Poisson Process vs Poisson distribution

Question

I am trying to understand the following Bold part

"The highway starts with zero traffic, and the vehicles are inserted randomly into lane 0; Vehicles arrival follows a Poisson distribution with an average rate of λ =2000veh/h which is chosen to be less than highway throughput of 2400 veh/h under default settings."

from http://www.sciencedirect.com/science/article/pii/S2214209615000145

I cannot understand how they did that and how it is different with Poisson Process. Does this mean that they used Poisson process and distribute the 2000 cars within an hour with the $\lambda=2000/3600(\text{seconds in an hour})$? In this case, the interarrival is fixed and should not follow Exponential distribution, Is this a correct conclusion?

I found some good explanation such as

Hiqmet's answer from:

https://www.researchgate.net/post/Difference_between_distribution_and_Process

and this nice explanation (1, and 2)

http://individual.utoronto.ca/zheli/poisson.pdf

and as explained below in the comments by Bruce.

Answer 1

As I understand your process, here is how I would simulate entry of $m = 10,000$ vehicles onto the highway. Let the interarrival times between entries be $X_i \sim \mathsf{Exp}(rate = \lambda),$ where $\lambda = 2000$ vehicles per hour. The first five interarrival times (in hours) are 1.250879e-03, 1.233794e-04, 3.290978e-06, 8.713730e-0, and 1.935913e-04.

Thus, starting at time $0$, the 'clock' times of the entries onto the highway of the $m$ vehicles are given by the cumulative sums of the interarrival times. The entry times of the first five vehicles are at 0.001250879, 0.001374259, 0.001377550, 0.002248923, and 0.002442514 hours; the last of the 10,000 vehicles entered 4.942467 hours after the start.

The number of cars entering in the first hour after the start is $Y = 1989,$ which is a realization of the random variable $Y \sim \mathsf{Pois}(\lambda= 2000).$ So it happens that slightly fewer than the the expected $E(Y) = 2000$ vehicles actually entered during the first hour. (The 1989th vehicle arrived just a bit before the end of the hour.)

Several other simulations using the same algorithm yielded slightly different numbers of vehicles entering within the first hour: 2013, 1999, 1997, and 2035.

I hope that thinking about the mechanics of this simulation will help you visualize the roles of the exponential and Poisson distributions in the Poisson process.

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The code in R statistical software for this simulation is shown below. [If you want to reproduce precisely the same simulation I did, then retain the first statement set.seed(1234); if you want to do a different simulation of your own; pick a seed other than 1234; or omit the set.seed statement, start R afresh and let it choose an unpredictable seed.]

set.seed(1234)
m = 10^4;  i = 1:m;  lam = 2000
x = rexp(m, lam);  x[1:5]
## 1.250879e-03 1.233794e-04 3.290978e-06 8.713730e-04 1.935913e-04
t = cumsum(x);  t[1:5]
## 0.001250879 0.001374259 0.001377550 0.002248923 0.002442514
t.max = t[m];  t.max     # arrival time of last of m vehicles
## 4.942467
y = max(i[t < 1]);  y    # number of vehicles arriving in first hour
## 1989