The Poisson process Random points

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1 12 The Poisso process I may radom pheomea we ecouter, it is ot just oe or two radom variables that play a role but a whole collectio. I that case oe ofte speaks of a radom process. The Poisso process is a simple kid of radom process, which models the occurrece of radom poits i time or space. There are umerous ways i which processes of radom poits arise: some examples are preseted i the first sectio. The Poisso process describes i a certai sese the most radom way to distribute poits i time or space. This is made more precise with the otios of homogeeity ad idepedece Radom poits Typical examples of the occurrece of radom time poits are: arrival times of messages at a server, the times at which asteroids hit the earth, arrival times of radioactive particles at a Geiger couter, times at which your computer crashes, the times at which electroic compoets fail, ad arrival times of people at a pump i a oasis. Examples of the occurrece of radom poits i space are: the locatios of asteroid impacts with earth (2-dimesioal), the locatios of imperfectios i a material (3-dimesioal), ad the locatios of trees i a forest (2-dimesioal). Some of these pheomea are better modeled by the Poisso process tha others. Loosely speakig, oe might say that the Poisso process model ofte applies i situatios where there is a very large populatio, ad each member of the populatio has a very small probability to produce a poit of the process. This is, for istace, well fulfilled i the Geiger couter example where, i a huge collectio of atoms, just a few will emit a radioactive particle (see [28]). A property of the Poisso process as we will see shortly is that poits may lie arbitrarily close together. Therefore the tree locatios are ot so well modeled by the Poisso process.

2 The Poisso process 12.2 Takig a closer look at radom arrivals A well-kow example that is usually modeled by the Poisso process is that of calls arrivig at a telephoe exchage the exchage is coected to a large umber of people who make phoe calls ow ad the. This will be our leadig example i this sectio. Telephoe calls arrive at radom times X 1,X 2,... at the telephoe exchage durig a time iterval [0,t]. 0 X 1 X 2 X 3 X 4 X 5 t Time.. The two basic assumptios we make o these radom arrivals are 1. (Homogeeity) Therateλ at which arrivals occur is costat over time: i a subiterval of legth u the expectatio of the umber of telephoe calls is λu. 2. (Idepedece) The umbers of arrivals i disjoit time itervals are idepedet radom variables. Homogeeity is also called weak statioarity. We deote the total umber of calls i a iterval I by N(I), abbreviatig N([0,t]) to N t. Homogeeity the implies that we require E[N t ]=λt. To get hold of the distributio of N t we divide the iterval [0,t]ito itervals of legth t/. Whe is large eough, every iterval I j, =((j 1) t/, j t/] will cotai either 0 or 1 arrival: For such a large (which also satisfies 0 t X 1 X 2 X 3 X 4 X 5 ( 1) t t Time.. >λt), let R j be the umber of arrivals i the time iterval I j,.sicer j is 0or1,R j has a Ber(p j ) distributio for some p j. Recall that for a Beroulli radom variable E[R j ] = 0 (1 p j )+1 p j = p j. By the homogeeity assumptio, for each j p j = λ legth of I j, = λt. Summig the umber of calls i the itervals gives the total umber of calls, hece N t = R 1 + R R.

3 12.2 Takig a closer look at radom arrivals 169 By the idepedece assumptio, the R j are idepedet radom variables, therefore N t has a Bi(, p) distributio, with p = λt/. Remark 12.1 (About this approximatio). The argumet just give seems pretty covicig, but actually R j does ot have a Beroulli distributio, whatever the value of. A way to see this is the followig. Every iterval I j, is a uio of the two itervals I 2j 1,2 ad I 2j,2. Hecethe probability that I j, cotais two calls is at least (λt/2) 2 = λ 2 t 2 /4 2, which is larger tha zero. Note however, that the probability of havig two arrivals is of smaller order tha the probability that R j takes the value 1. If we add a third assumptio, amely that the probability of two or more calls arrivig i a iterval I j, teds to zero faster tha 1/, the the coclusio below o the distributio of N t is valid. We have foud that (at least i first approximatio) P(N t = k) = ( k )( λt ) k ( 1 λt ) k for k =0,...,. I this aalysis is a rather artificial parameter, of which we oly kow that it should ot be too small. It therefore seems a good idea to get rid of by lettig go to ifiity, hopig that the probability distributio of N t will settle dow. Note that lim ( k ) 1 k = lim ad from calculus we kow that lim 1 k +1) ( 1 k! = 1 k!, ( 1 λt ) =e λt. Sice certaily ( lim 1 λt ) k =1, we obtai, combiig these three limits, that lim P(N t = k) = lim Sice ( k e λt ) 1 k (λt)k k=0 ( 1 λt ) ( 1 λt ) k = (λt)k e λt. k! (λt) k =e λt e λt =1, k! we have ideed ru ito a probability distributio o the umbers 0, 1, 2,.... Note that all these probabilities are determied by the sigle value λt. This motivates the followig defiitio.

