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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 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 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 initial value problem for the heat equation on the infinite interval is well-posed, that is, the solution to the partial differential equation
where the initial condition and the solution satisfy the following technical requirements:
exists for all time, is unique, and most importantly, can be represented as
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.
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.
Remark: The use of 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
with V (0,t) = 0, V (S,t) ~ S as S and
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:
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 - and S = Kex, and we will set
Remember, 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 Stochastic Calculus and Financial Applications by J. Michael Steele. Springer, New York, 2001 is appropriate here.
“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,) defined in terms of the old if we write the old V as a function of the new v and write the new 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.”
Following our advice, we first have to write
and the terminal condition
but V (S,T) = Kv(x, 0) so v(x, 0) = max(ex - 1, 0).
Now substitute all of the derivatives into the Black-Scholes equation to obtain:
Now the simplification begins:
What remains is the rescaled, constant coefficient equation:
We have made considerable progress, because
Instead, we will change the dependent variable scale yet again, by
where and are yet to be determined. Now using the product rule:
Put these into our constant coefficient partial differential equation, cancel the common factor of ex+ throughout and obtain:
Now gather like terms:
Then choose = -(k - 1)/2 so that the ux coefficient is 0, and then choose = 2 + (k - 1) - k = -(k + 1)2/4 so the u coefficient is likewise 0. With this choice, the equation is reduced to
We also need to transform the initial condition too. This would be
For future reference, we notice that this function is strictly positive when the argument x is strictly positive, that is u0(x) > 0 when x > 0, otherwise, u0(x) = 0 for x < 0.
Now we are in the final stage, since we are ready to apply the solution representation formula:
However, first we want to make a change of variable in the integration, by taking z = (s - x)/, (and thereby dz = (-1/) dx) so that the integration becomes:
We may as well only integrate over the domain where u0 > 0, that is for z > -x/. On that domain, u0 = e((k+1)/2)(x+z) - e((k-1)/2)(x+z) so we are down to:
Call the two integrals I1 and I2 respectively.
We will evaluate I1 ( 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
Now, change variables again on the integral, choosing y = z -(k + 1) so dy = dz, and all we need to change are the limits of integration:
The integral can be represented in terms of the cumulative distribution function of a normal random variable, typically denoted . That is,
where d1 = x/ + (k + 1) (note the use of the symmetry of the integral!) The calculation of I2 is identical, except that (k + 1) is replaced by (k - 1) throughout.
That, is the solution of the heat equation is
where d1 = x/ + (k + 1) and d2 = x/ + (k - 1).
Now, we must systematically unwind each of the changes of variables, from u, first v(x,) = e(-1/2)(k-1)x-(1/4)(k+1)2u(x,). (Notice how many of the exponentials neatly combine and cancel!) Then put x = log(S/K), = (1/2)2(T - t) and V (S,t) = Kv(x,).
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 :
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.
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 . The value of the call option at maturity plotted over a range of stock prices 70 < S < 130 surrounding the strike price is illustrated below:
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 < S < 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).
Notice a couple of trends in the value from this graph:
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:
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