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[0,1000] 随机游走：先到 0 的概率

[0,1000] 随机游走

专题
Probability / 概率
难度
L4

题目详情

Suppose we are doing a random walk on the interval [0, 1000], starting at 80. So, with probability 1/21 / 2 , this number increases or decreases by one at each step. We stop when one of the boundaries of the interval is reached. What is the probability that this boundary will be 0?

解析

对称随机游走在区间 {0,1,…,1000}\{0,1,\ldots,1000\} 上，从 80 开始，0 与 1000 为吸收边界。

位置过程 XtX_t 为鞅。令 τ\tau 为首次到达边界的停时，则

E[Xτ]=E[X0]=80.\mathbb{E}[X_\tau]=\mathbb{E}[X_0]=80.

设 p=P(Xτ=1000)p=\mathbb{P}(X_\tau=1000)，则

E[Xτ]=1000 p+0⋅(1−p)=1000p=80⇒p=0.08.\mathbb{E}[X_\tau]=1000\,p+0\cdot(1-p)=1000p=80\Rightarrow p=0.08.

故先到 0 的概率为

1−p=0.92.\boxed{1-p=0.92}.