Assume that customers arrive to a bus stop according to a homogenous Poisson process with rate $\alpha$ and that the arrival process of buses is an independent renewal process with interarrival distribution $F$. What is the long run percentage of customers who wait for a bus less than $x$ units of time?
I tried to solve this problem by considering a renewal reward process. In the case that $x$ is very large so that it imposes no restriction we may proceed as follows to solve a similar problem: we say that a new cycle of length $T$ begins each time a bus arrives. Further, we assume customers pay us money at a rate of $1$ per time unit while they wait for a bus. Then our "reward rate" at any time corresponds to the number of people who are waiting at that time. So, by the renewal reward theorem we have that: average number waiting = $E[R]/E[T]$, where $R$ is the total reward during a cycle. In that case, $E[R] = 1/2E[NT] = 1/2\alpha E[T^2]$, where $N$ is the number of arrival in a cycle, which follows from the definition of a Poisson process and since the points of the Poisson process are uniformly distributed in a cycle conditional on the total number of points in that cycle.
But I don't know how to define the reward with the restriction imposed by $x$. I tried the following conditional expectation: $E[R|T,N]=N x1_{T>x}/2+NT1_{T\leq x}/2$, but got stuck. Can someone please help me?
Best Answer
I ended up writing my comment as a full solution. Main ideas:
1.) Poisson process has zero probability of arrival at a particular time -- i.e. there is zero probability of a 'collision'
2.) Define a renewal reward process, where a new epoch begins each time a bus arrives, and a reward of 1 for each customer that waited less than $x$ in time. The long-run probability / percentage of customers with a sufficiently short wait ($\lt x$) is given by
$p=\lim_{t\to \infty}\frac{R(t)}{N(t)}$
where $N(t)$ of course is the r.v. for the Poisson counting process
3.) The Poisson process is memoryless, which is something we repeatedly exploit, e.g. allows the above renewal reward process to properly work, and the associated renewal function is quite simple $E[N(t)]=m(t)=\frac{t}{\mu}$, and so on.
The calculations
$p=\lim_{t\to \infty}\frac{R(t)}{N(t)}=\lim_{t\to \infty}\frac{\frac{R(t)}{t}}{\frac{N(t)}{t}}= \lim_{t\to \infty}\frac{R(t)}{t}\cdot \lim_{t\to \infty}\frac{t}{N(t)}$
i.) $\lim_{t\to \infty}\frac{t}{N(t)}=\mu$ by SLLN
ii.) $\lim_{t\to \infty}\frac{R(t)}{t}=\frac{E[R_1]}{E[T_1]}$ where $R_1$ and $T_1$ are rewards and time associated with first epoch. Using conditioning
$E\big[R_1\big]= E\big[E[R_1\vert T_1=t]\big]$ and in particular
if $t\in(0,x)$ then $E[R_1\vert T_1=t]=m(t)=\frac{t}{\mu}$
if $t\geq x$ then $E[R_1\vert T_1=t]=m(x)=\frac{x}{\mu}$
These both hold by memorylessness. In the latter case this reads: if the epoch length t is at least $x$ long what is the expected number of Poisson arrivals in the interval $(t-x,t)$, but by memorylessness this is the same as the expected number of Poisson arrivals in the interval $(0,x)$ and this is given by $m(x)=\frac{x}{\mu}$. For the former case it reads, what is the expected number of Poisson arrivals between the beginning and ending of the epoch, i.e. in the interval $(0,t)$ which is given by $m(t)=\frac{t}{\mu}$
Thus
$E\big[R_1\big]= E\big[E[R_1\vert T_1=t]\big]=\big(\int_{0}^x \frac{t}{\mu} dF(t)\big)+ \big(\int_{x}^\infty \frac{x}{\mu} dF(t)\big) = \frac{1}{\mu}\Big(\int_{0}^x t dF(t)\big)+ \big(\int_{x}^\infty x dF(t)\big)\Big)$
$\implies E\big[R_1\big]=\frac{1}{\mu}\cdot E\big[\min(T_1,x)\big]$
putting this all together
$p=\lim_{t\to \infty}\frac{R(t)}{t}\cdot \lim_{t\to \infty}\frac{t}{N(t)}=\Big(\frac{1}{\mu}\cdot E\big[\min(T_1,x)\big]\cdot\frac{1}{E[T_1]}\big)\cdot \big(\mu\big)= \frac{E[\min(T_1,x)]}{E[T_1]}$
$= \frac{\int_0^x Pr(T_1\gt t)dt}{E[T_1]}$