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\myheader \mytitle
\hr
\sectiontitle{Solution of the Black-Scholes Equation}
\hr
\usefirefox
\hr
\visual{Rating}{../../../../CommonInformation/Lessons/rating.png}
\section*{Rating} %one of
% Everyone: contains no mathematics.
% Student: contains scenes of mild algebra or calculus that may require guidance.
Mathematically Mature: may contain mathematics beyond calculus with
proofs. % Mathematicians Only: prolonged scenes of intense rigor.
\hr
\visual{QuestionofDay}{../../../../CommonInformation/Lessons/question_mark.png}
\section*{Question of the Day}
What is the solution method for the
Cauchy-Euler type of ordinary differential equation:
\[
x^2 \frac{d^2 v}{dx^2} + a x \frac{dv}{dx} + b v = 0\,?
\]
\hr
\visual{Key Concepts}{../../../../CommonInformation/Lessons/keyconcepts.png}
\section*{Key Concepts}
\begin{enumerate}
\item
We solve the Black-Scholes equation for the value of a European
call option on a security by judicious changes of variables that
reduce the equation to the heat equation. The heat equation has
a solution formula. Using the solution formula with the changes
of variables gives the solution to the Black-Scholes equation.
\item
Solving the Black-Scholes equation is an example of how to
choose and execute changes of variables to solve a partial
differential equation.
\end{enumerate}
\hr
\visual{Vocabulary}{../../../../CommonInformation/Lessons/vocabulary.png}
\section*{Vocabulary}
\begin{enumerate}
\item
A differential equation with auxiliary initial conditions and
boundary conditions, that is an initial value problem, is said
to be \defn{well-posed} if the solution exists, is unique, and
small changes in the equation parameters, the initial conditions
or the boundary conditions produce only small changes in the
solutions.%
\index{well-posed}
\end{enumerate}
\hr
\visual{Mathematical Ideas}{../../../../CommonInformation/Lessons/mathematicalideas.png}
\section*{Mathematical Ideas}
\subsection*{Conditions for Solution of the Black-Scholes Equation }
We have to start somewhere, and to avoid the problem of deriving
everything back to calculus, we will assert that the \emph{initial value
problem for the heat equation on the real line is well-posed}. That is,
consider the solution to the partial differential equation
\[
\frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial x^2}
\qquad -\infty < x < \infty, \quad \tau > 0.
\] We will take the initial condition
\[
u(x,0) = u_0(x).
\]%
\index{heat equation}
\index{heat equation!initial value problem}
We will assume the initial condition and the solution satisfy the
following technical requirements:
\begin{enumerate}
\item
\( u_0(x) \) has no more than a finite number of discontinuities
of the jump kind,
\item
\( \lim_{|x| \to \infty} u_0(x) e^{-a x^2} = 0 \) for any \( a
>0 \),
\item
\( \lim_{|x| \to \infty} u(x, \tau) e^{-a x^2} = 0 \) for any \(
a >0 \).
\end{enumerate}
Under these mild assumptions, the solution exists for all time and is
unique. Most importantly, the solution is represented as
\[
u(x, \tau) = \frac{1}{2\sqrt{\pi \tau}} \int_{-\infty}^{\infty} u_0(s)
e^{-(x-s)^2/4\tau} \, ds
\]
\begin{remark}
This solution can derived in several different ways, the easiest way
is to use Fourier transforms. The derivation of this solution
representation is standard in any course or book on partial
differential equations.
\end{remark}
\begin{remark}
Mathematically, the conditions above are unnecessarily restrictive,
and can be considerably weakened. However, they will be more than
sufficient for all practical situations we encounter in mathematical
finance.
\end{remark}
\begin{remark}
The use of \( \tau \) for the time variable (instead of the more
natural \( t \)) is to avoid a conflict of notation in the several
changes of variables we will soon have to make.
\end{remark}
The Black-Scholes terminal value problem for the value \( V(S,t) \) of a
European call option on a security with price \( S \) at time \( t \) is
\[
\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2
V}{\partial S^2} + r S \frac{\partial V}{\partial S} - r V = 0
\] with \( V(0,t) = 0 \), \( V(S,t) \sim S \) as \( S \to \infty \) and
\[
V(S,T) = \max( S- K, 0).
