The Black-Scholes PDE can be transformed into the heat equation: $$ u_t- u_{xx}=0, $$ with known boundary conditions and initial condition \(u(x,0)=f(x)\). The heat equation can be approximated using finite difference methods.
Let
- \(x\in [XS,XE]\), and \(t\in [0,\alpha]\), where \(\alpha=\frac{1}{2} r ~T~ \sigma^2\), \(XS << 1\) and \(XE>>1\).
- i.e. \(XS:=-10\) and \(XS:=2\), assuming the share price is bounded by the interval \(S_t=e^{-10} ~E\) and \(S_t=e^2 ~E\), during the lifetime of the option.
- Discretize \(x\) over \([XS,XE]\) with \(x_i=XS+i~h, ~\forall i \in[0,J]\), with an increment \(h=\frac{XE-XS}{J}\), where \(J\) is the number of points required over the \(x\)-space.
- Discretize \(t\) over \([0,\alpha]\) with \(t_n=n ~\Delta t, ~\forall n \in[0,N]\), with an increment \(\Delta t=\frac{\alpha}{N}\), where \(N\) is the number of points required over the \(t\)-space.
- For a put option, we have the initial condition of $$ u(x_i,0)=\max \left(e^{0.5(w-1)X_i}-e^{0.5(w+1)X_i},0 \right), $$ and boundary conditions \(u(XE,t_n)=0\) and \(V(x_1,t_n)=e^{0.5(w-1)x_1+(w+1)^2~t_n/4}\), where \(\displaystyle {w=\frac{2r}{\sigma^2}}\).
The equation of the forward method:
$$ \frac{u(x_i,t_{n+1})-u(x_i,t_n)}{\Delta t}-~ D^2_x \left[u(x_i,t_n) \right]=0, $$ where$$ D^2_x \left[u(x_i,t_n) \right]=\frac{u(x_{i+1},t_{n})-2 u(x_i,t_n)+u(x_{i-1},t_n)}{h^2}, $$ \(h=x_{i+1}-x_i\) and \(\Delta t=t_{n+1}-t_n\). This gives$$ u_i^{n+1}=\frac{\Delta t}{h^2} \left (u_{i+1}^n + u_{i-1}^n \right)+ \left(1-2 \frac{\Delta t}{h^2} \right) u_i^n. $$ It can be shown that the forward numerical scheme is only stable when$$ (1-2 \lambda)>0 \Longleftrightarrow N>\frac{r T \sigma^2 J^2}{(XE-XS)^2}, $$ where \(\lambda=\frac{\Delta t}{h^2}\).
The equation of the backward method:
$$ \frac{u(x_i,t_{n+1})-u(x_i,t_n)}{\Delta t}-~ D^2_x[u(x_i,t_{n+1})]=0, $$ where$$ D^2_x[u(x_i,t_{n+1})]=\frac{u(x_{i+1},t_{n+1})-2 u(x_i,t_{n+1})+u(x_{i-1},t_{n+1})}{h^2}, $$ \(h=x_{i+1}-x_i\) and \(\Delta t=t_{n+1}-t_n\). This gives$$ -\frac{\Delta t}{h^2} u_{i+1}^{n+1}+\left(1+2 \frac{\Delta t}{h^2} \right) u_i^{n+1}-\frac{\Delta t}{h^2} u_{i-1}^{n+1}=u_{i}^n. $$ It can be shown that the backward numerical scheme is always stable. Thus, we can find the solution by solving the system of linear equation
$$ M_{(J-1)\times (J-1)} ~U^{n+1}_{(J-1)\times 1} = V^{n}_{(J-1) \times 1}, $$ where \begin{align*} U^{n+1}_{(J-1)\times 1}= \left( \begin{array}{c} u_1^{n+1}\\ u_2^{n+1}\\ u_3^{n+1}\\ \vdots \\ u_{J-2}^{n+1} \\ u_{J-1}^{n+1} \end{array} \right),~ V^{n}_{(J-1)\times 1}= \left( \begin{array}{c} u_1^{n}+\lambda u_0^{n+1}\\ u_2^{n}\\ u_3^{n}\\ \vdots \\ u_{J-2}^{n}\\ u_{J-1}^{n}+\lambda u_{J}^{n+1} \end{array} \right) \end{align*}
\begin{align*} M_{(J-1)\times (J-1)} = \left( \begin{array}{ccccccc} 1+2\lambda & -\lambda& 0 &0&0&\cdots&0 \\ -\lambda&1+2\lambda & -\lambda& 0 &0&\cdots&0 \\ 0& -\lambda&1+2\lambda & -\lambda& 0 &\cdots&0 \\ 0 &\cdots&\cdots&\cdots&\cdots&\cdots&0 \\ 0 &\cdots&\cdots&\cdots&\cdots&\cdots&0 \\ 0 &\cdots&\cdots&\cdots& -\lambda&1+2\lambda & -\lambda \\ 0 &\cdots&\cdots&\cdots&0& -\lambda&1+2\lambda \\ \end{array} \right).~ \end{align*}
Gauss elimination can then be used to find \(u_i^{n+1}\) from \(u_i^{n}\), \(\forall n \in [1,N-1]\). Note that solving for \(U^{n+1}_{(J-1)\times 1} = M^{-1}_{(J-1)\times (J-1)} V^{n}_{(J-1) \times 1}\) is computationally expensive when calculating the inverse of a tridiagonal sparse matrix.
