Penalization for birth and death processes

Transcription

1 Penaization for birth and death processes Pierre Debs, Mihai Gradinaru To cite this version: Pierre Debs, Mihai Gradinaru Penaization for birth and death processes Journa of Theoretica Probabiity, Sprnger, 28, 21 3), pp <117/s > <ha > HAL Id: ha Submitted on 19 Apr 27 HAL is a muti-discipinary open access archive for the deposit and dissemination of scientific research documents, whether they are pubished or not The documents may come from teaching and research institutions in France or abroad, or from pubic or private research centers L archive ouverte puridiscipinaire HAL, est destinée au dépôt et à a diffusion de documents scientifiques de niveau recherche, pubiés ou non, émanant des étabissements d enseignement et de recherche français ou étrangers, des aboratoires pubics ou privés

2 Penaization for birth and death processes Pierre Debs and Mihai Gradinaru Institut Éie Cartan Nancy, BP 239, 5456 Vandœuvre-ès-Nancy Cede, France {PierreDebs, Abstract: In this paper we study a transient birth and death Markov process penaized by its sojourn time in Under the new probabiity measure the origina process behaves as a recurrent birth and death Markov process We aso show, in a particuar case, that an initiay recurrent birth and death process, behaves as an transient birth and death process after penaization with the event that it can reach zero in infinite time We iustrate some of our resuts with the Besse random wak eampe Key words: Birth and death Markov processes; penaization; sojourn time; Dynkin s formua; random wak; Brownian motion with drift; Besse chain and process; change of probabiity 2 Mathematics Subject Cassification: 6J8, 6J1, 6J27, 6J75, 6B1, 6G42, 6G44 1 Introduction Consider a transient Markov chain with positive states Is it possibe to force the chain to have an infinite number of visits at the state? Does one have a reasonabe manner to do this? Conditionay to this behaviour, is the resuting chain recurrent or transient? These are some natura questions when studying penaization for Markov chains, that is to modify their aws in order to ater some of their properties In the present paper by penaization one understands to condition by an event with probabiity zero, in our case, the event that the transient Markov chain returns infinitey often at The penaization of a probabiity measure for instance, the distribution of some Markov process) by some appropriate weights is the main too of some cassica probabiistic ideas as Doob s h-transform, Feynman-Kac transform or Girsanov transform For instance, in the atter case, an eponentia martingae occurs as weight factor and, under some natura conditions, a new probabiity measure is we defined Since, dividing a probabiity distribution by a famiy of weights does not create necessariy a probabiity famiy of distributions, one needs to made some imiting procedure often as the parameter goes to infinity) The main features of the penaization procedure have been pointed out in a series of papers by Roynette,Vaois,Yor see 12 for a survey) Essentiay, one considers continuous parameter Markov processes Brownian motion or Besse processes) and one penaizes by some function of oca time, maimum, one-sided maimum etc For instance, one considers Ω, {F t } t, {R t } t, ) the canonica Besse process of inde α 1, ), and one denotes {L t } t its oca time at eve Then one can be interested on the imit = Q h) A s ) = A s M s h), A s F s s fied), h : + +, im t A s hl t ) hl t ) 1

