We all know that the traditional BS equation is:

$$\frac{\partial \mathrm V}{ \partial \mathrm t } + \frac{1}{2}\sigma^{2} \mathrm S^{2} \frac{\partial^{2} \mathrm V}{\partial \mathrm S^2} + \mathrm r \mathrm S \frac{\partial \mathrm V}{\partial \mathrm S}\ - \mathrm r \mathrm V = 0$$

Suppose we want to write it is as:

$$\frac{\partial \mathrm V}{ \partial \mathrm t } + a(S, t)\frac{\partial^{2} \mathrm V}{\partial \mathrm S^2} + b(S, t) \frac{\partial \mathrm V}{\partial \mathrm S}\ + c(S, t)\mathrm V = 0$$

In order to emphasize the wide applicability of the finite difference methods.

Why do we have that the only constraint we must pose on the coefficients is that if we are solving a backward equation we must have $a > 0$ ?

  • $\begingroup$ a<0 requires you to be at-ease with the concept of negative variance (or, perhaps, complex-valued volatility). Neither has a clear physical meaning (at least to me!) $\endgroup$ Commented Nov 19, 2022 at 15:06


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