APPENDIX B

Solving for the Green’s Function of the Black-Scholes Equation

We want to solve

images

We can set the reference coordinates to zero because they can be added later by subtracting from the solution’s arguments. This is because the differential equation does not contain any explicit functions of the coordinates—that is, it is translation invarian. Thus

images

Take the 2-D Fourier transform of the equation, that is, (x, t) images (k, ω),

images

Then

images

The residue theorem from complex analysis means that the result is only nonzero for closing the integral over ω in the lower half of the complex ω-plane. This happens only if t < 0. Thus

images

By completing the square (recalling t < 0), we can factor out a function of x and t times an integral,

images

Finally substitute ttT, and xxxT to get

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