3 and one proves that the process {M s h) } s is a martingae Then one studies the canonica process {R t } t under the new probabiity Q h) and some interesting properties of this process are pointed out The discrete parameter counterpart of the same probems was performed very recenty by Debs 1 for random waks For instance, if Ω, {F n } n, {S n } n, ) is the canonica standard random wak and if one cas the oca time at the foowing sequence im p =, n+1 = n + {S n=}, n, then one can be interested on the imit A n hp ) {S p>} + g ) p ) {S p<}) hp ) {S p>} + g p ) {S p<} = Q h,g) A n ) = E A n M h,g) n, A n F n n fied), g, h : + Again, one proves that {M h,g) n } n is a martingae and one studies the canonica sequence {S n } n under the probabiity Q h,g) Let us notice that in amost of cases considered in the cited works, one uses as important toos precise informations on the processes : epicit aws of hitting times, epicit martingaes reated to considered processes and so on This aows to perform eact computations when proving certain properties In the present paper we consider the birth and death Markov chains Reca that these processes have the property that ony transitions to neighbouring states are possibe This mode comes from appications in bioogy, the deveoping in time of a popuation of partices: each partice ives for a random ength of time at the end of which it spits into two partices or dies The process is quite simpe and constitutes a generaisation of the standard random wak on in the discrete parameter setting) and have some simiarities with Besse processes in the continuous parameter setting) We address the previous simpe questions, for birth and death Markov chains with discrete and continuous parameter Precisey, we penaize the distribution of the transient birth and death process by the number of visits at the state which is ike oca time type penaization) When we force the chain to visit an infinitey often the state zero it is reasonabe to think that what we get is a recurrent chain Indeed, we prove that, under the new probabiity measure induced by penaization, the chain behaves as a recurrent birth and death chain The same procedure works in continuous parameter setting We point out that in 11 one studies the Besse process of dimension greater than 2 which is a transient process) penaized in some sense by its oca time at more precisey, by its sojourn time in an interva, see aso Remark 41 beow) Our approach is eementary in the discrete parameter setting, and it is based on Dynkin s formua for pure jump processes in the continuous parameter setting We woud ike to stress that for both situations a we need is the transient feature of the process, but we do not need neither epicit epressions of transition probabiities or rates), nor other epicit informations reated to our processes as in previous cited works In some sense this is one of the originaities of the present paper, besides the study of the penaization for birth and death chains Is it possibe to penaize a recurrent birth and death chain in order to obtain a transient one? We give a particuar eampe where we penaize a recurrent random wak by the event 2

4 to reach zero in infinite time and we get a recurrent birth and death chain Our proof for this case uses the epression of the aw of the hitting time of zero for a biased random wak and we obtain a particuar birth and death chain, the so-caed Besse random wak For this reason it seems that the case of a genera birth and death chain is more compicate We iustrate some other phenomena by further study of Besse random waks The paper is organised as foows : the main resuts concerning the penaization of transient birth and death chains, with discrete and aso with continuous parameter, are stated in the net section Section 2 aso contains an eampe of penaization of recurrent birth and death chain A the proofs are given in Section 3 The eampe of Besse random waks, but aso some other eampes are discussed in the ast section of the paper 2 Notations and main resuts 21 Discrete parameter setting Let {X n } n be a birth and death chain on the canonica probabiity space Ω, F, ) In other words, {X n } n is a Markov chain, with state space and having the transition probabiities given by: and p := X n+1 = + 1 X n = ), q := X n+1 = 1 X n = ), We wi denote r := X n+1 = X n = ) = 1 p q, 1, 21) p := X n+1 = 1 X n = ), q :=, r := 1 p 22) γ := 1, γ := and it can proved see for instance 6, p 77) the foowing cassica: k=1 q k p k, 1, 23) Proposition 21 The birth and death chain {X n } n is transient if and ony if γ <, 24) and in this case we wi denote S := γ 25) Set U n := n {X k =}, n 26) k= We are abe to state the first main resut: Theorem 21 Let {F n } n be the natura fitration associated to the transient birth and death chain {X n } n 3

5 1 For any integer n and any event A n F n, im κ A n {U κ} {U κ} = A n M n, 27) where M n := ) 1 1 p ) Un+ S S {Xn=} X n γ 28) 2 {M n } n is a positive {F n }-martingae which tends toward zero, amost surey Henceforth {M n } n is not uniformy integrabe 3 The equaity 27) induces a probabiity measure Q on Ω, F ): for any integer n and any event A n F n, Q A n ) := A n M n 29) Then, under Q, {X n } n is a recurrent birth and death chain with transition probabiities given by: and p = Q X n+1 = + 1 X n = ) := p P k +1 γ k)/ P k γ k), q = Q X n+1 = 1 X n = ) := q P k 1 γ k)/ P k γ k), r = Q X n+1 = X n = ) := r, 1, 21) p = Q X n+1 = 1 X n = ) := p S 1)/S p ), q :=, r := r S/S p ) 211) Remark 22 1 We suppose that the chain starts from zero ony to simpify the computations, but it is possibe to assume that the Markov chain {X n } n is starting from One can obtain a simiar resut with minor modifications: for instance in 28) one needs to repace the factor ) 1 1/S by γ 2 As a main eampe of birth and death chain, we wi consider see aso 41 beow) the so-caed Besse random wak with inde α > 1 or of dimension δ = 2α + 1) > see aso 7, p 448, or 8) The transition probabiities are given by p = 1, p = + 2α α + 1, q = 2 + 2α + 1, r =, 1 212) If α = 1 /2, {X n } n is the absoute vaue of the symmetric standard random wak Let us note that {X n } n is transient for α >, as we can easiy see by using Proposition 21 4

