The original pricing equation
All pricing equations satisfied by the price $u(t, x)$ of plain contracts in this article will have the form $$ \left\{\begin{align*} \partial_t u+\mathcal{A} u &=f \quad &\text{in} \quad &J \times G \\ u(0, x) &= u_0(x) \quad &\text{in} \quad &G \end{align*}\right., $$
where
- $\mathcal{A}$ is a linear second-order partial differential operator (w.r.t. $x$ )
- $u_0$ is the initial condition (the payoff)
- $f$ is the forcing term
- $G \subseteq \mathbb{R}^d$ (state space of admissible stock prices)
- $J=(0, T)$ ( $T>0$ the maturity)
The weak formulation
The equation above can be put into variational form. Given the Hilbert spaces $V$ and $H$ such that $V \hookrightarrow H \cong H^* \hookrightarrow V^*$, and assume that $u_0 \in H, \quad f \in L^2\left(J ; V^*\right)$, the abstract weak formulation of the equation above reads:
Find $u \in L^2(J ; V) \cap H^1 (J ; V^*)$ such that for $t \in J$, $$ \begin{align*} \frac{\mathrm{d}}{\mathrm{~d} t}\langle u(t), v\rangle_{V^*, V}+a(u(t), v)&=\langle f(t), v\rangle_{V^*, V}, \quad \forall v \in V \\ u(0)&=u_0 \quad \text { in } \quad H \end{align*}. $$
Here, $a(\cdot, \cdot): V \times V \rightarrow \mathbb{R}$ denotes the bilinear form associated with the operator $\mathcal{A} \in \mathcal{L}\left(V, V^*\right)$ via $a(u, v):=\langle\mathcal{A} u, v\rangle_{V^*, V}, \quad u, v \in V$, where $\langle\cdot, \cdot\rangle_{V^*, V}$ denotes the $V^* \times V$ duality pairing.
Check the well-posedness of the weak formulation
We need to check the existence and uniqueness of the solution to the weak formulation, which can be done with the following theorem:
Theorem (Well-posedness) Assume there exist constants $C_0, C_1>0, C_2 \geq 0$, such that $\forall u, v \in V$ there holds $$ \begin{align*} |a(u, v)| & \leq C_0||u||_V||v||_V \quad &\text { (continuity) } \\ a(u, u) & \geq C_1||u||_V^2-C_2||u||_H^2 \quad &\text { (Gårding inequality) } \end{align*}. $$ Assume further $u_0 \in H, f \in L^2\left(J ; V^*\right)$ and $0<T<\infty$. Then, problem (3.2) admits a unique solution $u \in L^2(J ; V) \cap H^1\left(J ; V^*\right)$.
Moreover, $u \in C^0(\bar{J} ; H)$, and there holds the a-priori estimate $$ ||u||_{C^0(\bar{J};H)} + ||u||_{L^2(J;V)}+||u||_{H^1(J ; V^*)} \leq C(||u_0||_H + ||f||_{L^2(J ; V^*)}) . $$
Discretization
For the discretization we use the method of lines where the weak formulation is only discretized in space in the first step to obtain a system of coupled ODEs. These ODEs will be solved in the second step.
Two steps
(1. Semi-Discretization in Space) Assume that $\left(V_N\right)_N$ is a family of subspaces $V_N \subset V$ with finite dimension $ N=\operatorname{dim} V_N<\infty$, and $u_{N, 0} \in V_N$ is an approximation of $u_0 \in H$.
Our goal is to approximate the solution $u(t)$ by an element $u_N(t) \in V_N$ for each fixed $t \in J$.
We first consider the semi-discrete formulation in space, which has the form as follows:
Find $u_N \in C^1\left(J ; V_N\right)$ such that $$ \begin{align*} \frac{\mathrm{d}}{\mathrm{~d} t}\left(u_N(t), v_N\right)_H+a\left(u_N(t), v_N\right) & =\left\langle f(t), v_N\right\rangle, \quad \forall v_N \in V_N, \quad \forall t \in J \\ u_N(0) & =u_{N, 0} \end{align*} $$
Assume that $V_N=\text{span}\left\{b_j: j=1, \ldots N\right\}$. For $t \in J$, let
$$ u_N(t)=\sum_{j=1}^N u_{N, j}(t) b_j, \quad \underline{u}_N(t):=\left(u_{N, 1}(t), \ldots, u_{N, N}(t)\right)^{\top} \in \mathbb{R}^N $$
Then, the semi-discrete formulation is equivalent to the matrix formulation as follows:
Find $\underline{u}_N \in C^1\left(J ; \mathbb{R}^N\right)$ such that $$ \begin{align*} \mathbf{M} \underline{\dot{u}}_N(t)+\mathbf{A} \underline{u}_N(t)&=\underline{f}(t) \quad \forall t \in J \\ \underline{u}_N(0)&=u_{N, 0} \end{align*}. $$
Here,
- $\underline{f}(t):=\left(f_1(t), \ldots, f_N(t)\right)^{\top} \in \mathbb{R}^N$ with $f_j(t):=\left\langle f(t), b_j\right\rangle$,
- $\mathbf{M}$ and $\mathbf{A}$ are the mass matrix and the stiffness matrix, respectively, with entries $\mathbf{M}_{i j}:=\left(b_j, b_i\right)_H, \mathbf{A}_{i j}:=a\left(b_j, b_i\right)$,
- $\underline{u}_{N, 0}$ is the coefficient vector of $u_{N, 0}$.
