每10和15分钟一班的两辆公交车中的第一辆公交车的等待时间的期望值

我碰到一个面试问题：

每隔10分钟就有一列红色火车驶来。每隔15分钟就有一列蓝色火车驶来。两者都是从随机时间开始的，因此您没有任何时间表。如果您在随机时间到达车站并乘坐第一班来往的火车，那么预计的等待时间是多少？

解决问题的一种方法是从生存功能开始。为了必须等待至少分钟，您必须为红色和蓝色火车都至少等待t分钟。因此，总体生存功能只是各个生存功能的乘积：$t$$t$

$$S(t) = \left( 1 - \frac{t}{10} \right) \left(1-\frac{t}{15} \right)$$

其中，对于，是，你必须等待至少概率牛逼分钟下一班车。这考虑到了对OP的澄清，该注释中的正确假设是，每列火车都在固定的时间表上，而不依赖于其他时间表以及与旅客的到达时间无关，并且两列火车的相位是均匀分布的，$0 \le t \le 10$$t$

然后得到pdf

$$p(t) = (1-S(t))' = \frac{1}{10} \left( 1- \frac{t}{15} \right) + \frac{1}{15} \left(1-\frac{t}{10} \right)$$

并以通常的方式获得期望值：

，$E[t] = \int_0^{10} t p(t) dt = \int_0^{10} \frac{t}{10} \left( 1- \frac{t}{15} \right) + \frac{t}{15} \left(1-\frac{t}{10} \right) dt = \int_0^{10} \left( \frac{t}{6} - \frac{t^2}{75} \right) dt$

算到分钟$\frac{35}{9}$

答案是 获取括号内的部分： ∫ý<XŶdŶ=ý2/2| X 0 =X2/2∫ý>XXdŶ=Xÿ| 15 X =15X-X2 所以，部分是： （。）=（∫ý<Xýdÿ

$$E[t]=\int_x\int_y \min(x,y)\frac 1 {10} \frac 1 {15}dx dy=\int_x\left(\int_{y<x}ydy+\int_{y>x}xdy\right)\frac 1 {10} \frac 1 {15}dx$$ $$\int_{y<x}ydy=y^2/2|_0^x=x^2/2$$ $$\int_{y>x}xdy=xy|_x^{15}=15x-x^2$$ 最后，ë[吨]= ∫ X（15X-X2/2） $$(.)=\left(\int_{y<x}ydy+\int_{y>x}xdy\right)=15x-x^2/2$$ $$E[t]=\int_x (15x-x^2/2)\frac 1 {10} \frac 1 {15}dx= (15x^2/2-x^3/6)|_0^{10}\frac 1 {10} \frac 1 {15}\\= (1500/2-1000/6)\frac 1 {10} \frac 1 {15}=5-10/9\approx 3.89$$

这是要模拟的MATLAB代码：

nsim = 10000000;
red= rand(nsim,1)*10;
blue= rand(nsim,1)*15;
nextbus = min([red,blue],[],2);
mean(nextbus)

Assuming each train is on a fixed timetable independent of the other and of the traveller's arrival time, the probability neither train arrives in the first $x$ minutes is $\frac{10-x}{10} \times \frac{15-x}{15}$ for $0 \le x \le 10$, which when integrated gives $\frac{35}9\approx 3.889$ minutes

Alternatively, assuming each train is part of a Poisson process, the joint rate is $\frac{1}{15}+\frac{1}{10}=\frac{1}{6}$ trains a minute, making the expected waiting time $6$ minutes

I am probably wrong but assuming that each train's starting-time follows a uniform distribution, I would say that when arriving at the station at a random time the expected waiting time for:

1. the $R$ed train is $\mathbb{E}[R] = 5$ mins
2. the $B$lue train is $\mathbb{E}[B] = 7.5$ mins
3. the train that comes the first is $\mathbb{E}[\min(R,B)] =\frac{15}{10}(\mathbb{E}[B]-\mathbb{E}[R]) = \frac{15}{4} = 3.75$ mins

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As pointed out in comments, I understood "Both of them start from a random time" as "the two trains start at the same random time". Which is a very limiting assumption.

Suppose that red and blue trains arrive on time according to schedule, with the red schedule beginning $\Delta$ minutes after the blue schedule, for some $0\le\Delta<10$. For definiteness suppose the first blue train arrives at time $t=0$.

Assume for now that $\Delta$ lies between $0$ and $5$ minutes. Between $t=0$ and $t=30$ minutes we'll see the following trains and interarrival times: blue train, $\Delta$, red train, $10$, red train, $5-\Delta$, blue train, $\Delta + 5$, red train, $10-\Delta$, blue train. Then the schedule repeats, starting with that last blue train.

If $W_\Delta(t)$ denotes the waiting time for a passenger arriving at the station at time $t$, then the plot of $W_\Delta(t)$ versus $t$ is piecewise linear, with each line segment decaying to zero with slope $-1$. So the average wait time is the area from $0$ to $30$ of an array of triangles, divided by $30$. This gives

$$\begin{align}\bar W_\Delta &:= \frac1{30}\left(\frac12[\Delta^2+10^2+(5-\Delta)^2+(\Delta+5)^2+(10-\Delta)^2]\right)\\&=\frac1{30}(2\Delta^2-10\Delta+125). \end{align}$$ Notice that in the above development there is a red train arriving $\Delta+5$ minutes after a blue train. Since the schedule repeats every 30 minutes, conclude $\bar W_\Delta=\bar W_{\Delta+5}$, and it suffices to consider $0\le\Delta<5$.

If $\Delta$ is not constant, but instead a uniformly distributed random variable, we obtain an average average waiting time of

$$\frac15\int_{\Delta=0}^5\frac1{30}(2\Delta^2-10\Delta+125)\,d\Delta=\frac{35}9.$$

This is a Poisson process. The red train arrives according to a Poisson distribution wIth rate parameter 6/hour.
The blue train also arrives according to a Poisson distribution with rate 4/hour. Red train arrivals and blue train arrivals are independent. Total number of train arrivals Is also Poisson with rate 10/hour. Since the sum of The time between train arrivals is exponential with mean 6 minutes. Since the exponential mean is the reciprocal of the Poisson rate parameter. Since the exponential distribution is memoryless, your expected wait time is 6 minutes.