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%Slide 1
\title{{\LARGE\bf Econ 102}\newline (Random walks\newline and high finance)\newline \newline}
\author{Tom Carter
\newline
\newline
\newline
\tthdump{\href{http://astarte.csustan.edu/\~tom/SFI-CSSS}{http://astarte.csustan.edu/\~\ tom/SFI-CSSS}}
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%%Complex Systems Summer School
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}
\date{Fall, 2008}
\maketitle

%Slide 2
\sectionhead{Our general topics:}
%%tth:\begin{itemize}
%%tth:\item
\tthdump{\hyperlink{Financial Modeling}
{\ \ $\circledcirc$ Financial Modeling\newline}}
%%tth:\makehyperlink{Financial Modeling}
%%tth:\item
\tthdump{\hyperlink{Some random (variable) background}
{\ $\circledcirc$ Some random (variable) background\newline}}
%%tth:\makehyperlink{Some random (variable) background}
%%tth:\item
\tthdump{\hyperlink{What is a random walk?}
{\ $\circledcirc$ What is a random walk?\newline}}
%%tth:\makehyperlink{What is a random walk?}
%%tth:\item
\tthdump{\hyperlink{Some Intuitive Derivations}
{\ $\circledcirc$ Some Intuitive Derivations\newline}}
%%tth:\makehyperlink{Some Intuitive Derivations}
%%tth:\end{itemize}

\pagedone

  \sectionhead{Financial Modeling}
  
       Let's use some of these ideas to do some financial modeling.  As an example,
       let's develop the (in)famous Black-Scholes model for options pricing.
  
       \begin{itemize}
       
            \item We can start with the simplest financial instrument, the fixed rate bond.
               If we  ``buy" amount $V_0$ of a rate $r$ bond at time $t=0$, then at time $t=1$
               we can redeem the for value $V_0(1 + r)$.  If we wait longer to redeem the bond, 
               then at some time in the future we can redeem the bond for
                  $$V(n, r) = V_0(1 + r)^n$$ 
               One can also think of this as a  ``savings account" with interest rate $r$.  In this
               case, we can ask the more general question, what is $V(t, r)$ for real values of
               $t$, rather than just integral values of $t$?
               
               This will depend on the specifics of the bond (savings account).  In its simplest form,
               the bond will have  ``coupons" that can be redeemed at specific times in the future, or
               in the case of a savings account, interest will be  ``compounded" on specific dates.
               
               Let's look at various possibilities for compounding.  Suppose the bond has (annual)
               interest rate $r$.  If interest is compounded $k$ times during the year ($k$ would be
               4 for quarterly compounding, 12 for monthly compounding, etc.), then the value
               at time $t$ would be
                  $$V(t, r, k) = V_0(1 + \frac{r}{k})^{kt}$$
               If we smooth this out, and let $k$ go to infinity (i.e.,  ``continuous compounding"),
               then we will have
                  $$V(t, r) = V_0 \lim_{k \to \infty}((1 + \frac{r}{k})^{kt}) = V_0 e^{rt}.$$
               We thus know how to set a price for a bond to be redeemed at some time $t$ in
               the future.
       \end{itemize}
  \pagedone
       \begin{itemize}
          \item Now let's generalize.  Suppose that the interest rate $r$, instead of being fixed,
               varied over time.  What price should we be willing to pay for such a financial
               instrument?  We can think of this financial instrument as a stock (share) in a corporation.
               Our  ``return on investment" will be uncertain, and will depend on the performance
               of the corporation (and also on the change in price of the stock).  We have the potential
               to make a large profit (if the price of the stock goes up), but we now also have the
               potential to lose money (if the price of the stock goes down).
               
               There is a difficulty here in that we don't really have the flexibility to buy the stock
               at whatever price we want today (depending on our calculation of the future value
               of the stock), but can only buy at today's price.
               
               One thing we can do (at least potentially, assuming there are sellers willing), is to purchase
               an  ``option" to buy the stock at some fixed price at some specific time in the future.
               Let's simplify things a bit, and assume that over the time period in question, the stock
               will not pay any dividends (in other words, our profit/loss will only depend on
               changes in the price of the stock).
               
               Once a market in  ``options" is developed, a variety of things become possible.  Not only
               can we purchase options to buy a stock at a given price at a given time in the future
               (a  ``call" option), but we can also purchase an option to sell a stock at a given price at
               a given time in the future (a  ``put" option).  Note that an option protects us against
               an adverse change in the price of the stock.  For example, if we purchase a  ``call"
               option, and the price of the stock goes down, we let the option expire, and we lose
               only the amount we paid for the option (the premium).  We are thus protected against
               large losses.  More generally, if we buy both  ``put" and  ``call" options, we can  ``hedge"
               our bets, and (we believe) protect ourselves against large losses, at the price of
               limiting the amount of gain we might make.
               
               Our task, then, is to develop a model that will allow us to determine (estimate?)
               the price we should be willing to pay for an option.
               
               In order to develop our model, we will have to make a variety of assumptions,
               many of them  ``simplifying" assumptions.  It is possible (likely?) that at least some
               or our assumptions will be unrealistic, but at least will allow us to do computations.
               This presents us with an interesting dilemma -- if our model is unrealistic, we may
               make very bad decisions if we depend on the model, but, if we make the model
                ``realistic," it may be useless to us because we can't do the computations.
               Apparently, such is life \ldots
               
\pagedone

          \item For this example, we're going to develop (a version of) the Black-Scholes
               option pricing model.
               
               We'll have to make a variety of assumptions.  These will include:
               \begin{enumerate}
                  \item There is a completely safe (e.g.,  ``FDIC insured savings account")
                     fixed rate asset available.
                  \item There are ``frictionless" markets (i.e., we can buy or sell any instrument at any
                     time in any amount).
                  \item There are no transaction costs.
                  \item No  ``arbitrage" (there are no financial instruments that provide  ``risk free"
                     profits above the fixed rate asset).
               \end{enumerate}
\pagedone
               The most critical assumptions we will have to make concern the form of variability of
               the  ``price" of the stock on which we will be buying our options.
                \begin{enumerate}
                  \item Variability is continuous (and possibly even smooth?).
                     This will allow us to work in continuous time.
                  \item The distribution is  ``stable" (i.e., the distribution of the variability
                     does not change over time).
                  \item Increments are independent (i.e., variability does not depend
                     on history -- there is no  ``memory" in the distribution).
                  \item The distribution has a finite mean and finite variance.
                  \item Variability is  ``independent of price" (i.e., the  ``value" of a change
                     in price does not depend on the specific current price).
               \end{enumerate}
\pagedone
               Putting all of this together, we will assume that the stock price $x(t)$ gives us a
               continuous random variable
               $$R(x, t) = \ln\left(\frac{x(t)}{x(0)}\right).$$
               This random variable $R(x, t)$ will be the return on the stock at time $t$, associated
               with stock price $x(t)$.  More specifically, we will assume that the price $x(t)$
               satisfies the stochastic differential equation
               $$dx(t) = \mu x(t)\,dt + \sigma x(t)\,dB_t$$
               where $B_t$ is Brownian motion, $\mu$ is the  ``drift," and $\sigma$ is the volatility.
               This is generally called {\em geometric  Brownian motion}.  This equation has
               the solution
               $$x(t) = x(0)e^{\left( \left(\mu - \frac{\sigma^2}{2} \right)t + \sigma B_t\right)},$$
               which is a log-normally distributed random variable with expected value
               $\expect{x(t)}= x(0)e^{\mu t}$, and variance
               $\operatorname{Var}(x(t)) = x(0)^2e^{2\mu t}  \left( e^{\sigma^2 t}-1\right)$.
                     
