventually ruined. 
\par
We begin by defining $q_k$ to be the probability that the gambler's stake reaches 0, i.e.,
she is ruined, before it reaches $M$, given that the initial stake is $k$.  We note that $q_0 =
1$ and $q_M = 0$.  The fundamental relationship among the $q_k$'s is the following:
$$q_k = pq_{k+1} + qq_{k-1}\ ,$$
where $1 \le k \le M-1$.  This holds because if her stake equals $k$, and she plays one game,
then her stake becomes $k+1$ with probability $p$ and $k-1$ with probability $q$.  In the first
case, the probability of eventual ruin is $q_{k+1}$ and in the second case, it is $q_{k-1}$. 
We note that since $p + q = 1$, we can write the above equation as
$$p(q_{k+1} - q_k) = q(q_k - q_{k-1})\ ,$$
or
$$q_{k+1} - q_k = {q\over p}(q_k - q_{k-1})\ .$$
From this equation, it is easy to see that
\begin{equation}
q_{k+1} - q_k = \biggl({q\over p}\biggr)^k(q_1 - q_0)\ .
\label{eq 12.2.2}
\end{equation}
We now use telescoping sums to obtain an equation in which the only unknown is $q_1$:
\begin{eqnarray*}
-1 &=& q_M - q_0 \\
   &=& \sum_{k = 0}^{M-1} (q_{k+1} - q_k)\ ,
\end{eqnarray*}
so
\begin{eqnarray*}
-1 &=& \sum_{k = 0}^{M-1} \biggl({q\over p}\biggr)^k(q_1 - q_0)\\
   &=& (q_1 - q_0) \sum_{k = 0}^{M-1} \biggl({q\over p}\biggr)^k\ .
\end{eqnarray*}
If $p \ne q$, then the above expression equals
$$(q_1 - q_0) {{(q/p)^M - 1}\over{(q/p) - 1}}\ ,$$
while if $p = q = 1/2$, then we obtain the equation
$$-1 = (q_1 - q_0) M\ .$$
For the moment we shall assume that $p \ne q$.   Then we have
$$q_1 - q_0 = -{{(q/p) - 1}\over{(q/p)^M - 1}}\ .$$
Now, for any $z$ with $1 \le z \le M$, we have
\begin{eqnarray*}
q_z - q_0 &=& \sum_{k = 0}^{z-1} (q_{k+1} - q_k)\\
          &=& (q_1 - q_0)\sum_{k = 0}^{z-1} \biggl({q\over p}\biggr)^k\\
          &=& -(q_1 - q_0){{(q/p)^z - 1}\over{(q/p) - 1}}\\
          &=& -{{(q/p)^z - 1}\over{(q/p)^M - 1}}\ .
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
q_z &=& 1 - {{(q/p)^z - 1}\over{(q/p)^M - 1}}\\
    &=& {{(q/p)^M - (q/p)^z}\over{(q/p)^M - 1}}\ .
\label{eq 12.2.3}
\end{eqnarray*}
Finally, if $p = q = 1/2$, it is easy to show that (see Exercise~\ref{exer 12.2.11})
$$q_z = {{M - z}\over M}\ .$$
We note that both of these formulas hold if $z = 0$.
\par
We define, for $0 \le z \le M$, the quantity $p_z$ to be the probability that the gambler's
stake reaches $M$ without ever having reached 0.  Since the game might continue indefinitely, it
is not obvious that $p_z + q_z = 1$ for all $z$.  However, one can use the same method as above
to show that if $p \ne q$, then
$$q_z = {{(q/p)^z - 1}\over{(q/p)^M - 1}}\ ,$$
and if $p = q = 1/2$, then
$$q_z = {z\over M}\ .$$
Thus, for all $z$, it is the case that $p_z + q_z = 1$, so the game ends with probability 1.

\subsection*{Infinitely Rich Adversaries}

We now turn to the problem of finding the probability of eventual ruin if the gambler is playing
against an infinitely rich adversary.  This probability can be obtained by letting $M$ go to
$\infty$ in the expression for $q_z$ calculated above.  If $q < p$, then the expression approaches
$(q/p)^z$, and if $q > p$, the expression approaches 1.  In the case $p = q = 1/2$, we recall
that $q_z = 1 - z/M$.  Thus, if $M \rightarrow \infty$, we see that the probability of eventual
ruin tends to 1.  