6 22 Continuous parameter setting Let {Xt)} t be a Markov process parametrized by t, ) and with the state space For instance, we can think that Xt) is the size of a popuation of partices at time t We foow 1, p 373, to describe the continuous time version of the birth and death process {Xt)} t The changes of states yieds as foows: when the process is in state, assume that are given two independent random variabes B) and D), which are independent from {Xt)} t, and eponentiay distributed with parameters b and d, respectivey A transition from to + 1 is made if B) D), which occurs with probabiity B) D) = b /b +d ); 213) otherwise a transition from to 1 is made The hoding time in state is B) D), which is eponentia with parameter b + d we assume that d = ) We may think of B) as the time unti a birth when the popuation is and simiary, D) is the time unti a death when the popuation is ; the popuation increases by one if a birth occurs prior to a death; otherwise the popuation decreases by one Usuay, the birth and death process in continuous time is described in terms of birth and death rates : in a popuation of size a partice is born at rate b and dies at rate d These refer to the infinitesima transition probabiities: b h + oh), if k = 1 Xt + h) = + k Xt) = ) = d h + oh), if k = 1 as h oh), if k > 1, It can be seen that these assumptions ead to a mode of jump Markov process with the hoding time parameter at λ := b + d and the jump from We wi denote { +1 with probabiity b /b ξ = +d ) 1 with probabiity d /b +d ) 214) J = and, for n 1, J n = inf{t J n 1 : Xt) XJ n 1 )} 215) Let us introduce the sequence of random variabes {Y n } n, given by Y n := XJ n ), n 216) The proof of the foowing proposition is cassica see, for instance 6, p and 9, p 115): Proposition 23 1 {Y n } n given by 216) is a birth and death discrete Markov chain with transition probabiities given by p = b /b +d ), q = d /b +d ), 1 and p = 1 We set as previousy q k d k γ = 1 and γ = =, 1 p k b k k=1 k=1 5

7 2 The continuous time birth and death process {Xt)} t is transient if and ony if the discrete Markov chain {Y n } n is transient, that is, if and ony if S := γ = k=1 d k b k < 217) Let us denote The second main resut is: V t) := {Xs)=}ds, t 218) Theorem 22 Let {F t } t be the natura fitration associated to the transient birth and death process {Xt)} t on the canonica probabiity space Ω, F, ) 1 For any t and any event A t F t, im κ A t {V ) κ} {V ) κ} = A t Mt), 219) where ) ) 1 b V t) Mt) := ep S S Xt) γ 22) 2 {Mt)} t is a positive {F t }-martingae which tends toward zero, amost surey Henceforth {Mt)} t is not uniformy integrabe 3 The equaity 219) induces a probabiity measure Q on Ω, F ): for any t and any event A t F t Q A t ) := AtMt) 221) Then, under Q, {Xt)} t is a continuous recurrent birth and death process with birth and death rates given by: b := b P k +1 γ k)/ P k γ k),, d := d P k 1 γ k)/ P k γ k), > 222) 23 Besse random wak as a penaized cassica random wak In this section we try to make the reverse work, more precisey, to penaize a recurrent birth and death Markov chain in order to obtain a transient birth and death Markov chain We present a particuar eampe of birth and death chain but the method of proof we have used in this case see 33 beow) seems more compicated for the genera situation Thus it remains an open question Let p, 1 /2) and we consider a recurrent random wak {Z n } n with transition probabiities p := Z n+1 = + 1 Z n = ) = p, q := Z n+1 = 1 Z n = ) = q = 1 p, 1 and p = Z n+1 = 1 Z n = ) = 1 223) It is a recurrent Markov chain by Proposition 21, since 1 γ = 1 q/p) = 6

