本章导读 Part II 开篇，构造关于布朗运动的随机积分。§13.1 随机微分方程与随机积分（Def 13.1 SDE \(dX_t=R_t dt+A_t dB_t\) (13.1)，理解为 \(X_t\) 在每一时刻以漂移 \(R_t\)、方差 \(A_t^2\) 演化；Rmk 13.1 \(A_t\)=下注、\(dB_t\)=赌局回报，故 \(dB_t\) 须在 \(A_t\) 之后；导数难定义故先定义积分 \(X_t=X_0+\int R_s ds+\int A_s dB_s\) (13.2)，只需处理随机部分 \(\int A_s dB_s\)）。§13.2 随机积分：13.2.1 简单过程（Def 13.2 简单过程、Def 13.3 其随机积分 (13.3)/(13.4)、Prop 13.1 四性质：线性、鞅、方差 \(\mathrm{Var}(Z_t)=\int_0^t\mathbb E[A_s^2]ds\)、连续；Rmk 13.5 只需 \(\mathrm{Var}(Z_t)<\infty\)）；13.2.2 连续过程（Lemma 13.1 用简单过程逼近 (13.8)、有界收敛定理 Thm 13.1；\(\{Z_t^{(n)}\}\) 是 \(\mathcal L^2\) Cauchy 列，取极限定义 \(Z_t\)；Prop 13.2 三性质保留但鞅性不再保证；Rmk 13.6 用"赌徒"反例说明）。无图。

定义 13.1（随机微分方程）与 (13.1)、(13.2) / Definition 13.1 (SDE) and (13.1), (13.2) 定义 13.1（随机微分方程）：设 \(\{S_t\}\) 是随机过程。称 \(dX_t=R_t dt+A_t dS_t\) 为随机微分方程 (SDE)，它与常微分方程相似，只是随机性通过 \(dS_t\) 嵌入。我们关心带布朗运动的 SDE：设 \(\{B_t\}\) 是标准一维布朗运动，则Definition 13.1 (Stochastic differential equation): let \(\{S_t\}\) be a stochastic process. The equation \(dX_t=R_t dt+A_t dS_t\) is a stochastic differential equation (SDE), similar to an ODE except that randomness is embedded through \(dS_t\). We are interested in the SDE with Brownian motion: let \(\{B_t\}\) be a standard one-dimensional Brownian motion, then

$$dX_t=R_t dt+A_t dB_t.\tag{13.1}$$

理解 (13.1)：\(\{X_t\}\) 是随机过程；在任意时刻 \(t\)，\(X_t\) 以漂移 \(R_t\)、方差 \(A_t^2\) 像布朗运动那样演化。由于 \(X_t\) 的导数难以定义，先定义随机积分 \(X_t=X_0+\int_0^t R_s ds+\int_0^t A_s dB_s\) (13.2)，并称 \(X_t\) 是 SDE (13.1) 的解当且仅当它满足 (13.2)。\(\int_0^t R_s ds\) 是普通积分，故只需理解随机部分 \(\int_0^t A_s dB_s\)。Understanding (13.1): \(\{X_t\}\) is a stochastic process; at any time \(t\), \(X_t\) evolves like a Brownian motion with drift \(R_t\) and variance \(A_t^2\). Since the derivative of \(X_t\) is hard to define, we first define the stochastic integral \(X_t=X_0+\int_0^t R_s ds+\int_0^t A_s dB_s\) (13.2), and say \(X_t\) solves the SDE (13.1) iff it satisfies (13.2). As \(\int_0^t R_s ds\) is an ordinary integral, we only need to make sense of the stochastic part \(\int_0^t A_s dB_s\).

