My friends and I were "thinking" yesterday in the pub about the following: if a person is standing on a bus stop that is served by a single bus which comes every p minutes, we would expect the average waiting time to be p/2 (which may or may not be correct). But we had no idea how to calculate the average waiting time if there is more than one bus. So let's assume there is n many buses serving the stop, and each comes once in m1, m2 … mn minutes. How would we go about calculating the average time a person has to wait for a bus? What is the theory behind it?
Thank you
Best Answer
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 $$ \mathbb E(T)=\frac{m_1}{n+1}. $$