SDE | 均方收敛阶、EM 均方收敛阶计算、一步逼近和基本均方收敛定理

2026-06-12 16:31:57 星期五

又要下雨，写不完了

从本节开始，我们介绍均方收敛阶的定义，并且给出EM的均方收敛阶的计算，最后我们介绍一步逼近，考察一步误差与全局误差的关系（基本均方收敛定理）.

均方收敛阶的定义

在确定情形的时候，我们有\(|X(t_n) - \overline{X}_n| \leq c (\Delta t)^p\)，其中 \(p\) 称为收敛阶.

在随机的情形，我们有两种类似的办法，但我们使用的是均方收敛阶和弱收敛阶. 我们先介绍均方收敛.

若下式成立，我们称该数值格式的均方收敛阶为 \(p\)：

\[\| X(t_n) - \bar{X}_n \|_{L^2(\Omega)} = \sqrt{ E | X(t_n) - \bar{X}_n |^2 } \leq C (\Delta t)^p, \quad n = 0,1,\dots, N \]

引理（离散gronwall不等式） If \(u_{n+1} \leq (1 + A\Delta t)u_n + k(\Delta t)^p\), \(n=0,1,\dots,N-1\), where \(\Delta t \cdot N = T\), \(A \geq 0\), \(k \geq 0\), \(p \geq 1\), \(u_n \geq 0\), then

\[u_n \leq e^{AT} u_0 + k \frac{e^{AT} - 1}{A} (\Delta t)^p. \]

---

EM method 均方收敛阶的计算

Thm The EM method has mean-square convergence order \(\dfrac{1}{2}\).

pf. Aim: \(\mathbb{E}|X(t_n)-\bar{X}_n|^2 \leq C \Delta t\).

\[\bar{X}_n = \bar{X}_{n+1} + f(\bar{X}_{n+1}) \Delta t + \sigma(\bar{X}_{n+1}) \Delta_n B \]

迭代得

\[\bar{X}_n = \bar{X}_0 + \sum_{i=0}^{n-1} f(\bar{X}_i) \Delta t + \sum_{i=0}^{n-1} \sigma(\bar{X}_i) \Delta_{i+1} B \]

即

\[\bar{X}_n = \bar{X}_0 + \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} f(\bar{X}_i) dt + \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} \sigma(\bar{X}_i) dB_t. \]

另一方面，

\[X(t_n) = X_0 + \int_0^{t_n} f(X(t)) dt + \int_0^{t_n} \sigma(X(t)) dB(t) \]

\[= X_0 + \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} f(X(t_i)) dt + \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} \sigma(X(t_i)) dB(t). \]

相减并整理得

\[\begin{aligned} X(t_n) - \overline{X}_n &= \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} \big[ f(X(t)) - f(X(t_i)) \big] dt \\ &\quad + \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} \big[ f(X(t_i)) - f(\overline{X}_i) \big] dt \\ &\quad + \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} \big[ \sigma(X(t)) - \sigma(X(t_i)) \big] dB_t \\ &\quad + \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} \big[ \sigma(X(t_i)) - \sigma(\overline{X}_i) \big] dB_t \\ &= \mathbf{(1)} + \mathbf{(2)} + \mathbf{(3)} + \mathbf{(4)}. \end{aligned} \]

两边取期望并利用 Cauchy-Schwarz 不等式，

\[\begin{aligned} \mathbb{E}\big| X(t_n) - \overline{X}_n \big|^2 &\leq 4 \left( \mathbb{E}\big|\mathbf{(1)}\big|^2 + \mathbb{E}\big|\mathbf{(2)}\big|^2 + \mathbb{E}\big|\mathbf{(3)}\big|^2 + \mathbb{E}\big|\mathbf{(4)}\big|^2 \right). \end{aligned} \]

分别估计各项：

\[\begin{aligned} \mathbb{E}|\mathbf{(1)}|^2 &= \sum_{i=0}^{n-1} \mathbb{E} \left[ \int_{t_i}^{t_{i+1}} |f(X(t)) - f(X(t_i))| dt \right]^2 \\ &\leq \sum_{i=0}^{n-1} (t_{i+1}-t_i) \mathbb{E} \int_{t_i}^{t_{i+1}} |f(X(t)) - f(X(t_i))|^2 dt \\ &\leq C \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} \mathbb{E}|X(t) - X(t_i)|^2 dt \\ &\leq C (1+\mathbb{E}|X_0|^2) \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} (t-t_i) dt \\ &\leq C (1+\mathbb{E}|X_0|^2) \Delta t. \end{aligned} \]