               The random variable $R(x,t)$ is normally distributed, with mean
               $(\mu - \sigma^2 / 2)t$ and variance $\sigma^2t$.
               
               We are interested in determining the current value of a financial instrument that will have
               a value in the future that depends on the price $x(t)$ (or return $R(x(t), t))$ of the underlying
               asset (stock).  Let us call the value of this derived instrument $w(x,t)$.  In the Black-Scholes
               case, $w(x,t)$ will be the value of a ``call' option.
               
               To understand this better, we will study ``portfolios" of various financial instruments.
               So, suppose we form a portfolio by putting one unit of our money into the secure
               interest bearing asset, and an amount $-1/w_1$ (where
               $w_1 = (\partial w / \partial x)$) in a ``call" option with
               strike price $K$ at time $T$ (i.e., at time $T$ we can buy the stock a price $K$).
               Then the value of the portfolio will be $p = x - w/w_1$.
               
               During a short period of time $\Delta t$, the value of the portfolio will change by
               $\Delta p = \Delta x - \Delta w / w_1$.  We want to expand this formula, and
               fortunately there is a nice method (part of the {\em Ito calculus}) that allows us
               to write
               $$\Delta w = \frac{\partial w}{\partial x} \Delta x + \frac{1}{2}\sigma^2 x^2
                    \frac{\partial^2 w}{\partial x \partial t} \Delta t + \frac{\partial w}{\partial t}\Delta t$$
               (this is essentially the {\em chain rule} for {\em stochastic differentials}).
               
               We then have
               $$\Delta p = -\frac{1}{w_1}\left(\frac{1}{2}\sigma^2 x^2w_{11} + w_2\right)\Delta t$$
               (where $w_1, w_{11},$ and $w_2$ are the appropriate partial derivatives, and
               $\sigma^2$ is a constant depending on $w$, essentially its variance or volatility).
               
               Now we will use the  ``no arbitrage" assumption.  Since $\Delta p$ is an assured return,
               it must be that
               $$\Delta p = r p \Delta t = r \left(x - \frac{w}{w_1}\right)\Delta t.$$
               If we equate these two expressions for $\Delta p$, we get
               $$ -\frac{1}{w_1}\left(\frac{1}{2}\sigma^2 x^2w_{11} + w_2\right)\Delta t
                   = r \left(x - \frac{w}{w_1}\right)\Delta t.$$

               Dividing through by $\Delta t$, we can simplify to
               $$ w_2 + r x w_1 + \frac{1}{2}\sigma^2 x^2 w_{11} -rw = 0.$$
               
               This is the Black-Scholes differential equation (B-S PDE) for $w(x, t)$.  It has boundary
               conditions:
               \begin{enumerate}
                  \item $w(0, t) = 0$ for all $t$.
                  \item $w(x, t) \sim x$ as $t \to \infty$.
                  \item $w(x, T) = \max(x - K, 0)$ (recall $K$ is the strike price at time $T$).
               \end{enumerate}
\pagedone
             \item Our next task is to develop a solution to the B-S equation.
               We'll start by making some changes of variables.  We'll reverse the order of time
               (with some normalization), since the  ``no arbitrage" rule allows us to use $t=T$ for
               a boundary condition.  We'll also move into a  ``logarithmic" mode (since the riskless
               asset grows exponentially), and normalize with respect to $K$ (the strike price) --
               we shouldn't worry about the units of the price (e.g., dollars or pounds):
               \begin{eqnarray*}
                  \tau & = & \frac{\sigma^2}{2}(T - t) \\
                  z & = & \ln(x/K) \\
                  v & = & \frac{w}{K}\\
               \end{eqnarray*}
               Now we do some calculations:
                 $$ \frac{\partial t}{\partial \tau} = - \frac{2}{\sigma^2}$$
               and
               $$\frac{\partial x}{\partial z}  = x.$$
               Then we have
                \begin{eqnarray*}
                  \frac{\partial v}{\partial \tau} & = & \frac{1}{K}\frac{\partial w}{\partial \tau}
                          =  \frac{1}{K}\left(\frac{\partial w}{\partial x}\frac{\partial x}{\partial \tau}
                                          + \frac{\partial w}{\partial t}\frac{\partial t}{\partial \tau}\right)\\
                         & = & \frac{1}{K}\left( - \frac{2}{\sigma^2}\right)w_2\\
               \end{eqnarray*}
               and
               \begin{eqnarray*}
                  \frac{\partial v}{\partial z} & = & \frac{1}{K}\frac{\partial w}{\partial z}
                          =  \frac{1}{K}\left(\frac{\partial w}{\partial x}\frac{\partial x}{\partial z}
                                          + \frac{\partial w}{\partial t}\frac{\partial t}{\partial z}\right)\\
                         & = & \frac{1}{K} x w_1. \\
               \end{eqnarray*}
                Using this, we get
                \begin{eqnarray*}
                  \frac{\partial^2 v}{\partial z^2} & = & \frac{1}{K}\frac{\partial xw_1}{\partial z}
                          =  \frac{1}{K}\left(\frac{\partial w_1}{\partial z}x 
                                          + w1\frac{\partial x}{\partial z}\right)\\
                         & = & \frac{1}{K}\left(\frac{\partial w_1}{\partial x}\frac{\partial x}{\partial z}x 
                                          + \frac{\partial w_1}{\partial t}\frac{\partial t}{\partial z}x
                                          + x w1\right)\\
                         & = & \frac{1}{K} (x^2 w_{11} + x w_1).\\
               \end{eqnarray*}
  \pagedone             
               
               Now we'll do some rearranging of the B-S equation, and then put in what
               we have gotten from the change of variables.  Starting from
               $$ w_2 + r x w_1 + \frac{1}{2}\sigma^2 x^2 w_{11} -rw = 0.$$
               we multiply by $\frac{-2}{K\sigma^2}$
               $$ \frac{-2w_2}{K\sigma^2} + \frac{-2}{K\sigma^2} r x w_1
                       - \frac{1}{K} x^2 w_{11} + \frac{2}{K\sigma^2} rw = 0.$$
               
