# Introduction:

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.

# Conclusion:

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.