\[\mathbb{E}|\mathbf{(2)}|^2 \leq C \sum_{i=0}^{n-1} \Delta t \mathbb{E}|X(t_i) - \overline{X}_i|^2. \]

\[\begin{aligned} \mathbb{E}|\mathbf{(3)}|^2 &= \sum_{i=0}^{n-1} \mathbb{E} \left[ \int_{t_i}^{t_{i+1}} (\sigma(X(t)) - \sigma(X(t_i))) dB(t) \right]^2 \\ &= \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} \mathbb{E}|\sigma(X(t)) - \sigma(X(t_i))|^2 dt \\ &\leq C \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} \mathbb{E}|X(t) - X(t_i)|^2 dt \\ &\leq C (1+\mathbb{E}|X_0|^2) \Delta t. \end{aligned} \]

\[\mathbb{E}|\mathbf{(4)}|^2 = \sum_{i=0}^{n-1} \int_{t_i}^{t_{i+1}} \mathbb{E}|\sigma(X(t_i)) - \sigma(\overline{X}_i)|^2 dt \leq C \sum_{i=0}^{n-1} \Delta t \mathbb{E}|X(t_i) - \overline{X}_i|^2. \]

综上，

\[\mathbb{E}|X(t_n) - \overline{X}_n|^2 \leq C \Delta t \sum_{i=0}^{n-1} \mathbb{E}|X(t_i) - \overline{X}_i|^2 + C (1+\mathbb{E}|X_0|^2) \Delta t. \]

记 \(V_n = \mathbb{E}|X(t_n) - \overline{X}_n|^2\)，则

\[V_n \leq C \Delta t \sum_{i=0}^{n-1} V_i + C (1+\mathbb{E}|X_0|^2) \Delta t. \]

设 \(U_n = \sum_{i=0}^{n-1} V_i\)，则

\[\begin{cases} V_n \leq C \Delta t U_n + C (1+\mathbb{E}|X_0|^2) \Delta t, \\ V_n = U_{n+1} - U_n. \end{cases} \]

由此得

\[U_{n+1} \leq (1 + C \Delta t) U_n + C (1+\mathbb{E}|X_0|^2) \Delta t. \]

递推得

\[U_n \leq e^{C T} U_0 + C (1+\mathbb{E}|X_0|^2) \frac{e^{C T} - 1}{C} \leq C (1+\mathbb{E}|X_0|^2). \]

因此

\[V_n \leq C (1+\mathbb{E}|X_0|^2) \Delta t. \]

这表明 EM 方法具有均方收敛阶 \(\frac{1}{2}\)。

一步逼近

我们定义一步逼近(one-step approximation):

\[\overline{X}^{t,x}(t + \Delta t) = x + A\bigl(t, x, \Delta t, B(r) - B(t), \; t \leq r \leq t + \Delta t \bigr) \]

基于此，我们可以重写数值格式：

\[\overline{X}_{n+1} = \overline{X}^{t_n, \overline{X}_n}(t_n + \Delta t) = \overline{X}_n + A\bigl(t_n, \overline{X}_n, \Delta t, B(r) - B(t_n), \; t_n \leq r \leq t_{n+1} \bigr) \qquad (1) \]

对于任意数值格式，我们可以写出其一步逼近：

1. EM method

   \[\overline{X}^{t,x}(t + \Delta t) = x+f(t,x) \Delta t+\sigma(t,x)(B(t+\Delta t )-B(t)) \]

2. Milstein method

   \[\overline{X}^{t,x}(t + \Delta t) = x + f(t,x) \Delta t + \sigma(t,x)\big(B(t+\Delta t )-B(t)\big) + \left( \frac{\partial \sigma}{\partial x} \sigma \right)(t, x) \int_{t}^{t + \Delta t} \int_{t}^{s} dB_r dB_s \]

3. \(\theta\)-EM method

   \[\overline{X}^{t,x}(t + \Delta t) = x + (1-\theta)f(t,x)\Delta t + \theta f(t+\Delta t, \overline{X}^{t,x}(t + \Delta t))\Delta t + \sigma(t,x)\big(B(t+\Delta t) - B(t)\big) \]

4. \(\theta\)-Milstein method

   \[\overline{X}^{t,x}(t + \Delta t) = x + (1-\theta)f(t,x)\Delta t + \theta f(t+\Delta t, \overline{X}^{t,x}(t + \Delta t))\Delta t + \sigma(t,x)\big(B(t+\Delta t) - B(t)\big) + \left( \frac{\partial \sigma}{\partial x} \sigma \right)(t, x) \int_{t}^{t + \Delta t} \int_{t}^{s} dB_r dB_s \]

