I currently have a sensor that picks up the inter-arrival times of vehicles. These inter-arrival times occur in aggregated bursts because there is a traffic light up-stream. So essentially, we have very short inter-arrival times clustered together with mean batch size $N$ which are separated by long inter-arrival times.

Problem Description

I am interested in two different perspectives that are taken to describe these batches of short inter-arrival times. They are as follows:

  1. We have a Poisson process with rate $\mu$ such that vehicles are separated on average by $\frac{1}{\mu}$ amount of time within this batch. This Poisson process is only allowed to be active for an Exponential distributed amount of time $\frac{1}{\lambda}$. This is like a continuous-time Markov process that dictates how long the active Poisson process may run for.
  2. We take a Renewal style approach to this. We have i.i.d. Exponential distributed inter-arrival times of length $\frac{1}{\mu}$. We observe on average $N$ of these inter-arrivals. We would observe $N(t)$ to be a Poisson Process with rate $\lambda$.

I would like to show that these two perspectives are the same.

My attempt:

To show that they are the same, it will suffice to show that $P(N(t)=k)$ is the same for both. Here $t$ is the duration of the batch/burst/activity and N(t) is the amount of vehicles or inter-arrivals that are found in the batch.

Case 1:

We have a Poisson process with rate $\mu$ that is only allowed to run for $t$ amount of time. However, $E[t] = \frac{1}{\lambda}$. Hence, $$ P(N(t)=k) = \frac{(\mu t)^k e^{-(\mu t)}}{k!} $$ $$\therefore P(N(t)=k) = \frac{(\frac{\mu}{\lambda})^k e^{-(\frac{\mu}{\lambda})}}{k!} = P \left( N \left( \frac{1}{\lambda} \right) =k \right) $$

We must note that $\mu > \lambda$ such that $t_{burst} > t_{arrival}$. We can thus say $N = \max\{n: n \leq \frac{\mu}{\lambda} \} $

Case 2:

This perspective is tricky for me. A Poisson process usually has to be observed for a fixed interval of time. We have an expected fixed interval of time which is the length of a burst. So our Poisson data/counts gives rise to a rate in the form of $n$ counts per $E[t]$ interval of time. Here $E[t]$ should be the sum of $N$ inter-arrivals each of which have mean length $\frac{1}{\mu}$. Intuitively, we then have $N \times \frac{1}{\mu}$ as the the length of the burst. But $E[N]=\lambda$ such that $E[t] = \frac{\lambda}{\mu}$. Hence the overall process is Poisson with rate $\frac{\mu}{\lambda}$ as in case 1.

I try to describe the intuitive explanation in more Math-like terms. Let $$ S_{N(t)} = \sum_{i=1}^{N(t)} X_i $$ $S_{N(t)}$ would then be a burst length. Furthermore, it has an Erlang probability distribution function.

Now we have

$$P(N(t)=k) = P(S_k \leq E[t]; X_{k+1} \geq E[t] - S_k)$$ $$ \therefore P(N(t)=k) = P(S_k \leq \frac{1}{\lambda}; X_{k+1} \geq \frac{1}{\lambda} - S_k)$$ $$ \therefore P(N(t)=k) = \int_0^{\frac{1}{\lambda}} \int_{\frac{1}{\lambda}-s}^{\infty} \left( \frac{\mu^k s^{k-1} e^{-\mu s}}{(k-1)!} \right) \mu e^{-\mu x} \, dx \, ds $$ $$ \therefore P(N(t)=k) = \frac{\mu^k}{(k-1)!} \int_0^{\frac{1}{\lambda}} (s^{k-1} e^{-\mu s}) e^{-\mu (\frac{1}{\lambda} - s)} \, ds $$ $$ \therefore P(N(t)=k) = \frac{(\frac{\mu}{\lambda})^k e^{-(\frac{\mu}{\lambda})}}{k!} $$

This is the same as in above.


Would you agree with what I have done? I am not too confident in the reasoning of what I have used. I've tried to solve this problem in between a lot of chaos hence the lack of confidence in what has been provided.

Please verify if you believe it to be correct and please correct me where wrong. Ideally, I would love for more explanations/solutions of a more elegant nature to be provided.

Thank you for your time



You must log in to answer this question.