Partial Integro-Differential Equation (PIDE), Merton Jump Diffusion
Posted: December 29th, 2020, 7:51 pm
The standard PIDE (before it is subjected to the 'usual' independent variable transformation and domain truncation) is
(A) \[\frac{\partial V}{\partial \tau} = \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + (r-\lambda k) S \frac{\partial V}{\partial S} - (r+\lambda) V +\lambda\int_{0}^{\infty} V(S\eta)\Gamma_\delta(\eta)\,d\eta = 0, \]
A big title and a big equation!
There's a few FDM schemes that are variations on a theme 1) [$]x = log(S)[$] and [$]y = log(\eta)[$] and 2) truncation of both the PDE domain and the integral domain and 3) usual suspect boundary conditions. Everyone does this but is very crude.
My question now is: not using (1), (2), (3) but transforming across the board [$]S,\eta = (0,\infty)[$] to [$]x,y = (0,1)[$] by [$]x = S/(S+1)[$], [$]y = \eta/(\eta + 1)[$].
Now I go from a pide on [$](0,\infty)[$] to [$](0,1)[$] for both the PDE and IE terms. NO errors incurred.
Question; Is this a good starting point, numerically? Is there a (magical) reason why this might not work?
// A '42' question: why do most articles go for tricks (1), (2), (3) (saying it's handy is not an answer). They work in practice but do they work in theory?
(A) \[\frac{\partial V}{\partial \tau} = \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + (r-\lambda k) S \frac{\partial V}{\partial S} - (r+\lambda) V +\lambda\int_{0}^{\infty} V(S\eta)\Gamma_\delta(\eta)\,d\eta = 0, \]
A big title and a big equation!
There's a few FDM schemes that are variations on a theme 1) [$]x = log(S)[$] and [$]y = log(\eta)[$] and 2) truncation of both the PDE domain and the integral domain and 3) usual suspect boundary conditions. Everyone does this but is very crude.
My question now is: not using (1), (2), (3) but transforming across the board [$]S,\eta = (0,\infty)[$] to [$]x,y = (0,1)[$] by [$]x = S/(S+1)[$], [$]y = \eta/(\eta + 1)[$].
Now I go from a pide on [$](0,\infty)[$] to [$](0,1)[$] for both the PDE and IE terms. NO errors incurred.
Question; Is this a good starting point, numerically? Is there a (magical) reason why this might not work?
// A '42' question: why do most articles go for tricks (1), (2), (3) (saying it's handy is not an answer). They work in practice but do they work in theory?