Why Backward Black–Scholes Equations Require Positive Diffusion

# Finite Difference Application


# Finite Difference Application












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 $a > 0$ ?

## Answer by Frido (score 1)

https://quant.stackexchange.com/a/81116

Actually quite an interesting question. Not sure if still relevant for the OP, but if the diffusion coefficient ($a(S,t)$ in the notation of the OP) is negative you end up with an ill-posed problem. See for instance these threads:

https://math.stackexchange.com/questions/952978/the-backwards-heat-equation-is-not-well-posed

https://math.stackexchange.com/questions/1674674/ill-posedness-and-well-posedness

Note, in the threads above the $t$ is actually $\tau := T-t$ in the OP's PDE.

Aside from the mathematical arguments, as already alluded to in the comment, there are physical arguments. In finance a complex-valued volatility makes little sense. In physics the $V$ in the OP's equation is for instance a probability density, which needs to decay as $\log S \to \pm \infty$, also known as normalizability. This will not be the case if the diffusion coefficient is negative.

## Answer by Andrea (score 0)

https://quant.stackexchange.com/a/81113

The answer to me is the fact that in the "natural" direction, the BS equation increases the smoothness of the solution.

Let's say one is given the BS price of a digital option at time 0 (or any negative time) and we know that the option expires at time 10. $V(S,10)$ is not even continuous.

Then by trial and error, one can go the "wrong" way round from $T=0$ to $T=10$, at which point the solution is the (digital) payoff.

One cannot go past this point because, assuming one could, then by executing the PDE the "natural" way round, from $T=11$ to $T=10$, one would generate the known digital payoff at $T=10$.

But this is impossible, because the solution of the equation is $C^2$ ($t<10$) and the payoff is not, so it cannot be generated from the same PDE started at some $T>10$.

So, I think you can try to solve it as an inverse problem, but up to a time point when the solution is not $C^2$, which means you have reached the payoff time.