Friday, February 15, 2019

Transformation from the Black-Scholes differential equation to the diffusion equation - and back


I know the derivation of the Black-Scholes differential equation and I understand (most of) the solution of the diffusion equation. What I am missing is the transformation from the Black-Scholes differential equation to the diffusion equation (with all the conditions) and back to the original problem.



All the transformations I have seen so far are not very clear or technically demanding (at least by my standards).


My question:
Could you provide me references for a very easily understood, step-by-step solution?



Answer



One starts with the Black-Scholes equation Ct+12σ2S22CS2+rSCSrC=0,(1) supplemented with the terminal and boundary conditions (in the case of a European call) C(S,T)=max C(0,t)=0,\qquad C(S,t)\sim S\ \mbox{ as } S\to\infty.\qquad\qquad\qquad\qquad\qquad\qquad The option value C(S,t) is defined over the domain 0< S < \infty, 0\leq t\leq T.


Step 1. The equation can be rewritten in the equivalent form \frac{\partial C}{\partial t}+\frac{1}{2}\sigma^2\left(S\frac{\partial }{\partial S}\right)^2C+\left(r-\frac{1}{2}\sigma^2\right)S\frac{\partial C}{\partial S}-rC=0. The change of independent variables S=e^y,\qquad t=T-\tau results in S\frac{\partial }{\partial S}\to\frac{\partial}{\partial y},\qquad \frac{\partial}{\partial t}\to - \frac{\partial}{\partial \tau}, so one gets the constant coefficient equation \frac{\partial C}{\partial \tau}-\frac{1}{2}\sigma^2\frac{\partial^2 C}{\partial y^2}-\left(r-\frac{1}{2}\sigma^2\right)\frac{\partial C}{\partial y}+rC=0.\qquad\qquad\qquad(3)


Step 2. If we replace C(y,\tau) in equation (3) with u=e^{r\tau}C, we will obtain that \frac{\partial u}{\partial \tau}-\frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial y^2}-\left(r-\frac{1}{2}\sigma^2\right)\frac{\partial u}{\partial y}=0.


Step 3. Finally, the substitution x=y+(r-\sigma^2/2)\tau allows us to eliminate the first order term and to reduce the preceding equation to the form \frac{\partial u}{\partial \tau}=\frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2} which is the standard heat equation. The function u(x,\tau) is defined for -\infty < x < \infty, 0\leq\tau\leq T. The terminal condition (2) turns into the initial condition u(x,0)=u_0(x)=\max(e^{\frac{1}{2}(a+1)x}-e^{\frac{1}{2}(a-1)x},0), where a=2r/\sigma^2. The solution of the heat equation is given by the well-known formula u(x,\tau)=\frac{1}{\sigma\sqrt{2\pi \tau}}\int_{-\infty}^{\infty} u_0(s)\exp\left(-\frac{(x-s)^2}{2\sigma^2 \tau}\right)ds.


Now, if we evaluate the integral with our specific function u_0 and return to the old variables (x,\tau,u)\to(S,t,C), we will arrive at the usual Black–Merton-Scholes formula for the value of a European call. The details of the calculation can be found e.g. in The Mathematics of Financial Derivatives by Wilmott, Howison, and Dewynne (see Section 5.4).


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