本章导读 Ito 公式是随机微积分的"基本定理"，有多个版本。先回顾普通微积分：\(C^k\) 函数的一阶 Taylor + 伸缩求和给出 \(f(1)=f(0)+\int_0^1 f'(t)dt\)；Ito 公式沿同一思路，但需保留 Taylor 展开的一阶与二阶项（因 \((dB_t)^2=dt\)）。§14.1 Ito 公式 1（Thm 14.1：\(df(B_t)=f'(B_t)dB_t+\frac12 f''(B_t)dt\) (14.3)）。§14.2 Ito 公式 2（Thm 14.2：含时间的 \(f(t,x)\)，多出 \(\dot f\,dt\) (14.7)；Ex 14.1 几何布朗运动）。§14.3 Ito 公式 3（Dummy's Guide：\((dt)^2=0\)、\((dB_t)^2=dt\)、\(dt\,dB_t=0\) Fact 14.1；Prop 14.1 二次变差 \(d\langle X\rangle_t=A_t^2 dt\)；Thm 14.3：一般 \(dX_t=R_t dt+A_t dB_t\) 下的 Ito (14.13)–(14.16)；Ex 14.2 指数 SDE 解 \(X_t=X_0 e^{\int A dB-\frac12\int A^2 ds}\)）。§14.4 乘积公式（Def 14.1 协变差、Prop 14.2 \(d\langle X,Y\rangle_t=A_t C_t dt\)、Thm 14.4 \(d(X_tY_t)=X_t dY_t+Y_t dX_t+d\langle X,Y\rangle_t\) (14.22)；多维 Fact 14.2、Ex 14.3）。§14.5 Ito 公式 4（Thm 14.5 多维向量形式 (14.23)；Ex 14.4 \(\nabla,\Delta\) 形式）。无图。

从普通微积分到 Ito / From ordinary calculus to Ito 设 \(f\) 是 \(C^k\) 函数（脚注 14.1：\(f\) 有直到 \(k\) 阶的连续导数）。一阶 Taylor：\(f(a+b)=f(a)+f'(a)b+o(b)\)（\(o(b)/b\to0\)）。伸缩求和 \(f(1)=f(0)+\sum_{j=1}^n\left[f\!\left(\tfrac jn\right)-f\!\left(\tfrac{j-1}n\right)\right]\) (14.1)，对每项一阶 Taylor 后取极限得 \(f(1)=f(0)+\int_0^1 f'(t)dt\)。Ito 公式沿同一思路，但需 Taylor 展开的一阶与二阶项。Let \(f\) be a \(C^k\) function (footnote 14.1: \(f\) has continuous derivatives up to order \(k\)). First-order Taylor: \(f(a+b)=f(a)+f'(a)b+o(b)\) (\(o(b)/b\to0\)). The telescoping sum \(f(1)=f(0)+\sum_{j=1}^n\left[f\!\left(\tfrac jn\right)-f\!\left(\tfrac{j-1}n\right)\right]\) (14.1), after a first-order Taylor on each term and taking the limit, gives \(f(1)=f(0)+\int_0^1 f'(t)dt\). Ito's formula follows the same idea but requires both the first- and second-order Taylor terms.

定理 14.1（Ito 公式 1）与注 14.1、14.2 / Theorem 14.1 (Ito's Formula 1) and Remarks 14.1, 14.2 设 \(f\) 是 \(C^2\) 函数，\(\{B_t\}\) 是标准布朗运动。则对 \(\forall t\)，Suppose \(f\) is a \(C^2\) function and \(\{B_t\}\) is a standard Brownian motion. Then for all \(t\),

$$f(B_t)=f(B_0)+\int_0^t f'(B_s)\,dB_s+\frac12\int_0^t f''(B_s)\,ds,\tag{14.2}$$

等价地 \(df(B_t)=f'(B_t)\,dB_t+\frac12 f''(B_t)\,dt\) (14.3)。注 14.1：(14.2) 与 (14.3) 表达同一件事，(14.3) 是 (14.2) 的简写。注 14.2：(14.3) 的直觉是——在每一时刻 \(t\)，\(f(B_t)\) 像漂移为 \(\frac12 f''(B_t)\)、方差为 \((f'(B_t))^2\) 的布朗运动那样演化。equivalently \(df(B_t)=f'(B_t)\,dB_t+\frac12 f''(B_t)\,dt\) (14.3). Remark 14.1: (14.2) and (14.3) state the same thing; (14.3) is shorthand for (14.2). Remark 14.2: the intuition of (14.3) is that at each time \(t\), \(f(B_t)\) evolves like a Brownian motion with drift \(\frac12 f''(B_t)\) and variance \((f'(B_t))^2\).

