Inhomogeneous Poisson process

Transcription

1 Chapter 22 Ihomogeeous Poisso process We coclue our stuy of Poisso processes with the case of o-statioary rates. Let us cosier a arrival rate, λ(t), that with time, but oe that is still Markovia. That is we assume the system has o memory (past arrivals are ot correlate with future arrivals) a isjoit time itervals are iepeet. For example, a football team may score goals at a higher rate attheeofagamethaatthebegiig if the opposig team tires more quickly. Aother example woul be that text messages are set more frequetly i the eveigs a afteroos tha i the morig, or the arrival of customers at a shop will epe o the time of ay. If we allow this flexibility i our moel, a let λ(t) vary,thewehaveamore powerful techique for moellig raom processes. We call this the ihomogeeous Poisso process to istiguish it from the staar (homogeeous) process i which the rate is costat at all times. ] Let us efie the ihomogeeous Poisso process as follows. It is a coutig process: {N(t),t }, sothatn has iteger values that ever ecrease over time, but jump up at raom times, a we specify the followig 4 coitios: (i) N() = (ii) icremets are iepeet (Markov) but ot statioary (iii) P (N(t + h) N(t) =1)=λ(t)h + o(h) (iv) P (N(t + h) N(t) > 1) = o(h) This is almost ietical to efiitio 2 for the homogeeous process. Of course, we shoul fi that ay results we obtai for the ihomogeeous process will automatically reuce, if we take λ costat, to the results for the homogeeous process. Our approach is to stuy how the probabilities chage with time, i the maer of calculus we take a ifiitesmal step forwar i time a cosier how the probabilities evolve urig this short time i the future. Let us first cosier P (N(t) =,theprobabilitythatafteratimet, o evets have occurre, a the cosier how this chages i time. We kow oe thig for certai: whe t =,P (N() = 1, that if we 153

2 154 CHAPTER 22. INHOMOGENEOUS POISSON PROCESS start coutig our evet from t =,thethecoutiszeroatthattime. P (N()) = 1, a P (N(t) =) =, =1, 2, 3,.... (22.1) Now cosier P (N(t) =)ap (N(t + h) =),thewecaiviethetimeiterval[,t+ h] itotwo isjoit time itervals [,t] a[t, t + h]. The, the probability that o evets occur i [,t+ h] meas that o evets occur i [,t] a[t, t + h]. I mathematical otatio: P (N(t + h) =)=P (N(t) = a N(t + h) N(t) =) (22.2) Sice these are isjoit time itervals (o overlap) they are iepeet. The we ca write that: P (N(t) = a N(t + h) N(t) =)=P (N(t) =) P (N(t + h) N(t) =). (22.3) Now usig our efiitio above, the probability of o evet happeig i a very short time iterval is: P (N(t + h) N(t) =)=1 λ(t)h + o(h). (22.4) So this gives us: P (N(t + h) =)=P (N(t) =) [1 λ(t)h + o(h)]. (22.5) Let s use some shortha, amely that P (N(t) =) p (t). The this equatio ca be rearrage to give: p (t + h) p (t) h = λ(t)p (t)+ o(h) h Now takig the limit h, gives the ifferetial equatio:. (22.6) p t = λ(t)p. (22.7) This equatio ca be solve irectly, sice the variable p a t are separable. We write this as: a ow itegrate to give: p p p = λ(t)t. (22.8) p = l p (t) l p () = But we kow that p () = 1, so that, this gives: l p (t) = λ(t)t. (22.9) λ(τ)τ. (22.1) λ(τ)τ (22.11) that is p (t) =e R t λ(τ )τ. (22.12) Now clearly if the process is homogeeous, that is λ is a costat (oes ot chage i time), the: λ(τ)τ = λt

3 155 a this gives the familiar expressio for the homogeeous case: P (N(t) =)=e λt,asitshoul. So cosier the case >, for P (N(t) =). This is slightly messier but repeats the same ieas. We seek P (N(t + h) = base o P (N(t) =), that is we look back i time (coitio o) the earlier values. P (N(t + h) =) = P (N(t + h) = N(t) =k) P (N(t) =k). (22.13) k= Now, the coitioal probabilities ca be simplifie: P (N(t + h) = N(t) =k) =P (N(t + h) N(t) = k). (22.14) Sice the cout is always icreasig i time we caot have (more evets i the past tha i the future). This traslates, mathematically, to the formula: P (N(t + h) N(t) = k) =, k >. (22.15) So the ifiite sum i (22.16) is actually fiite, sice there are zeros for all k>. By the same toke if the time step is extremely tiy h, there is the chace of (at most) oe evet happeig, a most likely that o evet will happe. I mathematical terms this meas (usig our rules) P (N(t + h) N(t) > 1) = o(h) P (N(t + h) N(t) =1) = λ(t)h + o(h) P (N(t + h) N(t) =) = 1 λ(t)h + o(h) So oly two (o-tiy) terms matter i the sum (22.16), correspoig to o extra evets or oe extra evet. This meas: P (N(t+h) =) =(1 λ(t)h+o(h)) P (N(t) =)+( λ(t)h+o(h)) P (N(t) = 1)+o(h). (22.16) Agai, switchig to a more coveiet shortha: p (t) P (N(t) =), this ca be writte as: p (t + h) p (t) h = λ(t)p (t)+λ(t)p 1 (t)+ o(h) h. (22.17) Now takig the limit h givestheifferetial equatio: t p (t) = λ(t)p (t)+λ(t)p 1 (t), >. (22.18) I fact, this is a set ( =1, 2, 3,...)ofcouplefirst-orerifferetial equatios. I cotiuous time we e up with ifferetial equatios rather tha iffierece equatios.

4 156 CHAPTER 22. INHOMOGENEOUS POISSON PROCESS 22.1 Solutio of the ifferetial equatios We have the ifferece-ifferetial equatio: a, for =: t p (t) = λ(t)p (t)+λ(t)p 1 (t), =1, 2, 3,.... (22.19) The iitial coitios, that efie the uique solutio are: t p (t) = λ(t)p (t). (22.2) p () = 1, p () = =1, 2, 3,.... (22.21) The equatio (22.19) has two variables, the iscrete (ifferece)variable,athecotiuous(ifferetial )variable,t. Wecarearrage(22.19),movigoetermfromrighttoleft-ha sie: t p (t)+λ(t)p (t) =λ(t)p 1 (t). (22.22) This is a first-orer oriary-ifferetial equatio i staar form, for which we have a few useful methos of solutio. For example, we ca use the itegratig factor techique, that is, we multiply both sies of the equatio by the term: e R t λ(τ )τ, (22.23) givig us: e R t λ(τ )τ t p (t)+λ(t)e R t λ(τ )τ p (t) =λ(t)e R t λ(τ )τ p 1 (t). (22.24) This leas to the simpler form: (e R ) t λ(τ )τ p (t) = λ(t)e R t λ(τ )τ p 1 (t) (22.25) t Now, for the sake of abbreviatio we efie, z (t) e R t λ(τ )τ p (t). The we have the simpler iffereceifferetial equatio: with the equatio for =,correspoigto(22.2): t z = λ(t)z 1 (t), =1, 2, 3,.... (22.26) t z =. (22.27) This equatio ca be solve by iteratio. Thatis,wesolve(22.27)firsttofiz (t), a the solve (??), for z 1 (t), usig the solutio from (22.27). The solve (22.26) for z 2 (t), a so o a ifiitum. The solutio of (22.27) with the bouary coitio, p () = 1, is, z =1,thatis p (t) =e R t λ(τ )τ, (22.28) as obtaie before. The, for =1, t z 1 = λ(t)z (t) =λ(t), (22.29)

5 22.1. SOLUTION OF THE DIFFERENTIAL EQUATIONS 157 This ca be itegrate irectly, recallig the iitial coitio z 1 () =. with solutio: The, accorig to (22.26) That is, substitutig for z 1 (t), we have: z 2 (t) = z 1 (t) = z 2 (t) = λ(τ)τ, (22.3) λ(τ)z 1 (τ)τ. (22.31) [ τ ] λ(τ) λ(τ )τ τ = 1 [ 2 λ(τ)τ]. (22.32) 2 You ca covice yourself that this is correct by calculatig z 2 /t, ( [ 1 t ] 2 ) [ ] [ t ] [ ] λ(τ)τ = λ(τ)τ λ(τ)τ = λ(t) λ(τ)τ t 2 t. (22.33) a showig that it satisfies (22.26) for =2. For =3wehave: z 3 (t) = λ(τ)z 2 (τ)τ = 1 [ 3 λ(τ)τ]. (22.34) 2 3 Cotiuig i the same maer, expressios ca be fou for all. Apattersooemergesawefi that, for ay value of we have: z (t) = 1 [ λ(τ)τ], (22.35) from which it follows that: R t p (t) = e λ(τ )τ [ t λ(τ)τ]. (22.36) Agai, i case of oubt oe ca verify that this is the solutio of(22.19)bysubstitutio. Ofcourse,i the special case of havig a homogeeous process we get: That is, we retrieve our efiitio 1 of a Poisso process. P (N(t) =) = e λt (λt). (22.37) Thus, accorig to (22.36) we have a Poisso istributio with mea (a variace) give by: E (N(t)) = λ(τ)τ (22.38) If istea, we start coutig the process at a time t 2 a stop coutig at a later time t 2 >,thethe bouary coitios woul mea that: p ( )=1, p ( )= =1, 2, 3,.... (22.39) That is, our lower limit of itegratio chages but ot the ifferetial equatios themselves. That is, we are simply resettig the zero of time to,butthisistheolychage,atheaalysisfollowsexactly

6 158 CHAPTER 22. INHOMOGENEOUS POISSON PROCESS the same lies. For example we get, for the probability of o evets betwee times a t 2, P (N(t 2 ) N( )=)=e R t 2 λ(τ )τ. (22.4) It the follows that, the probability of exactly evets occurrig betwee the limits is: P (N(t 2 ) N( )=) = e R t2 λ(τ )τ [2 ] λ(τ)τ. (22.41) Solutio by momet-geeratig fuctio Oe ca also arrive at the solutio (22.37), which we terme efiitio 1, through the momet-geeratig fuctio. It s worth presetig this iea for iactic reasosif othig else. Cosier the coutig process, N(t), for the homogeeous Poisso process. The total time iterval [,t] ca be ivie ito m (equal) pieces a the total cout will be the sum of the icremetal couts over each iterval. Let: t i be the start times of each iterval. The: ( ) i t, i m 1, (22.42) m N(t) ={N( ) N(t )} + {N(t 2 ) N( )} + + {N(t) N(t m 1 )}. (22.43) These sub-couts (icremets) will be iepeet of course, a sice the itervals are very small we ca use efiitio 2 to state that, for ay iteger 1 i m: P (N[t i+1 ] N[t i ]=1)=λ(t/m)+o(t/m), (22.44) a P (N[t i+1 ] N[t i ]=)=1 λ(t/m)+o(t/m). (22.45) The, the momet-geeratig fuctio takes the form: ( M N(t) (u) =E e un(t)) (22.46) The: M N(t) (u) =E (exp u [ + N(t i+1 ) N(t i )+ ]) (22.47) There are m terms (of iepeet variables) i the sum i the expoetial. This breaks ito the prouct of m expectatios. The geeral term has the form: E (exp[u (N(t i+1 ) N(t i ))]) = 1 λ.(t/m)+λ.(t/m)e u + o(t/m) (22.48) where we have use (22.44) a (22.45) to evaluate the expectatio. This simplifies to: E (exp[u (N(t i+1 ) N(t i ))]) = 1 + λ(t/m)(e u 1) + o(t/m) (22.49) So, with m terms of this form, multiplie together, a takig the limit m : M N(t) (u) = lim m (1 + λ(t/m)(eu 1)) m (22.5)

7 22.2. MARKOVIAN 159 This gives: M N(t) (u) =exp(λt(e u 1)) (22.51) This we recogise as the momet geeratio fuctio of a Poisso variable a hece, because of the uiqueess of the momet geeratig fuctio, we ca immeiately euce the probability mass fuctio for N(t): which is i agreemet with (22.37). λt (λt) P (N(t) =) =e, =, 1, 2..., (22.52) 22.2 Markovia So the solutio for the ihomogeous case is: p (t) =e R t λ(t)t ( λ(τ)τ ). (22.53) Sice the process is Markovia we ca start the clock at ay time, a reset the couter at ay time. So i geeral: P (N(t 2 )= N( )=)=e R t 2 λ(τ )τ ( 2 ) λ(τ)τ. (22.54) Of course, takig the special case of the homogeous process, λ costat, a startig from = a we regai the staar formula. λ(τ)τ = λt We ca easily verify that (22.53) is a solutio of (22.19). We ote that, usig the prouct rule: ( ) t t p (t) = [e R t λ(τ )τ] λ(τ)τ +e R t λ(τ )τ t t ( ) t λ(τ)τ. (22.55) The usig the result: we get: a which gives: t λ(τ)τ = λ(t) [e R t λ(t)t] = λ(t)e R t λ(t)τ (22.56) t [ t [ 1 λ(τ)τ] = λ(τ)τ] λ(t). (22.57) t ( ) t ( ) t p (t) = λ(t) [e R t λ(τ )τ] λ(τ)τ t 1 + λ(t)e R t λ(τ )τ λ(τ)τ ( 1)!. (22.58) That is: t p (t) = λ(t)p (t)+λ(t)p 1 (t). (22.59)

8 16 CHAPTER 22. INHOMOGENEOUS POISSON PROCESS which is what we wat. QUESTION The rate at which a ewly-qualife river has acciets is a Poisso process with rate: λ(t) =.1.2t per year over the first three years, t 3. Calculate the probability that the river has o acciets i hissecoyear(1 t 2), a the probability that he has exactly oe acciet i the seco year. ANSWER We use the formula: P (N(t 2 )= N( )=)=e R t 2 λ(τ )τ ( 2 ) λ(τ)τ (22.6) where =1at 2 =2. Foroacciets =aforoeacciet =1. Ieithercase,weeethe itegral: This gives, So 2 λ(τ)τ = 1 2 (.1.2τ)τ. (22.61) [.1τ.1τ 2 ] 2 =.16.9 =.7. (22.62) 1 P (N(2) = N(1) = ) = e (22.63) Similarly, the probability of exactly oe acciet ( = 1) is: P (N(2) = 1 N(1) = ) = e (22.64) 22.3 Time to arrival Cosier a ihomogeeous Poisso process, a the associate waitig times. Let T 1 eote the first arrival, the the waitig time is just the expoetial process, iscusse before: F T1 (t) =P (T 1 t) =1 e R t λ(τ )τ. (22.65) The correspoig probability esity, f T1 (t), is the erivative of this fuctio: f T1 (t) = t F T 1 (t) =λ(t)e R t λ(τ )τ. (22.66) The the expecte time for the first arrival was show to be: E (T )= + e R t λ(τ )τ t (22.67)