\subsection*{Historical Remarks}

In 1711, De Moivre\index{de MOIVRE, A.}, in his book  \emx{De Mesura Sortis}, gave an ingenious
derivation of the probability of ruin.  The following description of his argument is taken from
David.\footnote{F.~N. David, 
\emx {Games, Gods and Gambling} (London: Griffin, 1962).}\index{DAVID, F. N.}  The notation used
is as follows:  We imagine that there are two players, A and B, and the probabilities that they
win a game are $p$ and $q$, respectively.  The players start with $a$ and $b$ counters,
respectively. 
\begin{quote}
Imagine that each player starts with his counters before him in a pile, and that nominal values
are assigned to the counters in the following manner.  A's bottom counter is given the nominal
value $q/p$; the next is given the nominal value $(q/p)^2$, and so on until his top counter
which has the nominal value $(q/p)^a$.  B's top counter is valued $(q/p)^{a+1}$, and so on
downwards until his bottom counter which is valued $(q/p)^{a+b}$.  After each game the loser's
top counter is transferred to the top of the winner's pile, and it is always the top counter
which is staked for the next game.  Then \emx {in terms of the nominal values} B's stake is
always $q/p$ times A's, so that at every game each player's nominal expectation is nil.  This
remains true throughout the play; therefore A's chance of winning all B's counters, multiplied
by his nominal gain if he does so, must equal B's chance multiplied by B's nominal gain.  Thus,
\begin{equation}
P_a\biggl(\Bigl({q\over p}\Bigr)^{a+1} + \cdots +
\Bigl({q\over p}\Bigr)^{a+b}\biggr) =  
P_b\biggl(\Bigl({q\over p}\Bigr) + \cdots +
\Bigl({q\over p}\Bigr)^a\biggr)\ .
\label{eq 12.2.1}
\end{equation}
\end{quote}
\par
Using this equation, together with the fact that
$$P_a + P_b = 1\ ,$$
it can easily be shown that
$$P_a = {{(q/p)^a - 1}\over{(q/p)^{a+b} - 1}}\ ,$$
if $p \ne q$, and
$$P_a = {a\over{a+b}}\ ,$$
if $p = q = 1/2$.
\par
In terms of modern probability theory, de Moivre is changing the values of the counters to make
an unfair game into a fair game, which is called a martingale.  With the new values, the
expected fortune of player A (that is, the sum of the nominal values of his counters) after each
play equals his fortune before the play (and similarly for player B).  (For a simpler martingale
argument, see Exercise~\ref{exer 12.2.10}.)  De Moivre then uses the fact that when the game ends,
it is still fair, thus Equation~\ref{eq 12.2.1} must be true.  This fact requires proof, and is
one of the central theorems in the area of martingale theory.  

\exercises
\begin{LJSItem}
 
\i\label{exer 12.2.2} In the gambler's ruin problem, assume that the gambler initial stake
is 1 dollar, and assume that her probability of success on any one game is $p$.  Let $T$ be the
number of games until 0 is reached (the gambler is ruined).  Show that the  generating
function for~$T$ is  $$
h(z) = \frac{1 - \sqrt{1 - 4pqz^2}}{2pz}\ ,
$$
and that
$$
h(1) = \left \{ \begin{array}{ll}
              q/p, & \mbox{if $q \leq p$}, \\
                1, & \mbox{if $q \geq p,$} 
         \end{array}
         \right.    
$$
and
$$
h'(1) = \left \{ \begin{array}{ll}
               1/(q - p), & \mbox{if $q > p$}, \\
                  \infty, & \mbox{if $q = p.$}
         \end{array}
         \right. 
$$
Interpret your results in terms of the time $T$ to reach 0.  (See also 
Example~\ref{exam 10.1.7}.)

