SDE | Milstein method 的均方收敛阶计算

2026-06-14 14:21:33 星期日
嘻嘻，三天创造一个奇迹了，还有几十页笔记没看。

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本节讨论milstein格式的均方收敛阶.
首先对于SDE

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

对\(u(X_t)\) 使用 Ito 公式:

\[du(X_t) = \left[ u^{\prime}(X_t)f(X_t) + \frac{1}{2}u^{\prime\prime}(X_t)\sigma^2(X_t) \right] dt + u^{\prime}(X_t)\sigma(X_t) dB_t \]

定义算子 \(L\) 与 \(\Lambda\):

\[Lu(x) = u^{\prime}(x)f(x) + \frac{1}{2}u^{\prime\prime}(x)\sigma^2(x) \]

\[\Lambda u(x) = u^{\prime}(x)\sigma(x) \]

于是转化为了

\[du(X_t) = Lu(X_t) dt + \Lambda u(X_t) dB_t \]

借用上面的算子记号，我们之前介绍的milstein格式可以写成：

\[\overline{X}^{t,x}(t + \Delta t) = x + f(x) \, \Delta t + \sigma(x) \left( B(t + \Delta t) - B(t) \right) + \Lambda \sigma(x) \int_{t}^{t + \Delta t} \int_{t}^{s} dB_r dB_s, \quad \Lambda \sigma(x) = \sigma'(x) \sigma(x) \]

Thm Assume that \(|f(x)|+|f'(x)|+|\Lambda\sigma(x)|\leqslant C(1+|x|^2)^{1/2}\) and \(|\sigma(x)-\sigma(y)|\leqslant C|x-y|\). Then the Milstein method has mean-square convergence order 1.

Proof:

\[\begin{aligned} X^{t,x}(t+\Delta t)-\overline{X}^{t,x}(t+\Delta t) &= \int_{t}^{t+\Delta t}\left(f(X^{t,x}(s))-f(x)\right)ds \\ &\quad + \int_{t}^{t+\Delta t}\int_{t}^{s} \Lambda\sigma(X^{t,x}(r))dr dB_s \\ &\quad + \int_{t}^{t+\Delta t}\int_{t}^{s}\left(\Lambda\sigma(X^{t,x}(r))-\Lambda\sigma(x)\right)dB_r dB_s \\ &\triangleq I+II+III. \end{aligned} \]

\[\mathbb{E} \left[ \big| X^{t,x}(t + \Delta t) - \overline{X}^{t,x}(t + \Delta t) \big|^2 \right] \leq 3 \left( \mathbb{E} I^2 + \mathbb{E} II^2 + \mathbb{E} III^2 \right) \]

\[\begin{aligned} \mathbb{E} I^2 &= \mathbb{E} \left[ \int_t^{t+\Delta t} \left( f(X^{t,x}(s)) - f(x) \right) ds \right]^2 \\ &\leq \Delta t \int_t^{t+\Delta t} \mathbb{E} \left[ \big| f(X^{t,x}(s)) - f(x) \big|^2 \right] ds \quad \text{(利用 Cauchy-Schwarz 与 Fubini 定理)} \\ &\leq C \Delta t \int_t^{t+\Delta t} \mathbb{E} \left[ \big| X^{t,x}(s) - x \big|^2 \right] ds \quad \text{ ( Lipschitz 条件)} \\ &\leq C \Delta t \int_t^{t+\Delta t} (1 + |x|^2)(s - t) ds \quad \text{(利用 SDE 精确解性质)} \\ &\leq C (1 + |x|^2) (\Delta t)^3 \end{aligned} \]

\[\begin{aligned} \mathbb{E} II^2 &= \mathbb{E} \left[ \int_t^{t+\Delta t} \int_t^s \Lambda\sigma(X^{t,x}(r)) dB_r dB_s \right]^2 \\ &= \int_t^{t+\Delta t} \mathbb{E} \left| \int_t^s \Lambda\sigma(X^{t,x}(r)) dB_r \right|^2 ds \quad \text{(利用第一次 Itô 等距公式)} \\ &\leq \int_t^{t+\Delta t} (s-t) \int_t^s \mathbb{E} \left| \Lambda\sigma(X^{t,x}(r)) \right|^2 dr ds \quad \text{(利用 Cauchy-Schwarz 不等式榨出 } (s-t) \text{)} \\ &\leq C \int_t^{t+\Delta t} (s-t) \int_t^s (1 + \mathbb{E}|X^{t,x}(r)|^2) dr ds \quad \text{(利用 } \Lambda\sigma \text{ 的线性增长条件)} \\ &\leq C (1 + |x|^2) \int_t^{t+\Delta t} (s-t)^2 ds \quad \text{(利用二阶矩有界性与内层积分)} \\ &= C (1 + |x|^2) \cdot \frac{1}{3}(\Delta t)^3 \\ &\leq C (1 + |x|^2) (\Delta t)^3 \end{aligned} \]

