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泊松过程等车：前向等待与后向回溯期望

Property of Poisson Process

专题
Probability / 概率
难度
L4

题目详情

公交到达服从泊松过程，平均到达间隔为 10 分钟（λ=0.1\lambda=0.1/min）。

过程运行很久后，你在随机时刻到达车站：

  1. 你的期望等待时间是多少？
  2. 平均来说，上一班公交在多少分钟前离开？

You are waiting for a bus that arrives according to a Poisson process with an average arrival time of 10 minutes (λ=0.1\lambda=0.1/min). If the process has been running for a long time and you arrive at a random time, what is your expected waiting time? On average, how long ago did the previous bus leave?

解析

两者都为 10 分钟。

到达间隔服从参数 λ\lambda 的指数分布，指数分布具有无记忆性，因此从随机时刻起到下一次到达的剩余时间期望为 1/λ=101/\lambda=10。

同理向后回溯到上一次到达的期望时间也为 10。

Original Explanation

Interarrival times are exponentially distributed with parameter λ=0.1.\lambda=0.1. The exponential distribution is memoryless, so no matter when you arrive, you expect to wait 10 minutes. Similarly, looking backward, on average the previous bus left 10 minutes earlier.