本章导读 本章把布朗运动刻画为连续鞅。§3.1 滤波（Def 3.1 滤波、Def 3.2 由过程生成的滤波、Def 3.3 适应过程、Def 3.4 关于滤波的布朗运动、Def 3.5 滤波的"通常条件"）。§3.2 鞅：3.2.1 定义（Def 3.6 离散鞅、Def 3.7 连续鞅、Def 3.8 上鞅、Def 3.9 下鞅）；3.2.2 性质（命题 3.1 \(\mathbb E[M_T]=\mathbb E[M_0]\)；命题 3.2 鞅的平方是下鞅；注 3.2）；3.2.3 例（Ex 3.1 \(B_t^2-t\)、Ex 3.2 指数鞅 \(e^{\lambda B_t-\lambda^2 t/2}\)）。§3.3 布朗运动是鞅（命题 3.3 无漂移 BM 是连续鞅；注 3.3、3.4）。§3.4 可选采样定理：3.4.1 定理本身（Thm 3.1）；3.4.2 有界停时；3.4.3 无界停时（三个充分条件 + Ex 3.3 求 \(\mathbb P\{B_\tau=a\}\) 与 \(\mathbb E[\tau]=ab\)）。§3.5 Doob 极大不等式（Thm 3.2）。无图。

定义 3.1–3.3（滤波、生成滤波、适应过程）/ Definitions 3.1–3.3 定义 3.1（滤波）：滤波是一族递增的 \(\sigma\)-代数 \(\{\mathcal F_t:t\ge0\}\)，使得 \(\forall s带滤波的概率空间。直觉上 \(\mathcal F_t\) 是到 \(t\) 时刻为止可获得的信息，信息随时间只增不减。定义 3.2（由过程生成的滤波）：过程 \(\{X_t\}\) 生成的滤波为 \(\mathcal F_t^X\equiv\sigma\!\left(\bigcup_{0\le s\le t}\sigma(X_s)\right)\)，即由到 \(t\) 为止所有 \(X_s\) 生成的最小 \(\sigma\)-代数。定义 3.3（适应过程）：若 \(\forall t\)，\(X_t\) 关于 \(\mathcal F_t\) 可测，则称过程 \(\{X_t\}\) 适应于滤波 \(\{\mathcal F_t\}\)（脚注：\(\mathcal F_t\)-可测意指对任意 \(U_t\in\mathcal F_t\)，事件 \(\{X_t\in U_t\}\) 可由 \(\mathcal F_t\) 中信息判定）。Definition 3.1 (Filtration): a filtration is an increasing collection of \(\sigma\)-algebras \(\{\mathcal F_t:t\ge0\}\) such that for all \(sfiltered probability space. Intuitively \(\mathcal F_t\) is the information available up to time \(t\), and information only accumulates over time. Definition 3.2 (Filtration generated by a process): the filtration generated by a process \(\{X_t\}\) is \(\mathcal F_t^X\equiv\sigma\!\left(\bigcup_{0\le s\le t}\sigma(X_s)\right)\), the smallest \(\sigma\)-algebra generated by all \(X_s\) up to time \(t\). Definition 3.3 (Adapted process): a process \(\{X_t\}\) is adapted to a filtration \(\{\mathcal F_t\}\) if for all \(t\), \(X_t\) is \(\mathcal F_t\)-measurable (footnote: \(\mathcal F_t\)-measurable means that for any \(U_t\in\mathcal F_t\), the event \(\{X_t\in U_t\}\) is determined by the information in \(\mathcal F_t\)).

注 3.1 / Remark 3.1 任何过程都适应于它自己生成的滤波 \(\mathcal F_t^X\)——因为 \(X_t\) 按构造就是 \(\mathcal F_t^X\)-可测的。Any process is adapted to its own generated filtration \(\mathcal F_t^X\) — because \(X_t\) is by construction \(\mathcal F_t^X\)-measurable.