               We rearrange, and add/subtract appropriate terms:
                \begin{eqnarray*}
                    \frac{-2w_2}{K\sigma^2} & = & \frac{ x^2 w_{11}}{K}
                       +\frac{2 r x w_1}{K\sigma^2} - \frac{2 rw}{K\sigma^2}  \\
                    \frac{-2w_2}{K\sigma^2} & = & \frac{ x^2 w_{11}}{K}
                        + \frac{w_1 x}{K}
                       +\frac{2 r x w_1}{K\sigma^2}
                       + \frac{w_1 x}{K}
                       - \frac{2 rw}{K\sigma^2}  \\
                \end{eqnarray*}
                or,
                 \begin{eqnarray*}
                    \frac{-2w_2}{K\sigma^2} & = & \frac{1}{K}(x^2w_{11} + x w_1) \\
                         & \  & + \left(\frac{2r}{\sigma^2} - 1 \right)\frac{x w_1}{K}
                                - \frac{2 rw}{K\sigma^2}  \\
                \end{eqnarray*}
\pagedone
               Then, putting in the change of variables, we have:
               $$\frac{\partial v}{\partial \tau} = \frac{\partial^2 v}{\partial z^2}
                                                   + (k - 1) \frac{\partial v}{\partial z} - kv$$
               where $k = \frac{2r}{\sigma^2}$, and we have the
               boundary condition  $v(z, 0) = \max(e^z - 1, 0)$.
                                                   
               This is close to a heat equation, but we need one more change
               of variables.  If we let, for some constants $\alpha$, $\beta$,
               $$u(z, \tau) = e^{\alpha x + \beta \tau} v(z, \tau)$$
               we will have
               \begin{eqnarray*}
                  \beta u + \frac{\partial u}{\partial \tau} & = & \alpha^2 u
                                  + 2\alpha \frac{\partial u}{\partial z}
                                  + \frac{\partial^2 u}{\partial z^2} \\
                            & & + (k - 1)\left(\alpha u + \frac{\partial u}{\partial z}\right) - ku \\
               \end{eqnarray*}
              If we choose
                  $$ \alpha = - \frac{1}{2}(k - 1)$$
               and
                  $$ \beta = \alpha^2 + (k - 1) \alpha - k = - \frac{1}{4}(k + 1)^2,$$

\pagedone
              then we are left with the heat equation:
                 $$ \frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial z^2}\ \ . $$
                 
              At this point we'll just appeal to standard methods for solving the heat
              equation (e.g., Fourier series methods).  The general solution will
              be of the form:
              $$ u(z, \tau) =
                   \frac{1}{\sigma\sqrt{2\pi\tau}}\int_{-\infty}^{\infty}u_0(y)e^{-(z - y)^2/(2\sigma^2\tau)}\,dy.$$
                   
              Undoing the changes of variable, we get:
              $$w(x,t; K, T, r) = x\Phi(d_1) - Ke^{-r(T - t)}\Phi(d_2)\ ,$$
              where $\Phi$ is the standard normal cumulative distribution function
              $$\Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{(-\frac{u^2}{2})}du$$
              and
              \begin{eqnarray*}
                 d_1 & = & \frac{\ln(x/K) + (r + \frac{1}{2}\sigma^2)(T - t)}{\sigma\sqrt{T - t}}\\
                 d_2 & = & \frac{\ln(x/K) + (r - \frac{1}{2}\sigma^2)(T - t)}{\sigma\sqrt{T - t}}\\
              \end{eqnarray*}
                 
        \end{itemize}
        Notes for further discussion:
        
         c\`adl\`ag (French ``continue \`a droite, limit\'ee \`a gauche'') \ldots
         
         Functions of this sort provide sufficiently limited ``jumpiness'' to be amenable to these general
         methods.
         \vfill
         Following is some background material on random walks, which are central to the
         Black-Scholes (and related) models \ldots
\pagedone
\sectionhead{Some random (variable) background}

    This is a brief tour (peripatetic -- wandering about) of some topics in random
    walks.  It mostly consists of a few interesting ideas to start exploration
    of the general topic.  We'll start with a bit on random variables.  
    
\begin{itemize}

        \item A random variable is a view into a
           set of possible values.  Associated with each possible value is a probability.
           For a discrete random variable, there is a finite or countably infinite set
           of possible values and probabilities $\{(a_i, p_i)\}$, with the condition
           that $\sum_i p_i = 1$.  Thus, for example,
           we could talk about the random variable $\xi$ drawing values from the
           set of possible values $\{(1, 1/2), (-1, 1/2)\}$.  When we evaluate the
           random variable $\xi$, we get either 1 or -1, each with probability
           1/2.
           
\pagedone

        \item We can build new random variables.  For example, if $\xi_1$ and $\xi_2$
            are random variables over $\{(1, 1/2), (-1, 1/2)\}$, then $\xi_1 + \xi_2$
            is a random variable (but over the set $\{(2, 1/4), (0, 1/2), (-2, 1/4)\}$).
            We must be a bit careful sometimes -- for example, $\xi_1 + \xi_1$ is
            also a random variable (but this time over $\{(2, 1/2), (-2, 1/2)\}$).
            
        \item Given a random variable $X$ over $\{(a_i, p_i)\}$, we define the
            {\em expectation} (or {\em expected value}) of $X$ by:
            \[ \expect{X} = \sum_ip_ia_i. \]
            
            Note that the expectation is a linear operator:
            \[ \expect{\alpha X + \beta Y} = \alpha\expect{X} + \beta\expect{Y} \]
            for $\alpha, \beta$ real numbers, and $X, Y$ random variables.
            
\pagedone

        \item Note that the expectation of a constant $\alpha$ is that constant
            (a constant can be thought of as a random variable over $\{(\alpha, 1)\}$):
            \[ \expect{\alpha} = \alpha. \]

        \item Example: if $\xi_1$ and $\xi_2$ are random variables over
            $\{(1, 1/2), (-1, 1/2)\}$, then
            \begin{eqnarray*}
                \expect{\xi_1} & = & \frac{1}{2} * 1 + \frac{1}{2} * (-1)
                     = 0 = \expect{\xi_2}, \\
                \expect{\xi_1 + \xi_2} & = & \expect{\xi_1} + \expect{\xi_2} = 0 + 0 = 0, \\
                \expect{\xi_1\xi_2} & = & \frac{1}{2} * 1 + \frac{1}{2} * (-1) = 0,
            \end{eqnarray*}
            but
            \[\expect{\xi_1^{\ 2}} = \frac{1}{2} * 1^2 + \frac{1}{2} * (-1)^2 = 1, \]
            and
            \begin{eqnarray*}
                \expect{(\xi_1 + \xi_2)^2} & = & \expect{\xi_1^{\ 2} + 2 * \xi_1\xi_2
                    + \xi_2^{\ 2}} \\
                    & = & \expect{\xi_1^{\ 2}} + 2 * \expect{\xi_1\xi_2}
                    + \expect{\xi_2^{\ 2}} \\
                    & = & 1 + 0 + 1 \\
                    & = & 2.
            \end{eqnarray*}
            
\pagedone

        \item Given a random variable $X$, we define the {\em variance} of $X$ by:
            \[ V(X) = \expect{(X - \expect{X})^2}. \]
            We can also calculate this as:
            \begin{eqnarray*}
                V(X) & = & \expect{(X - \expect{X})^2} \\
                     & = & \expect{X^2 - 2 X \expect{X} + \expect{X}^2} \\
                     & = & \expect{X^2} - 2 \expect{X\expect{X}} + \expect{\expect{X}^2} \\
                     & = & \expect{X^2} - 2 \expect{X}^2 + \expect{X}^2 \\
                     & = & \expect{X^2} - \expect{X}^2.
             \end{eqnarray*}
             
         \item The {\em standard deviation} of a random variable $X$ is given by:
             \[ \sigma(X) = V(X)^{1/2}. \]
    \end{itemize}
\pagedone

