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(SK,0),(2)
C(0,t)=0,C(S,t)S  as S.
The option value C(S,t) is defined over the domain 0<S<, 0tT.


Step 1. The equation can be rewritten in the equivalent form Ct+12σ2(SS)2C+(r12σ2)SCSrC=0.

The change of independent variables S=ey,t=Tτ
results in SSy,tτ,
so one gets the constant coefficient equation Cτ12σ22Cy2(r12σ2)Cy+rC=0.(3)


Step 2. If we replace C(y,τ) in equation (3) with u=erτC, we will obtain that uτ12σ22uy2(r12σ2)uy=0.


Step 3. Finally, the substitution x=y+(rσ2/2)τ allows us to eliminate the first order term and to reduce the preceding equation to the form uτ=12σ22ux2

which is the standard heat equation. The function u(x,τ) is defined for <x<, 0τT. The terminal condition (2) turns into the initial condition u(x,0)=u0(x)=max(e12(a+1)xe12(a1)x,0),
where a=2r/σ2. The solution of the heat equation is given by the well-known formula u(x,τ)=1σ2πτu0(s)exp((xs)22σ2τ)ds.


Now, if we evaluate the integral with our specific function u0 and return to the old variables (x,τ,u)(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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