If the subway comes every 10 minutes on average, what is the expected wait time if I arrive at the station randomly? Can someone help me mathematically understand this problem?
[Math] Average wait time arriving at subway randomly
probabilitystatistics
Related Solutions
As mentioned in the comments, the answer depends very much on the model used to describe the passage times of the buses. The deterministic situation where the passage times of buses of type $k$ are $s_k+m_k\mathbb N$ for some initial passage time $s_k$ in $(0,m_k)$ is too unwieldy to be dealt with in full generality hence we now study two types of assumptions.
(1) Fully random passage times
Here the passage times of buses of type $k$ are a Poisson process of intensity $1/m_k$ and the passage times of buses of different types are independent. Then, starting at time $t_0$, the next bus of type $k$ arrives after a random time exponential with mean $m_k$ hence the waiting time $T$ is such that $$ \mathbb P(T\gt t)=\prod_k\mathbb P(\text{no bus of type}\ k\ \text{in}\ (t_0,t_0+t))=\prod_k\mathrm e^{-t/m_k}=\mathrm e^{-t/m}, $$ where $$ \frac1m=\sum_k\frac1{m_k}. $$ In particular, $T$ is exponentially distributed with parameter $1/m$, hence $$ \mathbb E(T)=m. $$ The case $m_1=m_2=\cdots=m_n$ yields $$ \mathbb E(T)=\frac{m_1}{n}. $$ (2) Fully periodic passage times with random uniform initializations
Here, buses of type $k$ pass at times in $S_k+m_k\mathbb N$ where $S_k$ is uniform on $(0,m_k)$ and the random variables $(S_k)$ are independent. Now, starting at time $t_0$, the next bus of type $k$ arrives after time $t_0+t$ if $t\leqslant m_k$ and if $S_k$ is not in a subinterval of $(0,m_k)$ of lenth $t/m_k$. Thus, $$ \mathbb P(T\gt t)=\prod_k\left(1-\frac{t}{m_k}\right),\qquad t\leqslant \bar m=\min\limits_km_k. $$ A consequence is that $$ \mathbb E(T)=\int_0^{+\infty}\mathbb P(T\gt t)\,\mathrm dt=\int_0^{\bar m}\prod_k\left(1-\frac{t}{m_k}\right)\,\mathrm dt. $$ Expanding the product yields $$ \mathbb E(T)=\sum_{i\geqslant0}(-1)^i\bar m^{i+1}\frac1{i+1}\sum_{|K|=i}\frac1{m_K}, $$ where, for every subset $K$, $$ m_K=\prod_{k\in K}m_k. $$ For example, time intervals $m_1$, $m_2$, $m_3$ with minimum $m_1$ yield $$ \mathbb E(T)=m_1-\frac{m_1^2}2\left(\frac1{m_1}+\frac1{m_2}+\frac1{m_3}\right)+\frac{m_1^3}{3}\left(\frac1{m_1m_2}+\frac1{m_2m_3}+\frac1{m_3m_1}\right)-\frac{m_1^4}{4m_1m_2m_3}, $$ which can be simplified a little bit (but not much) into $$ \mathbb E(T)=\frac{m_1}2-\frac{m_1^2}{6m_2}-\frac{m_1^2}{6m_3}+\frac{m_1^3}{12m_2m_3}. $$ The case $m_1=m_2=\cdots=m_n$ yields $$ \mathbb E(T)=\frac{m_1}{n+1}. $$
If the ones digit of the minutes of your arrival is 0 or 1, then you see the B train first. If the ones digit of the minutes of your arrival is 2, 3, 4, 5, 6, 7, 8, 9, then you see the A train first.
Your arrival time is uniformly distributed...
Best Answer
It depends on the shape of the distribution of train arrivals. To take two easy examples:
And there are many other possibilities.