由于任何数值格式都可以写出一步逼近，我们转而研究(1), 在研究(1)之前，我们首先需要解决一个问题：

\[\left. \begin{aligned} X^{t,x}(t+\Delta t)-\overline{X}^{t,x}(t+\Delta t) \quad &\text{（一步误差）} \\ X(t_n)-\overline{X}_n \quad &\text{（全局误差）} \end{aligned} \right\} \ \text{两者有什么关系？} \]

⭐Thm (Fundamental mean-square convergence theorem) If

\[\left| \mathbb{E}\big( X^{t,x}(t+\Delta t) - \overline{X}^{t,x}(t+\Delta t) \big) \right| \leqslant C(1+|x|^2)^{\frac{1}{2}}(\Delta t)^{p_1} \quad (2) \]

\[\left( \mathbb{E}\big| X^{t,x}(t+\Delta t) - \overline{X}^{t,x}(t+\Delta t) \big|^2 \right)^{\frac{1}{2}} \leqslant C(1+|x|^2)^{\frac{1}{2}}(\Delta t)^{p_2} \quad (3) \]

where \(p_2 \geqslant \frac{1}{2}\) and \(p_1 \geqslant p_2 + \frac{1}{2}\), then

\[\max_{0\leqslant n\leqslant N} \left( \mathbb{E}\big| X(t_n) - \overline{X}_n \big|^2 \right)^{\frac{1}{2}} \leqslant C\big(1+\mathbb{E}|X_0|^2\big)^{\frac{1}{2}}(\Delta t)^p \]

with \(p = p_2 - \frac{1}{2}\).

在证明之前，我们需要弄清三个问题：

• Q1: why can we replace \(\overline{X}_n / X(t_n)\) by \(X\) in the condition of thm?

• Q2: Can we bound \(X^{t_n, X(t_n)}(t_{n+1}) - X^{t_n, \overline{X}_n}(t_{n+1})\) by \(X(t_n) - \overline{X}_n\)?

• Q3: Can we have the boundedness of \(E|\overline{X}_n|^2\)?

(Q1): Recall the freezing lemma

设 \(\mathcal{F}_1, \mathcal{F}_2\) 相互独立，\(Y\) 是 \(\mathcal{F}_1\)-可测的随机变量。
\(\Psi : \mathbb{R} \times \Omega \to \mathbb{R}\) 是 \(\mathcal{B}(\mathbb{R}) \otimes \mathcal{F}_2\)-可测的函数。

\[\mathbb{E}[\Psi(Y, \cdot) \mid \mathcal{F}_1] = \phi(Y), \quad \text{其中 } \mathbb{E}[\Psi(x, \cdot)] = \phi(x) \]

条件 (2) 的条件期望与全期望版本

设 \(Y\) 是 \(\mathcal{F}_t\)-可测的随机变量：

\[\left| \mathbb{E}\big[ X^{t,Y}(t+\Delta t) - \overline{X}^{t,Y}(t+\Delta t) \mid \mathcal{F}_t \big] \right| \leqslant C(1 + |Y|^2)^{\frac{1}{2}} (\Delta t)^{p_1} \]

两边进行 平方：

\[\left| \mathbb{E}\big[ X^{t,Y}(t+\Delta t) - \overline{X}^{t,Y}(t+\Delta t) \mid \mathcal{F}_t \big] \right|^2 \leqslant C(1 + |Y|^2)(\Delta t)^{2p_1} \]

两边取期望得到：

\[\mathbb{E} \left| \mathbb{E}\big[ X^{t,Y}(t+\Delta t) - \overline{X}^{t,Y}(t+\Delta t) \mid \mathcal{F}_t \big] \right|^2 \leqslant C(1 + \mathbb{E}|Y|^2)(\Delta t)^{2p_1} \quad (2)' \]

条件 (3) 的条件期望与全期望版本

设 \(Y\) 是 \(\mathcal{F}_t\)-可测的随机变量：

\[\mathbb{E} \left[ \big| X^{t,Y}(t+\Delta t) - \overline{X}^{t,Y}(t+\Delta t) \big|^2 \mid \mathcal{F}_t \right] \leqslant C(1 + |Y|^2)(\Delta t)^{2p_2} \]

两边直接 取期望 得到：

\[\mathbb{E}\big| X^{t,Y}(t+\Delta t) - \overline{X}^{t,Y}(t+\Delta t) \big|^2 \leqslant C(1 + \mathbb{E}|Y|^2)(\Delta t)^{2p_2} \quad (3)' \]

