Part I 导读 / Part I overview Part I 布朗运动（Ch 1–12）系统建立布朗运动 (Brownian motion, BM) 的定义与性质：定义与存在性构造（Ch 1）、作为高斯过程（Ch 2）、作为连续鞅（Ch 3）、马尔可夫性（Ch 4）、停时（Ch 5）、二次变差（Ch 6）、零集（Ch 7）、重对数律（Ch 8）、多维 BM（Ch 9）、测度变换与 Girsanov 定理（Ch 10）、跳过程与 Lévy 过程（Ch 11）、对称稳定过程（Ch 12）。Part II（Ch 13–16）在此基础上发展随机积分与 Itô 微积分。

本章导读 §1.1 定义：随机过程（定义 1.1）、布朗运动（定义 1.2：原点出发、独立增量、平稳/同分布增量、路径连续）、定理 1.1（满足四条件 ⟹ \(B_t\sim\mathcal N(t\mu,t\sigma^2)\)）、标准布朗运动（定义 1.3 重定义、定义 1.4 标准、命题 1.1 \(Y_t=\mu t+\sigma B_t\)）。§1.2 标准 BM 的存在性：三步构造（在二进有理数集 \(\mathcal D\) 上定义并验证三条件、证一致连续、取极限延拓到 $[0,1]\(）。§1.3 **Hölder 连续**（定义 1.5/1.6，定理 1.2：BM 弱 Hölder 连续于阶 \)\tfrac12$、但非 Hölder 连续于阶 \(\tfrac12\)）。§1.4 不可微（定理 1.3：BM 几乎处处处处不可微）。§1.5 反射原理（定理 1.4：\(\mathbf P\{\max_{0\le s\le t}B_s\ge a\}=2\mathbf P\{B_t\ge a\}\)）。§1.6 标度律（定理 1.5：\(Y_t=\tfrac1a B_{at^2}\) 仍为标准 BM；例 1.1 用标度律 + 反射原理算"某区间内触零"概率 \(1-\tfrac2\pi\arctan\tfrac1{\sqrt t}\)）。

定义 1.1（随机过程 / Stochastic Process） 随机过程是概率空间 \((\Omega,\mathcal F,\mathbb P)\) 上、由 \(t\) 索引的随机变量的集合 \(\{X_t\}\)，离散情形 \(t\in\mathbb Z_+\)、连续情形 \(t\in\mathbb R_+\)。A stochastic process is a collection of random variables \(\{X_t\}\) on a probability space \((\Omega,\mathcal F,\mathbb P)\) indexed by \(t\), with \(t\in\mathbb Z_+\) in the discrete case and \(t\in\mathbb R_+\) in the continuous case.

定义 1.2（一维布朗运动 / One-dimensional Brownian Motion） 一维布朗运动是满足如下条件的随机过程 \(\{B_t\}\)：(1) 从原点出发：\(B_0=0\)；(2) 独立增量：若 \(r(3) 同分布（平稳）增量：若 \(s(4) 连续性：以概率 1，函数 \(t\to B_t\) 连续，即 \(\mathbf P\{\omega\in\Omega:B_t(\omega)\text{ continuous in }t\}=1\)。A one-dimensional Brownian motion is a stochastic process \(\{B_t\}\) satisfying: (1) Start from origin: \(B_0=0\); (2) Independent increments: if \(r(3) Identically distributed (stationary) increments: if \(s(4) Continuity: with probability one, the function \(t\to B_t\) is continuous, i.e. \(\mathbf P\{\omega\in\Omega:B_t(\omega)\text{ is continuous in }t\}=1\).

注 1.1–1.3 / Remarks 1.1–1.3 注 1.1：\(\Omega\) 中每个元素 \(\omega\) 给出一条确定性路径 \(B_t(\omega)\)；\(\mathcal F\) 是事件的集合（含一些确定性路径的集合），\(\mathbb P:\mathcal F\to[0,1]\) 为概率函数。注 1.2：(1)+(2)+(3) 给出更一般的一类随机过程——Lévy 过程；其中维纳过程是连续的 Lévy 过程，而 (1)+(2)+(3)+ 非(4) 称为跳过程 (jump process)，著名例子是泊松过程。注 1.3：若随机过程 \(\{X_t\}\) 满足 (2)+(3)（独立平稳增量），则它必可表为以下至多三者之和：(1) 确定性趋势（\(t\) 的确定性函数）；(2) 无漂移布朗运动；(3) 泊松过程的某种推广。细节见 Lévy–Khintchine 刻画（Ch 11）。Remark 1.1: each element \(\omega\in\Omega\) pins down a deterministic path \(B_t(\omega)\); \(\mathcal F\) is a set of events (containing some collections of deterministic paths), and \(\mathbb P:\mathcal F\to[0,1]\) is the probability function. Remark 1.2: (1)+(2)+(3) gives a more general class of stochastic processes, the Lévy process; the Wiener process is a continuous Lévy process, while (1)+(2)+(3)+not(4) is called a jump process, a famous example being the Poisson process. Remark 1.3: if a stochastic process \(\{X_t\}\) satisfies (2)+(3) (independent and stationary increments), then it must be expressible as a sum of at most: (1) a deterministic trend (a deterministic function of \(t\)); (2) a driftless Brownian motion; (3) some generalized version of the Poisson process. See the Lévy–Khintchine characterization (Ch 11) for details.

定理 1.1（满足四条件的过程为正态 / four conditions \(\Rightarrow\) normal） 若 \(\{B_t\}\) 满足定义 1.2 的条件 (1)-(4)，则对 \(\forall t\)，\(B_t\) 服从均值 \(t\mu\)、方差 \(t\sigma^2\) 的正态分布（\(\mu\in\mathbb R\)、\(\sigma^2\in\mathbb R_+\)），即 \(B_t\sim\mathcal N(t\mu,t\sigma^2)\)。If \(\{B_t\}\) satisfies conditions (1)-(4) of Definition 1.2, then for all \(t\), \(B_t\) is normally distributed with mean \(t\mu\) and variance \(t\sigma^2\) for some \(\mu\in\mathbb R\) and \(\sigma^2\in\mathbb R_+\), i.e. \(B_t\sim\mathcal N(t\mu,t\sigma^2)\).

