solving the Black-Scholes PDE by finite differences

This entry presents some examples of solving the Black-Scholes partial differential equation in one space dimension:

$rf={\displaystyle \frac{\partial f}{\partial t}}+rx{\displaystyle \frac{\partial f}{\partial x}}+\frac{1}{2}{\sigma }^2{x}^2{\displaystyle \frac{{\partial }^{2}f}{\partial x^2}},f=f(t,x),$

over the rectangle $0\le t\le T$, $X_{\mathrm{L}}\le x\le X_{\mathrm{U}}$, with various boundary conditions on the top, bottom, and right sides of the rectangle. The parameters $r$, $\sigma >0$ are arbitrary constants.

(Add diagram of domain here…)

The partial differential equation can be solved numerically using the basic methods based on approximating the partial derivatives with finite differences.

0.1 Finite-difference formulae

We summarize the equations for the finite differences below.

Let $n$, $m$, $k$ be some chosen positive integers, which determine the grid on which we are approximating the solution of the PDE.

Set $u_{i,j}$ to be the approximation to $f(T-i\mathrm{\Delta }t,j\mathrm{\Delta }x)$, for $0\le i\le m$ and $k\le j\le k+n+1$. (For convenience, we have made time move “backwards” as we increase $i$, because the original PDE is really a backwards heat equation, and evolves backwards in time.)

Also set:

$$\mathrm{\Delta }t=\frac{T}{m},\mathrm{\Delta }x=\frac{X_{\mathrm{U}}-X_{\mathrm{L}}}{n+1}$$

Explicit method.

${\displaystyle \frac{u_{i+1,j}-u_{i,j}}{\mathrm{\Delta }t}}$ $=-r{u}_{i,j}+rj\mathrm{\Delta }x{\displaystyle \frac{u_{i,j+1}-u_{i,j-1}}{2\mathrm{\Delta }x}}$

$\mathrm{\hspace{1em}\hspace{1em}}+{\displaystyle \frac{1}{2}}{\sigma }^2{(j\mathrm{\Delta }x)}^2{\displaystyle \frac{u_{i,j-1}-2u_{i,j}+u_{i,j+1}}{\mathrm{\Delta }{x}^2}}$

$u_{i+1,j}$ $=\left(\frac{1}{2}{(\sigma j)}^2\mathrm{\Delta }t-\frac{1}{2}rj\mathrm{\Delta }t\right)u_{i,j-1}$

$+\left(1-{(\sigma j)}^2\mathrm{\Delta }t-r\mathrm{\Delta }t\right)u_{i,j}$

$+\left(\frac{1}{2}{(\sigma j)}^2\mathrm{\Delta }t+\frac{1}{2}rj\mathrm{\Delta }t\right)u_{i,j+1}.$

Since the PDE to solve is parabolic and time-dependent, we can step through time to numerically approximate it. Given $u_{i,*}$, we can recursively compute $u_{i+1,*}$.

(Add stencil of numerical method here…)

Implicit method.

${\displaystyle \frac{u_{i+1,j}-u_{i,j}}{\mathrm{\Delta }t}}$ $=-r{u}_{i+1,j}+rj\mathrm{\Delta }x{\displaystyle \frac{u_{i+1,j+1}-u_{i+1,j-1}}{2\mathrm{\Delta }x}}$

$\mathrm{\hspace{1em}\hspace{1em}}+{\displaystyle \frac{1}{2}}{\sigma }^2{(j\mathrm{\Delta }x)}^2{\displaystyle \frac{u_{i+1,j-1}-2u_{i+1,j}+u_{i+1,j+1}}{\mathrm{\Delta }{x}^2}}$

$u_{i,j}$ $=\left(-\frac{1}{2}{(\sigma j)}^2\mathrm{\Delta }t+\frac{1}{2}rj\mathrm{\Delta }t\right)u_{i+1,j-1}$

$+\left(1+{(\sigma j)}^2\mathrm{\Delta }t+r\mathrm{\Delta }t\right)u_{i+1,j}$

$+\left(-\frac{1}{2}{(\sigma j)}^2\mathrm{\Delta }t-\frac{1}{2}rj\mathrm{\Delta }t\right)u_{i+1,j+1}.$

Crank-Nicolson method.

$$\begin{array}{c}\left(-\frac{1}{4}{(\sigma j)}^2\mathrm{\Delta }t+\frac{1}{4}rj\mathrm{\Delta }t\right)u_{i+1,j-1}\hfill \\ \hfill +\left(1+\frac{1}{2}{(\sigma j)}^2\mathrm{\Delta }t-\frac{1}{2}r\mathrm{\Delta }t\right)u_{i+1,j}+\left(-\frac{1}{4}{(\sigma j)}^2\mathrm{\Delta }t-\frac{1}{4}rj\mathrm{\Delta }t\right)u_{i+1,j+1}\\ \hfill =\left(\frac{1}{4}{(\sigma j)}^2\mathrm{\Delta }t-\frac{1}{4}rj\mathrm{\Delta }t\right)u_{i,j-1}+\left(1-\frac{1}{2}{(\sigma j)}^2\mathrm{\Delta }t-\frac{1}{2}r\mathrm{\Delta }t\right)u_{i,j}\\ \hfill +\left(\frac{1}{4}{(\sigma j)}^2\mathrm{\Delta }t+\frac{1}{4}rj\mathrm{\Delta }t\right)u_{i,j+1}\end{array}$$

0.2 Convergence of methods

(Briefly discuss convergence properties of these methods here…)

0.3 Example results

Figure 1: Basic stock call option price

Boundary conditions and parameters:

$r=0.10,\sigma =0.40,T=0.5,K=50.00.$

$X_{\mathrm{L}}=0,X_{\mathrm{U}}=100.00.$

$\mathrm{\hspace{1em}}m=100,n=200,\mathrm{\Delta }t=0.005,\mathrm{\Delta }x=0.4975.$

$f(T,x)$ $=\max (x-K,\mathrm{\hspace{0.25em}0}),$ $X_{\mathrm{L}}\le x\le X_{\mathrm{U}}$

$f(t,0)$ $=0,$ $0\le t\le T,$

$f(t,X_{\mathrm{U}})$ $=x-K{e}^{-r(T-t)},$ $0\le t\le T.$

(Describe analytic solution here…)

Figure 2: Price of call option with up-and-out barrier

$r=0.10,\sigma =0.40,T=0.5,K=50.00.$

$X_{\mathrm{L}}=0,X_{\mathrm{U}}=100.00.$

$\mathrm{\hspace{1em}}m=100,n=200,\mathrm{\Delta }t=0.005,\mathrm{\Delta }x=0.4975.$

$f(T,x)$ $=\max (x-K,\mathrm{\hspace{0.25em}0}),$ $X_{\mathrm{L}}\le x\le X_{\mathrm{U}}$

$f(t,0)$ $=0,$ $0\le t\le T,$

$f(t,X_{\mathrm{U}})$ $=0,$ $0\le t\le T.$

• http://svn.gold-saucer.org/math/PlanetMath/SolvingTheBlackScholesPDEByFiniteDifferences/bss.pyPython program that implements the finite-difference methods for the above two problems, and plots the results

Title	solving the Black-Scholes PDE by finite differences
Canonical name	SolvingTheBlackScholesPDEByFiniteDifferences
Date of creation	2013-03-22 16:30:59
Last modified on	2013-03-22 16:30:59
Owner	stevecheng (10074)
Last modified by	stevecheng (10074)
Numerical id	6
Author	stevecheng (10074)
Entry type	Example
Classification	msc 65M06
Classification	msc 91B28
Classification	msc 35K15