3
$\begingroup$

By the Black-Scholes operator I mean the following.

$$L_{BS}u(x) = \frac{1}{2}\sigma^2x^2\frac{\partial^2}{\partial x^2}u(x) + rx\frac{\partial}{\partial x}u(x) - ru(x)$$

Obviously, the domain of $L_{BS}$ must be a subset of twice continuously differentiable functions on some interval but making a definitive statement about the domain requires something specific about the application at hand (option pricing in this instance) and knowledge about the origin of the operator. So what is what I am looking for.

Feynman-Kac says something about which solutions of the Black-Scholes PDE satisfy the pricing formula, that is in the form of conditional expectation. The solution class is the set of twice continuously differentiable functions that are continuous at the boundary and whose growth is bounded by a function of the form $e^{\alpha x^2}$.

There is also the uniqueness class of the Black-Scholes PDE, which has a similar growth characteristic.

This paper mentions the Schwartz space as the domain of Black-Scholes operator but I have never seen this anywhere else.

I am looking for an answer that summarizes all the considerations that go into this, preferably with references. Thank you.

$\endgroup$
2
$\begingroup$

The domain would be twice differentiable in S, once differentiable in t functions of S and t defined on the Cartesian product of S in (0,infty) and t in (0,T) for finite T >0. To obtain a unique solution when boundary conditions are added, one must curtail the growth as S goes to infinity by adapting the corresponding condition for uniqueness of solutions to the heat equation.

$\endgroup$
  • $\begingroup$ Thank you very much for the explanation. What brought me to this question is a statement I read in some notes that the BS operator does not have an eigenvalue with positive real part. But if the domain is as you described then I cannot figure out (prove) why the BS operator should have this property. $\endgroup$ – Calculon May 29 '18 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.