本章导读 本章证明布朗运动增长速度的精确刻画——重对数律。§8.1 动机：BM 虽能增长很快，但不会任意快；存在渐近最小上包络 \(\psi(t)\) 使 \(\limsup_{t\to\infty}B_t/\psi(t)=1\)。§8.2 定理与证明（Thm 8.1：以概率 1，\(\limsup_{t\to\infty}\dfrac{B_t}{\sqrt{2t\ln(\ln t)}}=1\)；Rmk 8.1：由对称性 \(\liminf=-1\)）。证明分上下界两半：上界取 \(t_n=\rho^n\)，用反射原理 + Lemma 8.1（标准正态尾概率上下界）+ 缩放，证 \(\sum\mathbb P\{E_n\}<\infty\)（8.1），故 \(\mathbb P\{E_n\}\to0\)（反证）；下界取不相交增量 \(S_n\)，证 \(\sum\mathbb P\{S_n\}=\infty\)（用 \(\sum 1/(n\ln n)=\infty\)），由独立性推出无穷多 \(S_n\) 发生，再借已证上界控制 \(-B_{t_{n-1}}\)，令 \(\rho\to\infty\) 得 RHS\(\to1\)。无图。

上包络的思想 / The upper-envelope idea 布朗运动可能随时间增长得很快，但可以证明它不会任意快。事实上，对标准布朗运动 \(\{B_t\}\)，能找到一个渐近的最小上包络函数 \(\psi(t)\)，使得 \(\limsup_{t\to\infty}\dfrac{B_t}{\psi(t)}=1\)，其中 \(\limsup_{n\to\infty}x_n\equiv\lim_{n\to\infty}\left(\sup_{m\ge n}x_m\right)\)。有了这样的上包络函数，我们就能更好地理解布朗运动的增长速度。Brownian motion could grow very fast over time, but one can show it cannot grow arbitrarily fast. In fact, for a standard Brownian motion \(\{B_t\}\) we can find an asymptotic smallest upper envelope function \(\psi(t)\) such that \(\limsup_{t\to\infty}\dfrac{B_t}{\psi(t)}=1\), where \(\limsup_{n\to\infty}x_n\equiv\lim_{n\to\infty}\left(\sup_{m\ge n}x_m\right)\). With such an upper-envelope function we can better understand the growth speed of Brownian motion.

定理 8.1（重对数律）与注 8.1 / Theorem 8.1 (Law of Iterated Logarithm) and Remark 8.1 设 \(\{B_t\}\) 是标准布朗运动。则以概率 1，Let \(\{B_t\}\) be a standard Brownian motion. Then with probability one,

$$\limsup_{t\to\infty}\frac{B_t}{\sqrt{2t\ln(\ln t)}}=1.$$

注 8.1：由对称性，\(\liminf_{t\to\infty}\dfrac{B_t}{\sqrt{2t\ln(\ln t)}}=-1\)，其中 \(\liminf_{n\to\infty}x_n\equiv\lim_{n\to\infty}\left(\inf_{m\ge n}x_m\right)\)。Remark 8.1: by symmetry, \(\liminf_{t\to\infty}\dfrac{B_t}{\sqrt{2t\ln(\ln t)}}=-1\), where \(\liminf_{n\to\infty}x_n\equiv\lim_{n\to\infty}\left(\inf_{m\ge n}x_m\right)\).

引理 8.1（标准正态尾概率的上下界）/ Lemma 8.1 (Tail bounds for the standard normal) 设 \(Z\) 服从标准正态分布。则对 \(\forall x>0\)，Let \(Z\) have a standard normal distribution. Then for all \(x>0\),

$$\frac{x}{x^2+1}\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}\le\mathbb P\{Z>x\}\le\frac1x\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}.$$

引理 8.1 证明 / Proof of Lemma 8.1 右不等式：在被积函数中插入 \(z/x\ge1\)（当 \(z\ge x\)）放大：Right inequality: enlarge by inserting \(z/x\ge1\) (for \(z\ge x\)) into the integrand:

$$\mathbb P\{Z>x\}=\int_x^\infty\frac1{\sqrt{2\pi}}e^{-\frac12z^2}dz\le\int_x^\infty\frac1{\sqrt{2\pi}}\frac zx e^{-\frac12z^2}dz=\frac1{\sqrt{2\pi}}\left(-\frac1x e^{-\frac12z^2}\right)\Big|_x^\infty=\frac1x\frac1{\sqrt{2\pi}}e^{-\frac12x^2}.$$

