Chapter 2. Poisson Processes

Transcription

1 Chapter 2. Poisson Processes Prof. Ai-Chun Pang Graduate Institute of Networking and Multimedia, epartment of Computer Science and Information Engineering, National Taiwan University, Taiwan

2 utline Introduction to Poisson Processes Properties of Poisson processes Inter-arrival time distribution Waiting time distribution Superposition and decomposition Non-homogeneous Poisson processes (relaxing stationary) Compound Poisson processes (relaxing single arrival) Modulated Poisson processes (relaxing independent) Poisson Arrival See Average (PASTA) Prof. Ai-Chun Pang, NTU 1

3 ntroduction n ~ ( t ) Inter-arrival time ~ x 3 ~ x 2 ~ x 1 arrival epoch ~ ~ ~ ~ t ~ S 0 S 1 S 2 S 3 S 4 (i) n th arrival epoch S n is S n = x 1 + x x n = n i=1 x i S 0 = 0 (ii) Number of arrivals at time t is: ñ(t). Notice that: {ñ(t) n} iff { S n t}, {ñ(t) =n} iff { S n t and S n+1 >t} Prof. Ai-Chun Pang, NTU 2

4 ntroduction Arrival Process: X = { x i,i=1, 2,...}; x i s can be any S = { S i,i=0, 1, 2,...}; S i s can be any N = {ñ(t), t 0}; called arrival process Renewal Process: X = { x i,i=1, 2,...}; x i s are i.i.d. S = { S i,i=0, 1, 2,...}; S i s are general distributed N = {ñ(t), t 0}; called renewal process Poisson Process: X = { x i,i=1, 2,...}; x i s are iid exponential distributed S = { S i,i=0, 1, 2,...}; S i s are Erlang distributed N = {ñ(t), t 0}; called Poisson process Prof. Ai-Chun Pang, NTU 3

5 ounting Processes A stochastic process N = {ñ(t), t 0} is said to be a counting process if ñ(t) represents the total number of events that have occurred up to time t. From the definition we see that for a counting process ñ(t) must satisfy: 1. ñ(t) ñ(t) is integer valued. 3. If s<t,thenñ(s) ñ(t). 4. For s<t,ñ(t) ñ(s) equals the number of events that have occurred in the interval (s, t]. Prof. Ai-Chun Pang, NTU 4

6 efinition 1: Poisson Processes he counting process N = {ñ(t), t 0} is a Poisson process with rate λ λ>0), if: 1. ñ(0) = 0 2. Independent increments relaxed Modulated Poisson Process P [ñ(t) ñ(s) =k 1 ñ(r) =k 2,r s<t]=p [ñ(t) ñ(s) =k 1 ] 3. Stationary increments relaxed Non-homogeneous Poisson Process P [ñ(t + s) ñ(t) =k] =P [ñ(l + s) ñ(l) =k] 4. Single arrival relaxed Compound Poisson Process P [ñ(h) =1] = λh + o(h) P [ñ(h) 2] = o(h) Prof. Ai-Chun Pang, NTU 5

7 efinition 2: Poisson Processes he counting process N = {ñ(t), t 0} is a Poisson process with rate λ λ>0), if: 1. ñ(0) = 0 2. Independent increments 3. The number of events in any interval of length t is Poisson distributed with mean λt. That is, for all s, t 0 λt (λt)n P [ñ(t + s) ñ(s) =n] =e, n =0, 1,... n! Prof. Ai-Chun Pang, NTU 6

8 heorem: Definitions 1 and 2 are equivalent. roof. We show that Definition 1 implies Definition 2. To start, fix u 0 and let g(t) =E[e uñ(t) ] We derive a differential equation for g(t) as follows: g(t + h) = E[e uñ(t+h) ] = E {e uñ(t) e u[ñ(t+h) ñ(t)]} [ = E e uñ(t)] E = g(t)e { e u[ñ(t+h) ñ(t)]} by independent increments [ e uñ(h)] by stationary increments (1) Prof. Ai-Chun Pang, NTU 7

