EE126: Probability and Random Processes

Transcription

1 EE126: Probability and Random Processes Lecture 18: Poisson Process Abhay Parekh UC Berkeley March 17,

2 1 Review 2 Poisson Process 2

3 Bernoulli Process An arrival process comprised of a sequence of coin flips: X 1, X 2,... where the X i are iid Bernoulli rv s with probability of success p. 1 Let N(t) be the number of arrivals in t time slots. Then, N(t) = i X i is Binomial. 2 Let Y k be the time of the k t h arrival. Then Y 1 is the time to the first head i.e. Y 1 Geometric with parameter p. 3 T 1 = Y 1 is the time to the first arrival. T 2 = Y 2 Y 1 is the time between the first and second arrivals. Since coin flipping is memoryless, Y 2 Y 1 is also geometric and independent of Y 1. Definition 2 A sequence of iid geometric random variables with parameter p. 3

4 Distribution of the Time of the k th Arrival Y k is the sum of the first k interarrival times. The interarrival times are geometric rvs. (Geometric Mean: 1/p; Var: (1 p)/p 2.) E[Y k ] = var(y k ) = In the first t 1 time units we have to had exactly k 1 arrivals. The t th time unit must register an arrival. P(Y k = t) = ( ) t 1 p k 1 (1 p) t k p = k 1 ( ) t 1 p k (1 p) t k k 1 Called Pascal Distribution of order k. 4

5 Splitting Bernoulli Processes For each arrival, route it to A with prob q and to B with prob 1 q. Routing decisions are independent. A i : arrival to A at time slot i, and B i : arrival to B at slot i. { 1, if Xi = 1 and route to A; P(A i ) = 0, otherwise. So, A i is Bernoulli with prob pq and B i is Bernoulli with probability p(1 q). 5

6 Merging Bernoulli Processes A and B are independent Bernoulli Processes, with prob p A and p B respectively. Construct the arrival process C as follows: C i = 1 if A i = 1 or B i = 1 (includes if both have arrivals) and C i = 0 otherwise. Then P(C i = 0) = (1 p A )(1 p B ) and P(C i = 1) = p C = 1 (1 p A p B + p A p B ) = p A + p B p A p B. Thus C is Bernoulli Process with probability p C. 6

7 Poisson Process A continuous version of the Bernoulli Process, i.e. time is not in slots but defined on the non-negative reals. Immediate issue...how does p change as the slot size goes to zero? Slot size is δ: Fix λ > 0 and set p(δ) = λδ for λ > 0. Prob one arrival in a slot p(δ) = λδ Prob no arrival in a slot 1 λδ Expected Number of arrivals in in time τ: There are n = τ δ slots so number of arrivals in τ is Binomial (n,p). np = λτ Now what happens as we let δ 0 (but keep λ fixed)? Let N(t) the number of arrivals in t time units. P(N(t) = k) converges from a Binomial to a Poisson distribution. 7

8 Review: Poisson Random Variable Counting Successes with a Binomial rv can be computationally difficult: 1 Count misprints in a book. N is huge and p is very small. 2 What is the probability that exactly k out 500 randomly chosen people will share a birthday on New Year s Day? 3 A factory produces defective screws with prob What is the prob that a box of 100 screws has k defective items? When N is large and p is small an excellent approximation is: where λ = Np > 0. λ λk p X (k) = e k! 8

9 Approximation of Binomial by Poisson 9

10 10 Why? Fact:e x = lim n (1 + x n )n. Start with Binomial...set λ = np p = λ n p X (k) = = = n n n! (n k)!k! pk (1 p) n k n(n 1)... (n k + 1) n 1 n n k... n k + 1 n λ k k! (1 λ n )n k λ k k! (1 λ n ) k (1 λ n )n As n, p X (k) e λ λ k! when k << n.

11 11 Poison Process: Formal Definition Let N(t) be the number of arrivals in [0, t] and let P k (t) be the probability that N(t) = k. 1 Time Homogeneity: P k (τ) is the same for all intervals of length τ 2 Independence: The number of arrivals in any interval is independent of all arrival events outside that interval. 3 Small Interval Probabilities: P 0 (τ) = 1 λτ + o 0 (τ) P 1 (τ) = λτ + o 1 (τ) P k (τ) = o k (τ) for k = 2, 3,..., Note: o k (τ) lim = 0, k = 0, 1, 2, τ 0 τ

12 12 Occurrence 1 The number of goals in (90 minutes of) a soccer match 2 A renewal process is an arrival process in which the interarrival times are iid but not necessarily exponential. Lots of small independent renewal processes add up to a Poisson Process. Phone calls originating in a city The arrival of customers in a queue. Car accidents in a city Particle emissions from radioactive material The number of raindrops falling over an area

13 13 Poisson Process Let the number of arrivals of a Poisson process with rate λ in an interval of length t be N(t). Then for k = 0, 1,... Also: P(N(t) = k) = P k (t) = (λt)k e λt k! E[N(t)] = var(n(t)) = λt