注 13.1、注 13.2 / Remarks 13.1, 13.2 注 13.1：在 (13.1) 中，\(A_t\) 可理解为下注（投资额），\(dB_t\) 是 \(A_t\) 的赌局回报（投资收益）。故关键是让 \(dB_t\) 处于 \(A_t\) 的"未来"（后文随机积分讨论中将强调）。注 13.2：一般地，(13.1)、(13.2) 中的 \(\{X_t\},\{R_t\},\{A_t\},\{B_t\}\) 都是随机过程。Remark 13.1: in (13.1), \(A_t\) can be interpreted as the bet (amount invested), and \(dB_t\) is the gamble (investment return) for \(A_t\). So it is important to make \(dB_t\) lie in the future of \(A_t\) (emphasized later in the stochastic-integral discussion). Remark 13.2: in general, \(\{X_t\},\{R_t\},\{A_t\},\{B_t\}\) in (13.1) and (13.2) are all stochastic processes.

构造思路 / Construction strategy 要严格定义与 (13.2) 相关的过程 \(\{Z_t\}\)，其中 \(Z_t=\int_0^t A_s dB_s\)。设 \((\Omega,\mathcal F,\mathbb P)\) 是大概率空间，\(\{B_t\}\) 是其上的标准一维布朗运动且适应于滤波 \(\{\mathcal F_t\}\)。先对简单过程 \(\{A_t\}\) 定义 \(\{Z_t\}\)，再用此结果逼近连续过程 \(\{A_t\}\) 的 \(\{Z_t\}\)。To properly define the process \(\{Z_t\}\) related to (13.2), where \(Z_t=\int_0^t A_s dB_s\). Let \((\Omega,\mathcal F,\mathbb P)\) be a large probability space, \(\{B_t\}\) a standard one-dimensional Brownian motion on it adapted to a filtration \(\{\mathcal F_t\}\). We first define \(\{Z_t\}\) for a simple process \(\{A_t\}\), then use the result to approximate \(\{Z_t\}\) for a continuous process \(\{A_t\}\).

定义 13.2（简单过程）与注 13.3 / Definition 13.2 (Simple process) and Remark 13.3 定义 13.2（简单过程）：\(\{A_t\}\) 称为有界（或 \(\mathcal L^2\)）简单过程，若存在有限时刻集 \(0=t_0注 13.3：简单过程只在有限个时刻改变取值，两次改变之间保持常数。Definition 13.2 (Simple process): \(\{A_t\}\) is a bounded (or \(\mathcal L^2\)) simple process if there exist a finite collection of times \(0=t_0Remark 13.3: a simple process changes value only finitely often, staying constant between any two changes.

定义 13.3（简单过程的随机积分）与注 13.4 / Definition 13.3 and Remark 13.4 定义 13.3（简单过程的随机积分）：设 \(\{A_t\}\) 是有界简单过程。定义其随机积分过程 \(\{Z_t\}\) 使 \(Z_t=\int_0^t A_s dB_s\) (13.3)。由定义 13.2 可改写为Definition 13.3 (Stochastic integral for a simple process): let \(\{A_t\}\) be a bounded simple process. Define its stochastic integral process \(\{Z_t\}\) by \(Z_t=\int_0^t A_s dB_s\) (13.3). By Definition 13.2 this rewrites as

$$Z_t=\left[\sum_{i=1}^j Y_{i-1}(B_{t_i}-B_{t_{i-1}})\right]+Y_j(B_t-B_{t_j})\quad\text{for }t_j\le t

注 13.4：(13.4) 中 \(dB_t=B_{t_i}-B_{t_{i-1}}\) 被定义为处在 \(A_t=Y_{i-1}\) 的"未来"（对每个 \(t=t_i\)）。Remark 13.4: in (13.4), \(dB_t=B_{t_i}-B_{t_{i-1}}\) is defined to be in the future of \(A_t=Y_{i-1}\) for every \(t=t_i\).