\i\label{exer 12.2.3} Show that the Taylor series expansion for $\sqrt{1 - x}$ is
$$
 \sqrt{1 - x} = \sum_{n = 0}^\infty  {{1/2} \choose n} x^n\ ,
$$
where the binomial coefficient ${1/2} \choose n$ is
$$
{{1/2} \choose n} = \frac{(1/2)(1/2 - 1) \cdots (1/2 - n + 1)}{n!}\ .  
$$
Using this and the result of Exercise~\ref{exer 12.2.2}, show that the probability
that the gambler is ruined on the $n$th step is
$$
p_T(n) = \left \{ \begin{array}{ll}
 \frac{(-1)^{k - 1}}{2p} {{1/2} \choose k} (4pq)^k, & \mbox{if $n = 2k - 1$,} \\
                                                 0, & \mbox{if $n = 2k$.}
                  \end{array}
         \right.   
$$

\i\label{exer 12.2.4} For the gambler's ruin problem, assume that the
gambler starts with $k$ dollars.  Let $T_k$ be the time to reach 0 for the
first time.

\begin{enumerate}
\item Show that the generating function $h_k(t)$ for~$T_k$ is the $k$th
power of the generating function for the time $T$ to ruin starting at~1.  \emx{Hint}: Let $T_k
= U_1 + U_2 +\cdots+ U_k$, where $U_j$ is the time for the walk starting at~$j$ to reach
$j - 1$ for the first time.

\item Find $h_k(1)$ and $h_k'(1)$ and interpret your results.
\end{enumerate} 

\i\label{exer 12.2.5} (The next three problems come from Feller.\footnote{W.~Feller, op.\ cit.,
pg. 367.}) 
As in the text, assume that $M$ is a fixed positive integer. 

\begin{enumerate}
\item Show that if a gambler starts with an stake of 0 (and is allowed to have a negative
amount of money), then the probability that her stake reaches the value of $M$ before it returns
to 0 equals $p(1 - q_1)$.
\item Show that if the gambler starts with a stake of $M$ then the probability that her stake
reaches 0 before it returns to $M$ equals $qq_{M-1}$.
\end{enumerate}

\i\label{exer 12.2.6} Suppose that a gambler starts with a stake of 0 dollars.
\begin{enumerate} 
\item  Show that the probability that her stake never reaches $M$ before returning to 0 equals
$1 - p(1 - q_1)$.

\item  Show that the probability that her stake reaches the value $M$ exactly $k$ times before
returning to 0 equals $p(1-q_1)(1 - qq_{M-1})^{k-1}(qq_{M-1})$.  \emx {Hint}:  
Use Exercise~\ref{exer 12.2.5}.
\end{enumerate} 

\i\label{exer 12.2.7} In the text, it was shown that if $q < p$, there is a positive probability
that a gambler, starting with a stake of 0 dollars, will never return to the origin.  Thus, we
will now assume that $q \ge p$.  Using Exercise~\ref{exer 12.2.6}, show that if a gambler starts with a
stake of 0 dollars, then the expected number of times her stake equals $M$ before returning to
0 equals $(p/q)^M$, if $q > p$ and 1, if $q = p$.  (We quote from Feller:  ``The truly amazing
implications of this result appear best in the language of fair games.  A perfect coin is tossed
until the first equalization of the accumulated numbers of heads and tails.  The gambler receives
one penny for every time that the accumulated number of heads exceeds the accumulated number of
tails by $m$. \emx {The `fair entrance fee' equals 1 independent of $m$.}")

\i\label{exer 12.2.8} In the game in Exercise~\ref{exer 12.2.7}, let $p = q = 1/2$ and $M = 10$.  What
is the probability that the gambler's stake equals $M$ at least 20 times before it returns to 0?

\i\label{exer 12.2.9} Write a computer program which simulates the game in Exercise~\ref{exer 12.2.7}
for the case $p = q = 1/2$, and $M = 10$.  

\i\label{exer 12.2.10} In de Moivre's description of the game, we can modify the definition of
player A's fortune in such a way that the game is still a martingale (and the calculations are
simpler).  We do this by assigning nominal values to the counters in the same way as de Moivre,
but each player's current fortune is defined to be just the value of the counter which is being
wagered on the next game.  So, if player A has $a$ counters, then his current fortune is
$(q/p)^a$ (we stipulate this to be true even if $a = 0$).  Show that under this definition,
player A's expected fortune after one play equals his fortune before the play, if $p \ne q$. 
Then, as de Moivre does, write an equation which expresses the fact that player A's expected
final fortune equals his initial fortune.  Use this equation to find the probability of ruin of
player A.