定理 14.1 证明 / Proof of Theorem 14.1 设 \(t=1\)。伸缩求和 \(f(B_1)-f(B_0)=\sum_{j=1}^n\left[f(B_{\frac jn})-f(B_{\frac{j-1}n})\right]\) (14.4)。二阶 Taylor \(f(a+b)=f(a)+f'(a)b+\frac{f''(a)}2 b^2+o(b^2)\) (14.5)，应用于每项（\(\Delta_{j,n}=B_{\frac jn}-B_{\frac{j-1}n}\)）得 (14.6)。代入 (14.4)：Set \(t=1\). The telescoping sum \(f(B_1)-f(B_0)=\sum_{j=1}^n\left[f(B_{\frac jn})-f(B_{\frac{j-1}n})\right]\) (14.4). The second-order Taylor \(f(a+b)=f(a)+f'(a)b+\frac{f''(a)}2 b^2+o(b^2)\) (14.5), applied to each term (with \(\Delta_{j,n}=B_{\frac jn}-B_{\frac{j-1}n}\)), gives (14.6). Plug into (14.4):

$$f(B_1)-f(B_0)=\underbrace{\lim_{n\to\infty}\sum_{j=1}^n f'(B_{\frac{j-1}n})\Delta_{j,n}}_{\text{Part }A}+\underbrace{\lim_{n\to\infty}\sum_{j=1}^n\tfrac12 f''(B_{\frac{j-1}n})\Delta_{j,n}^2}_{\text{Part }B}+\underbrace{\lim_{n\to\infty}\sum_{j=1}^n o(\Delta_{j,n}^2)}_{\text{Part }C}.$$

Part A 是简单过程逼近的随机积分，\(=\int_0^1 f'(B_t)\,dB_t\)。Part C：BM 增量 \((\Delta_{j,n})^2\approx\frac1n\)，Part C 是 \(n\) 个比 \(\frac1n\) 更小阶的对象之和，故 $=0$。Part B：记 \(h(t)\equiv f''(B_t)\) 连续，用阶梯函数 \(h_\varepsilon\) 逼近，\(\lim_{n\to\infty}\sum h(\tfrac{j-1}n)(B_{\frac jn}-B_{\frac{j-1}n})^2=\int_0^1 h(t)dt\)，故 Part B \(=\frac12\int_0^1 f''(B_t)dt\)。于是 \(f(B_1)-f(B_0)=\int_0^1 f'(B_t)dB_t+\frac12\int_0^1 f''(B_t)dt\)，一般地即 (14.2)。\(\blacksquare\)Part A is the simple-process approximation of a stochastic integral, \(=\int_0^1 f'(B_t)\,dB_t\). Part C: the BM increment \((\Delta_{j,n})^2\approx\frac1n\), so Part C is a sum of \(n\) objects of order smaller than \(\frac1n\), hence $=0$. Part B: writing \(h(t)\equiv f''(B_t)\) (continuous), approximate by a step function \(h_\varepsilon\); \(\lim_{n\to\infty}\sum h(\tfrac{j-1}n)(B_{\frac jn}-B_{\frac{j-1}n})^2=\int_0^1 h(t)dt\), so Part B \(=\frac12\int_0^1 f''(B_t)dt\). Thus \(f(B_1)-f(B_0)=\int_0^1 f'(B_t)dB_t+\frac12\int_0^1 f''(B_t)dt\), which in general is (14.2). \(\blacksquare\)