9 heorem: Definitions 1 and 2 are equivalent. Conditioning on whether ñ(t) = 0orñ(t) = 1orñ(t) 2 yields [ E e uñ(h)] = 1 λh + o(h)+e u (λh + o(h)) + o(h) From (1) and (2), we obtain that implying that = 1 λh + e u λh + o(h) (2) g(t + h) =g(t)(1 λh + e u λh)+o(h) g(t + h) g(t) h = g(t)λ(e u 1) + o(h) h Prof. Ai-Chun Pang, NTU 8

10 heorem: Definitions 1 and 2 are equivalent. Letting h 0gives g (t) =g(t)λ(e u 1) or, equivalently, g (t) g(t) = λ(e u 1) Integrating, and using g(0) = 1, shows that or log(g(t)) = λt(e u 1) g(t) =e λt(e u 1) the Laplace transform of a Poisson r. v. Since g(t) is also the Laplace transform of ñ(t), ñ(t) is a Poisson r. v. Prof. Ai-Chun Pang, NTU 9

11 heorem: Definitions 1 and 2 are equivalent et P n (t) =P [ñ(t) =n]. e derive a differential equation for P 0 (t) in the following manner: P 0 (t + h) = P [ñ(t + h) =0] = P [ñ(t) =0, ñ(t + h) ñ(t) =0] = P [ñ(t) =0]P [ñ(t + h) ñ(t) =0] = P 0 (t)[1 λh +0(h)] ence P 0 (t + h) P 0 (t) = λp 0 (t)+ o(h) h h etting h 0 yields P 0(t) = λp 0 (t) ince P 0 (0) = 1, then P 0 (t) =e λt Prof. Ai-Chun Pang, NTU 10

12 heorem: Definitions 1 and 2 are equivalent imilarly, for n 1 P n (t + h) = P [ñ(t + h) =n] = P [ñ(t) =n, ñ(t + h) ñ(t) =0] +P [ñ(t) =n 1, ñ(t + h) ñ(t) =1] +P [ñ(t + h) =n, ñ(t + h) ñ(t) 2] = P n (t)p 0 (h)+p n 1 (t)p 1 (h)+o(h) = (1 λh)p n (t)+λhp n 1 (t)+o(h) hus P n (t + h) P n (t) = λp n (t)+λp n 1 (t)+ o(h) h h etting h 0, P n(t) = λp n (t)+λp n 1 (t) Prof. Ai-Chun Pang, NTU 11

13 he Inter-Arrival Time Distribution heorem. Poisson Processes have exponential inter-arrival time distribution, i.e., { x n,n=1, 2,...} are i.i.d and exponentially distributed with parameter λ (i.e., mean inter-arrival time = 1/λ). roof. x 1 : P ( x 1 >t)=p (ñ(t) =0)= e λt (λt) 0 x 1 e(t; λ) x 2 : P ( x 2 >t x 1 = s) 0! = P {0 arrivals in (s, s + t] x 1 = s} = e λt = P {0 arrivals in (s, s + t]}(by independent increment) = P {0 arrivals in (0,t]}(by stationary increment) = e λt x 2 is independent of x 1 and x 2 exp(t; λ). The procedure repeats for the rest of x i s. Prof. Ai-Chun Pang, NTU 12

14 he Arrival Time Distribution of the nth Event heorem. Thearrivaltimeofthen th event, S n (also called the waiting time until the n th event), is Erlang distributed with parameter (n, λ). roof. Method 1 : Method 2 : P [ S n t] =P [ñ(t) n] = f S n (t) = λe λt (λt) n 1 (n 1)! f Sn (t)dt = df Sn (t) =P [t < S n <t+ dt] k=n (exercise) e λt (λt) k = P {n 1 arrivals in (0,t] and 1 arrival in (t, t + dt)} + o(dt) = P [ñ(t) =n 1 and 1 arrival in (t, t + dt)] + o(dt) = P [ñ(t) =n 1]P [1 arrival in (t, t + dt)] + o(dt)(why?) k! Prof. Ai-Chun Pang, NTU 13

15 he Arrival Time Distribution of the nth Event = e λt (λt) n 1 (n 1)! lim dt 0 f S n (t)dt dt λdt + o(dt) = f Sn (t) = λe λt (λt) n 1 (n 1)! Prof. Ai-Chun Pang, NTU 14