证明（启发式 / heuristic）/ Proof 先构造二进有理数集 \(\mathcal D=\bigcup_{n=1}^\infty D_n\)，其中 \(D_n=\{\frac{j}{2^n}:j=0,1,\dots,2^n\}\)。对 \(t\in\mathbb R_+\) 定义 \(B_t=\lim_{s\to t}B_s\)（\(s\in\mathcal D_t\)）；由条件 (4)（连续路径），只需对 \(t\in\mathbb R_+\)、\(s\in\mathcal D_t\) 证明 \(B_s\sim\mathcal N(s\mu,s\sigma^2)\)。记 \(Z^{(n)}_{j,s}=B_{\frac{j+1}{2^n}s}-B_{\frac{j}{2^n}s}\)（\(j=0,\dots,2^n-1\)）。由条件 (2)-(3)，\(Z^{(n)}_{0,s},\dots,Z^{(n)}_{2^n-1,s}\) 是 i.i.d. 随机变量，记其均值、方差为 \(\frac{s\mu}{2^n}\)、\(\frac{s\sigma^2}{2^n}\)。由中心极限定理，\(\sqrt{2^n-1}\big[(\frac1{2^n}\sum_{j=0}^{2^n-1}Z^{(n)}_{j,s})-\frac{s\mu}{2^n}\big]\sim\mathcal N(0,\frac{s\sigma^2}{2^n-1})\)，整理（乘 \(2^n-1\)、求和）得 \(\big[\sum_{j=0}^{2^n-1}Z^{(n)}_{j,s}-s\mu\big]\sim\mathcal N(0,s\sigma^2)\)，即 \(B_s\sim\mathcal N(s\mu,s\sigma^2)\)。由于 \(s\in\mathcal D_t\) 任取，可取 \(s=1\)，于是 \(\mu,\sigma^2\) 可解读为 \(B_1\) 的均值与方差。最后由连续性 (4)，对 \(\forall t\in\mathbb R_+\) 有 \(B_t\sim\mathcal N(t\mu,t\sigma^2)\)。\(\blacksquare\)First construct the dyadic set \(\mathcal D=\bigcup_{n=1}^\infty D_n\), where \(D_n=\{\frac{j}{2^n}:j=0,1,\dots,2^n\}\). For \(t\in\mathbb R_+\) define \(B_t=\lim_{s\to t}B_s\) (\(s\in\mathcal D_t\)); by condition (4) (continuous paths), it suffices to show \(B_s\sim\mathcal N(s\mu,s\sigma^2)\) for \(t\in\mathbb R_+\), \(s\in\mathcal D_t\). Denote \(Z^{(n)}_{j,s}=B_{\frac{j+1}{2^n}s}-B_{\frac{j}{2^n}s}\) (\(j=0,\dots,2^n-1\)). By conditions (2)-(3), \(Z^{(n)}_{0,s},\dots,Z^{(n)}_{2^n-1,s}\) are i.i.d. random variables with mean and variance \(\frac{s\mu}{2^n}\) and \(\frac{s\sigma^2}{2^n}\). By the central limit theorem, \(\sqrt{2^n-1}\big[(\frac1{2^n}\sum_{j=0}^{2^n-1}Z^{(n)}_{j,s})-\frac{s\mu}{2^n}\big]\sim\mathcal N(0,\frac{s\sigma^2}{2^n-1})\); rearranging (multiplying by \(2^n-1\) and summing) gives \(\big[\sum_{j=0}^{2^n-1}Z^{(n)}_{j,s}-s\mu\big]\sim\mathcal N(0,s\sigma^2)\), i.e. \(B_s\sim\mathcal N(s\mu,s\sigma^2)\). Since \(s\in\mathcal D_t\) is arbitrary, we can choose \(s=1\), so \(\mu,\sigma^2\) can be interpreted as the mean and variance of \(B_1\). Finally, by continuity (4), \(B_t\sim\mathcal N(t\mu,t\sigma^2)\) for all \(t\in\mathbb R_+\). \(\blacksquare\)

定义 1.3、1.4 与命题 1.1 / Definitions 1.3, 1.4 and Proposition 1.1 定义 1.3（重定义：一维布朗运动）：带漂移 \(\mu\)、方差 \(\sigma^2\) 的一维布朗运动是满足下述条件的随机过程 \(\{B_t\}\)：(1) 从原点出发 \(B_0=0\)；(2) 独立增量；(3)* 正态性：若 \(s定义 1.4（标准一维布朗运动）：若 \(\mu=0\) 且 \(\sigma^2=1\)，则 \(\{B_t\}\) 为标准一维布朗运动。命题 1.1：若 \(\{B_t\}\) 是标准布朗运动、且 \(\{Y_t\}\) 满足 \(Y_t=\mu t+\sigma B_t\)（\(\forall t\)），则 \(\{Y_t\}\) 是带漂移 \(\mu\)、方差 \(\sigma^2\) 的布朗运动。注 1.4：命题 1.1 的逆也成立——若 \(\{Y_t\}\) 是带漂移 \(\mu\)、方差 \(\sigma^2\) 的 BM、\(\{B_t\}\) 满足 \(B_t=\frac1\sigma(Y_t-\mu t)\)，则 \(\{B_t\}\) 是标准 BM。Definition 1.3 (Redefine: One-dimensional Brownian Motion): a one-dimensional Brownian motion with drift \(\mu\) and variance \(\sigma^2\) is a stochastic process \(\{B_t\}\) satisfying: (1) start from origin \(B_0=0\); (2) independent increments; (3)* normality: if \(sDefinition 1.4 (Standard One-dimensional Brownian Motion): \(\{B_t\}\) is a standard one-dimensional Brownian motion if \(\mu=0\) and \(\sigma^2=1\). Proposition 1.1: if \(\{B_t\}\) is a standard Brownian motion and \(\{Y_t\}\) satisfies \(Y_t=\mu t+\sigma B_t\) (for all \(t\)), then \(\{Y_t\}\) is a Brownian motion with drift \(\mu\) and variance \(\sigma^2\). Remark 1.4: the reverse of Proposition 1.1 is also true — if \(\{Y_t\}\) is a BM with drift \(\mu\) and variance \(\sigma^2\) and \(\{B_t\}\) satisfies \(B_t=\frac1\sigma(Y_t-\mu t)\), then \(\{B_t\}\) is a standard BM.