8 We sha penaize this chain to reach the state zero in an infinite time As usua, we denote by T the hitting time of state zero by the Markov chain {Z n } n and it is not very difficut to get an equivaent for T κ), as κ see Lemma 35 beow) With this too in hand, we can prove: Proposition 24 Let {F n } n be the natura fitration associated to the recurrent birth and death chain {Z n } n with the transition probabiities given by 223) 1 For any integers 1, n and any event A n F n, im κ A n {T κ} {T κ} = A n M n, 224) where M n := Z n {T >n} ) Zn )/2 q 4pq) n /2 p is a positive {F n }-martingae which tends toward zero, amost surey 225) 2 The equaity 224) induces a probabiity measure Q on Ω, F ): for any integer n and any event A n F n, Q A n ) := A n M n 226) Then, under Q, {Z n } n is a transient birth and death chain with transition probabiities given by: Q Z n+1 = + 1 Z n = ) = + 1 2, Q Z n+1 = 1 Z n = ) = 1, 1, 2 and Q Z n+1 = Z n = 1) =, 227) that is a Besse random wak with inde 1 /2 or dimension 3) Now, a very natura probem is to make a simiar study for a recurrent birth and death process in continuous time setting Consider a continuous time random wak {Zt)} t : when the process is in a transition in + 1 or in 1 appears with probabiity b/b+d) or d/b+d), respectivey Here b, d > are the parameters of the eponentia times unti a birth or a death, respectivey Thus, the hoding times in any state are eponentia with parameter b + d According to Proposition 23, the process {Zt)} t is recurrent if and ony if d > b It seems ceary that, if we penaize this process to reach the state zero in an infinite time, we obtain a transient Markov process, at east at a heuristic eve Moreover, in the ight of Proposition 24, the underying Markov chain shoud be a Besse random wak with dimension 3 If T denotes the hitting time of state zero by {Zt)} t, it seems that the study of the equivaent of T κ), as κ, is more compicate as in the discrete time setting and it remains an open probem Let us point out that our intuition for the continuous time setting is reinforced by the study of the drifted recurrent Brownian motion Indeed, we get the same phenomenon as in the discrete time setting : when one penaizes the process to reach at an infinite time, the new process which we obtain is a Besse process of dimension 3 We state this resut in the foowing: 7

9 Proposition 25 Let {Bt)} t be a drifted Brownian motion Bt) := Bt) + µt, where µ < and {Bt)} t is a standard Brownian motion starting from We denote, as usuay, {F B,µ) t } t its natura fitration and T B,µ) the first hitting time of by B 1 For any t and any A t F B,µ) where im κ M B,µ) t) := t, A t {T B,µ) >κ} {T B,µ) >κ} {T B,µ) >t} Bt) = A t M B,µ) t), 228) ) ep µ Bt) ) ) + tµ2 2 is a positive F B,µ) -martingae which tends toward zero, amost surey 2 The equaity 228) induces a probabiity measure Q B,µ) A t F B,µ) t, Q B,µ) A t ) := A t M B,µ) t) 229) on: for any t and any and under Q B,µ), the infinitesima generator of {Bt)} t is L = 1 /2 d2 /d / d /d 3 Proofs 31 Proof of Theorem 21 a Proof of 27)-28) To begin with, we spit A n {U κ} into two terms 23) A n {U κ} = A n { {U n κ} + {U κ} {U n<κ}} = A n {U n κ} + A n {U κ} {U n<κ} and we note that {U n κ} vanishes if κ n /2 Consider the second term on the right hand side of the atter equaity and the Markov chain { X p := X n+p X n } p Then, for κ n /2, A n {U κ} {U n<κ} = A n {U κ} = A n {U θ n κ U n+ {Xn=}} = A n X n {Ũ κ Un+ {Xn=}} ) Here we denoted Ũ = p { X p=} and, as usuay, {θ n} n the famiy of shift operators At this eve we need the foowing: Lemma 31 For a integers n, nu κ) = R n/s)1 p /S) κ 1, 31) where R = S and R n := n γ, n 1 32) 8

