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\HCode{
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\section*{Key Concepts}
\begin{enumerate}
\item
The Black-Scholes equation for the value of a European call
option on a security can be solved by judicious changes of
variables that reduce the equation to the heat equation, for
which we can write down the solution.
\item
An example of how to choose and execute changes of variables to
solve a partial differential equation.
\end{enumerate}
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\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 \textbf{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.
\end{enumerate}
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\section*{Mathematical Ideas}
\subsection*{Analytic Solution of the Black-Scholes Equation }
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.
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 infinite interval is well-posed},
that is, the solution to the partial differential equation
$$
\frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial x^2}
\qquad -\infty < x < \infty, \quad \tau > 0
$$
with
$$
u(x,0) = u_0(x)
$$
where 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}
exists for all time, is unique, and most importantly, can be represented
as
$$
u(x, \tau) = \frac{1}{2\sqrt{\pi \tau}} \int_{-\infty}^{\infty} u_0(s)
e^{-(x-s)^2/4\tau} \; ds
$$
\emph{Remark:} This solution can derived in several different ways, I
think 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.
\emph{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.
\emph{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.
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
$$
V(S,T) = \max( S- K, 0).
$$
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, but 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.
\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 will eliminate each and every one of these objections with a suitable
change of variables. The plan is to change variables to reduce the
Black-Scholes terminal value problem to the heat equation, and then to
use the known solution of the heat equation to represent the solution,
and change variables back. This is a standard technique of solution in
partial differential equations, and none of the transformations we are
making are strange, unmotivated, or unknown.
First we will take $ t = T - \frac{\tau}{ (1/2)\sigma^2} $ and $ S = K e^x
$, and we will 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. 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 page 186 of \emph{Stochastic
Calculus and Financial Applications} by J. Michael Steele. Springer,
New York, 2001 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 our advice, we first have to write
$$
\tau = (1/2)\sigma^2( T - t)
$$
and
$$
x = \log(S/K)
$$
so
$$
\frac{\partial V}{\partial t} = K \frac{\partial v}{\partial \tau}
\cdot \frac{ d\tau}{dt} = K \frac{\partial v}{\partial \tau} \cdot
\frac{-\sigma^2}{2}
$$
and
$$
\frac{\partial V}{\partial S} = K \frac{\partial v}{\partial x}
\cdot \frac{dx}{dS} = K \frac{\partial v}{\partial x}\cdot\frac{1}{S}
$$
and
\begin{eqnarray*}
\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{eqnarray*}
and the terminal condition
$$
V(S,T) = \max(S-K, 0) = \max(K e^x - K, 0)
$$
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 the simplification begins:
\begin{enumerate}
\item
Factor 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
another (more or less hidden) rescaled time to expiry $ (1/2)
\sigma^2 T $ , not the original $ 4 $ dimensioned quantities,
namely $ 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 is now constant coefficient. In principle, we
could solve the equation directly now,
\end{enumerate}
Instead, we will change 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. Now using the
product rule:
$$
v_\tau = \beta e^{\alpha x + \beta \tau} u(x,\tau) + e^{\alpha x +
\beta \tau} u_\tau
$$
and
$$
v_x = \alpha e^{\alpha x + \beta \tau} u(x,\tau) + e^{\alpha x +
\beta \tau} u_x
$$
and
$$
v_{xx} = \alpha^2 e^{\alpha x + \beta \tau} u(x,\tau) + 2\alpha e^{\alpha
x + \beta \tau} u_x + e^{\alpha x + \beta \tau} u_{xx}
$$
Put these into our constant coefficient partial differential equation,
cancel the common factor of $ e^{\alpha x + \beta \tau} $ throughout and
obtain:
$$
\beta u + u_\tau = \alpha^2 u + 2\alpha u_x + u_{xx} + (k - 1) (\alpha
u + u_x) - k u
$$
Now gather like terms:
$$
u_\tau = u_{xx} + [ 2\alpha + (k-1)] u_x + [ \alpha^2 + (k-1) \alpha
- k - \beta] u.
$$
Then 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 also need to transform the initial condition too. This would be
\begin{eqnarray*}
u(x,0) &=& e^{-(-(k-1)/2)x - (-(k+1)^2/4)0} v(x,0) \\
&=& e^{((k-1)/2)x} \max( e^x - 1, 0) \\
&=& \max( e^{((k+1)/2)x} - e^{((k-1)/2)x}, 0 ).
\end{eqnarray*}
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 $.