第 1 步的构造与关键引理 / Step 1 construction and key lemma 从 \(B_0=0\)、\(B_1=N_1\) 开始（满足条件 (1)）。再定义 \(B_{1/2}=\frac12 B_1+\frac12 N_{1/2}\)，即 \(B_{1/2}=\frac12(B_1+B_0)+\frac12 N_{1/2}\)，从而 \(B_1-B_{1/2}=\frac12(B_1-B_0)-\frac12 N_{1/2}\)。引理 1.1：若 \(X,Y\) 独立同标准正态，\(Z=\frac{X+Y}{\sqrt2}\)、\(W=\frac{X-Y}{\sqrt2}\)，则 \(Z,W\) 独立同标准正态。断言 1.1：\(B_1-B_{1/2}\) 与 \(B_{1/2}\) 独立且同分布（均为 \(\mathcal N(0,\frac12)\)）。归纳地继续二分：对 \(n+1\)、\(q=\frac{2k+1}{2^{n+1}}\in D_{n+1}\setminus D_n\)，构造 \(B_q=B_{\frac{k}{2^n}}+\frac12\big(B_{\frac{k+1}{2^n}}-B_{\frac{k}{2^n}}\big)+\frac{N_q}{2^{(n+1)/2}}\)；可验证 \(B_q-B_{\frac{k}{2^n}}\) 与 \(B_{\frac{k+1}{2^n}}-B_q\) 独立且同分布 \(\mathcal N(0,\frac1{2^{n+1}})\)，故条件 (2)-(3) 满足。如此一直二分得到 \(\{B_t:t\in\mathcal D\}\)，始终满足条件 (1)-(3)。第 1 步完成。Start with \(B_0=0\) and \(B_1=N_1\) (which satisfies condition (1)). Then define \(B_{1/2}=\frac12 B_1+\frac12 N_{1/2}\), i.e. \(B_{1/2}=\frac12(B_1+B_0)+\frac12 N_{1/2}\), so that \(B_1-B_{1/2}=\frac12(B_1-B_0)-\frac12 N_{1/2}\). Lemma 1.1: if \(X,Y\) are independent standard normal, \(Z=\frac{X+Y}{\sqrt2}\) and \(W=\frac{X-Y}{\sqrt2}\), then \(Z,W\) are independent standard normal. Claim 1.1: \(B_1-B_{1/2}\) and \(B_{1/2}\) are independent and have the same distribution (both \(\mathcal N(0,\frac12)\)). Continue the bisection inductively: for \(n+1\) and \(q=\frac{2k+1}{2^{n+1}}\in D_{n+1}\setminus D_n\), construct \(B_q=B_{\frac{k}{2^n}}+\frac12\big(B_{\frac{k+1}{2^n}}-B_{\frac{k}{2^n}}\big)+\frac{N_q}{2^{(n+1)/2}}\); one verifies \(B_q-B_{\frac{k}{2^n}}\) and \(B_{\frac{k+1}{2^n}}-B_q\) are independent with the same distribution \(\mathcal N(0,\frac1{2^{n+1}})\), so conditions (2)-(3) are satisfied. Continuing the splitting yields \(\{B_t:t\in\mathcal D\}\), always satisfying conditions (1)-(3). Step 1 is done.

引理 1.1 与断言 1.1 的证明 / Proof of Lemma 1.1 and Claim 1.1 引理 1.1：\(Z,W\) 的任意线性组合都是 \(X,Y\) 的线性组合，故由引理 2.1（见 Ch 2）\(Z,W\) 联合正态。由多元正态的密度，\(Z,W\) 独立 \(\iff\mathrm{Cov}(Z,W)=0\)。而 \(\mathrm{Cov}(Z,W)=\mathrm{Cov}(\frac{X+Y}{\sqrt2},\frac{X-Y}{\sqrt2})=\frac12(\mathrm{Cov}(X,X)-\mathrm{Cov}(X,Y)+\mathrm{Cov}(Y,X)-\mathrm{Cov}(Y,Y))=0\)；又 \(\mathbb E[Z]=\mathbb E[W]=0\)、\(\mathrm{Var}(Z)=\mathrm{Var}(W)=\frac12\mathrm{Var}(X)+\frac12\mathrm{Var}(Y)=1\)。故 \(Z,W\) 独立同标准正态。\(\blacksquare\) 断言 1.1：\(N_1,N_{1/2}\) 独立标准正态，可写 \(B_1-B_{1/2}=\frac1{\sqrt2}\cdot\frac{N_1-N_{1/2}}{\sqrt2}\)、\(B_{1/2}=\frac1{\sqrt2}\cdot\frac{N_1+N_{1/2}}{\sqrt2}\)；由引理 1.1，\(\frac{N_1-N_{1/2}}{\sqrt2}\) 与 \(\frac{N_1+N_{1/2}}{\sqrt2}\) 独立标准正态，故 \(B_1-B_{1/2}\) 与 \(B_{1/2}\) 独立、同分布 \(\mathcal N(0,\frac12)\)。\(\blacksquare\)Lemma 1.1: any linear combination of \(Z,W\) is a linear combination of \(X,Y\), so by Lemma 2.1 (see Ch 2) \(Z,W\) are jointly normal. By the density of the multivariate normal, \(Z,W\) are independent \(\iff\mathrm{Cov}(Z,W)=0\). And \(\mathrm{Cov}(Z,W)=\mathrm{Cov}(\frac{X+Y}{\sqrt2},\frac{X-Y}{\sqrt2})=\frac12(\mathrm{Cov}(X,X)-\mathrm{Cov}(X,Y)+\mathrm{Cov}(Y,X)-\mathrm{Cov}(Y,Y))=0\); also \(\mathbb E[Z]=\mathbb E[W]=0\) and \(\mathrm{Var}(Z)=\mathrm{Var}(W)=\frac12\mathrm{Var}(X)+\frac12\mathrm{Var}(Y)=1\). So \(Z,W\) are independent standard normal. \(\blacksquare\) Claim 1.1: \(N_1,N_{1/2}\) are independent standard normal, and we can write \(B_1-B_{1/2}=\frac1{\sqrt2}\cdot\frac{N_1-N_{1/2}}{\sqrt2}\) and \(B_{1/2}=\frac1{\sqrt2}\cdot\frac{N_1+N_{1/2}}{\sqrt2}\); by Lemma 1.1, \(\frac{N_1-N_{1/2}}{\sqrt2}\) and \(\frac{N_1+N_{1/2}}{\sqrt2}\) are independent standard normal, so \(B_1-B_{1/2}\) and \(B_{1/2}\) are independent with the same distribution \(\mathcal N(0,\frac12)\). \(\blacksquare\)

第 2 步：一致连续性（概率 1）/ Step 2: uniform continuity (prob. one) 已得满足 (1)-(3) 的 \(\{B_t:t\in\mathcal D\}\)，现证以概率 1 函数 \(t\to B_t\)（\(t\in\mathcal D\)）一致连续：\(\forall\varepsilon>0\)，\(\exists\delta>0\)，使得 \(s,t\in\mathcal D\)、\(|s-t|\le\delta\) 时 \(|B_s-B_t|\le\varepsilon\)（概率 1）。定义 \(K_n=\sup\{|B_s-B_t|:s,t\in\mathcal D,|s-t|\le\frac1{2^n}\}\)，等价地只需证 \(\lim_{n\to\infty}K_n=0\)（概率 1）。命题 1.2：\(\forall\alpha<\frac12\)，\(\lim_{n\to\infty}2^{\alpha n}K_n=0\)（概率 1），特别地蕴含 \(\lim_{n\to\infty}K_n=0\)（概率 1）。Having obtained \(\{B_t:t\in\mathcal D\}\) satisfying (1)-(3), we now show that with probability one the function \(t\to B_t\) (\(t\in\mathcal D\)) is uniformly continuous: \(\forall\varepsilon>0\), \(\exists\delta>0\) such that for \(s,t\in\mathcal D\), \(|s-t|\le\delta\) implies \(|B_s-B_t|\le\varepsilon\) (with probability one). Define \(K_n=\sup\{|B_s-B_t|:s,t\in\mathcal D,|s-t|\le\frac1{2^n}\}\); equivalently it suffices to show \(\lim_{n\to\infty}K_n=0\) (prob. one). Proposition 1.2: \(\forall\alpha<\frac12\), \(\lim_{n\to\infty}2^{\alpha n}K_n=0\) (prob. one), which particularly implies \(\lim_{n\to\infty}K_n=0\) (prob. one).