定理 14.2（Ito 公式 2）/ Theorem 14.2 (Ito's Formula 2) 设 \(\{B_t\}\) 是标准布朗运动。若函数 \(f(t,x):[0,\infty)\times\mathbb R\to\mathbb R\) 关于 \(t\) 是 \(C^1\)、关于 \(x\) 是 \(C^2\)，则Let \(\{B_t\}\) be a standard Brownian motion. If \(f(t,x):[0,\infty)\times\mathbb R\to\mathbb R\) is \(C^1\) in \(t\) and \(C^2\) in \(x\), then

$$f(t,B_t)-f(0,B_0)=\int_0^t\left(\dot f(s,B_s)+\frac12 f''(s,B_s)\right)ds+\int_0^t f'(s,B_s)\,dB_s,\tag{14.7}$$

其中 \(\dot f=\dfrac{\partial f}{\partial t}\)、\(f'=\dfrac{\partial f}{\partial x}\)、\(f''=\dfrac{\partial^2 f}{(\partial x)^2}\)。简写 \(df(t,B_t)=\left(\dot f+\frac12 f''\right)dt+f'\,dB_t\)。where \(\dot f=\dfrac{\partial f}{\partial t}\), \(f'=\dfrac{\partial f}{\partial x}\), \(f''=\dfrac{\partial^2 f}{(\partial x)^2}\). Shorthand: \(df(t,B_t)=\left(\dot f+\frac12 f''\right)dt+f'\,dB_t\).

定理 14.2 证明 / Proof of Theorem 14.2 设 \(t=1\)，伸缩求和后拆为时间方向（Part 1）与空间方向（Part 2）：\(f(\tfrac jn,B_{\frac jn})-f(\tfrac{j-1}n,B_{\frac{j-1}n})=\left[f(\tfrac jn,B_{\frac jn})-f(\tfrac{j-1}n,B_{\frac jn})\right]+\left[f(\tfrac{j-1}n,B_{\frac jn})-f(\tfrac{j-1}n,B_{\frac{j-1}n})\right]\)。Part 1 对 \(\dot f\) 一阶 Taylor：\(\sum\left[\frac1n\dot f(\tfrac{j-1}n,B_{\frac jn})+o(\tfrac1n)\right]\to\int_0^1\dot f(s,B_s)ds\)。Part 2 与定理 14.1 同理（二阶 Taylor）\(\to\int_0^1 f'(s,B_s)dB_s+\frac12\int_0^1 f''(s,B_s)ds\)。合并即 (14.7)。\(\blacksquare\)Set \(t=1\); after telescoping, split into the time direction (Part 1) and the space direction (Part 2): \(f(\tfrac jn,B_{\frac jn})-f(\tfrac{j-1}n,B_{\frac{j-1}n})=\left[f(\tfrac jn,B_{\frac jn})-f(\tfrac{j-1}n,B_{\frac jn})\right]+\left[f(\tfrac{j-1}n,B_{\frac jn})-f(\tfrac{j-1}n,B_{\frac{j-1}n})\right]\). Part 1, a first-order Taylor in \(\dot f\): \(\sum\left[\frac1n\dot f(\tfrac{j-1}n,B_{\frac jn})+o(\tfrac1n)\right]\to\int_0^1\dot f(s,B_s)ds\). Part 2, as in Theorem 14.1 (second-order Taylor), \(\to\int_0^1 f'(s,B_s)dB_s+\frac12\int_0^1 f''(s,B_s)ds\). Combining gives (14.7). \(\blacksquare\)

例 14.1（几何布朗运动）/ Example 14.1 (Geometric Brownian Motion) 设 \(\{B_t\}\) 标准 BM，\(f(t,x)=e^{at+bx}\)（\(a,b\in\mathbb R\)），记 \(X_t=f(t,B_t)\)，求 \(dX_t\)。解：\(\dot f=aX_t\)、\(f'=bX_t\)、\(f''=b^2 X_t\)，由定理 14.2，Let \(\{B_t\}\) be a standard BM, \(f(t,x)=e^{at+bx}\) (\(a,b\in\mathbb R\)), \(X_t=f(t,B_t)\); solve for \(dX_t\). Solution: \(\dot f=aX_t\), \(f'=bX_t\), \(f''=b^2 X_t\), so by Theorem 14.2,

$$dX_t=\left(aX_t+\frac12 b^2 X_t\right)dt+bX_t\,dB_t=X_t\left[\left(a+\frac12 b^2\right)dt+b\,dB_t\right].\tag{14.11}$$

满足 (14.11) 的 \(\{X_t\}\) 是几何布朗运动。The \(\{X_t\}\) satisfying (14.11) is a geometric Brownian motion.