4 The Poisso process Defiitio. A discrete radom variable X has a Poisso distributio with parameter µ, where µ>0 if its probability mass fuctio p is give by p(k) =P(X = k) = µk k! e µ for k =0, 1, 2,... We deote this distributio by Pois(µ). Figure 12.1 displays the graphs of the probability mass fuctios of the Poisso distributio with µ = 0.9 (left) ad the Poisso distributio with µ = 5 (right) p(k) k p(k) Fig The probability mass fuctios of the Pois(0.9) ad the Pois(5) distributios. k Quick exercise 12.1 Cosider the evet exactly oe call arrives i the iterval [0, 2s]. The probability of this evet is P(N 2s =1)=λ 2s e λ 2s. But ote that this evet is the same as there is exactly oe call i the iterval [0,s) ad o calls i the iterval [s, 2s], or o calls i [0,s) ad exactly oe call i [s, 2s]. Verify (usig assumptios 1 ad 2) that you get the same aswer if you compute the probability of the evet i this way. We do have a hit 1 about what the expectatio ad variace of a Poisso radom variable might be: sice E[N t ]=λt for all, we aticipate that the limitig Poisso distributio will have expectatio λt. Similarly, sice N t has a Bi(, λt ) distributio, we aticipate that the variace will be 1 This is really ot more tha a hit: there are simple examples where the distributios of radom variables coverge to a distributio whose expectatio is differet from the limit of the expectatios of the distributios! (cf. Exercise 12.14).

5 12.3 The oe-dimesioal Poisso process 171 lim Var(N t) = lim λt ( 1 λt ) = λt. Actually, the expectatio of a Poisso radom variable X with parameter µ is easy to compute: E[X] = k=0 = µe µ k µk k! e µ =e µ k=1 k=1 µ k 1 (k 1)! = µe µ µ k (k 1)! j=0 µ j j! = µ. I a similar way the variace ca be determied (see Exercise 12.8), ad we arrive at the followig rule. The expectatio ad variace of a Poisso distributio. Let X have a Poisso distributio with parameter µ; the E[X] =µ ad Var(X) =µ The oe-dimesioal Poisso process We will derive some properties of the sequece of radom poits X 1,X 2,... that we cosidered i the previous sectio. What we derived so far is that for ay iterval (s, s + t] theumbern((s, s + t]) of poits X i i that iterval is a radom variable with a Pois(λt) distributio. Iterarrival times The differeces T i = X i X i 1 are called iterarrival times. Here we defie T 1 = X 1, the time of the first arrival. To determie the probability distributio of T 1, we observe that the evet {T 1 >t} that the first call arrives after time t isthesameastheevet {N t =0} that o calls have bee made i [0,t]. But this implies that P(T 1 t) =1 P(T 1 >t)=1 P(N t =0)=1 e λt. Therefore T 1 has a expoetial distributio with parameter λ. To compute the joit distributio of T 1 ad T 2, we cosider the coditioal probability that T 2 >t,givethatt 1 = s, ad use the property that arrivals i differet itervals are idepedet:

6 The Poisso process P(T 2 >t T 1 = s) = P(o arrivals i (s, s + t] T 1 = s) = P(o arrivals i (s, s + t]) =P(N((s, s + t]) = 0) = e λt. Sice this aswer does ot deped o s, we coclude that T 1 ad T 2 are idepedet, ad P(T 2 >t)=e λt, i.e., T 2 also has a expoetial distributio with parameter λ. Actually, although the coclusio is correct, the method to derive it is ot, because we coditioed o the evet {T 1 = s}, which has zero probability. This problem could be circumveted by coditioig o the evet that T 1 lies i some small iterval, but that will ot be doe here. Aalogously, oe ca show that the T i are idepedet ad have a Exp(λ) distributio. This ice property allows us to give a simple defiitio of the oe-dimesioal Poisso process. Defiitio. The oe-dimesioal Poisso process with itesity λ is a sequece X 1,X 2,X 3,... of radom variables havig the property that the iterarrival times X 1,X 2 X 1,X 3 X 2,... are idepedet radom variables, each with a Exp(λ) distributio. Note that the coectio with N t is as follows: N t is equal to the umber of X i that are smaller tha (or equal to) t. Quick exercise 12.2 We model the arrivals of messages at a server as a Poisso process. Suppose that o average 330 messages arrive per miute. What would you choose for the itesity λ i messages per secod? What is the expectatio of the iterarrival time? A obvious questio is: what is the distributio of X i? This has already bee aswered i Chapter 11: sice X i is a sum of i idepedet expoetially distributed radom variables, we have the followig. The poits of the Poisso process. For i = 1, 2,... the radom variable X i has a Gam(i, λ) distributio. The distributio of poits Aother iterestig questio is: if we kow that poits are geerated i a iterval, where do these poits lie? Sice the distributio of the umber of poits oly depeds o the legth of the iterval, ad ot o its locatio, it suffices to determie this for a iterval startig at 0. Let this iterval be [0,a]. We start with the simplest case, where there is oe poit i [0,a]: suppose that N([0,a]) = 1. The, for 0 <s<a:

7 12.4 Higher-dimesioal Poisso processes 173 P(X 1 s N([0,a]) = 1) = P(X 1 s, N([0,a]) = 1) P(N([0,a]) = 1) P(N([0,s]) = 1,N((s, a]) = 0) = P(N([0,a]) = 1) = λse λs e λ(a s) λae λa = s a. We fid that coditioal o the evet {N([0,a]) = 1}, the radom variable X 1 is uiformly distributed over the iterval [0,a]. Now suppose that it is give that there are two poits i [0,a]: N([0,a]) = 2. I a way similar to what we did for oe poit, we ca show that (see Exercise 12.12) P(X 1 s, X 2 t N([0,a]) = 2) = t2 (t s) 2 a 2. Now recall the result of Exercise 9.17: if U 1 ad U 2 are two idepedet radom variables, both uiformly distributed over [0,a], the the joit distributio fuctio of V =mi(u 1,U 2 )adz =max(u 1,U 2 )isgiveby P(V s, Z t) = t2 (t s) 2 a 2 for 0 s t a. Thus we have foud that, if we forget about their order, the two poits i [0,a] are idepedet ad uiformly distributed over [0,a]. With somewhat more work, this geeralizes to a arbitrary umber of poits, ad we arrive at the followig property. Locatio of the poits, give their umber. Give that the Poisso process has poits i the iterval [a, b], the locatios of these poits are idepedetly distributed, each with a uiform distributio o [a, b] Higher-dimesioal Poisso processes Our defiitio of the oe-dimesioal Poisso process, startig with the iterarrival times, does ot geeralize easily, because it is based o the orderig of the real umbers. However, we ca easily exted the assumptios of idepedece, homogeeity, ad the Poisso distributio property. To do this we eed a higher-dimesioal versio of the cocept of legth. We deote the k- dimesioal volume of a set A i k-dimesioal space by m(a). For istace, i the plae m(a) is the area of A, ad i space m(a) is the volume of A.

8 The Poisso process Defiitio. The k-dimesioal Poisso process with itesity λ is a collectio X 1,X 2,X 3,... of radom poits havig the property that if N(A) deotes the umber of poits i the set A, the 1. (Homogeeity) The radom variable N(A) has a Poisso distributio with parameter λm(a). 2. (Idepedece) For disjoit sets A 1,A 2,...,A the radom variables N(A 1 ),N(A 2 ),...,N(A ) are idepedet. Quick exercise 12.3 Suppose that the locatios of defects i a certai type of material follow the two-dimesioal Poisso process model. For this material it is kow that it cotais o average five defects per square meter. What is the probability that a strip of legth 2 meters ad width 5 cm will be without defects? I Figure 7.4 the locatios of the buildigs the architect wated to distribute over a 100-by-300-m terrai have bee geerated by a two-dimesioal Poisso process. This has bee doe i the followig way. Oe ca agai show that give the total umber of poits i a set, these poits are uiformly distributed over the set. This leads to the followig procedure: first oe geerates a value from a Poisso distributio with the appropriate parameter (λ times the area), the oe geerates times a poit uiformly distributed over the 100- by-300 rectagle. Actually oe ca geerate a higher-dimesioal Poisso process i a way that is very similar to the atural way this ca be doe for the oe-dimesioal process. Directly from the defiitio of the oe-dimesioal process we see that it ca be obtaied by cosecutively geeratig poits with expoetially distributed gaps. We will explai a similar procedure for dimesio two. For s>0, let M s = N(C s ), where C s is the circular regio of radius s, cetered at the origi. Sice C s has area πs 2, M s has a Poisso distributio with parameter λπs 2.LetR i deote the distace of the ith closest poit to the origi. This is illustrated i Figure Note that R i is the aalogue of the ith arrival time for the oe-dimesioal Poisso process: we have i fact that R i s if ad oly if M s i. I particular, with i =1ads = t, P ( R1 2 t ) ( =P R 1 ) t =P ( M t > 0) =1 e λπt. I other words: R1 2 is Exp(λπ) distributed. For geeral i, we ca similarly write P ( ( Ri 2 t) =P R i ) t =P ( M t i).