Now we are in the final stage, since we are ready to apply the solution
representation formula:
$$
u(x, \tau) = \frac{1}{2\sqrt{\pi \tau}} \int_{-\infty}^{\infty} u_0(s)
e^{-(x-s)^2/4\tau} \; ds.
$$
However, first we want to make a change of variable in the integration,
by taking $ z = (s-x)/\sqrt{2\tau} $, (and thereby $ dz = (-1/\sqrt{2\tau})
\; dx $) so that the integration becomes:
$$
u(x,\tau) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} u_0(z
\sqrt{2\tau} + x) e^{-z^2/2} \; dz
$$
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^{((k+1)/2)(x + z\sqrt
{2\tau})} - e^{((k-1)/2)(x + z\sqrt{2\tau}) } $ so we are down to:
$$
\frac{1}{\sqrt{2\pi}} \int_{-x/\sqrt{2\tau}}^{\infty} e^{\frac{k+1}{2}
(x + z\sqrt{2\tau})} e^{-z^2/2} \; dz - \frac{1}{\sqrt{2\pi}} \int_{-x/\sqrt
{2\tau}}^{\infty} e^{\frac{k-1}{2}(x + z\sqrt{2\tau})} e^{-z^2/2} \;
dz
$$
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, just completing the square in the exponent yields a standard,
tabulated integral. The exponent is
\begin{eqnarray*}
((k+1)/2)(x + z\sqrt{2\tau}) - z^2/2 &=& (-1/2)( z^2 -\sqrt{2\tau}(k+1)z)
+ ((k+1)/2) x \\
& = & (-1/2)(z^2 - \sqrt{2\tau}(k+1)z + \tau(k+1)^2/2) + ((k+1)/2)x
+ \tau(k+1)^2/4 \\
& = & (-1/2)(z - \sqrt{\tau/2}(k+1))^2 + (k+1)x/2 +\tau (k+1)^2/4
\end{eqnarray*}
Therefore
\[
\frac{1}{\sqrt{2\pi}} \int_{-x/\sqrt{2\tau}}^{\infty} e^{\frac{k+1}{2}
(x + z\sqrt{2\tau})} e^{-z^2/2} \; dz = \frac{ e^{(k+1)x/2 +\tau (k+1)^2/4}}
{\sqrt{2\pi}} \int_{-x/\sqrt{2\tau}}^\infty e^{\frac{-1}{2} (z -
\sqrt{\tau/2}(k+1))^2} \; dz
\]
Now, change variables again on the integral, choosing $ y = z - \sqrt{\tau/2}
(k+1) $ so $ dy = dz $, and all we need to change are the limits of
integration:
\[
\frac{ e^{ (k+1)x/2 +\tau (k+1)^2/4} }{ \sqrt{2\pi}} \int_{-x/\sqrt{2\tau}
- \sqrt{\tau/2}(k+1)}^\infty e^{(-1/2)y^2} \; dz
\]
The integral can be represented in terms of the cumulative distribution
function of a normal random variable, typically denoted $ \Phi $. That
is,
\[
\Phi(d) = (1/\sqrt{2\pi})\int_{-\infty}^{d} e^{-y^2/2} \; dy
\]
so
\[
I_1 = e^{ (k+1)x/2 +\tau (k+1)^2/4} \Phi(d_1)
\]
where $ d_1 = x/\sqrt{2\tau} + \sqrt{\tau/2}(k+1) $ (note the use of the
symmetry of the integral!) The calculation of $ I_2 $ is identical,
except that $ (k+1) $ is replaced by $ (k-1) $ throughout.
That, is the solution of the heat equation is
\[
u(x,\tau) = e^{ (k+1)x/2 +\tau (k+1)^2/4} \Phi(d_1) - e^{ (k-1)x/2 +\tau
(k-1)^2/4} \Phi(d_2)
\]
where $ d_1 = x/\sqrt{2\tau} + \sqrt{\tau/2}(k+1) $ and
$ d_2 = x/\sqrt{2\tau} + \sqrt{\tau/2}(k-1). $
Now, we must systematically unwind each of the changes of variables,
from $ u $, first $ v(x,\tau) = e^{(-1/2)(k-1)x - (1/4)(k+1)^2 \tau} u(x,\tau)
$. (Notice how many of the exponentials neatly combine and cancel!) Then put $
x = \log(S/K) $, $ \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 $:
\[
V(S,t) = S \Phi\left( \frac{\log(S/K) + ( r + \sigma^2/2)(T-t)}{\sigma
\sqrt{T-t}} \right) - K e^{-r(T-t)} \Phi\left( \frac{\log(S/K) + ( r
- \sigma^2/2)(T-t)}{\sigma \sqrt{T-t}} \right)
\]
Note, usually one doesn't see the solution as this full closed form
solution. Instead, 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.
\subsection*{Graphical Solution of the Black-Scholes Equation}
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 below:
\Picture[Value of the call option at maturity]{calloption.jpg}
Now, 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).
\Picture[Value of the call option at various times]{optionvalueS.jpg}
Notice a couple of trends in the value from this graph:
\begin{enumerate}
\item As the stock price increases, for a fixed time
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 can also plot the solution of the Black-Scholes equation as a
function of security price and the time to expiration as value
surface:
\Picture[Value surface from the Black-Scholes formula]{valuesurface.jpg}
\HCode{
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\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}
\HCode{
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\section*{Reading Suggestion:}
\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}
\HCode{
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\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, \EndLink 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 \EndLink Applied mathematicians transform the
Black-Scholes Equation into a simpler forward form and solve
it.It gives some examples and maple commands, Submitted by Chun
Fan, Nov. 12, 2002.
\item
\Link{http://www.blobek.com/black-scholes.html}{}\EndLink
An applet for calculating the option value based on
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 \EndLink A
sample spreadsheet based calculator for calculating the option
values, based on Black-Scholes model. Submitted by Yogesh
Makkar, November 19,2003
\end{enumerate}
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