命题 13.1（简单过程随机积分的性质）/ Proposition 13.1 \(\{Z_t\}\)（由 (13.3)/(13.4) 定义）满足：1. 线性：若 \(\{A_t\},\{C_t\}\) 是两个有界简单过程，\(\forall a,b\in\mathbb R\)，\(\int_0^t(aA_s+bC_s)dB_s=a\int_0^t A_s dB_s+b\int_0^t C_s dB_s\) (13.5)；2. 鞅：\(\{Z_t\}\) 是关于 \(\{\mathcal F_t\}\) 的鞅；3. 方差：\(\mathrm{Var}(Z_t)=\int_0^t\mathbb E[A_s^2]\,ds\)；4. 连续：以概率 1，\(t\mapsto Z_t\) 连续。\(\{Z_t\}\) (defined by (13.3)/(13.4)) satisfies: 1. Linearity: if \(\{A_t\},\{C_t\}\) are two bounded simple processes, then for all \(a,b\in\mathbb R\), \(\int_0^t(aA_s+bC_s)dB_s=a\int_0^t A_s dB_s+b\int_0^t C_s dB_s\) (13.5); 2. Martingale: \(\{Z_t\}\) is a martingale w.r.t. \(\{\mathcal F_t\}\); 3. Variance: \(\mathrm{Var}(Z_t)=\int_0^t\mathbb E[A_s^2]\,ds\); 4. Continuity: with probability 1, \(t\mapsto Z_t\) is continuous.

命题 13.1 证明 / Proof of Proposition 13.1 线性：取同时包含 \(\{A_t\},\{C_t\}\) 所有变点的更细网格 \(\{t_0,t_1^{A,C},\dots\}\)，在共同网格上 (13.4) 的求和逐项线性叠加即得 (13.5)。鞅：\(Z_t\) 关于 \(\mathcal F_t\) 可测、\(\mathbb E[|Z_t|]<\infty\)（\(\{A_t\}\) 有界）。对 \(s方差：\(\mathbb E[Z_t]=0\)（鞅），故 \(\mathrm{Var}(Z_t)=\mathbb E[Z_t^2]\)。展开 \(Z_t^2\)（13.6），交叉项 \(\mathbb E[Y_{i-1}(B_{t_i}-B_{t_{i-1}})Y_{k-1}(B_{t_k}-B_{t_{k-1}})]=0\)（\(i连续：\(dZ_t=A_t dB_t\)，\(A_t\) 有界，故 \(B_t\) 的连续性（概率 1）蕴含 \(Z_t\) 连续。\(\blacksquare\)Linearity: take a finer grid \(\{t_0,t_1^{A,C},\dots\}\) containing all change points of both \(\{A_t\},\{C_t\}\); on the common grid the sum in (13.4) superposes term-by-term linearly, giving (13.5). Martingale: \(Z_t\) is \(\mathcal F_t\)-measurable with \(\mathbb E[|Z_t|]<\infty\) (\(\{A_t\}\) bounded). For \(sVariance: \(\mathbb E[Z_t]=0\) (martingale), so \(\mathrm{Var}(Z_t)=\mathbb E[Z_t^2]\). Expanding \(Z_t^2\) (13.6), the cross terms \(\mathbb E[Y_{i-1}(B_{t_i}-B_{t_{i-1}})Y_{k-1}(B_{t_k}-B_{t_{k-1}})]=0\) (\(iContinuity: \(dZ_t=A_t dB_t\), \(A_t\) bounded, so the continuity of \(B_t\) (with probability 1) implies \(Z_t\) is continuous. \(\blacksquare\)

注 13.5 / Remark 13.5 \(\{A_t\}\) 不必有界，只需满足 \(\mathrm{Var}(Z_t)=\mathbb E[Z_t^2]=\int_0^t\mathbb E[A_s^2]\,ds<\infty\)，即足以保证 \(\mathbb E[|Z_t|]<\infty\)（二阶矩存在蕴含低阶矩存在），从而保证 \(\{Z_t\}\) 的鞅性。\(\{A_t\}\) need not be bounded; we only need \(\mathrm{Var}(Z_t)=\mathbb E[Z_t^2]=\int_0^t\mathbb E[A_s^2]\,ds<\infty\), which is enough to guarantee \(\mathbb E[|Z_t|]<\infty\) (the existence of a higher moment implies that of lower moments) and thus the martingale property of \(\{Z_t\}\).