\sectionhead{What is a random walk?}
     
    There are a variety of ways to define a random walk.  Here we
    start with a relatively simple version, which will allow us to develop some
    classical results on random walks.  Later, we can generalize.

    \begin{itemize}
    
        \item Let $\{\xi_i | i = 1, 2, 3, \ldots\}$ be a set of (independent) random variables
            over $\{(1, \frac{1}{2}), (-1, \frac{1}{2})\}$ (in particular, `observing' one
            of the random variables has no effect on observations of any of the rest of them).
            Then a {\em simple random walk} is a sequence $(S_n)$ where
            \begin{eqnarray*}
                S_0 & = & 0, \\
                S_n & = & \xi_1 + \xi_2 + \ldots + \xi_n.
            \end{eqnarray*}
            
        \item It is easy to see that
            \[ -n \le S_n \le n. \]
            
\pagedone
            
        \item  We have that
            \[ \expect{S_0} = 0, \]
            and thus
            \begin{eqnarray*}
                \expect{S_n} & = & \expect{S_{n-1} + \xi_n} \\
                             & = & \expect{S_{n-1}} + \expect{\xi_n} \\
                             & = & \expect{S_{n-1}} + 0 \\
                             & = & \expect{S_{n-1}} \\
                             & = & \expect{S_{n-2}} \\
                             & \vdots & \\
                             & = & \expect{S_0} \\
                             & = & 0.
            \end{eqnarray*}
            
        \item We also have
            \begin{eqnarray*}
                \expect{S_{n}^{\ 2}} & = & \expect{(S_{n-1} + \xi_n)^{\ 2}} \\
                       & = & \expect{S_{n-1}^{\ 2} + 2 * S_{n-1}\xi_n + \xi_n^{\ 2}} \\
                       & = & \expect{S_{n-1}^{\ 2}} + 2 * \expect{S_{n-1}\xi_n}
                              + \expect{\xi_n^{\ 2}} \\
                       & = & \expect{S_{n-1}^{\ 2}} + 2 * \sum_{i=1}^{n-1}\expect{\xi_i\xi_n}
                              + 1 \\
                       & = & \expect{S_{n-1}^{\ 2}} + 2 * \sum_{i=1}^{n-1}0
                              + 1 \\
                       & = & \expect{S_{n-1}^{\ 2}} + 1 \\
                       & = & \expect{S_{n-2}^{\ 2}} + 2 \\
                       & \vdots & \\
                       & = & \expect{S_0^{\ 2}} + n \\
                       & = & n,
            \end{eqnarray*}
            and thus
            \begin{eqnarray*}
                V(S_n) & = & \expect{S_{n}^{\ 2}} - \expect{S_{n}}^2 \\
                       & = & n - 0 \\
                       & = &  n.
            \end{eqnarray*}
            
            In other words, the variance of $S_n$ is $n$, and the standard deviation
            of $S_n$ is $\sqrt{n}$.
            
\pagedone

        \item  We know that $S_n$ ranges between $-n$ and $n$.  But how is it distributed
            across that range?  In other words, if $-n \le k \le n$, what is the probability
            that $S_n = k$ (i.e., $P(S_n = k)$)?  We can make a couple of observations.  First,
            by symmetry, $P(S_n = k) = P(S_n = -k)$, since each of the $\xi_i$ is over
            $\{(1, \frac{1}{2}), (-1, \frac{1}{2})\}$.  Second, by a parity argument,
            if $n$ is even and $k$ is odd, or if $n$ is odd and $k$ is even,
            then $P(S_n = k) = 0$.  Let us look, then, at the case $n$ is even and
            $k \ge 0$ is also even.
            
            We know that $S_n = \sum_i\xi_i$, and that each of the $\xi_i$ is either -1
            or 1.  Thus, $S_n = k$ when exactly ($\frac{n}{2} + \frac{k}{2}$) of the
            $\xi_i$ are +1 and the rest (i.e., $n - (\frac{n}{2} + \frac{k}{2})
            = (\frac{n}{2} - \frac{k}{2})$) are -1.  This can happen in
            $\binom{n}{\frac{n}{2} + \frac{k}{2}}$ different ways.  Each of these
            is equally likely, and there are $2^n$ total possibilities.
            
\pagedone

            Thus, we have that
            \begin{eqnarray*}
                P(S_n = k) & = & \binom{n}{\frac{n}{2} + \frac{k}{2}}\frac{1}{2^n} \\
                           & = & \frac{n!}{(\frac{n}{2} + \frac{k}{2})!
                                   \ (n - (\frac{n}{2} + \frac{k}{2}))!\ 2^n} \\
                           \\
                           & = & \frac{n!}{(\frac{n + k}{2})!
                                   \ (\frac{n - k}{2})!\ 2^n}
            \end{eqnarray*}
            
%%            We can also do the quick consistency check:
%%            $$\sum_{k=-n}^{n}P(S_n = k)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
%%             \ \ \ \ $$
%%            \begin{eqnarray*}
%%                & = & P(S_n = 0) + 2 \sum_{k=1}^{n}P(S_n = k) \\
%%                      & = & P(S_n = 0) + 2 \sum_{k=1}^{n/2}P(S_n = 2k) \\
%%                      & = & P(S_n = 0) + 2 \sum_{k=1}^{n/2}
%%                              \binom{n}{\frac{n}{2} + \frac{2k}{2}}\frac{1}{2^n}
%%            \end{eqnarray*}
%%            \begin{eqnarray*}
%%                      & = & P(S_n = 0) + \frac{1}{2^n}\ 2 \sum_{k=1}^{n/2}
%%                              \binom{n}{\frac{n}{2} + k} \\
%%                      & = & \frac{1}{2^n}\binom{n}{\frac{n}{2}}
%%                              + \frac{1}{2^n}\ 2 \sum_{k=1}^{n/2}
%%                              \binom{n}{\frac{n}{2} + k} \\
%%                      & = & \frac{1}{2^n}\left(\binom{n}{\frac{n}{2}}
%%                              + \sum_{k=1}^{n/2} \binom{n}{\frac{n}{2} + k}
%%                              + \sum_{k=1}^{n/2} \binom{n}{\frac{n}{2} + k}\right) \\
%%                      & = & \frac{1}{2^n}\left(\binom{n}{\frac{n}{2}}
%%                              + \sum_{k=1}^{n/2} \binom{n}{n - (\frac{n}{2} + k)}
%%                              + \sum_{k=1}^{n/2} \binom{n}{\frac{n}{2} + k}\right) \\
%%                      & = & \frac{1}{2^n}\left(\binom{n}{\frac{n}{2}}
%%                              + \sum_{k=0}^{n/2 - 1} \binom{n}{k}
%%                              + \sum_{k=n/2 + 1}^{n} \binom{n}{k}\right) \\
%%                      & = & \frac{1}{2^n}\left(\sum_{k=0}^{n} \binom{n}{k}\right) \\
%%                      & = & \frac{1}{2^n} * 2^n \\
%%                      \\
%%                      & = & 1.
%%            \end{eqnarray*}
%%            (ugh . . . :-)
%%            
%%\pagedone
%%
%%        \item Let's generalize slightly, and suppose that our random walk may have
%%            unequal probabilities of moving in the two directions.  In other words,
%%            suppose our random variables $\xi_i$ are over
%%            $\{(1, p), (-1, q)\}$, with $q = 1 - p$.
%%            
%%            In this case, given two such random variables $\xi_1$ and $\xi_2$, we have:
%%            \begin{eqnarray*}
%%                \expect{\xi_1} & = & p * 1 + q * (-1) = p - q \\
%%                      & = & 2p -1 = \expect{\xi_2}, \\
%%                \expect{\xi_1 + \xi_2} & = & \expect{\xi_1} + \expect{\xi_2} = 2 * (2p - 1) \\
%%                                      & = & 4p - 2, \\
%%                \expect{\xi_1\xi_2} & = & (p^2 + q^2) * 1 + 2pq * (-1) \\
%%                          & = & p^2 - 2pq + q^2 = (p - q)^2 \\
%%                          & = & (2p -1)^2 = 4p^2 - 4p + 1.
%%            \end{eqnarray*}
%%\pagedone
%%            We also have:
%%            \[\expect{\xi_1^{\ 2}} = p * 1^2 + q * (-1)^2 = p + q = 1, \]
%%            and
%%            \begin{eqnarray*}
%%                \expect{(\xi_1 + \xi_2)^2} & = & \expect{\xi_1^{\ 2} + 2 * \xi_1\xi_2
%%                    + \xi_2^{\ 2}} \\
%%                    & = & \expect{\xi_1^{\ 2}} + 2 * \expect{\xi_1\xi_2}
%%                    + \expect{\xi_2^{\ 2}} \\
%%                    & = & 1 + 2 * (p - q)^2 + 1 \\
%%                    & = & 2 + 2 * (p - q)^2 \\
%%                    & = & 2 + 2 * (4p^2 - 4p + 1) \\
%%                    & = & 2 + 8 p^2 - 8p + 2 \\
%%                    & = & 4 + 8 p^2 - 8p.
%%            \end{eqnarray*}