定义 3.4（关于滤波的布朗运动）/ Definition 3.4 (Brownian motion w.r.t. a filtration) 过程 \(\{B_t:t\ge0\}\) 称为关于滤波 \(\{\mathcal F_t:t\ge0\}\) 的标准布朗运动，若：(1) \(\{B_t\}\) 适应于 \(\{\mathcal F_t\}\)；(2) \(B_0=0\)；(3) 若 \(t>0\)，则 \(\mathcal D_t\equiv\sigma\{B_{s+t}-B_t:s\ge0\}\) 独立于 \(\mathcal F_t\)；(4) 对 \(sA process \(\{B_t:t\ge0\}\) is a standard Brownian motion with respect to a filtration \(\{\mathcal F_t:t\ge0\}\) if: (1) \(\{B_t\}\) is adapted to \(\{\mathcal F_t\}\); (2) \(B_0=0\); (3) if \(t>0\), then \(\mathcal D_t\equiv\sigma\{B_{s+t}-B_t:s\ge0\}\) is independent of \(\mathcal F_t\); (4) for \(s

定义 3.5（滤波的通常条件）/ Definition 3.5 (Usual conditions for a filtration) 出于技术原因，通常对滤波 \(\{\mathcal F_t:t\ge0\}\) 作两个假设：(1) 右连续，即 \(\forall t\)，\(\mathcal F_t=\bigcap_{\varepsilon>0}\mathcal F_{t+\varepsilon}\;(\equiv\mathcal F_{t+})\)；(2) \(\mathcal F\) 的每个零测集都属于每个 \(\mathcal F_t\)，从而所有零测集对 \(\forall t\) 都是 \(\mathcal F_t\)-可测的（脚注：\(A\) 为 \(\mathcal F\) 的零测集，若 \(\exists C\in\mathcal F\) 使 \(\mathbb P(C)=0\) 且 \(A\subseteq C\)）。For technical reasons, we usually make two assumptions about the filtration \(\{\mathcal F_t:t\ge0\}\): (1) right continuity, i.e. for all \(t\), \(\mathcal F_t=\bigcap_{\varepsilon>0}\mathcal F_{t+\varepsilon}\;(\equiv\mathcal F_{t+})\); (2) every null set of \(\mathcal F\) is in every \(\mathcal F_t\), so that all null sets are \(\mathcal F_t\)-measurable for all \(t\) (footnote: \(A\) is a null set of \(\mathcal F\) if there is \(C\in\mathcal F\) with \(\mathbb P(C)=0\) and \(A\subseteq C\)).

定义 3.6–3.9（离散鞅、连续鞅、上鞅、下鞅）/ Definitions 3.6–3.9 定义 3.6（离散鞅）：设滤波 \(\{\mathcal F_i\}\) 是由随机变量序列 \(M_0,M_1,\dots\) 生成的 \(\sigma\)-代数。序列 \(M_0,M_1,\dots\) 称为关于 \(\{\mathcal F_i\}\) 的（离散）鞅，若：(1) \(\forall t\)，\(M_t\) 关于 \(\mathcal F_t\) 可测且 \(\mathbb E[|M_t|]<\infty\)；(2) 对 \(s定义 3.7（连续鞅）：设 \(\{\mathcal F_t\}\) 是由连续随机过程 \(\{M_s:s\le t\}\) 生成的 \(\sigma\)-代数。\(\{M_t\}\) 称为关于 \(\{\mathcal F_t\}\) 的（连续）鞅，若：(1) \(\forall t\)，\(M_t\) 关于 \(\mathcal F_t\) 可测且 \(\mathbb E[|M_t|]<\infty\)；(2) 对 \(s定义 3.8（上鞅 super-martingale）：同上但 (2) 为 \(\mathbb E[M_t\mid\mathcal F_s]\le M_s\)。定义 3.9（下鞅 sub-martingale）：同上但 (2) 为 \(\mathbb E[M_t\mid\mathcal F_s]\ge M_s\)。Definition 3.6 (Discrete martingale): let the filtration \(\{\mathcal F_i\}\) be the \(\sigma\)-algebra generated by a sequence of random variables \(M_0,M_1,\dots\). The sequence \(M_0,M_1,\dots\) is called a (discrete) martingale w.r.t. \(\{\mathcal F_i\}\) if: (1) for each \(t\), \(M_t\) is \(\mathcal F_t\)-measurable with \(\mathbb E[|M_t|]<\infty\); (2) for \(sDefinition 3.7 (Continuous martingale): let \(\{\mathcal F_t\}\) be the \(\sigma\)-algebra generated by a continuous stochastic process \(\{M_s:s\le t\}\). The process \(\{M_t\}\) is called a (continuous) martingale w.r.t. \(\{\mathcal F_t\}\) if: (1) for each \(t\), \(M_t\) is \(\mathcal F_t\)-measurable with \(\mathbb E[|M_t|]<\infty\); (2) for \(sDefinition 3.8 (Super-martingale): same as above but (2) becomes \(\mathbb E[M_t\mid\mathcal F_s]\le M_s\). Definition 3.9 (Sub-martingale): same as above but (2) becomes \(\mathbb E[M_t\mid\mathcal F_s]\ge M_s\).