引理 13.1（用简单过程逼近）/ Lemma 13.1 (Approximation by simple processes) 基于上文简单过程 \(\{A_t\}\) 的讨论，现考虑连续过程 \(\{A_t\}\)（路径以概率 1 连续、适应于 \(\{\mathcal F_t\}\)，脚注 13.4：这蕴含 \(A_t\) 对 \(\forall t\) 是 \(\mathcal F_t\) 可测）的随机积分 \(Z_t=\int_0^t A_s dB_s\) (13.7)。引理 13.1：若 \(\{A_t\}\) 路径以概率 1 连续、有界、适应于 \(\{\mathcal F_t\}\)，则存在一列同界的简单过程 \(\left\{\{A_t^{(n)}\}_{t\ge0}\right\}_{n\in\mathbb N}\)，使得对每个 \(t\)，\(\lim_{n\to\infty}\int_0^t\mathbb E\!\left[(A_s^{(n)}-A_s)^2\right]ds=0\) (13.8)。Based on the discussion of the simple process \(\{A_t\}\) above, now consider the stochastic integral \(Z_t=\int_0^t A_s dB_s\) (13.7) for a continuous process \(\{A_t\}\) (continuous path with probability 1, adapted to \(\{\mathcal F_t\}\), footnote 13.4: this implies \(A_t\) is \(\mathcal F_t\)-measurable for all \(t\)). Lemma 13.1: if \(\{A_t\}\) has a continuous path with probability 1, is bounded, and is adapted to \(\{\mathcal F_t\}\), then there exists a sequence of simple processes \(\left\{\{A_t^{(n)}\}_{t\ge0}\right\}_{n\in\mathbb N}\) with the same bound such that for every \(t\), \(\lim_{n\to\infty}\int_0^t\mathbb E\!\left[(A_s^{(n)}-A_s)^2\right]ds=0\) (13.8).

