1. CIS 2033 based on Dekking et al. A Modern Introduction to Probability and Statistics. 2007 Instructor Longin Jan Latecki C12: The Poisson process

2. From Baron book: The number of rare events occurring within a fixed period of time has Poisson distribution. Essentially it means that two such events are extremely unlikely to occur simultaneously or within a very short period of time. Arrivals of jobs, telephone calls, e-mail messages, traffic accidents, network blackouts, virus attacks, errors in software, floods, and earthquakes are examples of rare events. This distribution bears the name of a famous French mathematician Siméon-Denis Poisson (1781–1840).

3. 12.2 – Poisson Distribution Definition: A discrete RV X has a Poisson distribution with parameter µ, where µ > 0 if its probability mass function is given by for k = 0,1,2…, where µ is the expected number of rare events, or number of successes, occurring in time interval [0, t], which is fixed for X. We can express µ = t λ, where t is the length of the interval, e.g., number of minutes. Hence λ = µ / t = number of events per time unite = probability of success. λ is also called the intensity or frequency of the Poisson process. We denote this distribution: Pois(µ) = Pois( t λ). Expectation E[X] = µ = tλ and variance Var(X) = µ = tλ

4. Dekking 12.6 A certain brand of copper wire has flaws about every 40 centimeters. Model the locations of the flaws as a Poisson process. What is the probability of two flaws in 1 meter of wire? The expected numbers of flaws in 1 meter is 100/40 = 2.5, and hence the number of flaws X has a Pois(2.5) distribution. The answer is P(X = 2): since

5. The inter-arrival times T 1 =X 1, T 2 =X 2 – X 1, T 3 =X 3 – X 2 … are independent RVs, each with an Exp(λ) distribution. Hence expected inter-arrival time is E(T i ) =1/λ. Since for Poisson λ = µ / t = (number of events) / (time unite) = intensity = probability of success, we have for the exponential distribution E(T i ) =1/λ = t / µ = (time unite) / (number of events) = wait time Let X 1, X 2, … be arrival times such that the probability of k arrivals in a given time interval [0, t] has a Poisson distribution Pois(tλ): The differences T i = X i – X i-1 are called inter-arrival times or wait times.

6. Quick exercise 12.2 We model the arrivals of email messages at a server as a Poisson process. Suppose that on average 330 messages arrive per minute. What would you choose for the intensity λ in messages per second? What is the expectation of the interarrival time? Because there are 60 seconds in a minute, we have λ = µ / t = (number of events) / (time unite) = 330 / 60 = 5.5 Since the interarrival times have an Exp(λ) distribution, the expected time between messages is 1/λ = 0.18 second, i.e., E(T) =1/λ = t / µ = (time unite) / (number of events) = 60/330=0.18

7. Each arrival time X i, is a random variable with Gam(i, λ) distribution for α=i : Let X 1, X 2, … be arrival times such that the probability of k arrivals in a given time interval [0, t] has a Poisson distribution Pois(λt): We also observe that Gam(1, λ) = Exp(λ):

8. a)It is reasonable to estimate λ with (nr. of cars)/(total time in sec.) = 0.192. b) 19/120 = 0.1583, and if λ = 0.192 then μ = 10 λ =1.92. Hence P(K = 0) = e -0.192*10 = 0.147 c) Again μ = 10 λ =1.92 and we have P(K = 10) = ((1.92 ) 10 / 10!) * e -1.92 = 2.71 * 10 -5.

9. 12.2 –Random arrivals  Example: Telephone calls arrival times  Calls arrive at random times, X 1, X 2, X 3 …  Homegeneity aka weak stationarity: is the rate lambda at which arrivals occur in constant over time: in a subinterval of length u the expectation of the number of telephone calls is λ u.  Independence: The number of arrivals in disjoint time intervals are independent random variables.  N(I) = total number of calls in an interval I  N t =N([0,t])  E[N t ] = t λ  Divide Interval [0,t] into n intervals, each of size t/n

10. 12.2 –Random arrivals  When n is large enough, every interval I j,n = ((j-1)t/n, jt/n] contains either 0 or 1 arrivals. Arrival: For such a large n ( n > λ t), R j = number of arrivals in the time interval I j,n, R j = 0 or 1  R j has a Ber(p) distribution for some p. Recall: (For a Bernoulli random variable) E[R j ] = 0 (1 – p) + 1 p = p  By Homogeneity assumption for each j p = λ length of I j,n = λ ( t / n)  Total number of calls: N t = R 1 + R 2 + … + R n.  By Independence assumption R j are independent random variables, so N t has a Bin(n,p) distribution, with p = λ t/n, hence λ = np/t  When n goes to infinity, Bin(n,p) converges to a Poisson distribution

11. Example form Baron Book:

12. Example 3.23 from Baron Book Baron uses λ for μ, hence and λ=np, where we have Bin(n, p).