\]%
\index{Black-Scholes equation}
\index{Black-Scholes
equation!terminal value problem}
Note that this looks a little like the heat equation on the infinite
interval in that it has a first derivative of the unknown with respect
to time and the second derivative of the unknown with respect to the
other (space) variable. On the other hand, notice:
\begin{enumerate}
\item
Each time the unknown is differentiated with respect to \( S \),
it also multiplied by the independent variable \( S \), so the
equation is not a constant coefficient equation.
\item
There is a first derivative of \( V \) with respect to \( S \)
in the equation.
\item
There is a zero-th order term \( V \) in the equation.
\item
The sign on the second derivative is the opposite of the heat
equation form, so the equation is of backward parabolic form.
\item
The data of the problem is given at the final time \( T \)
instead of the initial time \( 0 \), consistent with the
backward parabolic form of the equation.
\item
There is a \emph{boundary condition} \( V(0,t) = 0 \) specifying
the value of the solution at one sensible boundary of the
problem. The boundary is sensible since security values must
only be zero or positive. This boundary condition says that any
time the security value is 0, then the call value (with strike
price \( K \)) is also worth 0.%
\index{Black-Scholes
equation!boundary condition}
\item
There is another boundary condition \( V(S,t) \sim S \), as \( S
\to \infty \), but although this is financially sensible, (it
says that for very large security prices, the call value with
strike price \( K \) is approximately \( S \)) it is more in the
nature of a technical condition, and we will ignore it without
consequence.
\end{enumerate}
We eliminate each objection with a suitable change of variables. The
plan is to change variables to reduce the Black-Scholes terminal value
problem to the heat equation, then to use the known solution of the heat
equation to represent the solution, and finally change variables back.
This is a standard solution technique in partial differential equations.
All the transformations are standard, well-motivated, and well known.
\subsection*{Solution of the Black-Scholes Equation }
First we take \( t = T - \frac{\tau}{ (1/2)\sigma^2} \) and \( S = K e^x
\), and we set
\[
V(S,t) = K v(x,\tau).
\] Remember, \( \sigma \) is the volatility, \( r \) is the interest
rate on a risk-free bond, and \( K \) is the strike price. In the
changes of variables above, the choice for \( t \) reverses the sense of
time, changing the problem from backward parabolic to forward parabolic.
The choice for \( S \) is a well-known transformation based on
experience with the Euler equidimensional equation in differential
equations.%
\index{Euler equidimensional equation}
\index{Cauchy-Euler
equation|see{Euler equidimensional equation}}
In addition, the variables have been carefully scaled so as to make the
transformed equation expressed in dimensionless quantities. All of these
techniques are standard and are covered in most courses and books on
partial differential equations and applied mathematics.
Some extremely wise advice adapted from \emph{Stochastic Calculus and
Financial Applications} by J. Michael Steele,
\cite[page 186]{steele01}, is appropriate here.
\begin{quotation}
``There is nothing particularly difficult about changing variables
and transforming one equation to another, but there is an element of
tedium and complexity that slows us down. There is no universal
remedy for this molasses effect, but the calculations do seem to go
more quickly if one follows a well-defined plan. If we know that \(
V(S,t) \) satisfies an equation (like the Black-Scholes equation) we
are guaranteed that we can make good use of the equation in the
derivation of the equation for a new function \( v(x, \tau) \)
defined in terms of the old if we write the old \( V \) as a
function of the new \( v \) and write the new \( \tau \) and \( x \)
as functions of the old \( t \) and \( S \). This order of things
puts everything in the direct line of fire of the chain rule; the
partial derivatives \( V_t \), \( V_S \) and \( V_{SS} \) are easy
to compute and at the end, the original equation stands ready for
immediate use.''
\end{quotation}
Following the advice, write
\[
\tau = (1/2)\sigma^2(T - t)
\] and
\[
x = \log\left(\frac{S}{K}\right).
\] The first derivatives are
\[
\frac{\partial V}{\partial t} = K \frac{\partial v}{\partial \tau}
\cdot \frac{ d\tau}{dt} = K \frac{\partial v}{\partial \tau} \cdot
\frac{-\sigma^2}{2}
\] and
\[
\frac{\partial V}{\partial S} = K \frac{\partial v}{\partial x}
\cdot \frac{dx}{dS} = K \frac{\partial v}{\partial x}\cdot\frac{1}{S}.