命题 1.2 与引理 1.2 的证明 / Proof of Proposition 1.2 and Lemma 1.2 定义 \(Y(j,n)=\sup\{|B_q-B_{\frac{j-1}{2^n}}|:\frac{j-1}{2^n}\le q\le\frac{j}{2^n},q\in\mathcal D\}\)（第 \(j\) 个网格内 \(B_t\) 的局部最大差）、\(J_n=\max_{j=1,\dots,2^n}Y(j,n)\)（全局最大差）。由三角不等式，对 \(s,t\in\mathcal D\)、\(|s-t|\le\frac1{2^n}\) 有 \(K_n\le 3J_n\)，故 \(2^{\alpha n}K_n\le 3\cdot2^{\alpha n}J_n\)，只需证 \(\lim_{n\to\infty}3\cdot2^{\alpha n}J_n=0\)。对 \(\varepsilon>0\)：\(\mathbf P\{J_n\ge\varepsilon\}=\mathbf P\{\bigcup_{j=1}^{2^n}\{Y(j,n)\ge\varepsilon\}\}\le\sum_{j=1}^{2^n}\mathbf P\{Y(j,n)\ge\varepsilon\}=2^n\mathbf P\{Y(1,n)\ge\varepsilon\}\)（同增量分布）。由把 \(B_{\frac{j}{2^n}}\sim\mathcal N(0,\frac1{2^n})\) 放大 \(\sqrt{2^n}\) 倍构造 \(W^{(n)}_{q2^n}=2^{n/2}B_q\sim\mathcal N(0,1)\)，得 (1.1)：\(\sup\{|B_q|:0\le q\le\frac1{2^n},q\in\mathcal D\}=\frac1{2^{n/2}}\sup\{|W^{(n)}_q|\}\stackrel{d}{=}\frac1{2^{n/2}}\sup\{|B_q|:0\le q\le1,q\in\mathcal D\}\)。于是 \(\mathbf P\{J_n\ge\varepsilon\}\le2^n\mathbf P\{\frac1{2^{n/2}}Y(1,0)\ge\varepsilon\}=2^n\mathbf P\{Y(1,0)\ge2^{n/2}\varepsilon\}\)，其中 \(Y=Y(1,0)=\sup\{|B_q|:0\le q\le1,q\in\mathcal D\}\)。取 \(\varepsilon=\frac{C\sqrt n}{2^{n/2}}\)（\(C<\infty\)），则 \(3\cdot2^{\alpha n}\varepsilon=3\cdot2^{(\alpha-\frac12)n}C\sqrt n\to0\)（\(\alpha<\frac12\)），故只需证 \(\sum_{n=1}^\infty2^n\mathbf P\{Y(1,0)\ge C\sqrt n\}<\infty\)（它蕴含 \(2^n\mathbf P\{\dots\}\to0\)，从而 (1.2) \(3\cdot2^{\alpha n}J_n\to0\Rightarrow J_n\to0\)）。引理 1.2：\(\forall a>0\)，\(\mathbf P\{Y\ge a\}\le4\mathbf P\{B_1\ge a\}\)。[证：\(Y_n=\sup\{|B_{\frac{j}{2^n}}|:j=1,\dots,2^n\}\)，\(Y=\lim Y_n\) 递增，\(\mathbf P\{Y>a\}=\lim\mathbf P\{Y_n>a\}\)。令 \(A_{j,n}\) 为 \(\{B_{\frac{j}{2^n}}\}\) 首次在 \(j\) 处穿过 \(a\) 的事件，\(A_{j,n}\) 互斥、\(Y_n=\bigcup A_{j,n}\)。则 \(\mathbf P\{|B_1|>a\}\ge\sum\mathbf P\{A_{j,n}\cap\{|B_1|>a\}\}=\sum\mathbf P\{|B_1|>a\mid A_{j,n}\}\mathbf P\{A_{j,n}\}\ge\frac12\sum\mathbf P\{A_{j,n}\}=\frac12\mathbf P\{Y_n>a\}\)，故 \(\mathbf P\{Y_n>a\}\le2\mathbf P\{|B_1|>a\}\le4\mathbf P\{B_1>a\}\)。] 于是 \(\sum2^n\mathbf P\{Y(1,0)\ge C\sqrt n\}\le4\sum2^n\int_{C\sqrt n}^\infty\frac1{\sqrt{2\pi}}e^{-x^2/2}dx\le4\sum\frac1{\sqrt{2\pi}}\frac2{C\sqrt n}e^{n(\ln2-C^2/2)}\)。取 \(C>\sqrt{2\ln2}\)（使 \(\ln2-\frac{C^2}2<0\)），则该级数 \(\to0\)。一致连续性（概率 1）证毕。\(\blacksquare\)Define \(Y(j,n)=\sup\{|B_q-B_{\frac{j-1}{2^n}}|:\frac{j-1}{2^n}\le q\le\frac{j}{2^n},q\in\mathcal D\}\) (the local max difference of \(B_t\) within the \(j\)th grid) and \(J_n=\max_{j=1,\dots,2^n}Y(j,n)\) (the global max difference). By the triangle inequality, for \(s,t\in\mathcal D\), \(|s-t|\le\frac1{2^n}\) we have \(K_n\le3J_n\), so \(2^{\alpha n}K_n\le3\cdot2^{\alpha n}J_n\) and it suffices to show \(\lim_{n\to\infty}3\cdot2^{\alpha n}J_n=0\). For \(\varepsilon>0\): \(\mathbf P\{J_n\ge\varepsilon\}=\mathbf P\{\bigcup_{j=1}^{2^n}\{Y(j,n)\ge\varepsilon\}\}\le\sum_{j=1}^{2^n}\mathbf P\{Y(j,n)\ge\varepsilon\}=2^n\mathbf P\{Y(1,n)\ge\varepsilon\}\) (same increment distribution). Scaling \(B_{\frac{j}{2^n}}\sim\mathcal N(0,\frac1{2^n})\) up by \(\sqrt{2^n}\) to construct \(W^{(n)}_{q2^n}=2^{n/2}B_q\sim\mathcal N(0,1)\) gives (1.1): \(\sup\{|B_q|:0\le q\le\frac1{2^n},q\in\mathcal D\}=\frac1{2^{n/2}}\sup\{|W^{(n)}_q|\}\stackrel{d}{=}\frac1{2^{n/2}}\sup\{|B_q|:0\le q\le1,q\in\mathcal D\}\). So \(\mathbf P\{J_n\ge\varepsilon\}\le2^n\mathbf P\{\frac1{2^{n/2}}Y(1,0)\ge\varepsilon\}=2^n\mathbf P\{Y(1,0)\ge2^{n/2}\varepsilon\}\), where \(Y=Y(1,0)=\sup\{|B_q|:0\le q\le1,q\in\mathcal D\}\). Taking \(\varepsilon=\frac{C\sqrt n}{2^{n/2}}\) (\(C<\infty\)), then \(3\cdot2^{\alpha n}\varepsilon=3\cdot2^{(\alpha-\frac12)n}C\sqrt n\to0\) (\(\alpha<\frac12\)), so it suffices to show \(\sum_{n=1}^\infty2^n\mathbf P\{Y(1,0)\ge C\sqrt n\}<\infty\) (which implies \(2^n\mathbf P\{\dots\}\to0\), hence (1.2) \(3\cdot2^{\alpha n}J_n\to0\Rightarrow J_n\to0\)). Lemma 1.2: \(\forall a>0\), \(\mathbf P\{Y\ge a\}\le4\mathbf P\{B_1\ge a\}\). [Proof: \(Y_n=\sup\{|B_{\frac{j}{2^n}}|:j=1,\dots,2^n\}\), \(Y=\lim Y_n\) increasing, \(\mathbf P\{Y>a\}=\lim\mathbf P\{Y_n>a\}\). Let \(A_{j,n}\) be the event that \(\{B_{\frac{j}{2^n}}\}\) passes \(a\) for the first time at \(j\); the \(A_{j,n}\) are disjoint and \(Y_n=\bigcup A_{j,n}\). Then \(\mathbf P\{|B_1|>a\}\ge\sum\mathbf P\{A_{j,n}\cap\{|B_1|>a\}\}=\sum\mathbf P\{|B_1|>a\mid A_{j,n}\}\mathbf P\{A_{j,n}\}\ge\frac12\sum\mathbf P\{A_{j,n}\}=\frac12\mathbf P\{Y_n>a\}\), so \(\mathbf P\{Y_n>a\}\le2\mathbf P\{|B_1|>a\}\le4\mathbf P\{B_1>a\}\).] So \(\sum2^n\mathbf P\{Y(1,0)\ge C\sqrt n\}\le4\sum2^n\int_{C\sqrt n}^\infty\frac1{\sqrt{2\pi}}e^{-x^2/2}dx\le4\sum\frac1{\sqrt{2\pi}}\frac2{C\sqrt n}e^{n(\ln2-C^2/2)}\). Taking \(C>\sqrt{2\ln2}\) (so \(\ln2-\frac{C^2}2<0\)), the series \(\to0\). Uniform continuity (prob. one) is proved. \(\blacksquare\)