命题 3.1 / Proposition 3.1 若 \(\{M_t\}\) 是鞅，则 \(\mathbb E[M_T]=\mathbb E[M_0]\) 对 \(\forall T\ge0\) 成立。If \(\{M_t\}\) is a martingale, then \(\mathbb E[M_T]=\mathbb E[M_0]\) for all \(T\ge0\).

命题 3.1 证明 / Proof of Proposition 3.1 离散情形：用条件期望的塔性质（脚注 3.3：对 \(s连续情形：由定义直接有 \(\mathbb E[M_T\mid\mathcal F_0]=M_0\)，再用重期望律 \(\mathbb E[\mathbb E[M_T\mid\mathcal F_0]]=\mathbb E[M_0]\Rightarrow\mathbb E[M_T]=\mathbb E[M_0]\)。\(\blacksquare\)Discrete case: use the tower property of conditional expectation (footnote 3.3: for \(sContinuous case: by definition \(\mathbb E[M_T\mid\mathcal F_0]=M_0\) directly, and by the law of iterated expectation \(\mathbb E[\mathbb E[M_T\mid\mathcal F_0]]=\mathbb E[M_0]\Rightarrow\mathbb E[M_T]=\mathbb E[M_0]\). \(\blacksquare\)

命题 3.2 / Proposition 3.2 若 \(\{M_t\}\) 是鞅，则 \(\{M_t^2\}\) 是下鞅。If \(\{M_t\}\) is a martingale, then \(\{M_t^2\}\) is a sub-martingale.

命题 3.2 证明 / Proof of Proposition 3.2 要证 \(\{M_t^2\}\) 是下鞅。首先 \(\{M_t^2\}\) 适应于 \(\{M_t\}\) 生成的滤波，且 \(\mathbb E[M_t^2]<\infty\) 给定。只需证对 \(sTo show \(\{M_t^2\}\) is a sub-martingale: first \(\{M_t^2\}\) is adapted to the filtration generated by \(\{M_t\}\), and \(\mathbb E[M_t^2]<\infty\) is given. We only need to show that for \(s

$$\mathbb E\!\left[M_t^2\mid\mathcal F_s\right]\ge\left(\mathbb E[M_t\mid\mathcal F_s]\right)^2=M_s^2.$$

\(\blacksquare\)\(\blacksquare\)

注 3.2 / Remark 3.2 由命题 3.2，因 \(t\wedge\tau\le t\)，对鞅 \(\{M_t\}\) 有 \(\mathbb E[M_{t\wedge\tau}^2]\le\mathbb E[M_t^2]\)。By Proposition 3.2, since \(t\wedge\tau\le t\), for a martingale \(\{M_t\}\) we have \(\mathbb E[M_{t\wedge\tau}^2]\le\mathbb E[M_t^2]\).

例 3.1 与例 3.2 / Examples 3.1 and 3.2 例 3.1：设 \(\{B_t\}\) 是标准布朗运动（一个连续鞅）。证明 \(M_t=B_t^2-t\) 也是连续鞅。例 3.2：设 \(\{B_t\}\) 是标准布朗运动。证明对任意 \(\lambda\in\mathbb R\)，\(M_t=e^{\lambda B_t-\frac{\lambda^2 t}{2}}\) 也是连续鞅（指数鞅）。Example 3.1: let \(\{B_t\}\) be a standard Brownian motion (a continuous martingale). Show that \(M_t=B_t^2-t\) is also a continuous martingale. Example 3.2: let \(\{B_t\}\) be a standard Brownian motion. Show that for any \(\lambda\in\mathbb R\), \(M_t=e^{\lambda B_t-\frac{\lambda^2 t}{2}}\) is also a continuous martingale (the exponential martingale).