\] The second derivative is
\begin{align*}
\frac{\partial^2 V}{\partial S^2} &= \frac{\partial}{\partial S}
\left( \frac{\partial V}{\partial S} \right) \\
&= \frac{\partial}{\partial S} \left( K \frac{\partial v}{\partial x}
\frac{1}{S} \right) \\
&= K \frac{\partial v}{\partial x} \cdot \frac{-1}{S^2} + K \frac{\partial}
{\partial S} \left( \frac{\partial v}{\partial x} \right) \cdot
\frac{1}{S} \\
&= K \frac{\partial v}{\partial x} \cdot \frac{-1}{S^2} + K \frac{\partial}
{\partial x} \left( \frac{\partial v}{\partial x} \right) \cdot
\frac{dx}{dS} \cdot \frac{1}{S} \\
&= K \frac{\partial v}{\partial x} \cdot \frac{-1}{S^2} + K \frac{\partial^2
v}{\partial x^2} \cdot \frac{1}{S^2}.
\end{align*}
The terminal condition is
\[
V(S,T) = \max(S-K, 0) = \max(K e^x - K, 0)
\] but \( V(S,T) = K v(x,0) \) so \( v(x,0) = \max( e^x - 1, 0) \).
Now substitute all of the derivatives into the Black-Scholes equation to
obtain:
\[
K \frac{\partial v}{\partial \tau} \cdot \frac{-\sigma^2}{2} + \frac
{\sigma^2}{2} S^2 \left( K \frac{\partial v}{\partial x} \cdot \frac
{-1}{S^2} + K \frac{\partial^2 v}{\partial x^2} \cdot \frac{1}{S^2}
\right) + r S \left( K \frac{\partial v}{\partial x} \cdot \frac{1}{S}
\right) - r K v = 0.
\] Now begin the simplification:
\begin{enumerate}
\item
Isolate the common factor \( K \) and cancel.
\item
Transpose the \( \tau \)-derivative to the other side, and
divide through by \( \sigma^2/2 \)
\item
Rename the remaining constant \( r/(\sigma^2/2) \) as \( k \). \(
k \) measures the ratio between the risk-free interest rate and
the volatility.
\item
Cancel the \( S^2 \) terms in the second derivative.
\item
Cancel the \( S \) terms in the first derivative.
\item
Gather up like order terms.
\end{enumerate}
What remains is the rescaled, constant coefficient equation:
\[
\frac{\partial v}{\partial \tau} = \frac{\partial^2 v}{\partial x^2}
+ (k-1) \frac{\partial v}{\partial x} - k v.
\] We have made considerable progress, because
\begin{enumerate}
\item
Now there is only one dimensionless parameter \( k \) measuring
the risk-free interest rate as a multiple of the volatility and
a rescaled time to expiry \( (1/2) \sigma^2 T \), not the
original \( 4 \) dimensioned quantities \( K \), \( T \), \(
\sigma^2 \) and \( r \).
\item
The equation is defined on the interval \( -\infty < x < \infty \),
since this \( x \)-interval defines \( 0 < S < \infty \) through
the change of variables \( S = K e^x \).
\item
The equation now has constant coefficients.
\end{enumerate}
In principle, we could now solve the equation directly.
Instead, we will simplify further by changing the dependent variable
scale yet again, by
\[
v = e^{\alpha x + \beta \tau} u(x,\tau)
\] where \( \alpha \) and \( \beta \) are yet to be determined. Using
the product rule:
\[
v_\tau = \beta e^{\alpha x + \beta \tau} u + e^{\alpha x + \beta
\tau} u_\tau
\] and
\[
v_x = \alpha e^{\alpha x + \beta \tau} u + e^{\alpha x + \beta \tau}
u_x
\] and
\[
v_{xx} = \alpha^2 e^{\alpha x + \beta \tau} u + 2\alpha e^{\alpha x
+ \beta \tau} u_x + e^{\alpha x + \beta \tau} u_{xx}.
\] Put these into our constant coefficient partial differential
equation, cancel the common factor of \( e^{\alpha x + \beta \tau} \)
throughout and obtain:
\[
\beta u + u_\tau = \alpha^2 u + 2\alpha u_x + u_{xx} + (k - 1) (\alpha
u + u_x) - k u
\] Gather like terms:
\[
u_\tau = u_{xx} + [ 2\alpha + (k-1)] u_x + [ \alpha^2 + (k-1) \alpha
- k - \beta] u.
\] Choose \( \alpha = -(k-1)/2 \) so that the \( u_x \) coefficient is \(
0 \), and then choose \( \beta = \alpha^2 + (k-1) \alpha - k = -(k+1)^2/4
\) so the \( u \) coefficient is likewise \( 0 \). With this choice,
the equation is reduced to
\[
u_\tau = u_{xx}.
