对称随机游走：先到哪边与期望时间 The expected time 专题 Probability / 概率 难度 L4 来源 QuantQuestion 题目详情 What is the expected time for a symmetric random walk that starts at 0 to reach either aaa or bbb , where a<0<ba< 0< ba<0<b are two integers? What is the probability that it will reach bbb before aaa ? 解析 设对称随机游走从 0 出发，吸收边界为整数 a<0<ba<0<ba<0<b，令 T=inf{n≥0:Wn∈{a,b}}.T=\inf\{n\ge 0: W_n\in\{a,b\}\}.T=inf{n≥0:Wn∈{a,b}}. 先到 bbb 的概率：WnW_nWn 是鞅，对停时 TTT 用可选停止，有 0=E[WT]=a P(WT=a)+b P(WT=b).0=\mathbb{E}[W_T]=a\,\mathbb{P}(W_T=a)+b\,\mathbb{P}(W_T=b).0=E[WT]=aP(WT=a)+bP(WT=b). 解得 P(WT=b)=−ab−a,P(WT=a)=bb−a.\boxed{\mathbb{P}(W_T=b)=\frac{-a}{b-a}},\qquad \boxed{\mathbb{P}(W_T=a)=\frac{b}{b-a}}.P(WT=b)=b−a−a,P(WT=a)=b−ab. 期望吸收时间：Mn=Wn2−nM_n=W_n^2-nMn=Wn2−n 也是鞅，故 0=E[MT]=E[WT2]−E[T]⇒E[T]=E[WT2].0=\mathbb{E}[M_T]=\mathbb{E}[W_T^2]-\mathbb{E}[T]\Rightarrow \mathbb{E}[T]=\mathbb{E}[W_T^2].0=E[MT]=E[WT2]−E[T]⇒E[T]=E[WT2]. 代入 WT∈{a,b}W_T\in\{a,b\}WT∈{a,b} 及上述概率，得 E[T]=−ab.\boxed{\mathbb{E}[T]=-ab}.E[T]=−ab.