引理 13.1 证明（含有界收敛定理 Thm 13.1）/ Proof of Lemma 13.1 (with the Bounded Convergence Theorem 13.1) 给定 \(n\)，定义 \(A_s^{(n)}=\dfrac{\int_{(k-1)/n}^{k/n}A_r\,dr}{1/n}=n\int_{(k-1)/n}^{k/n}A_r\,dr\)（\(s\in[\tfrac kn,\tfrac{k+1}n)\)）。即 \(A_s^{(n)}\) 在每个小区间 \([\tfrac kn,\tfrac{k+1}n)\) 取 \(A_t\) 在前一小区间 \([\tfrac{k-1}n,\tfrac kn]\) 上的平均值，故 \(\{A_t^{(n)}\}\) 是简单过程。因 \(A_s^{(n)}\) 是小邻域上的平均且 \(t\mapsto A_t\) 以概率 1 连续，故 \(\lim_{n\to\infty}A_s^{(n)}=A_s\)。令 \(f_n(s)=(A_s^{(n)}-A_s)^2\)，则 \(f_n(s)\) 逐点收敛到 \(0\)。\(\{f_n(s)\}\) 一致有界（\(\{A_s\}\) 有界），由有界收敛定理（Thm 13.1：若 \(\{f_n\}\) 是有界测度空间 \((X,\Sigma,\mu)\) 上一致有界的可测函数列、逐点收敛到 \(f\)，则 \(\lim_{n\to\infty}\int_X f_n\,d\mu=\int_X f\,d\mu\)）在 Lebesgue 测度下 \(\lim_{n\to\infty}\int_0^t f_n(s)\,ds=\int_0^t f(s)\,ds=0\) (13.9)。再令 \(g_n(t)=\int_0^t(A_s^{(n)}-A_s)^2\,ds\)，由 (13.9) \(g_n(t)\to0\) 逐点收敛，且 \(\{g_n(t)\}\) 一致有界，再用有界收敛定理得 \(\lim_{n\to\infty}\mathbb E[g_n(t)]=0\)，即 \(\lim_{n\to\infty}\int_0^t\mathbb E\!\left[(A_s^{(n)}-A_s)^2\right]ds=0\)。\(\blacksquare\)Given \(n\), define \(A_s^{(n)}=\dfrac{\int_{(k-1)/n}^{k/n}A_r\,dr}{1/n}=n\int_{(k-1)/n}^{k/n}A_r\,dr\) (for \(s\in[\tfrac kn,\tfrac{k+1}n)\)). So \(A_s^{(n)}\) takes, on each little interval \([\tfrac kn,\tfrac{k+1}n)\), the average of \(A_t\) on the previous interval \([\tfrac{k-1}n,\tfrac kn]\), hence \(\{A_t^{(n)}\}\) is a simple process. Since \(A_s^{(n)}\) is an average over a small neighborhood and \(t\mapsto A_t\) is continuous with probability 1, \(\lim_{n\to\infty}A_s^{(n)}=A_s\). Let \(f_n(s)=(A_s^{(n)}-A_s)^2\); then \(f_n(s)\to0\) pointwise. \(\{f_n(s)\}\) is uniformly bounded (\(\{A_s\}\) bounded), so by the Bounded Convergence Theorem (Thm 13.1: if \(\{f_n\}\) is a uniformly bounded sequence of measurable functions on a bounded measure space \((X,\Sigma,\mu)\) converging pointwise to \(f\), then \(\lim_{n\to\infty}\int_X f_n\,d\mu=\int_X f\,d\mu\)), under Lebesgue measure \(\lim_{n\to\infty}\int_0^t f_n(s)\,ds=\int_0^t f(s)\,ds=0\) (13.9). Then let \(g_n(t)=\int_0^t(A_s^{(n)}-A_s)^2\,ds\); by (13.9) \(g_n(t)\to0\) pointwise, and \(\{g_n(t)\}\) is uniformly bounded, so by the Bounded Convergence Theorem again \(\lim_{n\to\infty}\mathbb E[g_n(t)]=0\), i.e. \(\lim_{n\to\infty}\int_0^t\mathbb E\!\left[(A_s^{(n)}-A_s)^2\right]ds=0\). \(\blacksquare\)

借 Cauchy 列定义连续过程的随机积分 / Defining the integral via a Cauchy sequence 由引理 13.1，对连续、有界、适应的 \(\{A_t\}\)，可这样定义 \(\{Z_t\}=\int_0^t A_s dB_s\)：找满足 (13.8) 的 \(\{A_t^{(n)}\}\)，定义 \(Z_t^{(n)}=\int_0^t A_s^{(n)}dB_s\)（由简单过程积分良定义）。考虑 \(Z_t^{(n)}-Z_t^{(m)}=\int_0^t(A_s^{(n)}-A_s^{(m)})dB_s\)，则 \(\lim_{n,m\to\infty}\mathbb E[(Z_t^{(n)}-Z_t^{(m)})^2]=\lim_{n,m\to\infty}\int_0^t\mathbb E[(A_s^{(n)}-A_s^{(m)})^2]\,ds=0\)（由方差公式与 (13.8)）。故 \(\left\{\{Z_t^{(n)}\}_{t\ge0}\right\}_{n\in\mathbb N}\) 是 \(\mathcal L^2\) 中的 Cauchy 列，蕴含 \(\mathcal L^2\) 极限 \(Z_t=\lim_{n\to\infty}Z_t^{(n)}\) 存在。记 \(\int_0^t A_s dB_s=Z_t\)，至此成功定义。By Lemma 13.1, for continuous, bounded, adapted \(\{A_t\}\), define \(\{Z_t\}=\int_0^t A_s dB_s\) as follows: find \(\{A_t^{(n)}\}\) satisfying (13.8) and set \(Z_t^{(n)}=\int_0^t A_s^{(n)}dB_s\) (well-defined via the simple-process integral). Consider \(Z_t^{(n)}-Z_t^{(m)}=\int_0^t(A_s^{(n)}-A_s^{(m)})dB_s\); then \(\lim_{n,m\to\infty}\mathbb E[(Z_t^{(n)}-Z_t^{(m)})^2]=\lim_{n,m\to\infty}\int_0^t\mathbb E[(A_s^{(n)}-A_s^{(m)})^2]\,ds=0\) (by the variance formula and (13.8)). So \(\left\{\{Z_t^{(n)}\}_{t\ge0}\right\}_{n\in\mathbb N}\) is a Cauchy sequence in \(\mathcal L^2\), implying the existence of the \(\mathcal L^2\) limit \(Z_t=\lim_{n\to\infty}Z_t^{(n)}\). Denote \(\int_0^t A_s dB_s=Z_t\); the integral is now defined.