注 1.5（路径连续性的含义 / Remark 1.5） 解读 \(t\to B_t\) 的连续性：\(B_t\) 是随机变量，谈连续性其实是谈其实现（realization）\(B_t(\omega)\)（\(\omega\in\Omega\)）的连续性。"以概率 1 连续"即 \(\mathbf P\{\omega:t\to B_t(\omega)\text{ continuous}\}=1\)，其中每个 \(\omega\) 对应一条可能路径、\(\Omega\) 是所有可能路径的集合。Interpreting the continuity of \(t\to B_t\): \(B_t\) is a random variable, so talking about continuity is actually talking about the continuity of its realization \(B_t(\omega)\) (\(\omega\in\Omega\)). "Continuous with probability one" means \(\mathbf P\{\omega:t\to B_t(\omega)\text{ continuous}\}=1\), where each \(\omega\) corresponds to a possible path and \(\Omega\) is the collection of all possible paths.

第 3 步：延拓到 $[0,1]$ 并逐条验证 / Step 3: extend to $[0,1]$ and verify conditions 在第 2 步基础上定义 \(B_s=\lim_{t\to s}B_t\)（\(t\in\mathcal D\)）(1.3)，\(s\in[0,1]\)，并按定义 1.3（带正态性的重定义）逐条验证 \(\{B_s:s\in[0,1]\}\) 为布朗运动。条件 (1)：\(B_0=0\) 平凡成立。条件 (2)（独立增量）：\(\{B_s:s\in[0,1]\}\) 由 \(\{B_t:t\in\mathcal D\}\) 逼近定义。若 \(s=\frac{j}{2^n}\)（某 \(j\in\mathbb N_+\)），增量独立性直接来自 \(\{B_t:t\in\mathcal D\}\)；若 \(s\in(\frac{j}{2^n},\frac{j+1}{2^n})\)（任意大 \(n\)），\(s\) 永不被 \(\mathcal D\) 中点真正取到，故对 \(m>s\)，\(B_m\) 由 \(s\) 右侧最近网格点逼近、对 \(h条件 (3)（正态/同分布增量）：记连续 BM 为 \(B\)、离散 BM 为 \(\hat B\)。对 \(s条件 (4)（连续性）：由 \(\{B_t:t\in\mathcal D\}\) 一致连续，连续情形下任意两点间有无穷多网格点，其距离可足够近以看到中间网格点，故连续性由离散情形成立。综上，标准布朗运动存在。\(\blacksquare\)On the basis of Step 2, define \(B_s=\lim_{t\to s}B_t\) (\(t\in\mathcal D\)) (1.3), \(s\in[0,1]\), and verify that \(\{B_s:s\in[0,1]\}\) is a Brownian motion by checking the conditions of Definition 1.3 (the redefinition with normality) one by one. Condition (1): \(B_0=0\) trivially. Condition (2) (independent increments): \(\{B_s:s\in[0,1]\}\) is defined by approximation of \(\{B_t:t\in\mathcal D\}\). If \(s=\frac{j}{2^n}\) (some \(j\in\mathbb N_+\)), increment independence follows directly from \(\{B_t:t\in\mathcal D\}\); if \(s\in(\frac{j}{2^n},\frac{j+1}{2^n})\) (for arbitrarily large \(n\)), \(s\) is never actually reached by points in \(\mathcal D\), so for \(m>s\), \(B_m\) is approximated by the nearest grid point to the right of \(s\), and for \(hCondition (3) (normal/identically distributed increments): denote the continuous BM by \(B\) and the discrete BM by \(\hat B\). For \(sCondition (4) (continuity): since \(\{B_t:t\in\mathcal D\}\) is uniformly continuous, any two points in the continuous case have infinitely many grid points between them, whose distance can be close enough to see grid points in between, so continuity holds from the discrete case. Hence a standard Brownian motion exists. \(\blacksquare\)

