1. The Poisson Process

2. Definition What is A Poisson Process? Examples:
The Poisson Process is a counting that counts the number of occurrences of some specific event through time.
It not only models many real-world phenomena, but the process allows for tractable mathematical analysis as well.
Examples:
- Number of customers arriving to a counter
- Number of calls received at a telephone exchange
- Number of packets entering a queue

3. The Poisson Process time
1st Event Occurs
2nd Event Occurs
3rd Event Occurs
4th Event Occurs X1 X2 X3 X4
time
t=0
X1, X2, … represent a sequence of +ve independent random variables with identical distribution
Xn depicts the time elapsed between the (n-1)th event and nth event occurrences
Sn depicts a random variable for the time at which the nth event occurs
Define N(t) as the number of events that have occurred up to some arbitrary time t.
The counting process { N(t), t>0 } is called a Poisson process if the inter-occurrence times X1, X2, … follow the exponential distribution

4. The Poisson Process: Example
For some reason, you decide everyday at 3:00 PM to go to the bus stop and count the number of buses that arrive. You record the number of buses that have passed after 10 minutes
Sunday
N (t=10 min) = 2
time
X1=5 min
X2=4 min
1st Bus Arrival
2nd Bus Arrival
X3=7 min
X4=2 min
3rd Bus Arrival
4th Bus Arrival
S1 = 5 min
t=0
S2 = 9 min
S3 = 16 min
S4 = 18 min

5. The Poisson Process: Example
For some reason, you decide everyday at 3:00 PM to go to the bus stop and count the number of buses that arrive. You record the number of buses that have passed after 10 minutes
Monday
N (t=10 min) =4
time
X1=1 min
X2=2 min
1st Bus Arrival
2nd Bus Arrival
X3=4 min
4th Bus Arrival
5th Bus Arrival
S1 = 1 min
t=0
S2 = 3 min
S3 = 7 min
S5 = 15 min
X4=2 min
3rd Bus Arrival
S4 = 9 min
X5=6 min

6. The Poisson Process: Example
For some reason, you decide everyday at 3:00 PM to go to the bus stop and count the number of buses that arrive. You record the number of buses that have passed after 10 minutes
Tuesday
N (t=10 min) =1
time
X1=10 min
2nd Bus Arrival
t=0
S1 = 10 min
S2 = 16 min
1st Bus Arrival
X2=6 min

7. The Poisson Process: Example
Given that Xi follow an exponential distribution then N(t=10) follows a Poisson Distribution

8. The Poisson Process
Defined as N(T) number of arrivals in time interval [0,t]
λ : arrival rate (average # of arrivals per time unit)
We divide [0,t] into n subintervals, each with duration delta δ = t/n.

9. The Poisson Distribution

10. Mean of the Poisson Distribution
where
On average the time between two consecutive events is 1/λ
This means that the event occurrence rate is λ
Consequently in time t, the expected number of events is λt

11. Variance of the Poisson Distribution

12. Inter-arrival time of Poisson process

13. The Exponential Distribution
The exponential distribution describes a continuous random variable. It models the arrival time
Cumulative Distribution Function (CDF) 1
Probability Density Function (PDF) λ

14. Mean of the Exponential Distribution

15. Variance of the Exponential Distribution

16. Moment Generating Function of Poisson Distribution
The Moment Generating Function of any PMF for a discrete random variable may be used to deduce different parameters and characteristics of the distribution
What could G(Z) be used for

17. Remaining Time of Exponential Distributions
Xk+1 is the time interval between the kth and k+1th arrivals
Condition:
T units have passed and the k+1th event has not occurred yet
Question:
Given that X*k+1 is the remaining time until the k+1th event occurs
What is Pr[X*k+1≤x]
kth Event Occurs
k+1th Event Occurs T
X*k+1

18. Remaining Time of Exponential Distributions
Xk+1 follows an exponential Distribution, i.e.,
Pr[Xk+1≤t]=1-e-λt
The remaining time X*k+1 follows an exponential distribution with the same mean 1/λ as that of the inter-arrival time Xk+1

19. The Memoryless Property of Exponential Distributions
The waiting time until the next arrival has the same exponential distribution as the original inter-arrival time regardless of long ago the last arrival occurred
Memoryless Property of Exponential Distribution and the Poisson Process
In the Poisson process, the number of arrivals within any time interval s follows a Poisson distribution with mean λs

20. Merging of Poisson Processes
{N1(t), t ≥ 0} and {N2(t), t ≥ 0} are two independent Poisson processes with respective rates λ1 and λ2,
{Ni (t)} corresponds to type i arrivals.
The merged process N(t) = N1(t) + N2(t), t ≥ 0. Then {N(t), t ≥ 0} is a Poisson process with rate λ = λ1 + λ2.
Zk is the inter-arrival time between the (k − 1)th and kth arrival in the merged process
Ik= i if the kth arrival in the merged process is a type i arrival,
For any k = 1, 2, ,
P{Ik = i | Zk = t} =λi/(λ1+λ2) , i= 1, 2, independently of t .

21. Splitting of Poisson Processes
{N(t), t ≥ 0} is a Poisson process with rate λ.
Each arrival of the process is classified as being a type 1 arrival or type 2 arrival with respective probabilities p1 and p2, independently of all other arrivals.
Ni (t) is the number of type i arrivals up to time t .
{N1(t)} and {N2(t)} are two independent Poisson processes having respective rates λp1 and λp2.

22. Example 1: A taxi problem
Group taxis are waiting for passengers at the central railway station.
Passengers for those taxis arrive according to a Poisson process with an average of 20 passengers per hour.
A taxi departs as soon as four passengers have been collected or ten minutes have expired since the first passenger got in the taxi.

23. Example 1: A taxi problem
(a) Suppose you get in the taxi as first passenger. What is the probability that you have to wait ten minutes until the departure of the taxi?
(b) Suppose you got in the taxi as first passenger and you have already been waiting for five minutes. In the meantime two other passengers got in the taxi. What is the probability that you will have to wait another five minutes until the taxi departs?

24. Example 1: A taxi problem
(a) Suppose you get in the taxi as first passenger. What is the probability that you have to wait ten minutes until the departure of the taxi?
You have to wait 10 if <3 passengers come.
P[wait 10 min] = P[N(10)=0]+ P[N(10)=1]+ P[N(10)=2]
Recall:
First, what is the arrival rate λ?
Average = 20 passengers/min

25. Example 1: A taxi problem
(b) Suppose you got in the taxi as first passenger and you have already been waiting for five minutes. In the meantime two other passengers got in the taxi. What is the probability that you will have to wait another five minutes until the taxi departs?
Let Y be r.v. for waiting time until next arrival. It follows from the memory less property, that Y has exponential PDF, so CDF F(Y) = 1-e-x/λ
P[Y > 5] = 1-P[Y<=5] = 1-(1-e-5/3) =

26. Example 2: A Waiting-time Problem
In the harbour of Amsterdam a ferry leaves every T minutes to cross the North Sea canal, where T is fixed.
Passengers arrive according to a Poisson process with rate λ.
The ferry has ample capacity. What is the expected total waiting time of all passengers joining a given crossing?