\i\label{exer 12.2.11} Assume in the gambler's ruin problem that $p = q = 1/2$.
\begin{enumerate}
\item  Using Equation~\ref{eq 12.2.2}, together with the facts that $q_0 = 1$ and $q_M = 0$,
show that for $0 \le z \le M$,
$$q_z = {{M - z}\over M}\ .$$
\item  In Equation~\ref{eq 12.2.3}, let $p \rightarrow 1/2$ (and since $q = 1 - p$, $q
\rightarrow 1/2$ as well).  Show that in the limit,
$$q_z = {{M - z}\over M}\ .$$  \emx {Hint}:  Replace $q$ by $1-p$, and use L'Hopital's rule.
\end{enumerate}

%We may want to move the next exercise to an earlier chapter.
\i\label{exer 12.2.12} In American casinos, the roulette wheels have the integers between 1 and
36, together with 0 and 00.  Half of the non-zero numbers are red, the other half are black, and
0 and 00 are green.  A common bet in this game is to bet a dollar on red.  If a red number
comes up, the bettor gets her dollar back, and also gets another dollar.  If a black or green
number comes up, she loses her dollar.
\begin{enumerate}
\item Suppose that someone starts with 40 dollars, and continues to bet on red until either her
fortune reaches 50 or 0.  Find the probability that her fortune reaches 50 dollars.
\item How much money would she have to start with, in order for her to have a 95\% chance of
winning 10 dollars before going broke?
\item A casino owner was once heard to remark that ``If we took 0 and 00 off of the roulette
wheel, we would still make lots of money, because people would continue to come in and play
until they lost all of their money.''  Do you think that such a casino would stay in business?
\end{enumerate}

\end{LJSItem}





 
\section{Arc Sine Laws\label{sec 12.3}}\index{arc sine laws}
\par
In Exercise~\ref{sec 12.1}.\ref{exer 12.1.6}, the distribution of the time
of the last equalization in the symmetric random walk was determined.  If we let $\alpha_{2k, 2m}$
denote the probability that a random walk of length $2m$ has its last equalization at time
$2k$, then we have
$$\alpha_{2k, 2m} = u_{2k}u_{2m-2k}\ .$$
We shall now show how one can approximate the distribution of the $\alpha$'s with a simple
function.  We recall that 
$$u_{2k} \sim {1\over{\sqrt {\pi k}}}\ .$$
Therefore, as both $k$ and $m$ go to $\infty$, we have
$$\alpha_{2k, 2m} \sim {1\over{\pi \sqrt{k(m-k)}}}\ .$$
This last expression can be written as
$${1\over{\pi m \sqrt{(k/m)(1 - k/m)}}}\ .$$
Thus, if we define 
$$f(x) = {1\over{\pi \sqrt{x(1-x)}}}\ ,$$
for $0 < x < 1$, then we have
$$\alpha_{2k, 2m} \approx {1\over m}f\biggl({k\over m}\biggr)\ .$$
The reason for the $\approx$ sign is that we no longer require that $k$ get large.  This means
that we can replace the discrete $\alpha_{2k, 2m}$ distribution by the continuous density $f(x)$ on
the interval $[0, 1]$ and obtain a good approximation.  In particular, if $x$ is a fixed real
number between 0 and 1, then we have
$$\sum_{k < xm}\alpha_{2k, 2m} \approx \int_0^x f(t)\,dt\ .$$
It turns out that $f(x)$ has a nice antiderivative, so we can write
$$\sum_{k < xm}\alpha_{2k, 2m} \approx {2\over \pi}\arcsin \sqrt x\ .$$  
\index{arc sine laws}
One can see from the graph of this last function that it has a minimum at $x = 1/2$ and is
symmetric about that point.  As noted in the exercise, this implies that half of the walks of
length $2m$ have no equalizations after time $m$, a fact which probably would not be guessed.
\par
It turns out that the arc sine density comes up in the answers to many other questions
concerning random walks on the line.  Recall that in Section~\ref{sec 12.1}, a random walk could be
viewed as a polygonal line connecting $(0,0)$ with $(m, S_m)$.  Under this interpretation, we
define $b_{2k, 2m}$ to be the probability that a random walk of length $2m$ has exactly $2k$ of
its $2m$ polygonal line segments above the $t$-axis.  
\par
The probability $b_{2k, 2m}$ is frequently interpreted in terms of a two-player game.  (The
reader will recall the game Heads or Tails, in Example~\ref{exam 1.3}.)  Player A is said to be in
the lead at time
$n$ if the random walk is above the
$t$-axis at that time, or if the random walk is on the $t$-axis at time $n$ but above the
$t$-axis at time
$n-1$.  (At time 0, neither player is in the lead.)  One can ask what is the most probable number
of times that player A is in the lead, in a game of length $2m$.  Most people will say that the
answer to this question is $m$.  However, the following theorem says that $m$ is the least likely
number of times that player A is in the lead, and the most likely number of times in the
lead is 0 or $2m$.
\begin{theorem}\label{thm 12.3.1}
If Peter and Paul play a game of Heads or Tails of length $2m$, the probability that Peter will
be in the lead exactly $2k$ times is equal to
$$
\alpha_{2k, 2m}\ .
$$
%\end{theorem}
\proof
To prove the theorem, we need to show that
\begin{equation}  
b_{2k, 2m} = \alpha_{2k, 2m}\ .
\label{eq 12.3.1}  
\end{equation}
Exercise~\ref{sec 12.1}.\ref{exer 12.1.7} shows that $b_{2m, 2m} = u_{2m}$ and $b_{0, 2m} = u_{2m}$,
so we only need to prove that Equation~\ref{eq 12.3.1} holds for $1 \le k \le m-1$.  We can obtain a
recursion involving the $b$'s and the $f$'s (defined in Section~\ref{sec 12.1}) by counting the
number of paths of length $2m$ that have exactly $2k$ of their segments above the $t$-axis, where $1
\le k \le m-1$.  To count this collection of paths, we assume that the first return occurs at time
$2j$, where
$1 \le j \le m-1$.  There are two cases to consider.  Either during the first $2j$ outcomes the
path is above the $t$-axis or below the $t$-axis.  In the first case, it must be true that the
path has exactly $(2k - 2j)$ line segments above the $t$-axis, between $t = 2j$ and $t = 2m$. 
In the second case, it must be true that the path has exactly $2k$ line segments above the
$t$-axis, between $t = 2j$ and $t = 2m$.  
\par We now count the number of paths of the various types
described above.  The number of paths of length $2j$ all of whose line segments lie above the
$t$-axis and which return to the origin for the first time at time $2j$ equals
$(1/2)2^{2j}f_{2j}$.  This also equals the number of paths of length $2j$ all of whose line
segments lie below the $t$-axis and which return to the origin for the first time at time $2j$. 
The number of paths of length $(2m - 2j)$ which have exactly $(2k - 2j)$ line segments above the
$t$-axis is $b_{2k-2j, 2m-2j}$.  Finally, the number of paths of length $(2m-2j)$ which have
exactly $2k$ line segments above the $t$-axis is $b_{2k,2m-2j}$.  Therefore, we have
$$b_{2k,2m} = {1\over 2} \sum_{j = 1}^k f_{2j}b_{2k-2j, 2m-2j} + 
{1\over 2}\sum_{j = 1}^{m-k} f_{2j}b_{2k, 2m-2j}\ .$$
\par
We now assume that Equation~\ref{eq 12.3.1} is true for $m < n$.  Then we have

