Chapter 5 - Euler scheme for stochastic differential equations - Exercises

Exercice 5.1 (strong convergence)
In Theorem 5.2.1, we have proved that for a SDE with globally Lipschitz coefficients $b, \sigma$, we have $$\mathbb{E}(\sup_{0\leq t \leq T}|X^{(h)}_t-X_t|^p)\leq O(h^{\frac p2})$$ for any $p>0$. The convergence order is 1/2. Show, that  if $\sigma$ is constant and $b$ is ${\cal C}^2$ in space and ${\cal C}^1$ in time with bounded derivatives, the convergence order is 1.

Exercice 5.2 (Milshtein scheme)
Denote by $(X_t)_{t\geq 0}$ the solution of the stochastic differential equation $$X_t=x+\int_0^t \sigma(X_s){\rm d}W_s+\int_0^t b(X_s){\rm d}s$$ where $\sigma,b:{\mathbb R}\rightarrow{\mathbb R}$ are bounded ${\cal C}^2$-functions with bounded derivatives.
  1. Show the short time $L_2$-approximation $$\mathbb{E}\left(\left(X_t-[x+b(x) t+\sigma(x)W_t]\right)^2\right)= \frac{(\sigma\sigma'(x))^2}{2}t^2+o(t^2).$$ The above estimate is instrumental to show the strong convergence of the Euler scheme at order $1/2$: essentially, the global quadratic error (of size $N^{-1}$) is the summation of the above local estimates (of order $N^{-2}$ with $t=h$) over $N$ times, to finally get the error bound for $p=2$. 
  2. Similarly, show $$\mathbb{E}\left(\left(X_t-[x+b(x)t+\sigma(x)W_t+\frac{1}{2}\sigma\sigma'(x)(W_t^2-t)]\right)^2\right)={\mathcal O}(t^3). $$ This estimate leads to a high-order scheme, called the Milshtein scheme, which is written \begin{equation} \begin{cases} X^{(h,M)}_0=&x,\\ X^{(h,M)}_{t_{i+1}}=&X^{(h,M)}_{t_{i}}+b(X^{(h,M)}_{t_{i}}) h +\sigma(X^{(h,M)}_{t_{i}}) (W_{t_{i+1}}-W_{t_{i}})\\ &+\frac{ 1 }{ 2}\sigma\sigma'(X^{(h,M)}_{t_{i}}) [(W_{t_{i+1}}-W_{t_{i}})^2-h]. \end{cases} \end{equation}
  3. Using the above estimate, prove that \begin{equation} \sup_{0\leq i \leq N}\mathbb{E}(|X^{(h,M)}_{t_{i}}-X_{ih}|^2)=O( h^{2}), \end{equation} i.e. a convergence rate at order $1$ for the strong convergence. 
This Milshtein scheme is effective, mainly for one-dimensional situations, because in the multidimensional case, it is written using the iterated stochastic integrals $\int_0^t W^i_s{\rm d}W^j_s$ for $1\leq i \neq j \leq d$, which cannot be easily simulated (in dimension 1, we simply have $\int_0^{t}W_s{\rm d}W_s=\frac{ 1 }{2 }(W_t^2-t)$).

Exercice 5.3 (convergence rate of weak convergence)
We consider the model of geometric Brownian motion: $$X_t=x+\int_0^t \sigma X_s {\rm d}W_s+\int_0^t \mu X_s {\rm d}s$$ with $x>0$.
  1. Compute $\mathbb{E}(X_T^2)$. 
  2. Let $X^h$ be the related Euler scheme with time step $h$. Set $y_i=\mathbb{E}((X^{(h)}_{t_i})^2)$. Find a relation between $y_{i+1}$ and $y_i$.
  3. Deduce that $\mathbb{E}((X^{(h)}_{T})^2)=\mathbb{E}(X_T^2)+O(h)$. 