事实 14.1（Dummy's Guide）与命题 14.1（二次变差）/ Fact 14.1 and Proposition 14.1 引入 Ito 公式 3 前，先给出随机微积分的"傻瓜指南"与二次变差。事实 14.1（Dummy's Guide to Stochastic Calculus）：\(\{B_t\}\) 标准 BM，则 \((dt)^2=0\)、\((dB_t)^2=dt\)、\((dt)(dB_t)=0\)。命题 14.1：\(\{B_t\}\) 标准 BM，\(\{X_t\}\) 满足 \(dX_t=R_t dt+A_t dB_t\)（\(X_t=X_0+\int_0^t R_s ds+\int_0^t A_s dB_s\)，\(\{A_t\},\{R_t\}\) 适应）。其二次变差 \(\langle X\rangle_t=\lim_{n\to\infty}\sum_{j:\frac jn\le t}\left(X_{\frac jn}-X_{\frac{j-1}n}\right)^2\) (14.12) 满足 \(\langle X\rangle_t=\int_0^t A_s^2\,ds\)，简写 \(d\langle X\rangle_t=A_t^2\,dt\)。Before Ito's Formula 3, we give the "dummy's guide" to stochastic calculus and the quadratic variation. Fact 14.1 (Dummy's Guide to Stochastic Calculus): for a standard BM \(\{B_t\}\), \((dt)^2=0\), \((dB_t)^2=dt\), \((dt)(dB_t)=0\). Proposition 14.1: let \(\{B_t\}\) be a standard BM and \(\{X_t\}\) satisfy \(dX_t=R_t dt+A_t dB_t\) (\(X_t=X_0+\int_0^t R_s ds+\int_0^t A_s dB_s\), with \(\{A_t\},\{R_t\}\) adapted). Its quadratic variation \(\langle X\rangle_t=\lim_{n\to\infty}\sum_{j:\frac jn\le t}\left(X_{\frac jn}-X_{\frac{j-1}n}\right)^2\) (14.12) satisfies \(\langle X\rangle_t=\int_0^t A_s^2\,ds\), shorthand \(d\langle X\rangle_t=A_t^2\,dt\).

命题 14.1 证明 / Proof of Proposition 14.1 考虑 (14.12) 每一项 \(\left(X_{\frac jn}-X_{\frac{j-1}n}\right)^2=\left[\left(\int_{\frac{j-1}n}^{\frac jn}R_s ds\right)+\left(\int_{\frac{j-1}n}^{\frac jn}A_s dB_s\right)\right]^2\)，展开为三部分：\(R\) 积分平方项 \(\sim\frac1{n^2}\)、\(A\) 积分平方项 \(\sim\frac1n\)、交叉项 \(\sim\frac1{n^{3/2}}\)。求和取极限后只有 \(A\) 积分平方项存活：\(\lim_{n\to\infty}\sum A_{\frac{j-1}n}^2\left(B_{\frac jn}-B_{\frac{j-1}n}\right)^2=\int_0^t A_s^2\,ds\)（或由 Dummy's Guide \((dB)^2=dt\) 直接得，亦类似定理 14.1 的 Part B）。\(\blacksquare\)Consider each term of (14.12), \(\left(X_{\frac jn}-X_{\frac{j-1}n}\right)^2=\left[\left(\int_{\frac{j-1}n}^{\frac jn}R_s ds\right)+\left(\int_{\frac{j-1}n}^{\frac jn}A_s dB_s\right)\right]^2\), expanding into three parts: the \(R\)-integral square \(\sim\frac1{n^2}\), the \(A\)-integral square \(\sim\frac1n\), the cross term \(\sim\frac1{n^{3/2}}\). After summing and taking the limit, only the \(A\)-integral square survives: \(\lim_{n\to\infty}\sum A_{\frac{j-1}n}^2\left(B_{\frac jn}-B_{\frac{j-1}n}\right)^2=\int_0^t A_s^2\,ds\) (or directly by the Dummy's Guide \((dB)^2=dt\), also analogous to Part B in Theorem 14.1). \(\blacksquare\)

