# Staying time of last arrived traveler before given time in Poisson Process

The arrival process of passengers to a bus station follows Poisson distribution with arrival rate $\lambda$. The inter-arrival time between passengers will then follow exponential distribution with Mean, $1/\lambda$.

Assume the first passenger always arrives at $t=0$ and starts a timer of duration 15 minutes. The bus will leave after the timer expires i.e., after exactly 15 minutes. What is the expected value of the staying/waiting time of the last passenger?

Some insights: The first passenger always arrive at $t=0$ (we always know). When the timer of 15 minutes duration is started, the rest of passengers follow Poisson distribution. If no (more) passenger arrives in this time with probability i.e., $p_{0}$ calculated according to Poisson PMF, it means the bus will take the only first passenger and the first is also the last passenger. The expected time of waiting of the last would be $p_{0} * 15$ min. Similarly, 1,2,3... more passengers can arrive in $(0,10$ min$]$ with probabilities $p_{1}$, $p_{2}$, $p_{3}$, etc., But I should multiply with which time as of course this should be less than 15 minutes, as I multiplied the $p_{0}$ with 15 minutes?

## 1 Answer

Let $L_t$ denote the arrival time of the last passenger to arrive before time $t$, then $0\leqslant L_t<t$ almost surely, and $t-L_t$ is distributed as $\min\{X,t\}$ where $X$ is exponential with parameter $\lambda$. Thus, the waiting time $W_t=t-L_t$ of the last passenger is such that $$E(W_t)=E(\min\{X,t\})=tP(X>t)+\int_0^txf_X(x)dx$$ that is, $$E(W_t)=te^{-\lambda t}+\int_0^tx \lambda e^{-\lambda x}dx=\frac{1-e^{-\lambda t}}{\lambda}$$ Now, plug in $t=15$ minutes.

Edit answering a new question asked in the comments below: The average waiting time of the passengers taking the bus can be computed as follows. The number of passengers taking the bus is $1+N_t$ where $N_t$ is Poisson with parameter $\lambda t$. The waiting time of one passenger equals $t$, the mean waiting time of every other passenger, if any, is $\frac12t$, thus, the mean waiting time $w_t$ of the passengers is $$w_t=E\left(\frac{t+\frac12tN_t}{1+N_t}\right)=\frac12t\left(1+E\left(\frac1{1+N_t}\right)\right)=\frac12t\left(1+\frac{1-e^{-\lambda t}}{\lambda t}\right)$$ Predictably, $\frac12t<w_t<t$ for every positive $\lambda$, $w_t\to t$ when $\lambda\to0$ (can you say why?) and $w_t\to\frac12t$ when $\lambda\to\infty$ (can you say why?).

• Thanks again @Did. May I ask you another question?, related to the above problem. What about the AVERAGE staying(waiting) time of the bus? Considering the same above conditions i.e., The first passenger arrival will trigger a clock for a duration of $15min$, and the bus will depart after $15 mint$? Is it simply $15 min/2$? – Hallian1990 Aug 13 '17 at 0:19
• Not quite (you are forgetting the influence of the first passenger), see edit. – Did Aug 13 '17 at 7:43
• Thank you @Did. Yes, you are right. if the $\lambda$ approaches $0$, there is no passenger except the first one, so the first has a staying time of $t$. In the other case, if $\lambda$ approaches $infinity$, the inter-arrival times between passenger approaches $0$. which leads the mean waiting time as half of the $t$. Hope its correct? – Hallian1990 Aug 14 '17 at 16:45
• Dear @Did, I went through the expressions and noticed that if $t$ is sufficiently large e.g., 15 minutes, the results match with my simulator. However, for the smaller value of $t$ such as 0.5sec, etc.,, there is always a deviation between simulation and mathematical expression in case of low value of $\lambda$. Mathematical results show higher values for low value of $\lambda$. I could not understand this deviation. – Hallian1990 Aug 25 '17 at 0:48
• Unless you show the details of your simulation, how do you suggest that I address your comment? :-) – Did Aug 25 '17 at 12:58