定义 1.5、1.6 与定理 1.2 / Definitions 1.5, 1.6 and Theorem 1.2 定义 1.5（Hölder 连续）：函数 \(f:[0,1]\to\mathbb R^d\) 是阶 \(\alpha\) 的 Hölder 连续，若 \(\exists C<\infty\) 使得 \(\forall s\le t\)，\(|f(t)-f(s)|\le C(t-s)^\alpha\)。注 1.6：\(\alpha>\beta>0\) 时，阶 \(\alpha\) 的 Hölder 连续蕴含阶 \(\beta\) 的 Hölder 连续。注 1.7：\(\alpha>1\) 时定义 1.5 变平凡（要求 \(f\) 为水平直线）：\(|\frac{f(t)-f(s)}{t-s}|\le C(t-s)^{\alpha-1}\)，\(t-s\to0\Rightarrow|\frac{f(t)-f(s)}{t-s}|\to0\)；\(\alpha=1\) 时即 Lipschitz 连续。定义 1.6（弱 Hölder 连续）：\(f:[0,1]\to\mathbb R^d\) 是阶 \(\alpha\) 的弱 Hölder 连续，若对 \(\forall\beta<\alpha\) 它都是阶 \(\beta\) 的 Hölder 连续。定理 1.2：以概率 1，布朗运动 \(\{B_t:t\in[0,1]\}\) 是阶 \(\frac12\) 的弱 Hölder 连续，但不是阶 \(\frac12\) 的 Hölder 连续。Definition 1.5 (Hölder Continuity): a function \(f:[0,1]\to\mathbb R^d\) is Hölder continuous of order \(\alpha\) if \(\exists C<\infty\) such that \(\forall s\le t\), \(|f(t)-f(s)|\le C(t-s)^\alpha\). Remark 1.6: for \(\alpha>\beta>0\), Hölder continuity of order \(\alpha\) implies Hölder continuity of order \(\beta\). Remark 1.7: for \(\alpha>1\), Definition 1.5 becomes trivial (it requires \(f\) to be a flat line): \(|\frac{f(t)-f(s)}{t-s}|\le C(t-s)^{\alpha-1}\), \(t-s\to0\Rightarrow|\frac{f(t)-f(s)}{t-s}|\to0\); for \(\alpha=1\) it is Lipschitz continuity. Definition 1.6 (Weakly Hölder Continuity): \(f:[0,1]\to\mathbb R^d\) is weakly Hölder continuous of order \(\alpha\) if it is Hölder continuous of order \(\beta\) for all \(\beta<\alpha\). Theorem 1.2: with probability one, Brownian motion \(\{B_t:t\in[0,1]\}\) is weakly Hölder continuous of order \(\frac12\) but not Hölder continuous of order \(\frac12\).

定理 1.2 证明 / Proof of Theorem 1.2 由命题 1.2，\(\lim_{n\to\infty}2^{\alpha n}K_n=0\) 对 \(\forall\alpha<\frac12\)（但不对 \(\alpha=\frac12\)，脚注：若 \(\alpha=\frac12\)，(1.2) 中 \(3\cdot2^{(\alpha-\frac12)n}C\sqrt n\to0\) 不成立）以概率 1 成立。(1.4) 等价于：以概率 1，对充分大 \(n\)、\(\forall\varepsilon\in(0,C)\)（某常数 \(C\)）有 \(2^{\alpha n}K_n\le\varepsilon\)，即 \(\sup\{|B_s-B_t|:s,t\in\mathcal D,|s-t|\le\frac1{2^n}\}\le(\frac1{2^n})^\alpha\varepsilon\)，对 \(t-s=\frac1{2^n}\) 特别成立：\(|B_s-B_t|\le(\frac1{2^n})^\alpha\varepsilon\le(t-s)^\alpha\varepsilon\Rightarrow|B_s-B_t|\le C(t-s)^\alpha\) (1.5)（某 \(C>0\)），证得 \(\mathcal D\) 上的近似 BM 弱 Hölder 连续于阶 \(\frac12\)。对连续 BM，(1.5) 两端在极限下等于连续情形，故 (1.5) 对连续 BM 也平凡成立。注意 \(\alpha\ge\frac12\) 时证明不成立，因为 (1.4) 对 \(\alpha\ge\frac12\) 不成立。\(\blacksquare\)By Proposition 1.2, \(\lim_{n\to\infty}2^{\alpha n}K_n=0\) for all \(\alpha<\frac12\) (but not for \(\alpha=\frac12\), footnote: if \(\alpha=\frac12\), \(3\cdot2^{(\alpha-\frac12)n}C\sqrt n\to0\) in (1.2) won't hold) with probability one. (1.4) is equivalent to: with probability one, for sufficiently large \(n\) and \(\forall\varepsilon\in(0,C)\) (some constant \(C\)), \(2^{\alpha n}K_n\le\varepsilon\), i.e. \(\sup\{|B_s-B_t|:s,t\in\mathcal D,|s-t|\le\frac1{2^n}\}\le(\frac1{2^n})^\alpha\varepsilon\), which holds in particular for \(t-s=\frac1{2^n}\): \(|B_s-B_t|\le(\frac1{2^n})^\alpha\varepsilon\le(t-s)^\alpha\varepsilon\Rightarrow|B_s-B_t|\le C(t-s)^\alpha\) (1.5) (some \(C>0\)), proving the approximate BM on \(\mathcal D\) is weakly Hölder continuous of order \(\frac12\). For the continuous BM, both sides of (1.5) equal those of the continuous case in the limit, so (1.5) also holds trivially for continuous BM. Note the proof won't hold if \(\alpha\ge\frac12\) since (1.4) doesn't hold for \(\alpha\ge\frac12\). \(\blacksquare\)

定理 1.3（BM 处处不可微 / nowhere differentiable） 以概率 1，布朗运动 \(\{B_t\}\) 处处不可微。With probability one, Brownian motion \(\{B_t\}\) is nowhere differentiable.

