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C12: The Poisson process

CIS 2033 based on Dekking et al. A Modern Introduction to Probability and Statistics. 2007 Instructor Longin Jan Latecki. C12: The Poisson process. From Baron book: The number of rare events occurring within a fixed period of time has Poisson distribution .

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C12: The Poisson process

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  1. CIS 2033 based onDekking 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(tλ). Expectation E[X] = µ = tλand 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. Let X1, X2, … 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 Ti = Xi – Xi-1 are called inter-arrival times or wait times. The inter-arrival times T1=X1, T2=X2 – X1, T3=X3 – X2 … are independent RVs, each with an Exp(λ) distribution. Hence expected inter-arrival time is E(Ti) =1/λ. Since for Poisson λ = µ / t= (number of events) / (time unite) = intensity = probability of success, we have for the exponential distribution E(Ti) =1/λ = t / µ = (time unite) / (number of events) = wait time

  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. Let X1, X2, … be arrival times such that the probability of k arrivals in a given time interval [0, t] has a Poisson distribution Pois(λt): Each arrival time Xi, is a random variable with Gam(i, λ) distribution for α=i : We also observe that Gam(1, λ) = Exp(λ):

  8. 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-1.92*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, X1, X2, X3… • 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 λu. • Independence: The number of arrivals in disjoint time intervals are independent random variables. • N(I) = total number of calls in an interval I • Nt=N([0,t]) • E[Nt] = 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 Ij,n = ((j-1)t/n , jt/n] contains either 0 or 1 arrivals.Arrival: For such a large n ( n > λ t), Rj = number of arrivals in the time interval Ij,n, Rj = 0 or 1 • Rj has a Ber(p) distribution for some p.Recall: (For a Bernoulli random variable)E[Rj] = 0 • (1 – p) + 1 • p = p • By Homogeneity assumption for each jp = λ• length of Ij,n = λ (t / n) • Total number of calls:Nt = R1 + R2 + … + Rn. • By Independence assumption Rj are independent random variables, so Nt 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, henceand λ=np, where we have Bin(n, p).

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