定理 14.3（Ito 公式 3）/ Theorem 14.3 (Ito's Formula 3) 设 \(\{B_t\}\) 标准 BM，\(f(t,x):[0,\infty)\times\mathbb R\to\mathbb R\) 关于 \(t\) 是 \(C^1\)、关于 \(x\) 是 \(C^2\)。\(\{X_t\}\) 由 \(dX_t=R_t dt+A_t dB_t\) 定义（\(\{R_t\},\{A_t\}\) 适应）。则Let \(\{B_t\}\) be a standard BM, \(f(t,x):[0,\infty)\times\mathbb R\to\mathbb R\) be \(C^1\) in \(t\) and \(C^2\) in \(x\), and \(\{X_t\}\) be defined by \(dX_t=R_t dt+A_t dB_t\) (\(\{R_t\},\{A_t\}\) adapted). Then

$$f(t,X_t)-f(0,X_0)=\int_0^t\left(\dot f+R_s f'+\frac12 A_s^2 f''\right)(s,X_s)\,ds+\int_0^t A_s f'(s,X_s)\,dB_s.\tag{14.13}$$

三种简写：\(df=\left(\dot f+R_t f'+\frac12 A_t^2 f''\right)dt+A_t f'\,dB_t\) (14.14)；\(df=\left(\dot f+\frac12 A_t^2 f''\right)dt+f'\,dX_t\) (14.15)；\(df=\left(\dot f+R_t f'\right)dt+\frac12 f''\,d\langle X\rangle_t+A_t f'\,dB_t\) (14.16)。Three shorthands: \(df=\left(\dot f+R_t f'+\frac12 A_t^2 f''\right)dt+A_t f'\,dB_t\) (14.14); \(df=\left(\dot f+\frac12 A_t^2 f''\right)dt+f'\,dX_t\) (14.15); \(df=\left(\dot f+R_t f'\right)dt+\frac12 f''\,d\langle X\rangle_t+A_t f'\,dB_t\) (14.16).

定理 14.3 证明 / Proof of Theorem 14.3 由 Ito 公式 2（定理 14.2）直接得 \(df(t,X_t)=\left(\dot f+\frac12 A_t^2 f''\right)dt+f'\,dX_t\) (14.17)（其中 \(A_t^2\) 来自 Taylor 二阶项的系数，即 \(d\langle X\rangle_t=A_t^2 dt\)）。把 \(dX_t=R_t dt+A_t dB_t\) 代入 (14.17)：\(df=\left(\dot f+\frac12 f''A_t^2\right)dt+f'(R_t dt+A_t dB_t)=\left(\dot f+f'R_t+\frac12 f''A_t^2\right)dt+f'A_t dB_t\)，即 (14.14)。(14.16) 由把 \(d\langle X\rangle_t=A_t^2 dt\) 代入 (14.14) 得。\(\blacksquare\)From Ito's Formula 2 (Theorem 14.2) directly, \(df(t,X_t)=\left(\dot f+\frac12 A_t^2 f''\right)dt+f'\,dX_t\) (14.17) (where \(A_t^2\) comes from the coefficient of the second-order Taylor term, i.e. \(d\langle X\rangle_t=A_t^2 dt\)). Plug \(dX_t=R_t dt+A_t dB_t\) into (14.17): \(df=\left(\dot f+\frac12 f''A_t^2\right)dt+f'(R_t dt+A_t dB_t)=\left(\dot f+f'R_t+\frac12 f''A_t^2\right)dt+f'A_t dB_t\), i.e. (14.14). (14.16) is obtained by plugging \(d\langle X\rangle_t=A_t^2 dt\) into (14.14). \(\blacksquare\)

例 14.2（指数随机微分方程）/ Example 14.2 (Exponential SDE) 设 \(\{B_t\}\) 标准 BM，\(\{X_t\}\) 由 \(dX_t=A_t X_t\,dB_t\) (14.18) 定义（\(\{A_t\}\) 适应）。证明 \(X_t=X_0 e^{\left[\left(\int_0^t A_s dB_s\right)-\frac12\left(\int_0^t A_s^2 ds\right)\right]}\)。Let \(\{B_t\}\) be a standard BM and \(\{X_t\}\) be defined by \(dX_t=A_t X_t\,dB_t\) (14.18) (\(\{A_t\}\) adapted). Show that \(X_t=X_0 e^{\left[\left(\int_0^t A_s dB_s\right)-\frac12\left(\int_0^t A_s^2 ds\right)\right]}\).