%% 
%%\pagedone

%%        \item  If we again let $S_0 = 0$ and $S_n = S_0 + \xi_1 + \ldots \xi_n$,
%%            we have that
%%            \[ \expect{S_0} = 0, \]
%%            and in general
%%            \begin{eqnarray*}
%%                \expect{S_n} & = & \expect{S_{n-1} + \xi_n} \\
%%                             & = & \expect{S_{n-1}} + \expect{\xi_n} \\
%%                             & = & \expect{S_{n-1}} + p - q \\
%%                             & = & \expect{S_{n-2}} + \expect{\xi_{n-1}} + p - q \\
%%                             & = & \expect{S_{n-2}} + 2 * (p - q)\\
%%                             & \vdots & \\
%%                             & = & \expect{S_0} + n * (p - q)\\
%%                             & = & n * (p - q).
%%            \end{eqnarray*}
%%            
%%        \item We also have
%%            \begin{eqnarray*}
%%                \expect{S_{n}^{\ 2}} & = & \expect{(S_{n-1} + \xi_n)^{\ 2}} \\
%%                       & = & \expect{S_{n-1}^{\ 2} + 2 * S_{n-1}\xi_n + \xi_n^{\ 2}} \\
%%                       & = & \expect{S_{n-1}^{\ 2}} + 2 * \expect{S_{n-1}\xi_n}
%%                              + \expect{\xi_n^{\ 2}} \\
%%                       & = & \expect{S_{n-1}^{\ 2}} + 2 * \sum_{i=1}^{n-1}\expect{\xi_i\xi_n}
%%                              + 1 \\
%%                       & = & \expect{S_{n-1}^{\ 2}} + 2 * \sum_{i=1}^{n-1}(p - q)^2
%%                              + 1 \\
%%                       & = & \expect{S_{n-1}^{\ 2}} + 2(n - 1)(p - q)^2 + 1 \\
%%                       & = & \expect{S_{n-2}^{\ 2}} + 2((n - 1) + (n - 2))(p - q)^2 + 2\\
%%                       & \vdots & \\
%%                       & = & \expect{S_0^{\ 2}} + 2\left(\sum_{i = 1}^{n - 1}i\right)(p - q)^2 + n \\
%%                       & = & 0 + n(n - 1)(p - q)^2 + n \\
%%                       & = & n + n(n - 1)(p - q)^2,
%%            \end{eqnarray*}
%%\pagedone
%%            and thus
%%            \begin{eqnarray*}
%%                V(S_n) & = & \expect{S_{n}^{\ 2}} - \expect{S_{n}}^2 \\
%%                       & = & n + n(n - 1)(p - q)^2 - n^2 * (p - q)^2 \\
%%                       & = & n - n(p - q)^2 \\
%%                       & = & n(p + q)^2 - n(p - q)^2 \\
%%                       & = & n((p + q)^2 - (p - q)^2) \\
%%                       & = & n(4pq) \\
%%                       & = & 4npq.
%%            \end{eqnarray*}
%%            
%%            In other words, the variance of $S_n$ is $4npq$, and the standard deviation
%%            of $S_n$ is $2\sqrt{npq}$.
     \end{itemize}
     
 \pagedone
 
 \sectionhead{Some Intuitive Derivations}
 
     Every so often, I like to be a physicist (or biologist) and cavalier about
     error bounds.
     \begin{itemize}
         \item Suppose $S_n = \sum_i\xi_i$ is a random walk, with $\xi_i$ random variables
             over $\{(1, \frac{1}{2}), (-1, \frac{1}{2})\}$.  Let us write
             $P(S_n = k)$ as $P(k, n)$.
             
             We can observe that
             $$P(k, n + 1) = \frac{1}{2} P(k-1, n) + \frac{1}{2} P(k+1, n).$$
             Now assume that $n$ and $k$ are large, let $\delta$ and $\tau$ be (small)
             real numbers, and then let $x = \delta k$ and $t = \tau n$.
             
             We then have $P(x, t) = P(\delta k, \tau n)$, and so
             
             $P(x, t + \tau) = P(\delta k, \tau(n + 1)) $
             \begin{eqnarray*}
                 & = & \frac{1}{2} P(\delta(k - 1), \tau n) + \frac{1}{2}P(\delta(k + 1), \tau n) \\
                 \\
                 & = & \frac{1}{2} P(x - \delta, t) + \frac{1}{2}P(x + \delta, t).
             \end{eqnarray*}
 \pagedone
         From this, we get
         
         $P(x, t + \tau) - P(x, t)$
         $$ = \frac{1}{2}(P(x - \delta, t) + P(x + \delta, t) - 2 P(x, t)).$$
         
         Now consider two approximations.  First, for small (infinitesimal) $\tau$,
         we have
         $$P(x, t + \tau) = P(x, t) + \tau * \frac{\partial P(x, t)}{\partial t},$$
         and for small (infinitesimal) $\delta$, we have
         
         $P(x + \delta, t) + P(x - \delta, t) $
         $$ = 2P(x, t) + \delta^2 * \frac{\partial^2P(x, t)} {\partial x^2}.$$
         Putting pieces together, we have:
         $$\frac{\partial P(x, t)}{\partial t} = \frac{\delta^2}{2\tau}
                 \frac{\partial^2P(x, t)} {\partial x^2}$$
         (i.e., the diffusion equation \ldots).
         