\begin{eqnarray*}
b_{2k, 2n} &=& {1\over 2} \sum_{j = 1}^k f_{2j}\alpha_{2k-2j, 2m-2j} + 
{1\over 2}\sum_{j = 1}^{m-k} f_{2j}\alpha_{2k, 2m - 2j}\\
&=& {1\over 2}\sum_{j = 1}^k f_{2j}u_{2k-2j}u_{2m-2k} + 
{1\over 2}\sum_{j = 1}^{m-k} f_{2j}u_{2k}u_{2m - 2j - 2k}\\
&=& {1\over 2}u_{2m-2k}\sum_{j = 1}^k f_{2j}u_{2k - 2j} +
{1\over 2}u_{2k}\sum_{j = 1}^{m-k} f_{2j}u_{2m - 2j - 2k}\\
&=& {1\over 2}u_{2m - 2k}u_{2k} + {1\over 2}u_{2k}u_{2m - 2k}\ ,
\end{eqnarray*}

where the last equality follows from Theorem~\ref{thm 12.1.2}.  Thus, we have
$$b_{2k, 2n} = \alpha_{2k, 2n}\ ,$$
which completes the proof.
\end{theorem}
%\putfig{4.5truein}{PSfig12-2}{The arc sine density and times in the lead.}{fig 12.2}
We illustrate the above theorem by simulating 10{,}000 games of Heads or Tails, with each game
consisting of 40 tosses.  The distribution of the number of times that Peter is in the lead is
given in Figure~\ref{fig 12.2}, together with the arc sine density.