(2. Full Discretization) Given a (not necessarily uniform) partition of $\bar{J}=[0, T]$,
$$ 0=t_0<t_1<\ldots<t_{M-1}<t_M=T, $$
define the time steps $k_m:=t_m-t_{m-1}$ for $m=1, \ldots, M$. We discretize the temporal ODE appearing in the semi-discrete formulation with the $\theta$-scheme for a fixed $\theta \in[0,1]$, i.e., for $t \in\left(t_{m-1}, t_m\right)$ we approximate
$$ \begin{array}{rll} \frac{\mathrm{d}}{\mathrm{~d} t} u_N(t) & \text { by } & \text { the difference quotient } \frac{u_N^m-u_N^{m-1}}{k_m} \\ u_N(t) & \text { by } & u_N^{m-1+\theta}:=\theta u_N^m+(1-\theta) u_N^{m-1} \\ f(t) & \text { by } & f^{m-1+\theta}:=\theta f\left(t_m\right)+(1-\theta) f\left(t_{m-1}\right) . \end{array} $$
This yields the fully discrete scheme
Find $u_N^m \in V_N$ such that for $m=1, \ldots, M$ and all $v_N \in V_N$ $$ \begin{align*} k_m^{-1}\left(u_N^m-u_N^{m-1}, v_N\right)_H+a\left(u_N^{m-1+\theta}, v_N\right)&=\left\langle f^{m-1+\theta}, v_N\right\rangle \\ u_N^0&=u_{N, 0} \end{align*} $$
or in matrix form
Find $\underline{u}_N^m \in \mathbb{R}^N$ such that for $m=1, \ldots, M$ $$ \begin{align*} k_m^{-1} \mathrm{M}\left(\underline{u}_N^m-\underline{u}_N^{m-1}\right)+\mathrm{A}\left(\theta \underline{u}_N^m+(1-\theta) \underline{u}_N^{m-1}\right)&=\underline{f}^{m-1+\theta} \\ \underline{u}_N^0&=\underline{u}_0 \end{align*} $$ where $\underline{f}^{m-1+\theta}:=\theta \underline{f}\left(t_m\right)+(1-\theta) \underline{f}\left(t_{m-1}\right)$.
Rearranging the terms in the equality, we obtain
Find $\underline{u}_N^m \in \mathbb{R}^N$ such that for $m=1, \ldots, M$ $$ \begin{gathered} \left(\mathbf{M}+k_m \theta \mathbf{A}\right) \underline{u}_N^m=\left(\mathbf{M}-k_m(1-\theta) \mathbf{A}\right) \underline{u}_N^{m-1}+k_m \underline{f}^{m-1+\theta} \ \underline{u}_N^0=\underline{u}_0 \end{gathered} $$
This corresponds to
- forward (explicit) Euler time-stepping for $\theta=0$,
- Crank-Nicolson time-stepping for $\theta=\frac{1}{2}$,
- backward (implicit) Euler time-stepping for $\theta=1$.
(#to be continued)
Stability
Discretization Error
Example
Theorem 1.2.6 We consider a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with filtration $\mathbb{F}$ and a Brownian motion $W$ on ( $\Omega, \mathcal{F}, \mathbb{P}$ ) adapted to $\mathbb{F}$. Assume there exists $C>0$ such that $b, \sigma: \mathbb{R}{+} \times \mathbb{R} \rightarrow \mathbb{R}$ in (1.2) satisfy $$ \begin{align} & |b(t, x)-b(t, y)|+|\sigma(t, x)-\sigma(t, y)| \leq C|x-y|, \quad x, y \in \mathbb{R}, t \in \mathbb{R}{+} \ & |b(t, x)|+|\sigma(t, x)| \leq C(1+|x|), \quad x \in \mathbb{R}, t \in \mathbb{R}_{+} \end{align} $$
Assume further $X_0=Z$ for a random variable which is $\mathcal{F}_0$-measurable and satisfies $\mathbb{E}\left[|Z|^2\right]<\infty$. Then, for any $T \geq 0$, (1.2) admits a $\mathbb{P}$-a.s. unique solution in $[0, T]$ satisfying $$ \mathbb{E}\left[\sup _{0 \leq t \leq T}\left|X_t\right|^2\right]<\infty $$
Let suppose the price $X$ is the solution of the SDE $$ \mathrm{d} X_t=b\left(t, X_t\right) \mathrm{d} t+\sigma\left(t, X_t\right) \mathrm{d} W_t, \quad X_0=Z. $$
Assume that the coefficients $b$, $\sigma$ are independent of time and satisfies the assumptions of 【Theorem】 (which ensures the existence and uniqueness of $X$). Further assume that $r(x)$ is a bounded and continuous function which stands for the riskless interest rate. Denote $g(\cdot)$ as the payoff of the function. The value of the option $V$ can be computed as the conditional expectation
$$ V(t, x)=\mathbb{E}\left[e^{-\int_t^T r\left(X_s\right) \mathrm{d} s} g\left(X_T\right) \mid X_t=x\right] . $$
Let $\mathcal{A}$ denote the differential operator which is, for functions $f \in C^2(\mathbb{R})$ with bounded derivatives, given by
$$ (\mathcal{A} f)(x)=\frac{1}{2} \sigma^2(x) \partial_{x x} f(x)+b(x) \partial_x f(x) $$
This $V(t,x)$ is a solution of
The variational fomulation of the BS equation can be expressed as:
Find $u \in L^2\left(J ; H^1(\mathbb{R})\right) \cap H^1\left(J ; L^2(\mathbb{R})\right)$ such that: $\left(\partial_t u, v\right)+a^{\mathrm{BS}}(u, v)=0, \forall v \in H^1(\mathbb{R})$, a.e. in $J$, $u(0)=u_0$,