9 12.4 Higher-dimesioal Poisso processes R.... R Fig The Poisso process i the plae, with the two circles of the two poits closest to the origi. So P ( Ri 2 t) i 1 =1 e λπt j=0 (λπt) j, j! which meas that Ri 2 has a Gam(i, λπ) distributio as we saw o page 157. Sice gamma distributios arise as sums of idepedet expoetial distributios, we ca also write Ri 2 = R2 i 1 + T i, where the T i are idepedet Exp(λπ) radom variables (ad where R 0 =0). Note that this is quite similar to the oe-dimesioal case. To simulate the two-dimesioal Poisso process from a sequece U 1,U 2,... of idepedet U(0, 1) radom variables, oe ca therefore proceed as follows (recall from Sectio 6.2 that (1/λ)l(U i ) has a Exp(λ) distributio): for i =1, 2,... put R i = Ri λπ l(u 2i); this gives the distace of the ith poit to the origi, ad the put the poit o this circle accordig to a agle value geerated by 2πU 2i 1.Thisisthe correct way to do it, because oe ca show that i polar coordiates the radius ad the agle of a Poisso process poit are idepedet of each other, ad the agle is uiformly distributed over [0, 2π]. The latter is called the isotropy property of the Poisso process.

10 The Poisso process 12.5 Solutios to the quick exercises 12.1 The probability of exactly oe call i [0,s) ad o calls i [s, 2s] equals P(N([0,s)) = 1,N([s, 2s]) = 0) = P(N([0,s)) = 1) P(N([s, 2s]) = 0) =P(N([0,s)) = 1) P(N([0,s]) = 0) = λse λs e λs, because of idepedece ad homogeeity. I the same way, the probability of exactly oe call i [s, 2s] ad o calls i [0,s)isequaltoe λs λse λs.ad ideed: λse λs e λs +e λs λse λs =2λse λ 2s Because there are 60 secods i a miute, we have 60λ = 330. It follows that λ = Sice the iterarrival times have a Exp(λ) distributio, the expected time betwee messages is 1/λ =0.18 secod The itesity of this process is λ =5perm 2. The area of the strip is 2 (1/20) = 1/10 m 2. Hece the probability that o defects occur i the strip is e λ (area of strip) =e 5 (1/10) =e 1/2 = Exercises 12.1 I each of the followig examples, try to idicate whether the Poisso process would be a good model. a. The times of bakruptcy of eterprises i the Uited States. b. The times a chicke lays its eggs. c. The times of airplae crashes i a worldwide registratio. d. The locatios of worgly spelled words i a book. e. The times of traffic accidets at a crossroad The umber of customers that visit a bak o a day is modeled by a Poisso distributio. It is kow that the probability of o customers at all is What is the expected umber of customers? 12.3 Let N have a Pois(4) distributio. What is P(N =4)? 12.4 Let X have a Pois(2) distributio. What is P(X 1)? 12.5 The umber of errors o a hard disk is modeled as a Poisso radom variable with expectatio oe error i every Mb, that is, i every 2 20 bytes. a. What is the probability of at least oe error i a sector of 512 bytes? b. The hard disk is a Gb disk drive with sectors. What is the probability of at least oe error o the hard disk?