\] We need to transform the initial condition too. This transformation
is
\begin{align*}
u(x,0) &= e^{-(-(k-1)/2)x - (-(k+1)^2/4) \cdot 0} v(x,0) \\
&= e^{((k-1)/2)x} \max( e^x - 1, 0) \\
&= \max \left( e^{\left(\left(k+1\right)/2\right)x} - e^{\left(\left(k-1\right)/2\right)x}, 0 \right).
\end{align*}
For future reference, we notice that this function is strictly positive
when the argument \( x \) is strictly positive, that is \( u_0(x) > 0 \)
when \( x > 0 \), otherwise, \( u_0(x) = 0 \) for \( x \le 0 \).
We are in the final stage since we are ready to apply the heat-equation
solution representation formula:
\[
u(x, \tau) = \frac{1}{2\sqrt{\pi \tau}} \int_{-\infty}^{\infty} u_0(s)
e^{-\left(x-s\right)^2/4\tau} \, ds.
\] However, first we want to make a change of variable in the
integration, by taking \( z = \left(s-x\right)/\sqrt{2\tau} \), (and thereby \( dz
= (-1/\sqrt{2\tau}) \, dx \)) so that the integration becomes:
\[
u(x,\tau) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} u_0\left(z
\sqrt{2\tau} + x\right) e^{-z^2/2} \, dz.
\] We may as well only integrate over the domain where \( u_0 > 0 \),
that is for \( z > -x/\sqrt{2\tau} \). On that domain, \( u_0 = e^{\left(\left(k+1\right)/2\right)
\cdot \left(x + z\sqrt {2\tau}\right)} - e^{\left(\left(k-1\right)/2\right) \cdot \left(x + z\sqrt{2\tau}\right) } \)
so we are down to:
\[
\frac{1}{\sqrt{2\pi}} \int_{-x/\sqrt{2\tau}}^{\infty} e^{\frac{k+1}{2}
\left(x + z\sqrt{2\tau}\right)} e^{-z^2/2} \, dz - \frac{1}{\sqrt{2\pi}} \int_{-x/\sqrt
{2\tau}}^{\infty} e^{\frac{k-1}{2}\left(x + z\sqrt{2\tau}\right)} e^{-z^2/2} \,
dz
\] Call the two integrals \( I_1 \) and \( I_2 \) respectively.
We will evaluate \( I_1 \) ( the one with the \( k+1 \) term) first.
This is easy, completing the square in the exponent yields a standard,
tabulated integral. The exponent is
\begin{align*}
\left(\left(k+1\right)/2\right)\left(x + z\sqrt{2\tau}\right) - z^2/2 &= \left(-1/2\right)\left( z^2 -\sqrt{2\tau}\left(k+1\right)z\right)
+ \left(\left(k+1\right)/2\right) x \\
& = \left(-1/2\right)\left(z^2 - \sqrt{2\tau}\left(k+1\right)z + \tau\left(k+1\right)^2/2\right) + \left(\left(k+1\right)/2\right)x +
\tau\left(k+1\right)^2/4 \\
& = \left(-1/2\right)\left(z - \sqrt{\tau/2}\left(k+1\right)\right)^2 + \left(k+1\right)x/2 +\tau \left(k+1\right)^2/4.
\end{align*}
Therefore
\[
\frac{1}{\sqrt{2\pi}} \int_{-x/\sqrt{2\tau}}^{\infty} e^{\frac{k+1}{2}
\left(x + z\sqrt{2\tau}\right)} e^{-z^2/2} \, dz = \frac{ e^{\left(k+1\right)x/2 +\tau \left(k+1\right)^2/4}}
{\sqrt{2\pi}} \int_{-x/\sqrt{2\tau}}^\infty e^{\frac{-1}{2} \left(z -
\sqrt{\tau/2}\left(k+1\right)\right)^2} \, dz.
\] Now, change variables again on the integral, choosing \( y = z -
\sqrt{\tau/2} \left(k+1\right) \) so \( dy = dz \), and all we need to change are
the limits of integration:
\[
\frac{e^{ \left(k+1\right)x/2 +\tau \left(k+1\right)^2/4} }{ \sqrt{2\pi}} \int_{-x/\sqrt{2\tau}
- \sqrt{\tau/2}\left(k+1\right)}^\infty e^{\left(-1/2\right)y^2} \, dz.