例 14.2 证明（猜测并验证）/ Solution to Example 14.2 (guess and verify) 令 \(Y_t=\left[\left(\int_0^t A_s dB_s\right)-\frac12\left(\int_0^t A_s^2 ds\right)\right]\)，等价于 \(dY_t=-\frac12 A_t^2 dt+A_t dB_t\)。验证 \(X_t=X_0 e^{Y_t}=f(t,Y_t)\) 解 (14.18)。由 Ito 公式 3（定理 14.3，此处 \(\dot f=0\)、\(f'=f''=X_0 e^{Y_t}=X_t\)，且 \(Y_t\) 的漂移 \(-\frac12 A_t^2\)、扩散 \(A_t\)）：Let \(Y_t=\left[\left(\int_0^t A_s dB_s\right)-\frac12\left(\int_0^t A_s^2 ds\right)\right]\), equivalently \(dY_t=-\frac12 A_t^2 dt+A_t dB_t\). Verify that \(X_t=X_0 e^{Y_t}=f(t,Y_t)\) solves (14.18). By Ito's Formula 3 (Theorem 14.3, here \(\dot f=0\), \(f'=f''=X_0 e^{Y_t}=X_t\), and \(Y_t\) has drift \(-\frac12 A_t^2\), diffusion \(A_t\)):

$$dX_t=\left[X_t\left(-\frac12 A_t^2\right)+\frac12 X_t A_t^2\right]dt+X_t A_t\,dB_t=A_t X_t\,dB_t,$$

与 (14.18) 一致，故 \(X_t=X_0 e^{\left[\left(\int A dB\right)-\frac12\left(\int A^2 ds\right)\right]}\) 是 (14.18) 的解。\(\blacksquare\)consistent with (14.18), so \(X_t=X_0 e^{\left[\left(\int A dB\right)-\frac12\left(\int A^2 ds\right)\right]}\) is the solution to (14.18). \(\blacksquare\)

定义 14.1（协变差过程）与命题 14.2 / Definition 14.1 (Covariation) and Proposition 14.2 设 \(\{B_t\}\) 标准 BM，\(\{R_t\},\{A_t\},\{S_t\},\{C_t\}\) 适应，\(dX_t=R_t dt+A_t dB_t\) (14.19)、\(dY_t=S_t dt+C_t dB_t\) (14.20)。定义 14.1（协变差过程）：\(\langle X,Y\rangle_t=\lim_{n\to\infty}\sum_{j:\frac jn\le t}\left[\left(X_{\frac jn}-X_{\frac{j-1}n}\right)\left(Y_{\frac jn}-Y_{\frac{j-1}n}\right)\right]\)。注 14.3：自身的协变差就是二次变差，\(\langle X,X\rangle_t=\langle X\rangle_t\)。命题 14.2：\(\langle X,Y\rangle_t=\int_0^t A_s C_s\,ds\)，简写 \(d\langle X,Y\rangle_t=A_t C_t\,dt\)（证明与命题 14.1 同）。Let \(\{B_t\}\) be a standard BM, \(\{R_t\},\{A_t\},\{S_t\},\{C_t\}\) adapted, \(dX_t=R_t dt+A_t dB_t\) (14.19), \(dY_t=S_t dt+C_t dB_t\) (14.20). Definition 14.1 (Covariation process): \(\langle X,Y\rangle_t=\lim_{n\to\infty}\sum_{j:\frac jn\le t}\left[\left(X_{\frac jn}-X_{\frac{j-1}n}\right)\left(Y_{\frac jn}-Y_{\frac{j-1}n}\right)\right]\). Remark 14.3: the covariation with self is the quadratic variation, \(\langle X,X\rangle_t=\langle X\rangle_t\). Proposition 14.2: \(\langle X,Y\rangle_t=\int_0^t A_s C_s\,ds\), shorthand \(d\langle X,Y\rangle_t=A_t C_t\,dt\) (proof same as Proposition 14.1).