左不等式：定义 \(f(x)=xe^{-\frac12x^2}-(x^2+1)\int_x^\infty e^{-\frac12z^2}dz\)。可验 \(f(0)=-\int_0^\infty e^{-\frac12z^2}dz<0\) 且 \(\lim_{x\to\infty}f(x)=0\)（Taylor 展开）。由 Leibniz 积分法则，Left inequality: define \(f(x)=xe^{-\frac12x^2}-(x^2+1)\int_x^\infty e^{-\frac12z^2}dz\). One checks \(f(0)=-\int_0^\infty e^{-\frac12z^2}dz<0\) and \(\lim_{x\to\infty}f(x)=0\) (by Taylor expansion). By the Leibniz integral rule,

$$f'(x)=\left(1-x^2+x^2+1\right)e^{-\frac12x^2}-\int_x^\infty 2xe^{-\frac12z^2}dz=-2x\underbrace{\left(\int_x^\infty e^{-\frac12z^2}dz-\frac{e^{-\frac12x^2}}{x}\right)}_{\le\,0\text{ by the right inequality}}\ge0.$$

故 \(\forall x>0\)，\(f(x)\le0\)，即 \(xe^{-\frac12x^2}\le(x^2+1)\int_x^\infty e^{-\frac12z^2}dz\)，整理即得左不等式。\(\blacksquare\)So for all \(x>0\), \(f(x)\le0\), i.e. \(xe^{-\frac12x^2}\le(x^2+1)\int_x^\infty e^{-\frac12z^2}dz\), which rearranges to the left inequality. \(\blacksquare\)

定理 8.1 证明（上界）/ Proof of Theorem 8.1 (Upper bound) 先证上界：\(\forall\varepsilon>0\)，\(\limsup_{t\to\infty}\dfrac{B_t}{\sqrt{2t\ln(\ln t)}}\le1+\varepsilon\)。取 \(t_n=\rho^n\)（\(\rho>1\)），记 \(E_n=\{B_t\ge(1+\varepsilon)\sqrt{2t_n\ln(\ln t_n)}\text{ for some }t_n\le t\le t_{n+1}\}\)，目标是 \(\sum_{n}\mathbb P\{E_n\}<\infty\)。用反射原理与缩放：First the upper bound: for all \(\varepsilon>0\), \(\limsup_{t\to\infty}\dfrac{B_t}{\sqrt{2t\ln(\ln t)}}\le1+\varepsilon\). Take \(t_n=\rho^n\) (\(\rho>1\)), denote \(E_n=\{B_t\ge(1+\varepsilon)\sqrt{2t_n\ln(\ln t_n)}\text{ for some }t_n\le t\le t_{n+1}\}\); the goal is \(\sum_{n}\mathbb P\{E_n\}<\infty\). Using the reflection principle and scaling:

$$\begin{aligned}\mathbb P\{E_n\}&\le\mathbb P\Big\{\max_{0\le t\le t_{n+1}}B_t\ge(1+\varepsilon)\sqrt{2t_n\ln(\ln t_n)}\Big\}\\&\overset{\text{reflection}}{=}2\,\mathbb P\Big\{B_{t_{n+1}}\ge(1+\varepsilon)\sqrt{2t_n\ln(\ln t_n)}\Big\}\\&\overset{\text{scaling}}{=}2\,\mathbb P\Big\{B_1\ge\underbrace{(1+\varepsilon)\sqrt{\tfrac2\rho\big(\ln n+\ln(\ln\rho)\big)}}_{\equiv\,x}\Big\}\overset{\text{Lemma 8.1}}{\le}\frac2{x\sqrt{2\pi}}e^{-\frac12x^2}.\end{aligned}$$

给定 \(\varepsilon>0\)，选 \(\rho>1\) 使 \(1+\delta\equiv\dfrac{1+\varepsilon}\rho>1\)。代入并整理（吸收常数为 \(M\)）：Given \(\varepsilon>0\), choose \(\rho>1\) such that \(1+\delta\equiv\dfrac{1+\varepsilon}\rho>1\). Substituting and simplifying (absorbing constants into \(M\)):

$$\mathbb P\{E_n\}\le M\exp\!\left\{-(1+\delta)^2\ln n\right\}=M\frac1{n^{(1+\delta)^2}}.$$

由 \((1+\delta)^2>1\)，得Since \((1+\delta)^2>1\),

$$\sum_{n=1}^\infty M\frac1{n^{(1+\delta)^2}}<\infty\;\Rightarrow\;\sum_{n=1}^\infty\mathbb P\{E_n\}<\infty.\tag{8.1}$$