例 3.1 解答 / Solution to Example 3.1 \(B_t\) 的连续性与有界性直接蕴含 \(M_t\) 的连续性与有界性，故只需验证对 \(sThe continuity and boundedness of \(B_t\) directly imply those of \(M_t\), so we only need to check that for \(s

$$\begin{aligned}\mathbb E[M_t\mid\mathcal F_s]&=\mathbb E\!\left[B_t^2-t\mid\mathcal F_s\right]=\mathbb E\!\left[(B_s+B_t-B_s)^2\mid\mathcal F_s\right]-t\\&=\mathbb E\!\left[B_s^2\mid\mathcal F_s\right]+2\,\mathbb E\!\left[B_s(B_t-B_s)\mid\mathcal F_s\right]+\mathbb E\!\left[(B_t-B_s)^2\mid\mathcal F_s\right]-t\\&=B_s^2+2B_s\,\underbrace{\mathbb E[B_t-B_s]}_{=\,0}+\underbrace{\mathbb E[(B_t-B_s)^2]}_{=\,t-s}-t\\&=B_s^2+(t-s)-t=B_s^2-s=M_s,\end{aligned}$$

其中用到增量独立性。\(\blacksquare\)where increment independence is used. \(\blacksquare\)

例 3.2 解答 / Solution to Example 3.2 同理只需验证 \(\mathbb E[M_t\mid\mathcal F_s]=M_s\)。Likewise we only need to verify \(\mathbb E[M_t\mid\mathcal F_s]=M_s\).

$$\begin{aligned}\mathbb E[M_t\mid\mathcal F_s]&=\mathbb E\!\left[e^{\lambda B_t-\frac{\lambda^2 t}{2}}\mid\mathcal F_s\right]=\mathbb E\!\left[e^{\lambda(B_s+B_t-B_s)}\mid\mathcal F_s\right]e^{-\frac{\lambda^2 t}{2}}\\&=e^{\lambda B_s}\,\mathbb E\!\left[e^{\lambda(B_t-B_s)}\right]e^{-\frac{\lambda^2 t}{2}}=e^{\lambda B_s}\,e^{\frac12\lambda^2(t-s)}\,e^{-\frac{\lambda^2 t}{2}}\\&=e^{\lambda B_s-\frac{\lambda^2 s}{2}}=M_s,\end{aligned}$$

其中用到增量独立性，以及 \(B_t-B_s\sim\mathcal N(0,t-s)\) 的矩母函数 \(\mathbb E[e^{\lambda(B_t-B_s)}]=e^{\frac12\lambda^2(t-s)}\)。\(\blacksquare\)where we used increment independence and the moment generating function \(\mathbb E[e^{\lambda(B_t-B_s)}]=e^{\frac12\lambda^2(t-s)}\) of \(B_t-B_s\sim\mathcal N(0,t-s)\). \(\blacksquare\)

命题 3.3 / Proposition 3.3 无漂移布朗运动 \(\{B_t\}\) 是连续鞅。Driftless Brownian motion \(\{B_t\}\) is a continuous martingale.

命题 3.3 证明 / Proof of Proposition 3.3 定义 3.7 的条件 (1) 平凡成立，只需证条件 (2)。对任意 \(sCondition (1) of Definition 3.7 trivially holds; we only need to show condition (2). For any \(s

$$\mathbb E[B_t\mid\mathcal F_s]=\mathbb E[B_s\mid\mathcal F_s]+\mathbb E[B_t-B_s\mid\mathcal F_s]=B_s+\mathbb E[B_t-B_s]=B_s,$$

其中第二个等号用到增量独立性，末项 \(\mathbb E[B_t-B_s]=0\)。\(\blacksquare\)where the second equality uses increment independence and the last term \(\mathbb E[B_t-B_s]=0\). \(\blacksquare\)

注 3.3 与注 3.4 / Remarks 3.3 and 3.4 注 3.3：由命题 3.3，可把布朗运动等价地重定义为一个连续鞅且其路径以概率 1 连续。注 3.4：易见若布朗运动带漂移 \(\mu\ne0\)，则 \(\mathbb E[B_t\mid\mathcal F_s]=B_s\) 在不去漂移（detrending）时不成立。Remark 3.3: by Proposition 3.3, Brownian motion can be equivalently redefined as a continuous martingale with paths continuous with probability one. Remark 3.4: it is easy to see that if Brownian motion has drift \(\mu\ne0\), then \(\mathbb E[B_t\mid\mathcal F_s]=B_s\) won't hold without detrending.