10 Let us postpone the proof of this emma and we finish the proof of 27) We deduce that, for κ n/2 : A n R Xn /S) 1 p /S) κ Un+ 1 {Xn=} A n {U κ} = and, when we combine with 31) for n =, we obtain, for κ n/2, A n {U κ} {U κ} = and the proof of the first part of Theorem 21 is done A n R Xn/S) 1 p /S) Un+ {Xn=} Proof of Lemma 31 Let us consider f : given by f) = and, for n 1 integer, fn) = n 1 = γ Then, it is a direct computation to prove that {fx n ) : n } is a F n -martingae We denote by T n the hitting time of n by the birth and death chain and we caim that, for a n 1, nt < ) = P n γ )/S 33) Indeed, if m n, by the optiona stopping theorem, we get and we deduce fn) = n fx Tm T ) = fm) n T m < T ) + fn 1)1 n T > T m ) nt m < T ) = fn) f)) /fm) f)) = P n 1 = γ )/ P m 1 = γ ) Since {X n } n is a transient Markov chain, by etting m we obtain nt = ) = P n 1 = γ )/S or nt < ) = P n γ )/S = Rn /S Let us introduce the foowing stopping times : By using 26), we can write T ) = T and T ) := inf{n T 1) : X n = }, if 1 U = ) = T 1) + + T 1) <, T 1) + + T ) = ) = T 1) < ) T 2) < T 1) < ) T 1) < T 1) + + T 1) < ) Therefore, on the one hand T ) = T 1) + + T 1) < ) = T 1) < ) 1 T 1) = ) = 1 T < ) 1 1T = ) = p /S)1 p /S) 1 U κ) = κ U = ) = 1 p /S) κ 1, whie, on the other hand, if n 1, nu κ) = n T < ) U κ) from which we get 31) and Lemma 31 is proved 9

11 Remark 32 Let us point out that in fact the imit in 27) is an eact equaity for κ > n /2 This is a somehow different situation with respect to those in 1 or 12 b Proof of the second part of Theorem 21 By using again the fact that {X n } n is a transient Markov chain, im n X n = amost surey, and by using 21) the remainder of the series 26) tends to zero Furthermore U n is amost surey finite, hence by 28) we obtain that im n M n =, amost surey Then, necessariy, by assuming that {M n } n is a martingae, it can not be uniformy integrabe, since M ) = 1 Hence the ony thing we need to prove is that {M n } n is a martingae Since, by 28) and 26), for each n, we see that M n is integrabe We can write: M n 1 p /S) Un 1 p /S) n, M n+1 F n = Mn+1 {X n+1 =X n+1} F n + Mn+1 {X n+1 =X n 1} {X n } F n + Mn+1 {X n+1 =X n} F n 34) By using 28) and 32), the first term in 34) can be written as foows 1/S)1 p /S) Un R Xn+1 {X n+1 =X n+1} F n = 1/S)1 p /S) Un R Xn+1 {X n+1 =X n+1} F n = 1/S)1 p /S) Un R Xn+1 p Xn For the second term in 34) we spit the epectation into two terms Mn+1 {X n+1 =X n 1} {X n,1} F n + Mn+1 {X n+1 =X n 1} {X n=1} F n = 1/S)1 p /S) Un R Xn 1 {X n+1 =X n 1} {X n,1} F n + 1/S)1 p /S) Un 1+1 R Xn 1 {X n+1 =X n 1} {X n=1} F n = 1 /S)1 p /S) Un R Xn 1 {X n } {X n 1} + {X n=1}) X n+1 =X n 1 F n = 1/S)1 p /S) Un R Xn 1 {X n }q Xn By a simiar reasoning we can prove that the third term in 34) can be written Mn+1 {X n+1 =X n} F n = 1/S)1 p /S) Un R Xn r Xn We repace in 34) and we can write on the one hand, Mn+1 {X n } F n = 1 /S)1 p /S) Un {X n } pxn R + q R + r R Xn+1 Xn Xn 1 Xn Xn = 1 /S)1 p /S) Un {X n } pxn + q + r )R + q γ p γ Xn Xn Xn Xn Xn 1 Xn Xn = M n {X n }, 1