(Q2)(lemma) There is a representation:

\[X^{t,x}(t+\Delta t) - X^{t,y}(t+\Delta t) = x - y + Z^{x,y} \]

where

\[\mathbb{E}\left| X^{t,x}(t+\Delta t) - X^{t,y}(t+\Delta t) \right|^2 \leqslant |x-y|^2 (1 + C \Delta t) \]

and

\[\mathbb{E}|Z^{x,y}|^2 \leqslant C \Delta t |x-y|^2 \]

(Q3) lemma Under the conditions of Thm, we have

\[\max_{0\leqslant n\leqslant N} \mathbb{E}|\overline{X}_n|^2 \leqslant C(1+\mathbb{E}|X_0|^2) \]

(限于时间，我们略去基本均方收敛定理的证明，需要掌握的是如何应用于实际的方程)

---

例子

\subsubsection{EM method in the additive noise case}

SDE:

\[dX_t = f(X_t) dt + \sigma dB_t \]

数值格式:

\[\overline{X}_{n+1} = \overline{X}_n + f(\overline{X}_n)\Delta t + \sigma\Delta_{n+1} B \]

一步逼近:

\[\overline{X}^{t,x}(t+\Delta t) = x + f(x)\Delta t + \sigma(B(t+\Delta t)-B(t)) \]

\[X^{t,x}(t+\Delta t) - \overline{X}^{t,x}(t+\Delta t) = \int_{t}^{t+\Delta t} \big( f(X^{t,x}(s)) - f(x) \big) ds \]

计算 \(p_2\):

\[\begin{aligned} \mathbb{E}\big| X^{t,x}(t+\Delta t) - \overline{X}^{t,x}(t+\Delta t) \big|^2 &= \mathbb{E}\left| \int_{t}^{t+\Delta t} \big( f(X^{t,x}(s)) - f(x) \big) ds \right|^2 \\ &\leq \Delta t \int_{t}^{t+\Delta t} \mathbb{E}\big| f(X^{t,x}(s)) - f(x) \big|^2 ds \quad \text{(Cauchy-Schwarz 与 Fubini)} \\ &\leq C \Delta t \int_{t}^{t+\Delta t} \mathbb{E}\big| X^{t,x}(s) - x \big|^2 ds \quad \text{(Lipschitz 条件)} \\ &\leq C (1+|x|^2) (\Delta t)^3 \quad \text{(由 } \mathbb{E}|X^{t,x}(s)-x|^2 \leq C(1+|x|^2)(s-t) \text{)} \end{aligned} \]

因此 \(p_2 = \frac{3}{2}\).

---

计算 \(p_1\):

\[\left| \mathbb{E}\big( X^{t,x}(t+\Delta t) - \overline{X}^{t,x}(t+\Delta t) \big) \right| = \left| \mathbb{E}\left( \int_{t}^{t+\Delta t} \big( f(X^{t,x}(s)) - f(x) \big) ds \right) \right| \]

对 \(f(X^{t,x}(s)) - f(x)\) 使用 Itô 公式:

\[f(X^{t,x}(s)) - f(x) = \int_{t}^{s} \left( f'(X^{t,x}(r)) f(X^{t,x}(r)) + \frac{\sigma^2}{2} f''(X^{t,x}(r)) \right) dr + \int_{t}^{s} \sigma f'(X^{t,x}(r)) dB_r \]

取期望：

\[\mathbb{E}\big( f(X^{t,x}(s)) - f(x) \big) = \mathbb{E}\left( \int_{t}^{s} \left( f'(X^{t,x}(r)) f(X^{t,x}(r)) + \frac{\sigma^2}{2} f''(X^{t,x}(r)) \right) dr \right) \]

代回原式：

\[\begin{aligned} LHS &= \left| \int_{t}^{t+\Delta t} \int_{t}^{s} \mathbb{E}\left[ f'(X^{t,x}(r)) f(X^{t,x}(r)) + \frac{\sigma^2}{2} f''(X^{t,x}(r)) \right] dr ds \right| \\ &\leq \int_{t}^{t+\Delta t} \int_{t}^{s} \mathbb{E}\left| f'(X^{t,x}(r)) f(X^{t,x}(r)) + \frac{\sigma^2}{2} f''(X^{t,x}(r)) \right| dr ds \\ &\leq C \int_{t}^{t+\Delta t} \int_{t}^{s} \mathbb{E}\big( 1 + |X^{t,x}(r)|^2 \big) dr ds \\ &\leq C \int_{t}^{t+\Delta t} \int_{t}^{s} \big( \mathbb{E}|1 + X^{t,x}(r)|^2 \big)^{1/2} dr ds \\ &\leq C (1 + |x|^2)^{1/2} (\Delta t)^2 \end{aligned} \]

因此 \(p_1 = 2\).