定理 3.1（可选采样定理）/ Theorem 3.1 (Optional Sampling Theorem) 设 \(\tau\) 是停时，\(\{M_t\}\) 是关于滤波 \(\{\mathcal F_t\}\) 的鞅。则由 \(Y_t=M_{t\wedge\tau}\)（\(\forall t\)）定义的过程 \(\{Y_t\}\) 也是鞅。Let \(\tau\) be a stopping time and \(\{M_t\}\) a martingale w.r.t. a filtration \(\{\mathcal F_t\}\). Then the process \(\{Y_t\}\) defined by \(Y_t=M_{t\wedge\tau}\) (for all \(t\)) is also a martingale.

定理 3.1 证明 / Proof of Theorem 3.1 需验证 \(\{Y_t\}\) 满足定义 3.7。设 \(\{\mathcal F_t\}\) 是 \(\{M_t\}\) 生成的滤波。先证条件 (1)：\(\forall t\)，\(Y_t\) 关于 \(\mathcal F_t\) 可测且 \(\mathbb E[|Y_t|]<\infty\)。当 \(t\ge\tau\) 时 \(Y_t=M_\tau\)（取某 \(s\in[0,t]\)），它 \(\mathcal F_t\)-可测且因 \(\{M_t\}\) 是鞅而 \(\mathbb E[|Y_t|]<\infty\)；当 \(t<\tau\) 时 \(Y_t=M_t\)，同样 \(\mathcal F_t\)-可测且 \(\mathbb E[|Y_t|]<\infty\)。再证条件 (2)：对 \(sWe verify \(\{Y_t\}\) satisfies Definition 3.7. Let \(\{\mathcal F_t\}\) be the filtration generated by \(\{M_t\}\). Condition (1): for all \(t\), \(Y_t\) is \(\mathcal F_t\)-measurable with \(\mathbb E[|Y_t|]<\infty\). If \(t\ge\tau\) then \(Y_t=M_\tau\) (for some \(s\in[0,t]\)), which is \(\mathcal F_t\)-measurable with \(\mathbb E[|Y_t|]<\infty\) since \(\{M_t\}\) is a martingale; if \(t<\tau\) then \(Y_t=M_t\), again \(\mathcal F_t\)-measurable with \(\mathbb E[|Y_t|]<\infty\). Condition (2): for \(s

注 3.5 / Remark 3.5 特别地，定理 3.1 蕴含对 \(\forall t\ge0\)，\(\mathbb E[M_{t\wedge\tau}]=\mathbb E[M_0]\)。In particular, Theorem 3.1 implies that for all \(t\ge0\), \(\mathbb E[M_{t\wedge\tau}]=\mathbb E[M_0]\).

有界停时 / Bounded stopping time 有界停时指存在 \(T<\infty\) 使 \(\mathbb P\{\tau\le T\}=1\)。要证A bounded stopping time means there exists \(T<\infty\) such that \(\mathbb P\{\tau\le T\}=1\). We want to show

$$\mathbb E[M_0]=\mathbb E[M_\tau]\tag{3.1}$$

在有界停时情形成立。对 \(\tau\le T\)，由命题 3.1 \(M_\tau\) 满足 (3.1)；对 \(\tau>T\)，有 \(\mathbb E[M_\tau]=\mathbb E[M_T]=\mathbb E[M_0]\)，同样满足 (3.1)。故停时有界时 (3.1) 恒成立。holds for a bounded stopping time. For \(\tau\le T\), by Proposition 3.1 \(M_\tau\) satisfies (3.1); for \(\tau>T\), \(\mathbb E[M_\tau]=\mathbb E[M_T]=\mathbb E[M_0]\), which also satisfies (3.1). So (3.1) always holds when the stopping time is bounded.