Exercice 5.4 (solving SDE using change of variables)
Consider an SDE of the form \begin{equation}X_t = x_0+\int_0^t b(X_s){\rm d}s + \int_0^t\sigma(X_s){\rm d}W_s. \end{equation} We study two transformations that lead to simpler equations and may help in the resolution of the initial SDE.
  1. Lamperti transformation. We assume that the SDE is defined on ${\mathbb R}$, and we assume that the coefficients $b:{\mathbb R}\to{\mathbb R}$ and $\sigma:{\mathbb R}\to(0,\infty)$ are of class ${\cal C}^1$ with bounded derivative, that the function $\frac1{\sigma(x)}$ is not integrable at $\pm \infty$, and that the function $b/\sigma - \sigma'/2$ is Lipschitz. 
    1. Verify that the function $f(x)=\int_{x_0}^x \frac{{\rm d}y}{\sigma(y)}$ is a bijection from ${\mathbb R}$ into ${\mathbb R}$.
    2. Show that the solution to the SDE can be put in the form $X_t=f^{-1}(Y_t)$, where the process $Y$ solves \[ Y_t = \int_0^t \tilde{b} (Y_s) {\rm d}s + W_t, \] for some function $\tilde{b}$ to explicit.
      Comment: The interest in Lamperti transformation is to retrieve an SDE with unit-diffusion coefficient; the approximation by Euler scheme of this SDE becomes more accurate (Exercise 5.1).
  2. Doss-Sussmann transformation. We consider now the case where the coefficients $b,\sigma:{\mathbb R}^d \to {\mathbb R}^d$ are Lipschitz, $\sigma$ is of class ${\cal C}^2_b$ (bounded with bounded derivatives), and $W_t$ is still a scalar Brownian motion. Denote by $F(\theta,x)$ the flow of the differential equation $x'_\theta=\sigma(x_\theta)$ in ${\mathbb R}^d$, i.e. \[ \partial_\theta F(\theta,x) = \sigma(F(\theta,x)), \qquad F(0,x)=x, \qquad \forall \: (\theta,x) \in {\mathbb R} \times {\mathbb R}^d, \] and by $\partial_x F(\theta,x)$ the Jacobian matrix of $F$. Using that $F$ is of class ${\cal C}^2$ on ${\mathbb R}\times{\mathbb R}^d$ and that $\partial_x F(\theta,x)$ is invertible for any $(\theta,x)$, show that the solution $X_t$ to the SDE is written in the form \[ X_t(\omega)=F(W_t(\omega),Z_t(\omega)) \] for any $t$, where the differentiable process $Z_t$ solves --- $\omega$ by $\omega$ --- the Ordinary Differential Equation (ODE) \[ Z_t = x_0+\int_0^t (\partial_x F(W_s,Z_s))^{-1} \Bigl(b - \frac12 (\partial_x \sigma) \sigma \Bigr)(F(W_s,Z_s)) {\rm d}s, \] with the matrix $(\partial_x F(\theta,x))^{-1}$ being the inverse of $\partial_x F(\theta,x)$. 

Exercice 5.5 (simulation of exit time)
Write a simulation program to compute ${\mathbb E}(\mathbb{1}_{\sup_{0\leq t\leq T}S_t\leq U}(K-S_T)_+)$ where $S$ is a geometric Brownian motion. Compare the three schemes (discrete time approximation, Brownian bridge method, boundary shifting method).

Exercice 5.6 (exit time, non convergence)
In the theoretical results of Section 5.4, the assumption of non-degeneracy (i.e. $\sigma(x)$ invertible) plays an important role in the validity of the convergence results. Otherwise, we may face some pathological situations, as illustrated below.
  1. In dimension 1, consider the model with $b(y)=y$, $\sigma(y)=0$, $x_0 =1$ and $D=(-\infty,\exp(1))$, $T=1$. Prove that $\tau^{(h)}_{disc.}>1$ and $\tau=1$, so that $$\mathbb{P}(\tau^{(h)}_{disc.}> T)-\mathbb{P}(\tau> T)=1$$ does not converge to 0 as $h\to0$. 
  2. In dimension 1, consider the model $b(x) = \cos(x), \sigma(x) = \sin(x)$, $x_0=\pi/2$ with $D =(-\pi,2\pi)$. Prove that $\tau=+\infty$ almost surely and $\tau^{(h)}_{disc.}<+\infty$ almost surely.

No comments:

Post a Comment