16 onditional Distribution of the Arrival Times heorem. Given that ñ(t) =n, then arrival times S 1, S 2,..., S n have the same distribution as the order statistics corresponding to n i.i.d. uniformly distributed random variables from (0,t).... rder Statistics. Let x 1, x 2,..., x n be n i.i.d. continuous random variables having common pdf f. Define x (k) as the k th smallest value among all x i s, i.e., x (1) x (2) x (3)... x (n),then x (1),..., x (n) are known as the order statistics corresponding to random variables x 1,..., x n. We have that the joint pdf of x (1), x (2),..., x (n) is f x(1), x (2),..., x (n) (x 1,x 2,...,x n )=n!f(x 1 )f(x 2 )...f(x n ), where x 1 <x 2 <...<x n (check the textbook [Ross]).... Prof. Ai-Chun Pang, NTU 15

17 onditional Distribution of the Arrival Times roof. Let 0 <t 1 <t 2 <... <t n+1 = t and let h i be small enough so that t i + h i <t i+1,i=1,...,n. P [t i < S i <t i + h i,i=1,...,n ñ(t) =n] exactly one arrival in each [t P i,t i + h i ] i =1, 2,...,n, andnoarrivalelsewherein[0,t] = P [ñ(t) =n] = (e λh 1 λh 1 )(e λh 2 λh 2 )...(e λh n λh n )(e λ(t h 1 h 2... h n ) ) e λt (λt) n /n! = n!(h 1h 2 h 3...h n ) t n P [t i < S i <t i + h i,i=1,...,n ñ(t) =n] = n! h 1 h 2...h n t n Prof. Ai-Chun Pang, NTU 16

18 onditional Distribution of the Arrival Times Taking lim ( h i 0,i=1,...,n ), then f S1, S 2,..., S n ñ(t) (t 1,t 2,...,t n n) = n! t n, 0 <t 1 <t 2 <...<t n. Prof. Ai-Chun Pang, NTU 17

19 onditional Distribution of the Arrival Times xample (see Ref [Ross], Ex. 2.3(A) p.68). Suppose that travellers arrive at a train depot in accordance with a Poisson process with rate λ. If the train departs at time t, what is the expected sum of the waiting times of travellers arriving in (0,t)? That is, E[ ñ(t) i=1 (t S i )] =? ~ ~ ~ ~ S 1 S 2 S 3 S n ~(t ) t Prof. Ai-Chun Pang, NTU 18

20 onditional Distribution of the Arrival Times nswer. Conditioning on ñ(t) = n yields ñ(t) E[ (t S i ) ñ(t) =n] =nt E[ i=1 = nt E[ = nt E[ n ũ (i) ] i=1 n ũ i ] ( i=1 n i=1 S i ] (by the theorem) n ũ (i) = i=1 n ũ i ) = nt t 2 n = nt ( E[ũ i ]= t 2 2 ) ñ(t) To find E[ (t S i )], we should take another expectation ñ(t) E[ i=1 i=1 (t S i )] = t 2 E[ñ(t)] }{{} =λt i=1 = λt2 2 Prof. Ai-Chun Pang, NTU 19

21 uperposition of Independent Poisson Processes heorem. Superposition of independent Poisson Processes N (λ i,i=1,...,n), is also a Poisson process with rate λ i. 1 Poisson λ1 Poisson λ2 Poisson Poisson λn rate = N 1 λi Homework> Prove the theorem (note that a Poisson process must satisfy Definitions 1 or 2). Prof. Ai-Chun Pang, NTU 20

22 ecomposition of a Poisson Process heorem. Given a Poisson process N = {ñ(t),t 0}; If ñ i (t) represents the number of type-i events that occur by time t, i =1, 2; Arrival occurring at time s is a type-1 arrival with probability p(s), and type-2 arrival with probability 1 p(s) then ñ 1, ñ 2 are independent, ñ 1 (t) P (k; λtp), and ñ 2 (t) P (k; λt(1 p)), where p = 1 t t 0 p(s)ds Prof. Ai-Chun Pang, NTU 21

23 ecomposition of a Poisson Process Poisson λ p(s) Poisson ( k; λ p( s) ds) 1-p(s) Poisson ( k; λ [1 p( s)] ds) pecial case: If p(s) = p is constant, then Poisson λ p Poisson rate λp 1-p Poisson rate λ( 1 p) Prof. Ai-Chun Pang, NTU 22