无界停时的充分条件 / Sufficient conditions for unbounded stopping time 实际中常没有有界停时，只有 \(\mathbb P\{\tau<\infty\}=1\)，此时需附加条件才能保证 (3.1)。首先注意：由定理 3.1 与命题 3.1，\(\lim_{t\to\infty}\mathbb E[M_{t\wedge\tau}]=\mathbb E[M_0]\)（3.2）；由 \(\mathbb P\{\tau<\infty\}=1\)，\(\mathbb E[\lim_{t\to\infty}M_{t\wedge\tau}]=\mathbb E[M_\tau]\)（3.3）。只要能证In practice we often lack a bounded stopping time and only have \(\mathbb P\{\tau<\infty\}=1\), in which case extra conditions are needed for (3.1). First note: by Theorem 3.1 and Proposition 3.1, \(\lim_{t\to\infty}\mathbb E[M_{t\wedge\tau}]=\mathbb E[M_0]\) (3.2); by the assumption \(\mathbb P\{\tau<\infty\}=1\), \(\mathbb E[\lim_{t\to\infty}M_{t\wedge\tau}]=\mathbb E[M_\tau]\) (3.3). As long as we can show

$$\lim_{t\to\infty}\mathbb E[M_{t\wedge\tau}]=\mathbb E\!\left[\lim_{t\to\infty}M_{t\wedge\tau}\right]\tag{3.4}$$

则由 (3.2)(3.3) 得到重要结论 \(\mathbb E[M_0]=\mathbb E[M_\tau]\)。使 (3.4)（从而 (3.1)）成立的若干条件：条件 1：\(\{M_t\}\) 一致有界；条件 2：存在可积随机变量 \(Y\) 使 \(|M_{t\wedge\tau}|\le Y\) 对 \(\forall t\) 成立；条件 3：存在 \(C<\infty\) 使 \(\mathbb E[M_{t\wedge\tau}^2]\le C\) 对 \(\forall t\) 成立。then by (3.2)(3.3) we obtain the important conclusion \(\mathbb E[M_0]=\mathbb E[M_\tau]\). Conditions under which (3.4) (hence (3.1)) holds: Condition 1: \(\{M_t\}\) is uniformly bounded; Condition 2: there exists an integrable random variable \(Y\) with \(|M_{t\wedge\tau}|\le Y\) for all \(t\); Condition 3: there exists \(C<\infty\) such that \(\mathbb E[M_{t\wedge\tau}^2]\le C\) for all \(t\).

注 3.6 / Remark 3.6 这三个条件只是三个例子，可能还有更多。但它们共同的特征都是保证 \(M_t\) 一致可积——这是关键，只要它成立，对 \(\forall t\) 都有 \(M_t\) 可积，于是交换取极限与取期望的次序就是合法的。These three conditions are just three examples; there may be more. But they all share the same feature: ensuring \(M_t\) is uniformly integrable — which is the key, since whenever it holds, \(M_t\) is integrable for all \(t\) and so interchanging the order of taking limit and taking expectation is legitimate.

例 3.3（可选采样定理的应用）/ Example 3.3 (Application of Theorem 3.1) 设 \(\{B_t\}\) 是标准布朗运动，\(a,b>0\)，\(\tau\) 是 \(\{B_t\}\) 的停时 \(\tau=\inf\{t:B_t=a\text{ or }B_t=-b\}\)。由停时性质有 \(\mathbb P\{\tau<\infty\}=1\)，且 \(B_{t\wedge\tau}\) 被 \(\max\{|a|,|-b|\}\) 一致有界。于是可用 (3.1) 求出 \(\mathbb P\{B_\tau=a\}\) 与 \(\mathbb P\{B_\tau=-b\}\)，并进一步求出期望停时 \(\mathbb E[\tau]=ab\)。Let \(\{B_t\}\) be a standard Brownian motion, \(a,b>0\), and \(\tau\) the stopping time for \(\{B_t\}\) given by \(\tau=\inf\{t:B_t=a\text{ or }B_t=-b\}\). By the nature of the stopping time, \(\mathbb P\{\tau<\infty\}=1\) and \(B_{t\wedge\tau}\) is uniformly bounded by \(\max\{|a|,|-b|\}\). So we can use (3.1) to solve for \(\mathbb P\{B_\tau=a\}\) and \(\mathbb P\{B_\tau=-b\}\), and further for the expected stopping time \(\mathbb E[\tau]=ab\).