 \pagedone
 
         More generally, if we have a biased random walk (over $\{(1, p), (-1, q)\}$),
         then
             $$ P(x, t + \tau) = p * P(x - \delta, t) + q * P(x + \delta, t),$$
         and, using the approximations, we have
             \begin{eqnarray*}
                 &   &\tau * \frac{\partial P(x, t)}{\partial t} \\
                 \\
                 & = & P(x, t + \tau) - P(x, t) \\
                 \\
                 & = & p * P(x - \delta, t) + q * P(x + \delta, t) - P(x, t) \\
                 \\
                 & = & p * \left(P(x, t) - \delta * \frac{\partial P(x, t)}{\partial x}
                        + \frac{\delta^2}{2} * \frac{\partial^2 P(x, t)}{\partial x^2}\right) \\
                 \\
                 &   &  + q * \left(P(x, t) + \delta * \frac{\partial P(x, t)}{\partial x}
                        + \frac{\delta^2}{2} * \frac{\partial^2 P(x, t)}{\partial x^2}\right) \\
                 \\
                 &   &  - P(x, t) \\
                 & = & (p + q - 1) P(x, t) + (q - p) * \delta
                         * \frac{\partial P(x, t)}{\partial x} \\
                 &   &  + \frac{(p + q)\delta^2}{2} * \frac{\partial^2 P(x, t)}{\partial x^2} \\
                 \\
                 & = & (q - p) * \delta * \frac{\partial P(x, t)}{\partial x}
                        + \frac{\delta^2}{2} * \frac{\partial^2 P(x, t)}{\partial x^2}.
             \end{eqnarray*}
 \pagedone
         Writing this in a slightly different form, 
         we have
             $$\frac{\partial P(x, t)}{\partial t}
                  = D * \frac{\partial^2P(x, t)} {\partial x^2}
                    + D * \beta * \frac{\partial P(x, t)}{\partial x},$$
                    
         where
             $$D = \frac{\delta^2}{2 \tau}\ \mathrm{and}\ \beta = \frac{2(1 - 2p)}{\delta}$$
         
         (i.e., diffusion with drift \ldots).
             
\pagedone
        \item Let's look at another approach to continuous versions of these issues.
            Instead of looking at random variables over a discrete set, let the
            random variables draw their values from a probability distribution.
            In particular, let
                $$w: \Re \to [0, 1]$$
            be an integrable (measurable)
            function, with
                $$\int_{-\infty}^\infty w(s) ds = 1.$$
            
            Then a random variable $\xi$ over $w(s)$ gives the value $s$ with
            probability $w(s)$ (i.e., $P(\xi = s) = w(s)$).
            
            Note that we can use the Dirac delta function $\delta(x - x_0)$ to recover
            the discrete examples if we want to.
\pagedone
            Recall that the Dirac delta function has the
            properties
            \begin{eqnarray*}
                \delta(x - x_0) dx & = & 0\ \ \mathrm{if}\ |x - x_0| > \frac{dx}{2} \\
                \delta(x - x_0) dx & = & 1\ \ \mathrm{if}\ |x - x_0| \le \frac{dx}{2}
            \end{eqnarray*}
            and
            $$ \int_{-\infty}^\infty \delta(x - x_0) dx = 1.$$
            
            Then, if we let
            $$ w(x) = \sum_i p_i * \delta(x - a_i),$$
            we are back in the discrete case.
            
            Let's mention here also that the delta function has the property:
            $$\delta(x - x_0) = \frac{1}{2 \pi} \int_{-\infty}^\infty
                e^{it(x_0 - x)} dt.$$
\pagedone
  
            Given a random variable $\xi$ over a probability distribution $w(s)$,
            we can look at the expected value of the random variable
            $$\expect{\xi} = \int_{-\infty}^\infty s * w(s) ds,$$
            the mean square
            $$\expect{\xi^2} = \int_{-\infty}^\infty s^2 * w(s) ds,$$
            the variance
            $$V(\xi) = \expect{\xi^2} - \expect{\xi}^2,$$
            and so on.
            
            Now suppose we have a probability distribution $w(s)$ with mean $\mu$
            and standard deviation $\sigma$ (i.e., if $\xi$ is a random variable
            over $w(s)$, then $\mu = \expect{\xi}$, and $\sigma^2 = V(\xi)$).  Note that
            there is no guarantee for a given distribution $w(s)$ that either the mean
            $\mu$ or standard deviation $\sigma$ exist -- the integrals could diverge.  In
            this example, we are assuming they do exist.
            
\pagedone
   
            Let $\{\xi_i\}$ be a collection of (independent) random variables over
            the distribution $w(s)$, and let $S_n = S_0 + \sum_1^n\xi_i$ (with $S_0 = 0$)
            be a random walk.  What can we say about the distribution of the values of
            $\frac{1}{n}S_n$?  In other words, what can we say about the distribution
            of the average of $n$ (identically distributed) random variables?
            
            We want to find the probability that $\frac{1}{n}S_n = x$ (let's write this as
            $P_n(x)$).  We will have $\frac{1}{n}S_n = x$ if $\xi_i = s_i$ and
            $\frac{1}{n}\sum_is_i = x$.
            The probability of this happening is $\prod_iw(s_i)$, since the $\xi_i$
            are independent of each other.  We need to add up the probabilities over
            all possible ways that $\frac{1}{n}\sum_is_i = x$ (as we did in the discrete case).
            In other words, we will have
            $$P_n(x) = \underset{\frac{1}{n}\sum_is_i = x}{\int\int\cdots\int}w(s_1)\cdots w(s_n)
                        ds_1\cdots ds_n.$$
            The limits of integration are fairly messy, so we will use the Dirac
            delta function.
            
\pagedone

            We will then have
            
            $P_n(x)$
            $$ = \int_{-\infty}^\infty\cdots\int_{-\infty}^\infty\delta(x - \frac{1}{n}\sum_js_j)
                    w(s_1)\cdots w(s_n)ds_1\cdots ds_n.$$
            Using the (fourier transform) property of the delta function, we have
            
            $2\pi P_n(x)$
            \begin{eqnarray*}
                &= & \int_{-\infty}^\infty...\int_{-\infty}^\infty
                    e^{it(\frac{1}{n}\sum_js_j - x)}
                    w(s_1)...w(s_n)dtds_1...ds_n \\
                \\
                &= & \int_{-\infty}^\infty\cdots\int_{-\infty}^\infty
                    e^{-itx}
                    \prod_j(e^{it\frac{s_j}{n}}w(s_j))dtds_1\cdots ds_n \\
                &= & \int_{-\infty}^\infty e^{-itx}
                    \left(\prod_j\int_{-\infty}^\infty(e^{it\frac{s_j}{n}}w(s_j))ds_j\right)dt.
            \end{eqnarray*}
            If we now let $Q(t) = \int_{-\infty}^\infty e^{\frac{its}{n}}w(s)ds$, we have
            $$P_n(x) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{-itx}Q^n(t) dt.$$
            
