solving the Black-Scholes PDE by finite differences

This entry presents some examples of solvingthe Black-Scholes partial differential equationin 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 rightsides of the rectangle. The parameters $r$, $\sigma >0$are arbitrary constants.

(Add diagram of domain here…)

The partial differential equation can be solvednumerically using the basic methodsbased on approximating the partial derivativeswith 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 solutionof 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 PDEis 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