定理 14.4（乘积公式）/ Theorem 14.4 (Product Formula) 设 \(\{B_t\}\) 标准 BM，\(\{X_t\},\{Y_t\}\) 由 (14.19)、(14.20) 定义。则 \(d(X_tY_t)=X_t dY_t+Y_t dX_t+A_t C_t\,dt\) (14.21)，等价地 \(d(X_tY_t)=X_t dY_t+Y_t dX_t+d\langle X,Y\rangle_t\) (14.22)。Let \(\{B_t\}\) be a standard BM and \(\{X_t\},\{Y_t\}\) be defined by (14.19), (14.20). Then \(d(X_tY_t)=X_t dY_t+Y_t dX_t+A_t C_t\,dt\) (14.21), equivalently \(d(X_tY_t)=X_t dY_t+Y_t dX_t+d\langle X,Y\rangle_t\) (14.22).

定理 14.4 证明 / Proof of Theorem 14.4 分解 \(d(X_tY_t)=X_{t+dt}Y_{t+dt}-X_tY_t=(X_{t+dt}-X_t)(Y_{t+dt}-Y_t)+X_t(Y_{t+dt}-Y_t)+(X_{t+dt}-X_t)Y_t=dX_t dY_t+X_t dY_t+Y_t dX_t\)。其中 \(dX_t dY_t=(R_t dt+A_t dB_t)(S_t dt+C_t dB_t)=A_t C_t dt\)（由 Dummy's Guide），或直接 \(dX_t dY_t=d\langle X,Y\rangle_t\)。\(\blacksquare\)Decompose \(d(X_tY_t)=X_{t+dt}Y_{t+dt}-X_tY_t=(X_{t+dt}-X_t)(Y_{t+dt}-Y_t)+X_t(Y_{t+dt}-Y_t)+(X_{t+dt}-X_t)Y_t=dX_t dY_t+X_t dY_t+Y_t dX_t\). Here \(dX_t dY_t=(R_t dt+A_t dB_t)(S_t dt+C_t dB_t)=A_t C_t dt\) (by the Dummy's Guide), or directly \(dX_t dY_t=d\langle X,Y\rangle_t\). \(\blacksquare\)

事实 14.2（多维协变差）与例 14.3 / Fact 14.2 (Multi-dim covariation) and Example 14.3 事实 14.2：设 \(\mathbf B_t=(B_{1,t},\dots,B_{d,t})\)，\(\{B_{1,t}\},\dots,\{B_{d,t}\}\) 独立一维标准 BM。\(\mathbf X_t=(X_{1,t},\dots,X_{m,t})\)，\(dX_{i,t}=R_{i,t}dt+\sum_{j=1}^d A_{ij,t}dB_{j,t}\)。由命题 6.1，\(\langle B_i,B_i\rangle_t=t\)；由独立性 \(i\neq k\) 时 \(\langle B_i,B_k\rangle_t=0\)。故 \(\langle X_1,X_2\rangle_t=\int_0^t\left(\sum_{j=1}^d A_{1j,s}A_{2j,s}\right)ds\)，简写 \(d\langle X_1,X_2\rangle_t=\left(\sum_{j=1}^d A_{1j,t}A_{2j,t}\right)dt\)。例 14.3：\(dX_t=R_t dt+\sum_{j=1}^d A_{j,t}dB_{j,t}\)、\(dY_t=S_t dt+\sum_{j=1}^d C_{j,t}dB_{j,t}\)，由乘积公式 (14.22)，\(d(X_tY_t)=\left(X_t S_t+Y_t R_t+\sum_{j=1}^d A_{j,t}C_{j,t}\right)dt+\left[\sum_{j=1}^d\left(X_t C_{j,t}+Y_t A_{j,t}\right)dB_{j,t}\right]\)。Fact 14.2: let \(\mathbf B_t=(B_{1,t},\dots,B_{d,t})\) with \(\{B_{1,t}\},\dots,\{B_{d,t}\}\) independent one-dimensional standard BMs, and \(\mathbf X_t=(X_{1,t},\dots,X_{m,t})\) with \(dX_{i,t}=R_{i,t}dt+\sum_{j=1}^d A_{ij,t}dB_{j,t}\). By Proposition 6.1, \(\langle B_i,B_i\rangle_t=t\); by independence, for \(i\neq k\), \(\langle B_i,B_k\rangle_t=0\). So \(\langle X_1,X_2\rangle_t=\int_0^t\left(\sum_{j=1}^d A_{1j,s}A_{2j,s}\right)ds\), shorthand \(d\langle X_1,X_2\rangle_t=\left(\sum_{j=1}^d A_{1j,t}A_{2j,t}\right)dt\). Example 14.3: for \(dX_t=R_t dt+\sum_{j=1}^d A_{j,t}dB_{j,t}\), \(dY_t=S_t dt+\sum_{j=1}^d C_{j,t}dB_{j,t}\), by the product formula (14.22), \(d(X_tY_t)=\left(X_t S_t+Y_t R_t+\sum_{j=1}^d A_{j,t}C_{j,t}\right)dt+\left[\sum_{j=1}^d\left(X_t C_{j,t}+Y_t A_{j,t}\right)dB_{j,t}\right]\).

