A new restaurant in my hometown gets most of its income from a few important clients. I want to model the probability that a client comes to my restaurant.
Let P(i) be the probability that a client shows up on day i as a function of the total number of clients who visited the restaurant the previous day, n. Assume each client’s decision to go to the restaurant is independent.
Simple case:
- There are 3 clients in the hometown.
- The probability function is defined as f(n) = 1 – n/3 when n = 1, 2, or 3.
- f(0) = 0.
Question 1: What is the probability of exactly one client in the restaurant if there were 2 customers at the restaurant the previous day?
Question 2: Calculate the expected number of days until no people show up.
Question 3: Now instead of f(0) = 0, suppose f(0) is split evenly between 0 people showing up and 1 person showing up. What is the expected number of clients dining in my restaurant far in the future?
这道题本质上是一个小状态空间的概率递推与马尔可夫链期望分析题。由于总共只有 3 个客户，每个人是否到店相互独立，并且次日到店概率只由前一天到店人数决定，因此可以把状态定义为“当天到店人数”0、1、2、3，并写出状态转移概率。第一问直接根据前一天有 2 人时的单人到店概率 f(2)=1-2/3=1/3 来求“恰好 1 人到店”的概率。第二问要求“直到无人到店”为止的期望天数，可将 0 视为吸收态，设 E[n] 表示从 n 人状态出发到达 0 的期望步数，建立线性方程组求解。第三问把 0 状态改成以 50% 概率留在 0、50% 概率转到 1，系统不再必然吸收到 0，而是需要求长期稳态下的期望到店人数，通常通过稳态分布或解平稳方程得到。
