随机过程英语讲义-8

2.2 Properties of Poisson processesExample Suppose that people immigrate into a territory at a Poisson rateλ=1 per day. (a) What is the expected time until the tenth immigrant arrives? (b) What is the probability that the elapsed time between the tenth and the eleventh arrival exceeds two days? Solution: (a) E[S10]=10/λ= 10 days (b) P{X11>2}= e -2λ= e-2≈ 0.1333

2.2 Properties of Poisson processesArrival time distribution Proposition 2.2.2: The arrival time of the nth event Sn follows aΓ distribution with parameter (n,λ). (λt ) n 1 f (t )=λe λ t Proof: (n 1)!{Sn≤ t} {N(t)≥n} P{Sn≤ t}= P{N(t)≥n}=

∑k=n

∞

e λ t

(λ t ) k k!

differentiating the two sides of equation with respect to t:

2.2 Properties of Poisson processes

2.2 Properties of Poisson processesSuppose we are told that exactly one event of a Poisson process has taken place by time t, and we are asked to determine the distribution of the time at which the event occurred. Since a Poisson process possesses stationary and independent increments, it seems reasonable that each interval in[0,t] of equal length should have the same probability of containing the event. In other words, the time of the event should be uniformly distributed over[0,t].

2.2 Properties of Poisson processes

This result may be generalized, but before doing so we need to introduce the concept of order statistics.5

2.2 Properties of Poisson processesOrder statistics Order statistics Let Y1, Y2……Yn are n random variables, if we arrange these random variables from small to big, note Y(1)= y1 is the smallest in the sequence, Y(2)= y2 is the second smallest,…. Y(n)= yn is the biggest in the sequence. Y(1)< Y(2)……< Y(n), Y(1)……Y(n) or y1……yn are the order statistics of Y1…Yn.

2.2 Properties of Poisson processes,

Let,

f is density of distribution of Yi, if f follows the uniform density over (0,t), the joint density of{Y(i)} is:Y(1) .....Y( n )

f

( y1,..., yn )= n !Ci=1

n

n! f ( yi )= n, t

0< y1< ...< yn< t

2.2 Properties of Poisson processesPast arrival times given– Joint density of past arrival times Proposition 2.2.3: Given that N(t)=n, the n arrival times S1… Sn have the same distribution as the order statistics corresponding to the n i.i.d. samples from U(0,t). that is,

n! f S1 .....S n N ( t )(t1,..., t n n)= n, tProof:

0< t1< ...< t n< t

2.2 Properties of Poisson processesP{ti≤ Si≤ t+hi, i=1,…n|N(t)= n}P{one event in[ti, ti+ hi], 1≤ i≤ n, no events elsewhere in[0,t]}= P{N (t )= n}

=

λh1e

λh1

n!= n h1 ...hn t

...λhn e e e λt (λ t ) n n!

λhn1

λ ( t h1 ...hn )

P{N (t )= n}= e

λt

(λ t ) n, n= 0,1, 2,...... n!

2.2 Properties of Poisson processesn! P{ti≤ Si≤ t+hi, i=1,…n|N(t)= n}= n h1 ...hn t P{ti≤ S i≤ ti+ hi, i= 1...n N (t )= n} n!= n h1....hn ttaking the limit

s as hi→ 0 for all i, we obtain n! f S1 ..... S n N ( t )(t1,..., t n n)= n t

2.2 Properties of Poisson processesExample: A cable TV company collects$1/unit time from each subscriber. Subscribers sign up in accordance with a Poisson process with rateλ. What is the expected total revenue received in (0,t]? Solution: (Depends on the total number of subscribers and their arriving time)

2.2 Properties of Poisson processesLet N(t) denote the number of subscribers, and Si denote the收益 arrival time of the ith customer. The revenue generated by this customer in (0,t] is t-Si. Adding the revenues generated by all arrivals in (0,t]

N (t ) ∑ (t Si ), E ∑ (t S i ) i=1 i=1 find the previous expectation by conditioning on N(t)N (t )

N (t ) n n E ∑ (t S i ) N (t )= n = E ∑ (t S i ) N (t )= n = nt E ∑ S i N (t )= n i=1 i=1 i=1

2.2 Properties of Poisson processesLet U1,…Un be iid random variables which follow U(0,t). sot n n n n E ∑ Si N (t )= n = E ∑ U (i ) = E ∑ U i =∑ E[U i]= n 2 i=1 i=1 i=1 i=1

so

N (t ) t t E ∑ (t Si ) N (t )= n = nt n= n 2 2 i=1

Calculate the expectation by conditional expectation: N (t ) E[ N (t )]t 1 2=λt E ∑ (t S i ) = 2 2 i=1

2.2 Properties of Poisson processesDecomposition of Poisson process (an important application of Proposition 2.2.3) A Poisson process N={N(t),t≥0} with rateλ. We consider the case in which if an arrival occurs at time s, it is a type-1 arrival with probability P(s) and a type-2 arrival with probability 1-P(s). The type of arrival depends on the epoch of arrival. By using Proposition 2.2.3 we can prove the following propositon.

2.2 Properties of Poisson processesProposition 2.2.4 Let Ni={Ni(t), t≥0}, i=1 and 2, where Ni(t) denotes the number of type-i arrivals in (0,t]. N1(t) and N2(t) are two independent Poisson random variables with meansλpt andλqt, where1 t p=∫ P ( s )ds and q= 1 p t 0 λ pt (λpt ) n λ qt (λqt ) m P{N1 (t )= n, N 2 (t )= m}= e e n! m!

2.2 Properties of Poisson processes

2.2 Properties of Poisson processes

The importance of the above proposition is illustrated by the following example.17

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