\putfig{3.5truein}{PSfig12-2}{Times in the lead.}{fig 12.2}

We end this section by stating two other results in which the arc sine density appears. 
Proofs of these results may be found in Feller.\footnote{W.~Feller, op.\ cit., pp. 93--94.}
\begin{theorem}\label{thm 12.3.2}
Let $J$ be the random variable which, for a given random walk of length $2m$, gives the
smallest subscript $j$ such that $S_{j} = S_{2m}$.  (Such a subscript $j$ must be even,
by parity considerations.)  Let $\gamma_{2k, 2m}$ be the probability that $J = 2k$.  Then we
have
$$\gamma_{2k, 2m} = \alpha_{2k, 2m}\ .$$
\end{theorem}
The next theorem says that the arc sine density is applicable to a wide range of
situations.  A continuous distribution function $F(x)$ is said to
be \emx {symmetric} if $F(x) = 1 - F(-x)$.  (If $X$ is a continuous random variable
with a symmetric distribution function, then for any real $x$, we have $P(X \le x) = P(X
\ge -x)$.)  We imagine that we have a random walk of length $n$ in which each summand
has the distribution
$F(x)$, where $F$ is continuous and symmetric.  The subscript of the \emx {first
maximum}\index{first maximum of a random walk} of such a walk is the unique subscript $k$ such that 
$$S_k > S_0,\ \ldots,\ S_k > S_{k-1},\ S_k \ge S_{k+1},\ \ldots,\ S_k \ge S_n\ .$$
We define the random variable $K_n$ to be the subscript of the
first maximum.  We can now state the following theorem concerning the random variable $K_n$. 
\begin{theorem}\label{thm 12.3.3}
Let $F$ be a symmetric continuous distribution function, and let $\alpha$ be a fixed real number
strictly between 0 and 1.  Then as $n \rightarrow \infty$, we have
$$P(K_n < n\alpha) \rightarrow {2\over \pi} \arcsin\sqrt \alpha\ .$$
\end{theorem}
\par
A version of this theorem that holds for a symmetric random walk can also be found in Feller.

\exercises
\begin{LJSItem}

\i\label{exer 12.3.1} For a random walk of length $2m$, define $\epsilon_k$ to equal 1 if
$S_k > 0$, or if $S_{k-1} = 1$ and $S_k = 0$.   Define $\epsilon_k$ to equal -1 in all other
cases.  Thus, $\epsilon_k$ gives the side of the $t$-axis that the random walk is on during the
time interval $[k-1, k]$.  A ``law of large numbers" for the sequence $\{\epsilon_k\}$ would say
that for any $\delta > 0$, we would have
$$P\biggl(-\delta < {{\epsilon_1 + \epsilon_2 + \cdots + \epsilon_n}\over{n}} < \delta \biggr)
\rightarrow 1$$
as $n \rightarrow \infty$.  Even though the $\epsilon$'s are not independent, the above assertion
certainly appears reasonable.  Using Theorem~\ref{thm 12.3.1}, show that if $-1 \le x \le 1$, then
$$\lim_{n \rightarrow \infty} P\biggl({{\epsilon_1 + \epsilon_2 + \cdots + \epsilon_n}\over{n}}
< x\biggr) = {2\over{\pi}} \arcsin\sqrt{{{1 + x}\over{2}}}\ .$$
 
\i\label{exer 12.3.2} Given a random walk $W$ of length $m$, with summands 
\[ \{X_1, X_2, \ldots,X_m\}\ , \]  
\noindent define the \emx {reversed} random walk to be the walk $W^*$ with
summands 
\[ \{X_m, X_{m-1}, \ldots, X_1\}\ . \]  
\begin{enumerate}
\item  Show that the $k$th partial sum $S^*_k$ satisfies the equation
$$S^*_k = S_m - S_{n-k}\ ,$$
where $S_k$ is the $k$th partial sum for the random walk $W$.
\item Explain the geometric relationship between the graphs of a random walk and its
reversal.  (It is not in general true that one graph is obtained from the other by reflecting in a
vertical line.)
\item Use parts (a) and (b) to prove Theorem~\ref{thm 12.3.2}.
\end{enumerate}

\end{LJSItem}
\vfill
\pagebreak