定理 1.3 证明 / Proof of Theorem 1.3 设 \(B_t:[0,1]\to\mathbb R\)（脚注：仍聚焦 $[0,1]$ 上的 BM，易推广到 \([0,+\infty)\)）在某 \(t\in[0,1]\) 可微，则 \(\exists M<\infty\)、\(\delta>0\) 使 \(\forall|s-t|\le\delta\) 有 \(|B_t-B_s|\le M|t-s|\)，从而 \(\exists N>0\) 使 \(\forall n>N\)（脚注：大 \(N\) 保证 \(\frac{j}{n},\frac{j+1}{n},\frac{j+2}{n},\frac{j+3}{n}\) 落入 \(\delta\) 范围）至少能找到某 \(j\in\{0,\dots,n\}\) 使 (1.6)：\(|B_{\frac{j+1}{n}}-B_{\frac{j}{n}}|\le\frac Mn\)、\(|B_{\frac{j+2}{n}}-B_{\frac{j+1}{n}}|\le\frac Mn\)、\(|B_{\frac{j+3}{n}}-B_{\frac{j+2}{n}}|\le\frac Mn\)。记 \(A_{n,M}\) 为"至少存在某 \(j\) 使 (1.6) 成立"的事件（\(M<\infty\)、\(n\) 大），只需证 \(\mathbf P\{A_{n,M}\}=0\)（\(\forall M>0\)）。则 \(\mathbf P\{A_{n,M}\}\le n\,\mathbf P\{|B_{\frac1n}|\le\frac Mn,|B_{\frac2n}-B_{\frac1n}|\le\frac Mn,|B_{\frac3n}-B_{\frac2n}|\le\frac Mn\}=n(\mathbf P\{|B_{\frac1n}|\le\frac Mn\})^3=n(\mathbf P\{|B_1|\le\frac M{\sqrt n}\})^3\le n(\frac{2M}{\sqrt{2\pi}\sqrt n})^3=(\frac{2M}{\sqrt{2\pi}})^3\frac1{\sqrt n}\to0\)（\(n\to\infty\)，\(\forall M>0\)）。其中：第一行有 \(n\) 个 bin、\(n\) 次机会满足 \(A_{n,M}\)；第二行由增量独立；第三行由 BM 定义 \(B_1\stackrel{d}=\sqrt n B_{\frac1n}\sim\mathcal N(0,1)\)；第四行由 \(\mathbf P\{|B_1|\le\frac M{\sqrt n}\}=2\int_0^{M/\sqrt n}\frac1{\sqrt{2\pi}}e^{-x^2/2}dx\le2\int_0^{M/\sqrt n}\frac1{\sqrt{2\pi}}dx=\frac{2M}{\sqrt{2\pi}\sqrt n}\)。\(\blacksquare\)Suppose \(B_t:[0,1]\to\mathbb R\) (footnote: still focusing on BM on $[0,1]\(, easily generalized to \)[0,+\infty)\() is differentiable at some \)t\in[0,1]\(, then \)\exists M<\infty$, \(\delta>0\) such that \(\forall|s-t|\le\delta\), \(|B_t-B_s|\le M|t-s|\), which implies \(\exists N>0\) such that \(\forall n>N\) (footnote: large \(N\) ensures \(\frac{j}{n},\frac{j+1}{n},\frac{j+2}{n},\frac{j+3}{n}\) fall into the range of \(\delta\)) we can find at least some \(j\in\{0,\dots,n\}\) such that (1.6): \(|B_{\frac{j+1}{n}}-B_{\frac{j}{n}}|\le\frac Mn\), \(|B_{\frac{j+2}{n}}-B_{\frac{j+1}{n}}|\le\frac Mn\), \(|B_{\frac{j+3}{n}}-B_{\frac{j+2}{n}}|\le\frac Mn\). Denote \(A_{n,M}\) the event of "at least having some \(j\) such that (1.6) holds" (\(M<\infty\), large \(n\)); it suffices to show \(\mathbf P\{A_{n,M}\}=0\) (for all \(M>0\)). Then \(\mathbf P\{A_{n,M}\}\le n\,\mathbf P\{|B_{\frac1n}|\le\frac Mn,|B_{\frac2n}-B_{\frac1n}|\le\frac Mn,|B_{\frac3n}-B_{\frac2n}|\le\frac Mn\}=n(\mathbf P\{|B_{\frac1n}|\le\frac Mn\})^3=n(\mathbf P\{|B_1|\le\frac M{\sqrt n}\})^3\le n(\frac{2M}{\sqrt{2\pi}\sqrt n})^3=(\frac{2M}{\sqrt{2\pi}})^3\frac1{\sqrt n}\to0\) (\(n\to\infty\), for all \(M>0\)). Here: the first line has \(n\) bins and \(n\) opportunities to satisfy \(A_{n,M}\); the second line is by increment independence; the third line is by the BM definition \(B_1\stackrel{d}=\sqrt n B_{\frac1n}\sim\mathcal N(0,1)\); the fourth line is by \(\mathbf P\{|B_1|\le\frac M{\sqrt n}\}=2\int_0^{M/\sqrt n}\frac1{\sqrt{2\pi}}e^{-x^2/2}dx\le2\int_0^{M/\sqrt n}\frac1{\sqrt{2\pi}}dx=\frac{2M}{\sqrt{2\pi}\sqrt n}\). \(\blacksquare\)

注 1.8（不可微的直觉 / Remark 1.8） 布朗运动处处不可微相当直观：BM 的核心思想是每一刻的随机游走，意味着下一刻的运动不可预测；若其路径在任一点可微，则该点处下一刻的运动就可预测，与随机性矛盾。Nowhere differentiability of Brownian motion is quite intuitive: the core idea of BM is a random walk at each moment, which means the next movement is unpredictable; if its path were differentiable at any point, then the next movement at that point would be predictable, which contradicts randomness.

定理 1.4（反射原理 / Reflection Principle） 若 \(\{B_t\}\) 是标准布朗运动，对 \(a>0\) 有 \(\mathbf P\{\max_{0\le s\le t}B_s\ge a\}=2\mathbf P\{B_t\ge a\}\)。If \(\{B_t\}\) is a standard Brownian motion, then for \(a>0\), \(\mathbf P\{\max_{0\le s\le t}B_s\ge a\}=2\mathbf P\{B_t\ge a\}\).