24 ecomposition of a Poisson Process roof. It is to prove that, for fixed time t, P [ñ 1 (t) =n, ñ 2 (t) =m] = P [ñ 1 (t) =n]p [ñ 2 (t) =m] = e λpt (λpt) n n! e λ(1 p)t [λ(1 p)t] m m!... P [ñ 1 (t) =n, ñ 2 (t) =m] = P [ñ 1 (t) =n, ñ 2 (t) =m ñ 1 (t)+ñ 2 (t) =k] P [ñ 1 (t)+ñ 2 (t) =k] k=0 = P [ñ 1 (t) =n, ñ 2 (t) =m ñ 1 (t)+ñ 2 (t) =n + m] P [ñ 1 (t)+ñ 2 (t) =n + m] Prof. Ai-Chun Pang, NTU 23

25 ecomposition of a Poisson Process From the condition distribution of the arrival times, any event occurs at some time that is uniformly distributed, and is independent of other events. Consider that only one arrival occurs in the interval [0,t]: = = P [type - 1 arrival ñ(t) =1] t 0 t 0 P [type - 1 arrival arrival time S 1 = s, ñ(t) =1] P (s) 1 t ds = 1 t t 0 P (s)ds = p f S1 ñ(t) (s ñ(t) =1)ds Prof. Ai-Chun Pang, NTU 24

26 ecomposition of a Poisson Process P [ñ 1 (t) =n, ñ 2 (t) =m] = P [ñ 1 (t) =n, ñ 2 (t) =m ñ 1 (t)+ñ 2 (t) =n + m] P [ñ 1 (t)+ñ 2 (t) =n + m] = n + m p n (1 p) m e λt (λt) n+m n (n + m)! = (n + m)! p n (1 p) m e λt (λt) n+m n!m! (n + m)! = e λpt (λpt) n n! e λ(1 p)t [λ(1 p)t] m m! Prof. Ai-Chun Pang, NTU 25

27 ecomposition of a Poisson Process xample (An Infinite Server Queue, textbook [Ross]). G Poisson λ departure G s (t) =P ( S t), where S = service time G s (t) is independent of each other and of the arrival process ñ 1 (t): the number of customers which have left before t; ñ 2 (t): the number of customers which are still in the system at time t; ñ 1 (t)? andñ 2 (t)? Prof. Ai-Chun Pang, NTU 26

28 ecomposition of a Poisson Process nswer. ñ 1 (t): the number of type-1 customers ñ 2 (t): the number of type-2 customers type-1: P (s) = P (finish before t) type-2: 1 P (s) = Ḡ s(t s) ñ 1 (t) P ñ 2 (t) P = P ( S t s) =G s (t s) ( k; λt 1 t ( k; λt 1 t t 0 t 0 ) G s (t s)ds ) Ḡ s (t s)ds Prof. Ai-Chun Pang, NTU 27

29 ecomposition of a Poisson Process E[ñ 1 (t)] = λt 1 t = λ = λ 0 t t 0 t 0 G(t s)ds G(y)( dy) G(y)dy t s = y s = t y As t,wehave lim E[ñ 2(t)] = λ t t 0 Ḡ(y)dy = λe[ S] (Little s formula) Prof. Ai-Chun Pang, NTU 28

30 on-homogeneous Poisson Processes The counting process N = {ñ(t),t 0} is said to be a non-stationary or non-homogeneous Poisson Process with time-varying intensity function λ(t),t 0, if: 1. ñ(0) = 0 2. N has independent increments 3. P [ñ(t + h) ñ(t) 2] = o(h) 4. P [ñ(t + h) ñ(t) =1]=λ(t) h + o(h) Define integrated intensity function m(t) = Theorem. t 0 λ(t )dt. P [ñ(t + s) ñ(t) =n] = e [m(t+s) m(t)] [m(t + s) m(t)] n Proof. < Homework >. n! Prof. Ai-Chun Pang, NTU 29