\] The integral can be represented in terms of the cumulative
distribution function of a normal random variable, usually denoted \(
\Phi \). That is,
\[
\Phi(d) = (1/\sqrt{2\pi})\int_{-\infty}^{d} e^{-y^2/2} \, dy
\] so
\[
I_1 = e^{ \left(k+1\right)x/2 +\tau \left(k+1\right)^2/4} \Phi(d_1)
\] where \( d_1 = x/\sqrt{2\tau} + \sqrt{\tau/2}\left(k+1\right) \). Note the use
of the symmetry of the integral! The calculation of \( I_2 \) is
identical, except that \( \left(k+1\right) \) is replaced by \( \left(k-1\right) \)
throughout.
The solution of the transformed heat equation initial value problem is
\[
u(x,\tau) = e^{ \left(k+1\right)x/2 +\tau \left(k+1\right)^2/4} \Phi(d_1) - e^{ \left(k-1\right)x/2 +\tau
\left(k-1\right)^2/4} \Phi(d_2)
\] where \( d_1 = x/\sqrt{2\tau} + \sqrt{\tau/2}\left(k+1\right) \) and \( d_2 = x/\sqrt
{2\tau} + \sqrt{\tau/2}\left(k-1\right). \)
Now we must systematically unwind each of the changes of variables, from
\( u \). First, \( v(x,\tau) = e^{\left(-1/2\right)\left(k-1\right)x - \left(1/4\right)\left(k+1\right)^2 \tau} u(x,\tau)
\). Notice how many of the exponentials neatly combine and cancel!
Next put \( x = \log\left(S/K\right) \), \( \tau = (1/2)\sigma^2(T-t) \) and \( V(S,t)
= Kv(x,\tau) \).
The final solution is the Black-Scholes formula for the value of a
European call option at time \( T \) with strike price \( K \), if the
current time is \( t \) and the underlying security price is \( S \),
the risk-free interest rate is \( r \) and the volatility is \( \sigma \):
\[
V(S,t) = S \Phi\left( \frac{\log(S/K) + ( r + \sigma^2/2)(T-t)}{\sigma
\sqrt{T-t}} \right) - K e^{-r(T-t)} \Phi\left( \frac{\log(S/K) + ( r
- \sigma^2/2)(T-t)}{\sigma \sqrt{T-t}} \right).
\]
Usually one doesn't see the solution as this full closed form solution.
Most versions of the solution write intermediate steps in small pieces,
and then present the solution as an algorithm putting the pieces
together to obtain the final answer. Specifically,
let
\begin{align*}
d_1 &= \frac{\log(S/K) + ( r + \sigma^2/2)(T-t)}{\sigma \sqrt
{T-t}} \\
d_2 &= \frac{\log(S/K) + ( r - \sigma^2/2)(T-t)}{\sigma \sqrt
{T-t}}
\end{align*}
so that
\[
V_C(S, t) = S \cdot \Phi\left( d_1 \right)
- K e^{-r(T-t)} \cdot \Phi\left( d_2 \right).
\]
\index{Black-Scholes pricing formula!call option}
\subsection*{Solution of the Black-Scholes Equation Graphically}
Consider for purposes of graphical illustration the value of a call
option with strike price \( K = 100 \). The risk-free interest rate per
year, continuously compounded is 12\%, so \( r = 0.12 \), the time to
expiration is \( T = 1 \) measured in years, and the standard deviation
per year on the return of the stock, or the volatility is \( \sigma =
0.10 \). The value of the call option at maturity plotted over a range
of stock prices \( 70 \le S \le 130 \) surrounding the strike price is
illustrated in~%
\ref{fig:calloption}
\begin{figure}
\centering
\includegraphics{calloption.png}
\caption{Value of the call option at maturity%
\label{fig:calloption}}
\end{figure}
We use the Black-Scholes formula above to compute the value of the
option prior to expiration. With the same parameters as above the value
of the call option is plotted over a range of stock prices \( 70 \le S
\le 130 \) at time remaining to expiration \( t = 1 \) (red), \( t=0.8 \),
(orange), \( t=0.6 \) (yellow), \( t=0.4 \) (green), \( t = 0.2 \) (blue)
and at expiration \( t = 0 \) (black).
\begin{figure}
\centering
\includegraphics[scale=0.50]{optionvalueS.png}
\caption{Value of the call option at various times%
\label{fig:optionvalueS}}
\end{figure}
Using this graph notice two trends in the option value:
\begin{enumerate}
\item
For a fixed time, as the stock price increases the option value
increases,
\item
As the time to expiration decreases, for a fixed stock value
price the value of the option decreases to the value at
expiration.