定理 1.4 证明 / Proof of Theorem 1.4 令 \(\tau=\min\{t:B_t\ge a\}=\min\{t:B_t=a\}\)（首次触 \(a\) 的时刻）。则 \(\{\max_{0\le s\le t}B_s\ge a\}=\{\tau\le t\}\)。又 \(\mathbf P\{\tau\le t\}-\underbrace{\mathbf P\{\tau=t\}}_{=0\text{, measure }0}\le\mathbf P\{\tauLet \(\tau=\min\{t:B_t\ge a\}=\min\{t:B_t=a\}\) (the first time to hit \(a\)). Then \(\{\max_{0\le s\le t}B_s\ge a\}=\{\tau\le t\}\). Also \(\mathbf P\{\tau\le t\}-\underbrace{\mathbf P\{\tau=t\}}_{=0\text{, measure }0}\le\mathbf P\{\tau

定理 1.5（标度规则 / The Scaling Rule） 设 \(\{B_t\}\) 是标准布朗运动，\(a\in\mathbb R\)、\(a\ne0\)。定义 \(\{Y_t\}\) 为 \(Y_t=\frac1a B_{at^2}\) (1.7)，则 \(\{Y_t\}\) 也是标准布朗运动。Let \(\{B_t\}\) be a standard Brownian motion, \(a\in\mathbb R\), \(a\ne0\). Define \(\{Y_t\}\) by \(Y_t=\frac1a B_{at^2}\) (1.7), then \(\{Y_t\}\) is also a standard Brownian motion.

定理 1.5 证明 / Proof of Theorem 1.5 需证 \(\{Y_t\}\) 满足定义 1.2 的 (1)-(4) 且无漂移、方差为一。\(B_0=0\Rightarrow Y_0=\frac1a B_0=0\)，满足 (1)。对 \(rWe must show \(\{Y_t\}\) satisfies (1)-(4) of Definition 1.2 and is driftless with variance one. \(B_0=0\Rightarrow Y_0=\frac1a B_0=0\), satisfying (1). For \(r

例 1.1（用标度律 + 反射原理算触零概率 / hitting-zero probability） 设 \(\{B_t\}\) 是标准布朗运动。对 \(0Let \(\{B_t\}\) be a standard Brownian motion. For \(0

例 1.1 解 / Solution 由标度规则 (1.7)，\(X_r=\sqrt{\frac1s}B_{sr}\) 也是标准 BM。于是 \(\mathbf P\{B_r=0\text{ for some }s\le r\le a\}=\mathbf P\{B_{sr}=0\text{ for some }1\le r\le\frac as\}=\mathbf P\{\sqrt{\frac1s}B_{sr}=0\text{ for some }1\le r\le\frac as\}=\mathbf P\{X_r=0\text{ for some }1\le r\le\frac as\}=\mathbf P\{B_r=0\text{ for some }1\le r\le\frac as\}\)（末行因 \(\{B_t\},\{X_t\}\) 均为标准 BM）。记 \(1+t=\frac as\)（\(t>0\)），定义事件 \(A\equiv\{B_r=0\text{ for some }1\le r\le1+t\}\)。则By the scaling rule (1.7), \(X_r=\sqrt{\frac1s}B_{sr}\) is also a standard BM. So \(\mathbf P\{B_r=0\text{ for some }s\le r\le a\}=\mathbf P\{B_{sr}=0\text{ for some }1\le r\le\frac as\}=\mathbf P\{\sqrt{\frac1s}B_{sr}=0\text{ for some }1\le r\le\frac as\}=\mathbf P\{X_r=0\text{ for some }1\le r\le\frac as\}=\mathbf P\{B_r=0\text{ for some }1\le r\le\frac as\}\) (last line since \(\{B_t\},\{X_t\}\) are both standard BM). Denote \(1+t=\frac as\) (\(t>0\)) and define the event \(A\equiv\{B_r=0\text{ for some }1\le r\le1+t\}\). Then

$$\mathbf P\{A\}=\int_{-\infty}^\infty\mathbf P\{A\mid B_1=x\}\,d\mathbf P\{B_1=x\}=\int_{-\infty}^\infty\mathbf P\{A\mid B_1=x\}\frac1{\sqrt{2\pi}}e^{-\frac12 x^2}dx=2\int_0^\infty\mathbf P\{A\mid B_1=x\}\frac1{\sqrt{2\pi}}e^{-\frac12 x^2}dx\tag{1.8}$$

而由对称性、反射原理、标度律：And by symmetry, the reflection principle, and the scaling rule:

$$\mathbf P\{A\mid B_1=x\}=\mathbf P\{B_s\le-x\text{ for some }0\le s\le t\}=\mathbf P\{B_s\ge x\text{ for some }0\le s\le t\}=2\mathbf P\{B_t\ge x\}=2\mathbf P\{\sqrt t B_1\ge x\}=2\mathbf P\{B_1\ge\tfrac x{\sqrt t}\}\tag{1.9}$$

把 (1.9) 代入 (1.8)：Plugging (1.9) into (1.8):

$$\mathbf P\{A\}=4\int_0^\infty\mathbf P\{B_1\ge\tfrac x{\sqrt t}\}\frac1{\sqrt{2\pi}}e^{-\frac12 x^2}dx=\frac2\pi\int_0^\infty\int_{x/\sqrt t}^\infty e^{-\frac12(x^2+y^2)}\,dy\,dx\tag{1.10}$$

化为极坐标 \(x=r\cos\theta\)、\(y=r\sin\theta\)，记 \(\alpha=\arctan(\frac1{\sqrt t})\)：\(\mathbf P\{A\}=\frac2\pi\int_\alpha^{\pi/2}\int_0^\infty e^{-\frac12 r^2}r\,dr\,d\theta=\frac2\pi(\frac\pi2-\alpha)(-e^{-\frac12 r^2}|_0^\infty)=\frac2\pi(\frac\pi2-\alpha)=1-\frac2\pi\arctan(\frac1{\sqrt t})\)。故对 \(0Transforming to polar coordinates \(x=r\cos\theta\), \(y=r\sin\theta\), and denoting \(\alpha=\arctan(\frac1{\sqrt t})\): \(\mathbf P\{A\}=\frac2\pi\int_\alpha^{\pi/2}\int_0^\infty e^{-\frac12 r^2}r\,dr\,d\theta=\frac2\pi(\frac\pi2-\alpha)(-e^{-\frac12 r^2}|_0^\infty)=\frac2\pi(\frac\pi2-\alpha)=1-\frac2\pi\arctan(\frac1{\sqrt t})\). So for \(0

注 1.9（极限直觉 / Remark 1.9） (1) \(t\to0\) 时 \(\arctan(\frac1{\sqrt t})\to\frac\pi2\)，故 \(\mathbf P\{B_r=0\text{ for some }1\le r\le1+t\}\to0\)；(2) \(t\to\infty\) 时 \(\arctan(\frac1{\sqrt t})\to0\)，故概率 \(\to1\)。结论很直观：BM 在太短时间内不可能触零、但在长时间内必然触零。(1) When \(t\to0\), \(\arctan(\frac1{\sqrt t})\to\frac\pi2\), so \(\mathbf P\{B_r=0\text{ for some }1\le r\le1+t\}\to0\); (2) when \(t\to\infty\), \(\arctan(\frac1{\sqrt t})\to0\), so the probability \(\to1\). The result is very intuitive: BM is impossible to cross 0 in too short a time, but will surely cross 0 in a long time.