\[\begin{aligned} \mathbb{E} III^2 &= \mathbb{E} \left[ \int_t^{t+\Delta t} \int_t^s \left( \Lambda\sigma(X^{t,x}(r)) - \Lambda\sigma(x) \right) dB_r dB_s \right]^2 \\ &= \int_t^{t+\Delta t} \mathbb{E} \left[ \int_t^s \left( \Lambda\sigma(X^{t,x}(r)) - \Lambda\sigma(x) \right) dB_r \right]^2 ds \quad \text{(利用第一次 Itô 等距公式)} \\ &= \int_t^{t+\Delta t} \int_t^s \mathbb{E} \left[ \big| \Lambda\sigma(X^{t,x}(r)) - \Lambda\sigma(x) \big|^2 \right] dr ds \quad \text{(利用第二次 Itô 等距公式)} \\ &\leq C \int_t^{t+\Delta t} \int_t^s \mathbb{E} \left[ \big| X^{t,x}(r) - x \big|^2 \right] dr ds \quad \text{(利用 } \Lambda\sigma \text{ 的 Lipschitz 条件)} \\ &\leq C \int_t^{t+\Delta t} \int_t^s (1 + |x|^2)(r - t) dr ds \quad \text{(利用精确解性质)} \\ &= C (1 + |x|^2) \int_t^{t+\Delta t} \frac{1}{2} (s - t)^2 ds \\ &\leq C (1 + |x|^2) (\Delta t)^3 \end{aligned} \]

综合 \(I, II, III\) 三项的放缩结果：

\[\Rightarrow p_2 = \frac{3}{2} \]

下面计算\(p_1\)

\[\begin{aligned} \left| \mathbb{E} \big( X^{t,x}(t + \Delta t) - \overline{X}^{t,x}(t + \Delta t) \big) \right| &= | \mathbb{E} I | \\ &= \left| \mathbb{E} \left[ \int_t^{t+\Delta t} \big( f(X^{t,x}(s)) - f(x) \big) ds \right] \right| \end{aligned} \]

对 \(f(X^{t,x}(s)) - f(x)\) 使用 Itô 公式：

\[\begin{aligned} &= \left| \mathbb{E} \left[ \int_t^{t+\Delta t} \int_t^s Lf(X^{t,x}(r)) dr ds + \int_t^{t+\Delta t} \int_t^s \Lambda f(X^{t,x}(r)) dB_r ds \right] \right| \\ &= \left| \int_t^{t+\Delta t} \int_t^s \mathbb{E} \left[ Lf(X^{t,x}(r)) \right] dr ds + \int_t^{t+\Delta t} \int_t^s \mathbb{E} \left[ \Lambda f(X^{t,x}(r)) dB_r \right] ds \right| \quad \text{(Fubini 定理)} \\ &= \left| \int_t^{t+\Delta t} \int_t^s \mathbb{E} \left[ Lf(X^{t,x}(r)) \right] dr ds \right| \quad \text{(Itô 积分项期望为 0)} \\ &\leq \int_t^{t+\Delta t} \int_t^s \mathbb{E} \left[ \big| Lf(X^{t,x}(r)) \big| \right] dr ds \quad \text{(绝对值移入积分内)} \\ &\leq C \int_t^{t+\Delta t} \int_t^s \mathbb{E} \left[ (1 + |X^{t,x}(r)|^2)^{1/2} \right] dr ds \quad \text{(利用 } Lf \text{ 的多项式增长条件)} \\ &\leq C \int_t^{t+\Delta t} \int_t^s \left( \mathbb{E} \left[ 1 + |X^{t,x}(r)|^2 \right] \right)^{1/2} dr ds \quad \text{(Cauchy-Schwarz 不等式)} \\ &\leq C (1 + |x|^2)^{1/2} \int_t^{t+\Delta t} \int_t^s 1 \, dr ds \quad \text{(真实解的二阶矩均匀有界性)} \\ &= C (1 + |x|^2)^{1/2} \cdot \frac{1}{2} (\Delta t)^2 \\ &\leq C (1 + |x|^2)^{1/2} (\Delta t)^2 \end{aligned} \]

综上，根据基本均方收敛定理，该数值方法的均方收敛阶是1.