12 by using 21), 23) and 32) On the other hand Mn+1 {X n=} F n = 1/S)1 p /S) Un {X n=} p γ + r = 1 /S)1 p /S) Un {X n=}s1 p /S) = 1 /S)1 p /S) Un+1 {X n=} 1 γ = M n {X n=}, where we used the fact that 1 γ = S 1 and r = 1 p Hence {M n } n is a martingae c Proof of the third part of Theorem 21 Assuming that, under Q, the sequence {X n } n is a birth and death Markov chain, with transition probabiities given by 21), it is no difficut to prove that, under Q, the chain is recurrent Indeed, it suffices to use the resut of Proposition 21 Set γ := 1, γ := k=1 Therefore, by using 23), 21) and 32), we can write γ = 1 1 k=1 q k R k 1 = γ p k R k+1 1 = SR 1 k=1 1 R k 1 = R R 1 γ R k q k p k, 1 35) 1 /+1 1 / ) = SR 1 im N 1 /R N+1 1 /R 1 ) =, 36) since im N R N+1 =, by Proposition 21 We prove that, under Q, {X n } n is a birth and death Markov chain with transition probabiities given by 21) Consider, for a integers n 1 and a integers a,, a n 1, the event A n 1 := {X = a,, X n 1 = a n 1 } We can write, for 1 and a integers n 1, {X p = Q X n+1 = + 1 X n =, A n 1 = n+1 =+1,X n=} A n 1 M n+1, by 29) {X n=} A n 1 M n = +1 = = +1 {X n+1 =+1,X n=} A n 1 1 p /S) U n+1 +1, by 28) {X n=} A n 1 1 p /S) Un {X n+1 =+1,X n=} A n 1 1 p /S) U n+1 F n {X n=} A n 1 1 p /S) Un {X n=} A n 1 1 p /S) Un X n+1 = X n + 1 F n ) {X n=} A n 1 1 p /S) Un, since 1 = +1 {X n=} A n 1 1 p /S) Un p Xn R +1 = p {X n=} A n 1 1 p /S) Un 11

13 With a simiar computation we get that, for 1, r = r Assume now that = Then, by using a simiar reasoning, for a a,, a n 1, p = Q X n+1 = 1 X n =, A n 1 {X = n+1 =1,X n=} A n 1 1 p /S) U n+1 R 1, by 29) and 28) {X n=} A n 1 1 p /S) Un+ {Xn=} R Furthermore: = R 1 S {X n=} A n 1 1 p /S) Un p Xn {X n=} A n 1 1 p /S) Un+1 = p R 1 S r = Q X n+1 = X n =, A n 1 {X n+1 =,X n=} A n 1 1 p /S) U n+1+ {X n+1 =} R = {X n=} A n 1 1 p /S) Un+ {Xn=} R = 32 Proof of Theorem 22 1 p S ) 1 {X n+1 =,X n=} A n 1 1 p /S) Un 1+1 {X n=} A n 1 1 p /S) Un+1 = r 1 p ) 1 S The proof contains severa simiar ideas as in the proof of Theorem 21 We sha emphasize the main differences a Proof of 219)-22) This part is quite simiar to the discrete case We spit A t {V ) κ} = A t {V t) κ} + A t {V ) κ} {V t)<κ} and we note that {V t) κ} vanishes, if κ > t Set { Xs) := Xt + s) Xt)} s and note that V ) = V t) + V ) θ t Again and, as usuay, {θ t } t denotes the famiy of shift operators Then, for κ > t: A t {V ) κ} {V t)<κ} = ) A t {V ) θ t κ V t)} = A t Xt) {Ṽ ) κ V t), where Ṽ ) := { Xt)=} ds As for the discrete case we can state: Lemma 33 For a integers n, and for any t, where R n is given by 32) Let us finish the proof of 219) For κ > t: nv ) > t) = R n/s)e b t/s, 37) A t {V ) κ} = A t R Xt)/S)e b t V t))/s 12

14 Xt) e 1 e 2 e 3 J 1 J 4 J 5 J 6 J 7 t Figure 1: Xt) conditionnay to the event U = 3 and using the preceding emma again, we obtain for κ > t: that is 219)-22) A t {V ) κ} {V ) κ} = A t R Xt)/S)e b V t) /S, Proof of Lemma 33 Reca that {Y n } n is given by 216) and we sha denote U Y ) := n {Y n=} It is no difficut to see that, on the event {U Y ) = k}, V ) is the sum of k independent random variabes e 1,, e k, the hoding times at, having eponentia distribution of same parameter λ = b, and being independent from U Y ) see Figure 1) Therefore, by using 31), for a integers n, nv ) > t) = ) n U Y ) = k, e e k > t k = ) n U Y ) = k e e k > t) = k k = R n S 1 S e b t b t) i i! i k i+1 The proof of Lemma 33 is done R n S 1 1 ) k 1 1 S 1 1 S 1 1 ) k 1 = R n 1 S S S e b t b t) i i! i S e b t k 1 i= b t) i i! ) i S = R n S e bt S Remark 34 We point out again that in fact the imit in 219) is an eact equaity for κ > t, as was noted in the discrete parameter setting b Proof of the second part of Theorem 22 As in the discrete case we point out a usefu martingae At this eve we need to reca Dynkin s formua for pure jump processes 13