\pagedone

            Now let's look at $Q(t)$.  We can expand the exponential to get
            \begin{eqnarray*}
                Q(t) & = & \int_{-\infty}^\infty e^{\frac{its}{n}}w(s)ds \\
                     & = & \int_{-\infty}^\infty w(s)(1 + \frac{its}{n}
                            - \frac{1}{2}\frac{t^2s^2}{n^2} \cdots)ds \\
                     & = & \int_{-\infty}^\infty w(s)ds
                            + \frac{it}{n}\int_{-\infty}^\infty s * w(s)ds \\
                     &   & - \frac{t^2}{2n^2} \int_{-\infty}^\infty s^2w(s)ds + \cdots \\
                     & = & 1 + \frac{it}{n}\expect{s} - \frac{1}{2n^2}t^2\expect{s^2} + \cdots
            \end{eqnarray*}
            Now we take the log, and use the expansion $\ln(1 + y) = y - \frac{1}{2}y^2 + \cdots$
            to get
            \begin{eqnarray*}
                \ln(Q^n(t)) & = & n \ln(1 + \frac{it}{n}\expect{s}
                            - \frac{t^2}{2n^2}\expect{s^2} + \cdots) \\
                     & = & n * (\frac{it}{n}\expect{s} - \frac{1}{2n^2}t^2\expect{s^2}
                         - \frac{1}{2}(\frac{it}{n}\expect{s})^2 + \cdots)\\
                     & = & (it\expect{s}
                         - \frac{1}{2n}t^2(\expect{s^2} - \expect{s}^2) + \cdots) \\
                     & = & (it\mu - \frac{1}{2n}t^2\sigma^2 + \cdots)
            \end{eqnarray*}
            where $\mu$ and $\sigma$ are the mean and standard deviation of the distribution
            $w(s)$.
  
\pagedone
 
            Discarding all the higher order terms, and taking antilogs, we get
            $$Q^n(t) = e^{it\mu - \frac{1}{2n}t^2\sigma^2},$$
            and then that
            $$P_n(x) = \frac{1}{2\pi}\int_{-\infty}^\infty
                    e^{it(\mu -x) - \frac{1}{2n}t^2\sigma^2}dt.$$
            Now we use the formula
            $$\int_{-\infty}^\infty e^{at - bt^2}dt = \sqrt{\frac{\pi}{b}}
                    * e^{\left(\frac{a^2}{4b}\right)}$$
            with $a = i(\mu - x)$, $b = \frac{1}{2n}\sigma^2$, to finally get 
            \begin{eqnarray*}
                P_n(x) & = & \frac{1}{2\pi}\int_{-\infty}^\infty
                              e^{it(\mu -x) - \frac{1}{2n}t^2\sigma^2}dt \\
                       & = & \frac{1}{2\pi}\sqrt{\frac{\pi}{\frac{1}{2n}\sigma^2}}
                              * e^{\left(-\frac{(\mu - x)^2}
                                  {4 * \frac{1}{2n}\sigma^2}\right)} \\
                       & = & \frac{1}{\frac{\sigma}{\sqrt{n}}\sqrt{2 \pi}}
                              * e^{\left(-\frac{(x - \mu)^2}
                                 {2\left(\frac{\sigma^2}{n}\right)}\right)}.
            \end{eqnarray*}
            In other words, it is a normal distribution with mean $\mu$ and standard deviation
            $\frac{\sigma}{\sqrt{n}}$.
            
\pagedone

            Note that we made no assumptions about the distribution $w(s)$ except that
            it actually has a mean ($\mu$) and a standard deviation ($\sigma$) (and,
            of course, that it goes to zero fast enough for large $|s|$ that the
            approximations work out right \ldots).
            
            What this says is that if we average a bunch of identically distributed
            independent random variables, the result is a normal distribution, whether
            or not the original distribution was normal.
            
            This is usually called the {\em Central Limit Theorem}.
     \end{itemize}
\pagedone

  
 \pagedone
 
 \footnotesize
 \bibliographystyle{plain}
 \tthdump{\hypertarget{References}{}\hyperlink{Our general topics:}{\hfil \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
   \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ To top $\leftarrow$}}
%%tth:{\special{html: <A NAME="References"></A><a href="\#Top of file">       Top</a>}}
\vspace{-1.0 in}

\begin{thebibliography}{12}
%%tth:{\special{html: <font size="+0">}}

\bibitem{bachelier}
Bachelier, Louis, {\em Theory of \$peculation} - The Origins of Modern Finance,
Translated and with Commentary by Mark Davis and Alison Etheridge,
Princeton University Press, Princeton, 2006.

 \bibitem{spitzer}
 Spitzer, Frank,
 {\em Principles of Random Walk}, 2nd ed.,
Springer, New York, 1976.

\bibitem{denny}
Denny, Mark and Gaines, Steven,
{\em Chance in Biology} - Using Probability to Explore Nature,
Princeton University Press, Princeton, 2000.


\end{thebibliography}

\tthdump{\hyperlink{Our general topics:}{\hfil To top $\leftarrow$}}
%%tth:{\special{html: <a href="\#Top of file">       Back to top of file</a>}}