31 on-homogeneous Poisson Processes xample. The output process of the M/G/ queue is a non-homogeneous Poisson process having intensity function λ(t) = λg(t), where G is the service distribution. int. Let D(s, s + r) denote the number of service completions in the interval (s, s + r] in(0,t]. If we can show that D(s, s + r) follows a Poisson distribution with mean λ s+r s G(y)dy, and the numbers of service completions in disjoint intervals are independent, then we are finished by definition of a non-homogeneous Poisson process. Prof. Ai-Chun Pang, NTU 30

32 on-homogeneous Poisson Processes nswer. An arrival at time y is called a type-1 arrival if its service completion occurs in (s, s + r]. Consider three cases to find the probability P (y) that an arrival at time y is a type-1 arrival: Case 1: y s. Case 1 Case 2 Case 3 s s+r t y y y P (y) =P {s y< S <s+ r y} = G(s + r y) G(s y) Case 2: s<y s + r. P (y) =P { S <s+ r y} = G(s + r y) Prof. Ai-Chun Pang, NTU 31

33 on-homogeneous Poisson Processes Case 3: s + r<y t. P (y) =0 Based on the decomposition property of a Poisson process, we conclude that D(s, s + r) follows a Poisson distribution with mean λpt, wherep =(1/t) t 0 P (y)dy. t 0 P (y)dy = = = s 0 s+r 0 s+r 0 [G(s + r y) G(s y)]dy + + t s+r (0)dy G(s + r y)dy G(z)dz s 0 s 0 G(z)dz = s+r s G(s y)dy s+r s G(z)dz G(s + r y)dy Prof. Ai-Chun Pang, NTU 32

34 on-homogeneous Poisson Processes Because of the independent increment assumption of the Poisson arrival process, and the fact that there are always servers available for arrivals, the departure process has independent increments Prof. Ai-Chun Pang, NTU 33

35 ompound Poisson Processes A stochastic process { x(t),t 0} is said to be a compound Poisson process if it can be represented as x(t) = ñ(t) i=1 ỹ i, t 0 {ñ(t),t 0} is a Poisson process {ỹ i,i 1} is a family of independent and identically distributed random variables which are also independent of {ñ(t),t 0} The random variable x(t) is said to be a compound Poisson random variable. E[ x(t)] = λte[ỹ i ]andvar[ x(t)] = λte[ỹ 2 i ]. Prof. Ai-Chun Pang, NTU 34

36 ompound Poisson Processes Example (Batch Arrival Process). Consider a parallel-processing system where each job arrival consists of a possibly random number of tasks. Then we can model the arrival process as a compound Poisson process, which is also called a batch arrival process. Let ỹ i be a random variable that denotes the number of tasks comprising a job. We derive the probability generating function P x(t) (z) as follows: [ P x(t) (z) = E z x(t)] [ [ ]] [ [ ]] = E E z x(t) ñ(t) = E E zỹ1+ +ỹñ(t) ñ(t) [ ]] [zỹ1+ +ỹñ(t) = E E (by independence of ñ(t) and{ỹ i }) [ [ ] [ ]] = E E zỹ1 E zỹñ(t) (by independence of ỹ 1,, ỹñ(t) ) [ = E (Pỹ(z))ñ(t)] = Pñ(t) (Pỹ(z)) Prof. Ai-Chun Pang, NTU 35

37 odulated Poisson Processes Assume that there are two states, 0 and 1, for a modulating process. 0 1 When the state of the modulating process equals 0 then the arrive rate of customers is given by λ 0, and when it equals 1 then the arrival rate is λ 1. The residence time in a particular modulating state is exponentially distributed with parameter μ and, after expiration of this time, the modulating process changes state. The initial state of the modulating process is randomly selected and is equally likely to be state 0 or 1. Prof. Ai-Chun Pang, NTU 36

38 odulated Poisson Processes Foragivenperiodoftime(0,t), let Υ be a random variable that indicates the total amount of time that the modulating process has been in state 0. Let x(t) be the number of arrivals in (0,t). Then, given Υ, the value of x(t) is distributed as a non-homogeneous Poisson process and thus P [ x(t) =n Υ =τ] = (λ 0τ + λ 1 (t τ)) n e (λ 0τ+λ 1 (t τ)) As μ 0, the probability that the modulating process makes no transitions within t seconds converges to 1, and we expect for this case that P [ x(t) =n] = 1 { (λ0 t) n e λ 0t + (λ 1t) n e λ } 1t 2 n! n! n! Prof. Ai-Chun Pang, NTU 37

