# Poisson Process vs Poisson distribution

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."

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

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.

• Interarrival times of a Poisson Process are exponentially distributed. Jun 25, 2017 at 16:07
• Thank you, what about Poisson distribution? what distribution does the interarrival times of Poisson distribution have? Jun 25, 2017 at 17:04
• The Poisson distribution $\mathsf{Pois}(\lambda)$ has PDF $P(X =k) = e^{-\lambda}\lambda^k/k!,$ for $k = 0, 1, 2, \dots .$ The number of events in a Poisson process within interval $(t, t +s)$ has a Poisson distribution with mean $\lambda s,$ where $\lambda$ is the average number of events in an interval of length $1$ time unit. // Suggest you look carefully at definitions of Poisson distribution and Poisson process in your text. Jun 25, 2017 at 17:44
• I have updated the question based on your comment, thank you. Jun 25, 2017 at 19:50

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.

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

• Thank you @BruceET, Very helpful. So, this means Poisson distribution shows the $\lambda$ and the number of events in a fixed interval one hour, but the interarrival follows Exp. Jun 26, 2017 at 2:03
• Just an additional point the sum of independent exponentially distributed random variables is Gamma-distributed from here: math.stackexchange.com/q/467915 and stat.ucla.edu/~nchristo/statistics100B/… Jun 26, 2017 at 17:36
• Also, as I recall, my $T_1, \dots T_k$ in $(0, s)$ are distributed as order statistics of $k$ observations from $\mathsf{Unif}(0,s).$ You are wise to so be looking carefully at this example. It is much studied, of theoretical interest, and very useful in reliability and queueing applications. Interesting confluence of Poisson, exponential, gamma, and uniform distributions. Wish you best success in your studies. Jun 27, 2017 at 3:17
• It is indeed a great example, much appreciated. Jun 27, 2017 at 8:49