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% 
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% 
% %\bibitem{abrams2}
% Abrams D S and Lloyd S, 
% Non-Linear Quantum Mechanics implies Polynomial Time 
% solution for NP-complete and $\#$P problems,
% %in {\it LANL e-print} quant-ph/9801041, http://xxx.lanl.gov (1998)
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9801041}{}{}
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\end{document}
%%         \item Oh, all right.  Since we're here, let's use Stirling's approximation
%%             $$n! \approx   \sqrt{2\pi}\ n^ne^{-n}\sqrt{n}.$$
%%             to calculate:
%%             
%%             $$P(S_n = k) = \frac{n!}{(\frac{n + k}{2})!
%%                                    \ (\frac{n - k}{2})!\ 2^n} $$
%%                                    
%%             \begin{eqnarray*} & \approx & \frac{\sqrt{2\pi}\ n^ne^{-n}\sqrt{n}}
%%                                  {\sqrt{2\pi}\ (\frac{n + k}{2})^{\frac{n + k}{2}}
%%                                       e^{-\frac{n + k}{2}}\sqrt{\frac{n + k}{2}}
%%                                    \ \sqrt{2\pi}\ (\frac{n - k}{2})^{\frac{n - k}{2}}
%%                                    e^{-\frac{n - k}{2}}\sqrt{\frac{n - k}{2}}\ 2^n} \\
%%                               \\
%%                               & = & \frac{n^n\sqrt{n}}
%%                                  {\sqrt{2\pi}\ (\frac{n + k}{2})^{\frac{n + k}{2}}
%%                                       \sqrt{\frac{n + k}{2}}
%%                                    \ \ (\frac{n - k}{2})^{\frac{n - k}{2}}
%%                                    \sqrt{\frac{n - k}{2}}\ 2^n} \\
%%                               \\
%%                               & = & \frac{n^n\sqrt{n}}
%%                                  {\sqrt{2\pi}\ (n + k)^{\frac{n + k}{2}}
%%                                       \sqrt{\frac{n + k}{2}}
%%                                    \ \ (n - k)^{\frac{n - k}{2}}
%%                                    \sqrt{\frac{n - k}{2}}} \\
%%                               \\
%%                               & = & \frac{n^n\sqrt{\frac{4n}{(n+k)(n-k)}}}
%%                                  {\sqrt{2\pi}\ (n + k)^{\frac{n + k}{2}}
%%                                    \ (n - k)^{\frac{n - k}{2}}}
%%             \end{eqnarray*}
%%             
%% \pagedone
%% 
%%             \begin{eqnarray*}
%%                               & = & \frac{e^{n \ln(n)}\sqrt{\frac{4n}{(n + k)(n - k)}}}
%%                                  {\sqrt{2\pi}\ e^{\frac{n + k}{2}\ln(n + k)}
%%                                    \ e^{\frac{n - k}{2}\ln(n - k)}} \\
%%                               \\    
%%                               & = & \frac{1}{\sqrt{2\pi}}e^{n \ln(n)-\frac{n + k}{2}\ln(n + k)
%%                                     - \frac{n - k}{2}\ln(n - k)} \\
%%                               &   & * \sqrt{\frac{4n}{(n + k)(n - k)}}                                
%%             \end{eqnarray*}
%%             so
%%             
%%             $\ln(P(s_n = k)) $
%%             \begin{eqnarray*}
%%                               & \approx & \ln(\frac{1}{\sqrt{2\pi}}) 
%%                                      + n \ln(n)-\frac{n + k}{2}\ln(n + k)\\
%%                               &   & - \frac{n - k}{2}\ln(n - k)
%%                                     + \frac{1}{2}\ln\left(\frac{4n}{(n + k) (n - k)}\right) \\
%%                               \\
%%                               & = & \ln(\frac{1}{\sqrt{2\pi}}) 
%%                                      + n \ln(n)-\frac{n + k}{2}\ln(n + k)\\
%%                               &   & - \frac{n - k}{2}\ln(n - k)
%%                                     + \frac{1}{2}(\ln(4)) + \frac{1}{2}\ln(n)\\
%%                               &   & - \frac{1}{2}\ln((n + k)(n - k))                               
%%             \end{eqnarray*}
%%             
%% \pagedone
%%             \begin{eqnarray*}
%%                               & = & \ln(\frac{2}{\sqrt{2\pi}})
%%                                      + \frac{2n + 1}{2} \ln(n)-\frac{n + k + 1}{2}\ln(n + k)\\
%%                               &   & - \frac{n - k + 1}{2}\ln(n - k) \\
%%                               \\
%%                               & = & \ln(\frac{2}{\sqrt{2\pi}})
%%                                      + \frac{2n + 1}{2}(\ln(n)
%%                                      - \frac{n+k+1}{2n + 1}\ln(n + k) \\
%%                               &   &  - \frac{n-k+1}{2n + 1}\ln(n - k)) \\
%%                               \\
%%                               & = & \ln(\frac{2}{\sqrt{2\pi}})
%%                                      + \frac{1}{\frac{2}{2n + 1}}(\ln(n)
%%                                      - \frac{n+k+1}{2n + 1}\ln(n + k) \\
%%                               &   &  - \frac{n-k+1}{2n + 1}\ln(n - k)) \\                           
%%             \end{eqnarray*}
%%             
%%             so $ \lim_{n \to \infty} \ln(P(S_n = k)) $
%%             \begin{eqnarray*}
%%                               & = & \ln(\frac{2}{\sqrt{2\pi}})
%%                                     - \lim_{n \to \infty}\frac{1}{\frac{4}{(2n+1)^2}}
%%                                          (\frac{1}{n} \\
%%                               &   &  + \frac{2k + 1}{(2n + 1)^2}\ln(n+k)
%%                                      - \frac{n + k + 1}{(2n + 1)(n + k)} \\
%%                               &   &  - \frac{2k - 1}{(2n + 1)^2}\ln(n-k)
%%                                      - \frac{n - k + 1}{(2n + 1)(n - k)})
%%             \end{eqnarray*}
%%             
%% \pagedone
%% 
%%             \begin{eqnarray*}
%%                               & = & \ln(\frac{2}{\sqrt{2\pi}})
%%                                     + \lim_{n \to \infty}\frac{1}{4}
%%                                          (\frac{(2n+1)^2}{n} \\
%%                               &   &  + (2k + 1)\ln(n+k)
%%                                      - \frac{(n + k + 1)(2n + 1)}{(n + k)} \\
%%                               &   &  - (2k - 1)\ln(n-k)
%%                                      - \frac{(n - k + 1)(2n + 1)}{(n - k)}) \\
%%                               \\
%%                               & = & \ln(\frac{2}{\sqrt{2\pi}})
%%                                     + \lim_{n \to \infty}\frac{1}{4}
%%                                          (\frac{(2n+1)^2}{n} \\
%%                               &   &  - \frac{(n - k + 1)(2n + 1)}{(n - k)}
%%                                      - \frac{(n + k + 1)(2n + 1)}{(n + k)} \\
%%                               &   &  + (2k + 1)\ln(n+k) - (2k - 1)\ln(n-k)) \\
%%                               \\
%%                               & = & \ln(\frac{2}{\sqrt{2\pi}})
%%                                     + \lim_{n \to \infty}\frac{2n+1}{4}
%%                                          (\frac{(2n+1)}{n} \\
%%                               &   &  - \frac{n - k + 1}{(n - k)}
%%                                      - \frac{n + k + 1}{(n + k)}) \\
%%                               \\
%%                               & = & \ln(\frac{2}{\sqrt{2\pi}})
%%                                     + \lim_{n \to \infty}\frac{2n+1}{4}
%%                                       (\frac{(2n+1)(n-k)(n+k)}{n(n-k)(n+k)} \\
%%                               &   & - \frac{n(n-k+1)(n+k) + n(n-k)(n+k+1)}{n(n-k)(n+k)})
%%             \end{eqnarray*}
%%             
%% \pagedone
%% 
%%             \begin{eqnarray*}
%%                               & = & \ln(\frac{2}{\sqrt{2\pi}})
%%                                     + \lim_{n \to \infty}\frac{2n+1}{4}
%%                                       (\frac{(2n+1)(n^2 - k^2)}{n^3 - nk^2} \\
%%                               &   & - \frac{(n-k+1)(n^2+nk) + (n^2-nk)(n+k+1)}{n^3 - nk^2}) \\
%%                               \\
%%                               & = & \ln(\frac{2}{\sqrt{2\pi}})
%%                                     + \lim_{n \to \infty}\frac{2n+1}{4} \\
%%                               &   &   (\frac{2n^3 - 2nk^2 + n^2 - k^2}{n^3 - nk^2} \\
%%                               &   & - \frac{n^3 +n^2k - kn^2 -nk^2 +n^2 +nk}{n^3 - nk^2} \\
%%                               &   & - \frac{n^3 +n^2k + n^2 -n^2k -nk^2 -nk}{n^3 - nk^2}) \\
%%                               \\
%%                               & = & \ln(\frac{2}{\sqrt{2\pi}})
%%                                     + \lim_{n \to \infty}\frac{2n+1}{4}
%%                                     (\frac{- n^2 - k^2 -2nk}{n^3 - nk^2}) \\
%%                               \\
%%                               & = & \ln(\frac{2}{\sqrt{2\pi}}) - \frac{1}{2}
%%             \end{eqnarray*}
%%             
%%         