例 3.3 解答（命中概率与期望停时）/ Solution to Example 3.3 命中概率：由本例条件，(3.1) 成立，即 \(0=\mathbb E[B_0]=\mathbb E[B_\tau]\)。由停止规则的定义显式展开：Hitting probabilities: by the conditions of this example, (3.1) holds, i.e. \(0=\mathbb E[B_0]=\mathbb E[B_\tau]\). Expand explicitly by the definition of the stopping rule:

$$\begin{aligned}\mathbb E[B_\tau]&=a\cdot\mathbb P\{B_\tau=a\}+(-b)\cdot\mathbb P\{B_\tau=-b\}\\&=a\cdot\mathbb P\{B_\tau=a\}-b\,(1-\mathbb P\{B_\tau=a\})=0\\\Rightarrow\;\mathbb P\{B_\tau=a\}&=\frac{b}{a+b},\qquad\mathbb P\{B_\tau=-b\}=\frac{a}{a+b}.\end{aligned}$$

（上式即 (3.5) 与 (3.6)。）期望停时：令 \(M_t\equiv B_t^2-t\)。注意 \(M_{t\wedge\tau}=B_{t\wedge\tau}^2-(t\wedge\tau)\) 不再一致有界（因 \(t\wedge\tau\) 不一致有界，\(\tau\) 是随机变量），但仍有 \(|M_{t\wedge\tau}|\le(a^2+b^2)+\tau\)（\(\forall t\)）。假设 \(\mathbb E[\tau]<\infty\)（最后验证），则 \(\tau\) 可积，由条件 2 知 (3.1) 成立，即 \(\mathbb E[M_\tau]=\mathbb E[M_0]=0\)。整理：(These are (3.5) and (3.6).) Expected stopping time: let \(M_t\equiv B_t^2-t\). Note \(M_{t\wedge\tau}=B_{t\wedge\tau}^2-(t\wedge\tau)\) is no longer uniformly bounded (since \(t\wedge\tau\) is not, \(\tau\) being a random variable), but still \(|M_{t\wedge\tau}|\le(a^2+b^2)+\tau\) for all \(t\). Assume \(\mathbb E[\tau]<\infty\) (verified at the end); then \(\tau\) is integrable, and by Condition 2 (3.1) holds, i.e. \(\mathbb E[M_\tau]=\mathbb E[M_0]=0\). Rearranging:

$$\begin{aligned}0=\mathbb E[M_\tau]&=\mathbb E[B_\tau^2]-\mathbb E[\tau]=\mathbb P\{B_\tau=a\}\,a^2+\mathbb P\{B_\tau=-b\}\,b^2-\mathbb E[\tau]\\&=\frac{b}{a+b}a^2+\frac{a}{a+b}b^2-\mathbb E[\tau]\\\Rightarrow\;\mathbb E[\tau]&=\frac{ab(a+b)}{a+b}=ab,\end{aligned}$$

这验证了 \(\mathbb E[\tau]=ab<\infty\)。\(\blacksquare\)which verifies \(\mathbb E[\tau]=ab<\infty\). \(\blacksquare\)

定理 3.2（Doob 极大不等式）/ Theorem 3.2 (Doob's Maximal Inequality) 若 \(\{M_t\}\) 是连续鞅，\(a>0\)，则If \(\{M_t\}\) is a continuous martingale and \(a>0\), then

$$\mathbb P\!\left\{\max_{0\le s\le t}M_s\ge a\right\}\le\frac{\mathbb E[M_t^2]}{a^2}.$$

定理 3.2 证明 / Proof of Theorem 3.2 取停时 \(\tau=\min\{s:M_s\ge a\}\)。则Take the stopping time \(\tau=\min\{s:M_s\ge a\}\). Then

$$\mathbb P\!\left\{\max_{0\le s\le t}M_s\ge a\right\}=\mathbb P\{M_{t\wedge\tau}=a\}\le\mathbb P\{M_{t\wedge\tau}\ge a\}\le\frac{\mathbb E[M_{t\wedge\tau}^2]}{a^2}\le\frac{\mathbb E[M_t^2]}{a^2},$$

其中第三个不等号用 Markov 不等式，最后一个不等号由命题 3.2（\(\{M_t^2\}\) 是下鞅）且 \(t\wedge\tau\le t\) 得到。\(\blacksquare\)where the third inequality is Markov's inequality and the last holds since by Proposition 3.2 \(\{M_t^2\}\) is a sub-martingale and \(t\wedge\tau\le t\). \(\blacksquare\)