14 Interarrival times Starting from time τ let τ + T be the time of the next arrival. P(T t) = P(N(t) > 0) = 1 P(N(t) = 0) = 1 (λt)0 0! = 1 e λt e λt This is the CDF of an exponential r.v. T k : k th inter-arrival time. Then the T k are all identically distributed exponential rvs. They describe non-overlapping intervals they are independent. Given a sequence of iid exponential random variables T 1, T 2..., we can construct a Poisson process: Y 1 = T 1, Y k = Y k Y k 1, k = 2, 3,... where Y k is the time of the k th arrival. Thus a Poisson Process is a process where the interarrival times are iid exponential random variables. 14

15 Time to k th arrival Y k is the time the k th arrival. f Yk (y)δ P(k th arrival in [y, y + δ]) The k t h arrival occurs in the interval with prob λδ The other k 1 arrivals occurred before y. I.e. with probability: (λy) k 1 (k 1)! e λy. The two events are independent f Yk (y)δ λδ (λy)k 1 (k 1)! e λy. f Yk (y) = λk y k 1 (k 1)! e λy. This is the Erlang PDF of order k. 15

16 16 Erlang Distribution The distribution of Y k, the k th arrival time is given by f Yk (y) = λk y k 1 (k 1)! e λy. The k-order Erlang is the sum of k iid exponentials.

17 Video Game Alice is playing a game in which she must move her player from between two points. Machine gun blasts that annihilate everything between A and B arrive according to a Poisson process with rate λ. Alice decides to move her player when no machine gun blast has arrived for τ seconds. Let N be the number of blasts she sees before moving her player. 1 E[N]: If N = n, there are n interarrivals less than τ and then one greater than τ. The interarrival times are independent and exponentially distributed with λ. So N + 1 is geometric with parameter e λτ. So E[N] + 1 = e λτ E[N] = e λτ 1. 2 It takes τ 1 seconds to move the player. Find P(player is annihilated): At the time she decides to move the player, residual time of the interarrival time is exponentially distributed so 1 e λτ 1. 17

18 18 Example: Fishing Bob catches fish according to a Poisson Process with rate λ = 0.6 per hour. If catches as at least one in the first two hours he quits. Else he continues until he has caught the first fish. 1 Prob Bob fishes for > 2 hours: e Prob Bob fishes for time in [2, 5] hours: e 1.2 (1 e 1.8 ). 3 Prob Bob catches at least 2 fish: 1 e e Expected number of fish caught: e Expected total fishing time given it is > 4 hours: = hours.

19 19 Merging of two Independent Poisson Processes Two processes, A and B with parameters λ A and λ B respectively. The merged process, C, has an arrival whenever either A or B has an arrival. It is clearly Time Homogeneous and Independent. In a small interval of length δ: 1 P(0 arrivals in C) (1 λ A δ)(1 λ B δ) = 1 λ A δ λ B δ λ A λ B δ 2 P(0 arrivals in C) 1 (λ A + λ B )δ 2 P(1 arrival in C) λ A δ(1 λ B δ) + λ B δ(1 λ A δ) = (λ A + λ B )δ. Process C is Poisson with parameter λ A + λ B.

20 20 Merging of two Independent Poisson Processes Two processes, A and B with parameters λ A and λ B respectively. The merged process, C, is Poisson with parameter λ A + λ B. Suppose a single arrival occurs in a small interval of length δ: 1 P(1 type A arrival in [0, δ] 1 arrival in [0, δ]): λ A δ (λ A + λ B )δ 2 Since C is time homogeneous, this is true of any arrival in C: The probability that a particular arrival is a type A arrival is λ A (λ A + λ B )

21 21 Splitting of Poisson Processes Same idea as with Bernouli: Each arrival of a PP with rate λ is routed up with probability p and down with prob 1 p. Upper is Poisson with rate λp Lower is Poisson with rate λ(1 p) Are two processes independent? Yes! Let N u (t): number of arrivals to the upper process; N l (t): number of arrivals to the lower process P(N u (t) = m, N l (t) = n) = P(N u (t) = m, N l (t) = n P(N(t) = m + n)p(n(t) = ( m + n = )p m (1 p) n (λt)m+n m (m + n)! e λt (m + n)! = p m (1 p) n (λt)m+n m!n! (m + n)! e λt = (λt)m (λt) n (m)! e λpt (n)! e λ(1 p)t

22 22 Tasks on Processors There n tasks on n processors. Two stages to each stage. Time for each stage is iid exponential with parameter 1. Find: Prob k half done tasks when the first task is done. There are exactly n iid competing exponentials at all times. Merge them to get a Poisson Process of rate n. Each arrival in the merged process is tagged i with prob 1 n for i = 1, 2,.., n. Prob k + 1 successive distinct tags and k + 2 nd duplicate tag. This is 1 n 1 n n 2 n...n k n ( n k+1) (k + 1)! k + 1 n k n k + 1 n