再证 \(\lim_{n\to\infty}\mathbb P\{E_n\}=0\)（反证）：若不然，则 \(\forall N>0,\exists n>N\) 使 \(\mathbb P\{E_n\}=\xi>0\)；如此可取出 \(k_10\)，于是 \(\sum_n\mathbb P\{E_n\}\ge\sum\xi_{\min}=\infty\)，与 (8.1) 矛盾。上界得证。\(\blacksquare\)Then \(\lim_{n\to\infty}\mathbb P\{E_n\}=0\) (by contradiction): if not, then for all \(N>0\) there is \(n>N\) with \(\mathbb P\{E_n\}=\xi>0\); one extracts \(k_10\), so \(\sum_n\mathbb P\{E_n\}\ge\sum\xi_{\min}=\infty\), contradicting (8.1). The upper bound is proved. \(\blacksquare\)

定理 8.1 证明（下界）/ Proof of Theorem 8.1 (Lower bound) 同样取 \(t_n=\rho^n\)（\(\rho>1\)），记不相交增量事件 \(S_n=\{B_{t_n}-B_{t_{n-1}}\ge\sqrt{2(t_n-t_{n-1})\ln(\ln(t_n-t_{n-1}))}\}\)，目标是 \(\sum_n\mathbb P\{S_n\}=\infty\)。由 \(B_{t_n}-B_{t_{n-1}}\sim\mathcal N(0,t_n-t_{n-1})\)，令 \(Z\equiv\dfrac{B_{t_n}-B_{t_{n-1}}}{\sqrt{t_n-t_{n-1}}}\sim\mathcal N(0,1)\)。则Again take \(t_n=\rho^n\) (\(\rho>1\)) and denote the disjoint-increment events \(S_n=\{B_{t_n}-B_{t_{n-1}}\ge\sqrt{2(t_n-t_{n-1})\ln(\ln(t_n-t_{n-1}))}\}\); the goal is \(\sum_n\mathbb P\{S_n\}=\infty\). Since \(B_{t_n}-B_{t_{n-1}}\sim\mathcal N(0,t_n-t_{n-1})\), let \(Z\equiv\dfrac{B_{t_n}-B_{t_{n-1}}}{\sqrt{t_n-t_{n-1}}}\sim\mathcal N(0,1)\). Then

$$\mathbb P\{S_n\}=\mathbb P\Big\{Z\ge\underbrace{\sqrt{2\ln(\ln(t_n-t_{n-1}))}}_{\equiv\,x}\Big\}\overset{\text{Lemma 8.1}}{\ge}\frac{x}{x^2+1}e^{-\frac12x^2}\overset{\text{large }x}{\ge}\frac1{\sqrt2\,c\ln\rho}\frac1{n\ln n}.$$

由于Since

$$\sum_{n=2}^\infty\frac1{n\ln n}\ge\int_2^\infty\frac1{x\ln x}dx=\ln(\ln x)\Big|_2^\infty=\infty,$$

故 \(\sum_n\mathbb P\{S_n\}=\infty\)。因 \(\mathbb P\{S_n\}\le1\) 且增量独立，必有无穷多 \(n\) 使 \(S_n\) 发生，即 \(B_{t_n}\ge B_{t_{n-1}}+\sqrt{2(t_n-t_{n-1})\ln(\ln(t_n-t_{n-1}))}\)。再用已证上界控制 \(-B_{t_{n-1}}\le2\sqrt{2t_{n-1}\ln(\ln t_{n-1})}\)，于是对充分大 \(n\)，so \(\sum_n\mathbb P\{S_n\}=\infty\). Since \(\mathbb P\{S_n\}\le1\) and the increments are independent, infinitely many \(n\) must have \(S_n\) occur, i.e. \(B_{t_n}\ge B_{t_{n-1}}+\sqrt{2(t_n-t_{n-1})\ln(\ln(t_n-t_{n-1}))}\). Controlling \(-B_{t_{n-1}}\le2\sqrt{2t_{n-1}\ln(\ln t_{n-1})}\) via the already-proved upper bound, for \(n\) large enough,

$$\frac{B_{t_n}}{\sqrt{2t_n\ln(\ln t_n)}}\ge-2\sqrt{\frac{t_{n-1}}{t_n}}+\frac{t_n-t_{n-1}}{t_n}=-\frac2{\sqrt\rho}+\frac{\rho^n-\rho^{n-1}}{\rho^n}=1-\frac1\rho-\frac2{\sqrt\rho}.$$

故 \(\limsup_{t\to\infty}\dfrac{B_t}{\sqrt{2t\ln(\ln t)}}\ge1-\dfrac1\rho-\dfrac2{\sqrt\rho}\)，其 RHS 在 \(\rho\to\infty\) 时趋于 \(1\)，证毕。\(\blacksquare\)So \(\limsup_{t\to\infty}\dfrac{B_t}{\sqrt{2t\ln(\ln t)}}\ge1-\dfrac1\rho-\dfrac2{\sqrt\rho}\), whose RHS tends to \(1\) as \(\rho\to\infty\), and we are done. \(\blacksquare\)