The equation of Crank-Nickolson method:
$$ \frac{u(x_i,t_{n+1})-u(x_i,t_n)}{\Delta t}-~ D^2_x[\frac{u(x_i,t_{n})+u(x_i,t_{n+1})}{2}]=0, $$ where$$ D^2_x[\frac{u(x_i,t_{n})+u(x_i,t_{n+1})}{2}]=\frac{1}{2} \left(D^2_x[u(x_i,t_{n})]+D^2_x[u(x_i,t_{n+1})] \right). $$ It can be shown that Crank-Nickolson numerical scheme is always stable. Thus, we can find the solution by solving the system of linear equation
$$ M_{(J-1)\times (J-1)} ~U^{n+1}_{(J-1)\times 1} = V^{n}_{(J-1) \times 1}, $$ where \begin{align*} U^{n+1}_{(J-1)\times 1}= \left( \begin{array}{c} u_1^{n+1}\\ u_2^{n+1}\\ u_3^{n+1}\\ \vdots \\ u_{J-2}^{n+1} \\ u_{J-1}^{n+1} \end{array} \right),~ V^{n}_{(J-1)\times 1}= \left( \begin{array}{c} f_1^{n}+\lambda u_0^{n+1}\\ f_2^{n}\\ f_3^{n}\\ \vdots \\ f_{J-2}^{n}\\ f_{J-1}^{n}+\lambda u_{J}^{n+1} \end{array} \right), \end{align*}
\begin{align*} M_{(J-1)\times (J-1)} = \left( \begin{array}{ccccccc} 1+2\lambda & -\lambda& 0 &0&0&\cdots&0 \\ -\lambda&1+2\lambda & -\lambda& 0 &0&\cdots&0 \\ 0& -\lambda&1+2\lambda & -\lambda& 0 &\cdots&0 \\ 0 &\cdots&\cdots&\cdots&\cdots&\cdots&0 \\ 0 &\cdots&\cdots&\cdots&\cdots&\cdots&0 \\ 0 &\cdots&\cdots&\cdots& -\lambda&1+2\lambda & -\lambda \\ 0 &\cdots&\cdots&\cdots&0& -\lambda&1+2\lambda \\ \end{array} \right),~ \end{align*}
where \(f_i^n:=\lambda u_{i+1}^n+(1-2\lambda) u_i^n+\lambda u_{i-1}^n\). Gauss elimination can then be used to find \(u_i^{n+1}\) from \(u_i^{n}\), \(\forall n \in [1,N-1]\).
Ameican Options.
For American options, we have to include the property of the early exercise of the option. For a Put option, the early exercise has a value of $$ g_i^{n+1}=e^{\frac{t_{n+1}}{4}(w+1)^2} \mathrm{max} \left(e^{\frac{w-1}{2}x_i}-e^{\frac{w+1}{2}x_i},0 \right)$$.
Taking the forward method as an example, we get $$ u_i^{n+1}= \mathrm{max} \left( g_i^{n+1}, \frac{\Delta t}{h^2} \left (u_{i+1}^n + u_{i-1}^n \right)+ \left(1-2 \frac{\Delta t}{h^2} \right) u_i^n \right). $$
 |
| Forward Finite Differences |
 |
| Backward Finite Differences |
 |
| Crank-Nicolson Finite Differences |
References
- Burden, R. L., & Faires, J. D. (2005).
Numerical analysis
. Thomson Brooks/Cole. - Haidar, H. (2008).
Pricing options for special prices movement
,” MSc dissertation, University of Sussex.
No comments:
Post a Comment