39 odulated Poisson Processes As μ, then the modulating process makes an infinite number of transitions within t seconds, and we expect for this case that P [ x(t) =n] = (βt)n e βt Example (Modeling Voice). n!, where β = λ 0 + λ 1 2 A basic feature of speech is that it comprises an alternation of silent periods and non-silent periods. The arrival rate of packets during a talk spurt period is Poisson with rate λ 1 and silent periods produce a Poisson rate with λ 0 0. The duration of times for talk and silent periods are exponentially distributed with parameters μ 1 and μ 0, respectively. The model of the arrival stream of packets is given by a modulated Poisson process. Prof. Ai-Chun Pang, NTU 38

40 oisson Arrivals See Time Averages (PASTA) PASTA says: as t Fraction of arrivals who see the system in a given state upon arrival (arrival average) = Fraction of time the system is in a given state (time average) = The system is in the given state at any random time after being steady Counter-example (textbook [Kao]: Example 2.7.1) 1/2 1/2 service time = 1/2 inter-arrival time = Prof. Ai-Chun Pang, NTU 39

41 oisson Arrivals See Time Averages (PASTA) Arrival average that an arrival will see an idle system = 1 Time average of system being idle = 1/2 Mathematically, Let X = { x(t),t 0} be a stochastic process with state space S, and B S Define an indicator random variable 1, if x(t) B ũ(t) = 0, otherwise Let N = {ñ(t),t 0} be a Poisson process with rate λ denoting the arrival process then, Prof. Ai-Chun Pang, NTU 40

42 oisson Arrivals See Time Averages (PASTA) lim t t 0 ũ(s)dñ(s) ñ(t) (arrival average) t ũ(s)ds 0 = lim t t (time average) Condition For PASTA to hold, we need the lack of anticipation assumption (LAA): for each t 0, the arrival process {ñ(t + u) ñ(t),u 0} is independent of { x(s), 0 s t} and {ñ(s), 0 s t}. Application: To find the waiting time distribution of any arriving customer Given: P[system is idle] = 1 ρ; P[system is busy] = ρ Prof. Ai-Chun Pang, NTU 41

43 oisson Arrivals See Time Averages (PASTA) Case 1: system is idle Poisson Case 2: system is busy Poisson P ( w t) = P ( w t idle) P (idle upon arrival) + P ( w t busy) P (busy upon arrival) Prof. Ai-Chun Pang, NTU 42

44 emoryless Property of the Exponential Distribution A random variable x is said to be without memory, or memoryless, if P [ x>s+ t x >t]=p [ x>s] for all s, t 0 (3) The condition in Equation (3) is equivalent to or P [ x>s+ t, x >t] P [ x>t] = P [ x>s] P [ x>s+ t] =P [ x>s]p [ x>t] (4) Since Equation (4) is satisfied when x is exponentially distributed (for e λ(s+t) = e λs e λt ), it follows that exponential random variable are memoryless. Not only is the exponential distribution memoryless, but it is the unique continuous distribution possessing this property. Prof. Ai-Chun Pang, NTU 43

45 omparison of Two Exponential Random Variables uppose that x 1 and x 2 are independent exponential random variables ith respective means 1/λ 1 and 1/λ 2.WhatisP[ x 1 < x 2 ]? P [ x 1 < x 2 ] = = = = = λ 1 λ 1 + λ 2 P [ x 1 < x 2 x 1 = x]λ 1 e λ1x dx P [x < x 2 ]λ 1 e λ1x dx e λ2x λ 1 e λ1x dx λ 1 e (λ 1+λ 2 )x dx Prof. Ai-Chun Pang, NTU 44

46 inimum of Exponential Random Variables uppose that x 1, x 2,, x n are independent exponential random variables, ith x i having rate μ i,i=1,,n. It turns out that the smallest of the x i s exponential with a rate equal to the sum of the μ i. P [min( x 1, x 2,, x n ) >x] = P [ x i >xfor each i =1,,n] n = P [ x i >x] (by independence) = i=1 n i=1 ow about max( x 1, x 2,, x n )? (exercise) e μ ix { ( n ) } = exp μ i x i=1 Prof. Ai-Chun Pang, NTU 45