定理 14.5（Ito 公式 4）与注 14.4 / Theorem 14.5 (Ito's Formula 4) and Remark 14.4 设函数 \(f(t,\mathbf x)\)（\(\mathbf x=(x_1,\dots,x_m)'\)）关于 \(t\) 是 \(C^1\)、关于 \(\mathbf x\) 是 \(C^2\)。\(\{X_{1,t}\},\dots,\{X_{m,t}\}\) 构成 \(\mathbf X_t=(X_{1,t},\dots,X_{m,t})'\)。则Let \(f(t,\mathbf x)\) (\(\mathbf x=(x_1,\dots,x_m)'\)) be \(C^1\) in \(t\) and \(C^2\) in \(\mathbf x\), and let \(\{X_{1,t}\},\dots,\{X_{m,t}\}\) form \(\mathbf X_t=(X_{1,t},\dots,X_{m,t})'\). Then

$$df(t,\mathbf X_t)=\dot f(t,\mathbf X_t)\,dt+\sum_{j=1}^m\partial_j f(t,\mathbf X_t)\,dX_{j,t}+\frac12\sum_{j=1}^m\sum_{k=1}^m\partial_{jk}f(t,\mathbf X_t)\,d\langle X_j,X_k\rangle_t,\tag{14.23}$$

其中 \(\partial_j f=\dfrac{\partial f}{\partial x_j}\)、\(\partial_{jk}f=\dfrac{\partial^2 f}{\partial x_j\partial x_k}\)（证明与定理 14.2 同理）。注 14.4：当 \(\mathbf X_t\) 退化为一维 \(X_t\) 时，(14.23) 与 (14.16)、(14.14) 完全相同。where \(\partial_j f=\dfrac{\partial f}{\partial x_j}\), \(\partial_{jk}f=\dfrac{\partial^2 f}{\partial x_j\partial x_k}\) (proof analogous to Theorem 14.2). Remark 14.4: when \(\mathbf X_t\) degenerates to the one-dimensional \(X_t\), (14.23) is exactly the same as (14.16) and (14.14).

例 14.4（梯度与 Laplace 形式）/ Example 14.4 (gradient and Laplace form) 设 \(\mathbf B_t\) 如事实 14.2，对 (14.23) 取 \(\mathbf X_t=\mathbf B_t\)：因 \(d\langle B_j,B_k\rangle_t=0\)（\(j\neq k\)）、\(d\langle B_j\rangle_t=dt\)，故Let \(\mathbf B_t\) be as in Fact 14.2; applying \(\mathbf X_t=\mathbf B_t\) to (14.23): since \(d\langle B_j,B_k\rangle_t=0\) (\(j\neq k\)) and \(d\langle B_j\rangle_t=dt\),

$$df(t,\mathbf B_t)=\dot f(t,\mathbf B_t)\,dt+\nabla f(t,\mathbf B_t)\cdot d\mathbf B_t+\frac12\Delta f(t,\mathbf B_t)\,dt,$$

其中 \(\nabla f=(\partial_1 f,\dots,\partial_m f)'\) 是梯度，\(\Delta f=\sum_{j=1}^m\partial_{jj}f\) 是 Laplace 算子。where \(\nabla f=(\partial_1 f,\dots,\partial_m f)'\) is the gradient and \(\Delta f=\sum_{j=1}^m\partial_{jj}f\) is the Laplace operator.