\end{enumerate}
We predicted both trends from our intuitive analysis of options. The
Black-Scholes option pricing formula makes the intuition precise.
We can also plot the solution of the Black-Scholes equation as a
function of security price and the time to expiration as value surface:
\begin{figure}
\centering
\includegraphics{valuesurface.png}
\caption{Value surface from the Black-Scholes formula%
\label{fig:valuesurface}}
\end{figure}
This value surface shows both trends.
\subsection*{Sources}
This discussion is drawn from Section 4.2, pages 59--63; Section 4.3,
pages 66--69; Section 5.3, pages 75--76; and Section 5.4, pages 77--81
of \emph{The Mathematics of Financial Derivatives: A Student
Introduction} by P. Wilmott, S. Howison, J. Dewynne, Cambridge
University Press, Cambridge, 1995. Some ideas are also taken from
Chapter 11 of \emph{Stochastic Calculus and Financial Applications} by
J. Michael Steele, Springer, New York, 2001.
\nocite{wilmott95}
\nocite{steele01}
\hr
\visual{Problems to Work}{../../../../CommonInformation/Lessons/solveproblems.png}
\section*{Problems to Work for Understanding}
\begin{enumerate}
\item
Explicitly evaluate the integral \( I_2 \) in terms of the
c.d.f. \( \Phi \) and other elementary functions as was done for
the integral \( I_1 \).
%%\HCode{} Solution \HCode{}
\item
What is the price of a European call option on a
non-dividend-paying stock when the stock price is \$52, the
strike price is \$50, the risk-free interest rate is 12\% per
annum (compounded continuously), the volatility is 30\% per
annum, and the time to maturity is 3 months?
%% adapted from Hull, Problem 10.21, page 241.
%%\HCode{} Solution \HCode{}
\item
What is the price of a European call option on a non-dividend
paying stock when the stock price is \$30, the exercise price is
\$29, the risk-free interest rate is 5\%, the volatility is 25\%
per annum, and the time to maturity is 4 months?
% adapted from Hull, Problem 10.23 (a) page 241.
%%\HCode{} Solution \HCode{}
\item
Show that the Black-Scholes formula for the price of a call
option tends to \( \max( S-K, 0) \) as \( t \to T \).
%%\HCode{} Solution \HCode{}
\end{enumerate}
\hr
\visual{Books}{../../../../CommonInformation/Lessons/books.png}
\section*{Reading Suggestion:}
\bibliography{../../../../CommonInformation/bibliography}
% \begin{enumerate}
% \item
% Section 4.2, pages 59--63; Section 4.3, pages 66--69; Section
% 5.3, pages 75--76; and Section 5.4, pages 77--81 of \emph{The
% Mathematics of Financial Derivatives: A Student Introduction }
% by P. Wilmott, S. Howison, J. Dewynne, Cambridge University
% Press, Cambridge, 1995.
% \item
% Chapter 11 of \emph{Stochastic Calculus and Financial
% Applications } by J. Michael Steele. Springer, New York, 2001.
% \end{enumerate}
\hr
\visual{Links}{../../../../CommonInformation/Lessons/chainlink.png}
\section*{Outside Readings and Links:}
\begin{enumerate}
\item
\link{http://www.cs.cornell.edu/Info/Courses/Spring-98/CS522/content/lecture2.math.pdf}
{Cornell University, Department of Computer Science, Prof. T.
Coleman Rhodes and Prof. R. Jarrow} Numerical Solution of
Black-Scholes Equation, Submitted by Chun Fan, Nov. 12, 2002.
\item
\link{http://http://www.maths.monash.edu.au/mth3251/Lectures/NumericalSol.pdf}
{Monash University, Department of Mathematical Science, Eric. W.
Chu} This link gives some examples and maple commands, Submitted
by Chun Fan, Nov. 12, 2002.
\item
\link{http://www.blobek.com/black-scholes.html}{An applet for
calculating the option value} based on the Black-Scholes model.
Also contains tips on options, business news and literature on
options. Submitted by Yogesh Makkar, November 19, 2003.
\item
\link{www.xleverywhere.com/samples/bs/bs.htm}{ExcelEverywhere},
a commercial application for spreadsheets on the Web. A sample
spreadsheet based calculator for calculating the option values,
based on Black-Scholes model. Submitted by Yogesh Makkar,
November 19,2003
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