命题 13.2（连续过程随机积分的性质）与注 13.6 / Proposition 13.2 and Remark 13.6 \(\{Z_t\}\)（由 (13.7) 定义）满足：1. 线性：\(\int_0^t(aA_s+bC_s)dB_s=a\int_0^t A_s dB_s+b\int_0^t C_s dB_s\) (13.10)；2. 方差：\(\mathrm{Var}(Z_t)=\int_0^t\mathbb E[A_s^2]\,ds\)；3. 连续：以概率 1，\(t\mapsto Z_t\) 连续。证明：三条性质都由简单过程 \(\{Z_t^{(n)}\}\) 的相同性质与收敛 \(Z_t^{(n)}\to Z_t\) 得到。但对连续 \(\{A_t\}\)，不再保证鞅性。注 13.6：要看为何 \(\{Z_t\}\) 可能不是鞅，把 \(Z_t=\int_0^t A_s dB_s\) 看作赌局——\(A_s\) 可被适当选取（在某点任意大），使得（概率 1）\(Z_1\ge1\)，则 \(\mathbb E[Z_t]\ge1\neq0\)，故 \(\{Z_t\}\) 不可能是鞅。（脚注：设目标 \(Z_1=1\)；若某 \(t\in(0,1)\) 已 \(Z_t>1\) 则保持微调；若 \(Z_t<1\) 则在 \(B_{t+\Delta}-B_t>0\) 时大幅加仓 \(A_t\) 使 \(Z_{t+\Delta t}>1\)；若运气不好继续向下，则进一步加大 \(A_{t+\Delta t}\) 重复赌局；最终以概率 1，\(Z_1>1\) 成立。）\(\{Z_t\}\) (defined by (13.7)) satisfies: 1. Linearity: \(\int_0^t(aA_s+bC_s)dB_s=a\int_0^t A_s dB_s+b\int_0^t C_s dB_s\) (13.10); 2. Variance: \(\mathrm{Var}(Z_t)=\int_0^t\mathbb E[A_s^2]\,ds\); 3. Continuity: with probability 1, \(t\mapsto Z_t\) is continuous. Proof: the three properties follow from the same properties of the simple-process \(\{Z_t^{(n)}\}\) and the convergence \(Z_t^{(n)}\to Z_t\). However, for continuous \(\{A_t\}\), we no longer have the martingale property for sure. Remark 13.6: to see why \(\{Z_t\}\) could fail to be a martingale, view \(Z_t=\int_0^t A_s dB_s\) as a gamble — \(A_s\) can be properly chosen (arbitrarily large at some point) so that (with probability 1) \(Z_1\ge1\), hence \(\mathbb E[Z_t]\ge1\neq0\), so \(\{Z_t\}\) cannot be a martingale. (Footnote: suppose the target is \(Z_1=1\); if at some \(t\in(0,1)\) already \(Z_t>1\), keep fine-tuning; if \(Z_t<1\), load \(A_t\) heavily so that \(Z_{t+\Delta t}>1\) when \(B_{t+\Delta}-B_t>0\); if bad luck pushes it further down, increase \(A_{t+\Delta t}\) even more to repeat the gamble; finally, with probability one, \(Z_1>1\).)