11 12.6 Exercises A certai brad of copper wire has flaws about every 40 cetimeters. Model the locatios of the flaws as a Poisso process. What is the probability of two flaws i 1 meter of wire? 12.7 The Poisso model is sometimes used to study the flow of traffic ([15]). If the traffic ca flow freely, it behaves like a Poisso process. A 20-miute time iterval is divided ito 10-secod time slots. At a certai poit alog the highway the umber of passig cars is registered for each 10-secod time slot. Let j be the umber of slots i which j cars have passed for j =0,...,9. Suppose that oe fids j j Note that the total umber of cars passig i these 20 miutes is 230. a. What would you choose for the itesity parameter λ? b. Suppose oe estimates the probability of 0 cars passig i a 10-secod time slot by 0 divided by the total umber of time slots. Does that (reasoably) agree with the value that follows from your aswer i a? c. What would you take for the probability that 10 cars pass i a 10-secod time slot? 12.8 Let X be a Poisso radom variable with parameter µ. a. Compute E[X(X 1)]. b. Compute Var(X), usig that Var(X) =E[X(X 1)] + E[X] (E[X]) Let Y 1 ad Y 2 be idepedet Poisso radom variables with parameter µ 1, respectively µ 2. Show that Y = Y 1 + Y 2 also has a Poisso distributio. Istead of usig the additio rule i Sectio 11.1 as i Exercise 11.2, you ca prove this without doig ay computatios by cosiderig the umber of poits of a Poisso process (with itesity 1) i two disjoit itervals of legth µ 1 ad µ Let X be a radom variable with a Pois(µ) distributio. Show the followig. If µ<1, the the probabilities P(X = k) are strictly decreasig i k. Ifµ > 1, the the probabilities P(X = k) are first icreasig, the decreasig (cf. Figure 12.1). What happes if µ =1? Cosider the oe-dimesioal Poisso process with itesity λ. Show that the umber of poits i [0,t], give that the umber of poits i [0, 2t] is equal to, hasabi(, 1 2 ) distributio. Hit: write the evet {N([0,s]) = k, N([0, 2s]) = } as the itersectio of the (idepedet!) evets {N([0,s]) = k} ad {N((s, 2s]) = k}.

12 The Poisso process We cosider the oe-dimesioal Poisso process. Suppose for some a>0 it is give that there are exactly two poits i [0,a], or i other words: N a = 2. The goal of this exercise is to determie the joit distributio of X 1 ad X 2, the locatios of the two poits, coditioal o N a =2. a. Prove that for 0 <s<t<a P(X 1 s, X 2 t, N a =2) b. Deduce from a that =P(X 2 t, N a =2) P(X 1 >s,x 2 t, N a =2). P(X 1 s, X 2 t, N a =2)=e λa ( λ 2 t 2 c. Deduce from b that for 0 <s<t<a 2! λ2 (t s) 2 ). 2! P(X 1 s, X 2 t N a =2)= t2 (t s) 2 a Walkig through a meadow we ecouter two kids of flowers, daisies ad dadelios. As we walk i a straight lie, we model the positios of the flowers we ecouter with a oe-dimesioal Poisso process with itesity λ. It appears that about oe i every four flowers is a daisy. Forgettig about the dadelios, what does the process of the daisies look like? This questio will be aswered with the followig steps. a. Let N t be the total umber of flowers, X t the umber of daisies, ad Y t be the umber of dadelios we ecouter durig the first t miutes of our walk. Note that X t + Y t = N t. Suppose that each flower is a daisy with probability 1/4, idepedet of the other flowers. Argue that ( ) + m ( 1 ( 3 ) m. P(X t =, Y t = m N t = + m) = 4) 4 b. Show that P(X t =, Y t = m) = ) ( 3 ) me! m!( λt (λt) +m, 4 4 by coditioig o N t ad usig a. c. By writig e λt =e (λ/4)t e (3λ/4)t ad summig over m, show that P(X t = ) = 1! e (λ/4)t( λt ). 4 Sice it is clear that the umbers of daisies that we ecouter i disjoit time itervals are idepedet, we may coclude from c that the process (X t )is agai a Poisso process, with itesity λ/4. Oe ofte says that the process (X t ) is obtaied by thiig the process (N t ). I our example this correspods to pickig all the dadelios.

13 12.6 Exercises I this exercise we look at a simple example of radom variables X that have the property that their distributios coverge to the distributio of a radom variable X as, while it is ot true that their expectatios coverge to the expectatio of X. Let for = 1, 2,... the radom variables X be defied by P(X =0)=1 1 ad P(X =7) = 1. a. Let X be the radom variable that is equal to 0 with probability 1. Show that for all a the probability mass fuctios p X (a) ofthex coverge to the probability mass fuctio p X (a) ofx as. Note that E[X]=0. b. Show that oetheless E[X ] = 7 for all.

Infinite Sequences and Series

Infinite Sequences and Series Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet