Numerical Quadrature for a Lévy PIDE Integral
Summary
The document presents a spatial discretization approach for the jump integral in a pricing integro-differential equation with an exponentially damped Lévy density. It first changes variables from the asset price to its logarithm, turning multiplicative jumps into shifts of the log-price grid. Near zero, where the density has a singularity, a Taylor expansion approximates the payoff-function difference by its first derivative; integrating this term against the density gives a finite expression over the first grid cell.
For intervals away from zero, the response linearly interpolates the function between neighboring grid points and integrates the resulting terms against the Lévy kernel. This reduces the computation to exponential-integral terms such as the integral of a damped reciprocal over each cell, for which it notes fast methods are available. The exposition sketches quadrature rather than comparing methods or providing a complete implementation, and it focuses on the positive-jump contribution; the negative side requires corresponding treatment.
Key ideas
- Changing to log-price coordinates converts multiplicative jumps into additive shifts.
- Near the singularity, a Taylor expansion isolates a finite first-derivative contribution.
- Away from zero, piecewise linear interpolation approximates the shifted function on each grid cell.
- The cell integrals reduce partly to damped reciprocal integrals that can be evaluated with specialized methods.
- The presented derivation covers the positive-jump side and leaves implementation details open.
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# Numerical Solutions for PIDE
# Numerical Solutions for PIDE
I want to solve an exotic options of PIDE by Numerical Methods.I just focus on the integral part of PIDE and want to underestand some tips on numerical solution of how to numerically solve it. Exactly I just consider $$\int_{-\infty}^{\infty}(F(e^yS_{t^{-}},t)-F(S_{t^{-}},t))k(y)dy$$ I am going to solve numerically for the given Lévy density $$k(y)=\frac{e^{-\lambda_p y}}{v\,y}1_{y>0}+\frac{e^{-\lambda_n |y|}}{v\,|y|}1_{y<0}$$ Thanks.
## Answer by user16891 (score 2)
https://quant.stackexchange.com/a/18994
Without loss of generality, we can assume $y>0$.let $x =\ln S_t$ and defining $\tilde{F}(x, t)=F(S, t)$ we have $$\int_{0}^{\infty}(\,{F}(e^yS_{t^{-}},t)-{F}(S_{t^{-}},t)\,)k(y)\,dy=\int_{0}^{\infty}(\,\tilde F(x+y,t)-\tilde F(x,t)\,)k(y)\,dy$$ Beginning,it is natural to look at the following interval first $$\int_{0}^{\Delta x}(\,\tilde{F}(x+y,t)-\tilde{F}(x,t)\,)\,k(y)dy$$ for some small $\Delta x > 0$. Using Taylor expansion, we can write $\tilde {F} (x + y, t)$ as follows: $$\tilde{F}(x+y,t)=\tilde{F}(x,t)+y\frac{\partial\tilde{F}}{\partial x}(x,t)+\frac{1}{2}y^2\frac{\partial^2\tilde{F}}{\partial x^2}(\xi,t)$$ or equivalently $$\tilde{F}(x+y,t)-\tilde{F}(x,t)=y\frac{\partial\tilde{F}}{\partial x}(x,t)+O(y^2)$$ Substituting and we get \begin{align} &\int_{0}^{\Delta x}(\,\tilde{F}(x+y,t)-\tilde{F}(x,t)\,)\,k(y)dy=\int_{0}^{\Delta x}y\frac{\partial\tilde{F}}{\partial x}(x,t)\,k(y)dy\\ &\\ &\hspace{7.25cm}=\frac{\partial\tilde{F}}{\partial x}(x,t) \int_{0}^{\Delta x}y\,\frac{e^{-\lambda_p y}}{v\,y}1_{y>0}\,dy\\ &\\ &\hspace{7.25cm}=\frac{1}{v}\frac{\partial\tilde{F}}{\partial x}(x,t) \int_{0}^{\Delta x}e^{-\lambda_p y}\,dy\\ &\\ &\hspace{7.25cm}=\frac{1}{v\,\lambda_p}\frac{\partial\tilde{F}}{\partial x}(x,t)(1-e^{-\lambda_p\Delta x}) \\ \end{align} Away from zero, we can look at some arbitrary small subinterval. For $y\in (k\Delta x\,, (k + 1)\Delta x)$, forsome integer k we should evaluate the following integral $$\int_{k\Delta x}^{(k + 1)\Delta x}(\,\tilde{F}(x+y,t)-\tilde{F}(x,t)\,)\,k(y)dy$$ Using linear approximation yields we can write $\tilde{F}(x+y,t)$ on $(k\Delta x \, (k + 1)\Delta x)$ as follow $$\tilde{F}(x+y,t)=\tilde{F}(x+k\Delta x,t)+(y-k\Delta x)\frac{\tilde{F}(x+(k+1)\Delta x,t)-\tilde{F}(x+k\Delta x,t)}{\Delta x}+O(\Delta x^2)$$ Substituting it would yield \begin{align} &\int_{k\Delta x}^{(k + 1)\Delta x}\left[\,\tilde{F}(x+k\Delta x,t)+(y-k\Delta x)\frac{\tilde{F}(x+(k+1)\Delta x,t)-\tilde{F}(x+k\Delta x,t)}{\Delta x}-\tilde{F}(x,t)\,\right]\frac{e^{-\lambda_p y}}{v\,y}1_{y>0}\,dy\\ &=\frac{(1-k)\tilde{F}(x+k\Delta x,t)-\tilde{F}(x,t)-k\tilde{F}(x+(k+1)\Delta x,t)}{v}\times\int_{k\Delta x}^{(k + 1)\Delta x}\frac{e^{-\lambda_py}}{y}dy+\frac{\tilde{F}(x+(k+1)\Delta x,t)-\tilde{F}(x+k\Delta x,t)}{\lambda_p\,v\,\Delta x}(e^{-\lambda_pk\Delta x}-e^{-\lambda_p(k+1)\Delta x}) \end{align} So the only thing you have to do is calculate $$\int_{k\Delta x}^{(k + 1)\Delta x}\frac{e^{-\lambda_py}}{y}dy$$ there are several Methods for fast calculation of it.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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