The Poisson Process

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

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

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

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

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

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

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.

The Poisson Distribution

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

Variance of the Poisson Distribution

Inter-arrival time of Poisson process

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

Mean of the Exponential Distribution

Variance of the Exponential Distribution

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

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

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

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

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 .

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.

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.

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?

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

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) = 0.1889

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?