15 see, for instance 5, p 262): if f : is a bounded function such that {fxt))} t is a pure jump process, then M f t) := fxt)) fx)) is a oca martingae, where we denoted LfXs))ds 38) Lf) := b f + 1) f) + d f 1) f) 39) If we take in 39) f) = γ = R, we obtain: Lf) = b R +1 R ) + d R 1 R ) = b γ + d γ 1 = b =1 1 d d + d = b γ {=} = b b b {=} =1 Hence {M f t)} t given by 38) is a oca martingae Therefore, we can appy the integration by parts formua see, for instance 5, p 22) to the semimartingae R Xt) = M f t) + S b {Xs)=} ds and to the finite variation process At) := ep b V t)/s) We note that dr Xt) = dm f t) b {Xt)=} dt, dat) = b /S)e b V t) /S {Xt)=} dt and R X ), A = We obtain, by using 22), S Mt) = R Xt) e b V t) /S = S + = S + R Xs) das) + R Xs) b S eb V s)/s {Xs)=} ds + e b V s) /S As) dr Xs) + R X ), A t dm f s) b = S + {Xs)=} ds ) e b V s) /S dm f s) and consequenty, {Mt)} t is a oca martingae In fact, it is a true) martingae since, for a T, sup t T Mt) ) ep b T/S) b Proof of the third part of Theorem 22 First, we prove that, under Q, {Xt)} t is a Markov process For any n 1 and arbitrary sequences t < t 1 < < t n 1 and a,, a n 1, we consider the event A n 1 = {Xt ) = a,, Xt n 1 ) = a n 1 } 14

16 Then, for a integers k,, Q Xt n+1 ) = Xt n ) = k, A n 1 ) = = R k {Xt n+1 )=,Xt n)=k} A n 1 Mt n+1 ) {Xt n)=k} A n 1 Mt n ) = = R k = R k {Xt n+1 )=,Xt n)=k} A n 1 e b V t n+1 ) /S {Xt n)=k} A n 1 R k e b V tn) /S {Xt n+1 )=,Xt n)=k} A n 1 ep b V t n)+v t n+1 t n) θ tn )/S) {Xt n)=k} A n 1 e b V tn) /S {Xt n)=k} A n 1 e b V tn) /S {Xt n+1 )=} ep b V t n+1 t n) θ tn )/S) F tn {Xt n)=k} A n 1 e b V tn) /S {Xt n)=k} A n 1 e b V tn) /S k {Xt n)=k} A n 1 e b V tn) /S { Xt n+1 t n)=} eb Ṽ t n+1 tn) /S = R k k { Xt n+1 t n)=} eb Ṽ t n+1 tn) /S, which does not depend on A n 1 This proves that, under Q, {Xt)} t is a Markov process Let us study the hoding times, under Q, of the Markov process {Xt)} t, for instance we sha compute the distribution of J 1 For any integer 1 and for any α : Q e αj 1 = = S Q e αj1 {X)=} Q X) = = {B) D)} e αb) MB)) = 1 {X)=} e αj 1 MJ 1 ) M) = S R {X)=} + {B) D)} e αb) +1 + e αj 1 MJ 1 ) {D) B)} e αd) MD)) ) {D) B)} e αd) 1 ), since {B) D)} V J 1) =, Q e αj 1 = +1 -as and by using 22) and the stopping theorem Hence {B) D)} e αb) + 1 {D) B)} e αd) By using 213) we can compute : {B) D)}e αb) = Therefore, we obtain: d e α e b b dy e dy d = Q e αj 1 = b +1 + d 1 b + d + α) = b + d b + d + α b b + d + α We concude that the hoding time at 1 is an eponentia random variabe with parameter λ = b + d We stress that for 1 the distribution of the hoding time is the same under the probabiities and Q 15

17 The reasoning is simiar for =, : Q e αj 1 = {X)=} e αj 1 MJ 1 ) M) = e αb) R 1 S eb V B))/S {X)=} = R 1 S e αb) e b B) S = S 1 S e B) α+ b /S) = b b /S b b /S + α, and we deduce that the hoding time at is an eponentia random variabe of parameter b S 1)/S Now, we study the underying Markov chain {Y n } n under Q For any positive Bore function g : and 1, as previousy: Q gy 1 ) = Q gxj 1 )) = = Q {X)=} gxj 1 )) Q {X)=} {B) D)} g + 1)+1 Q {X)=} {X)=} + {X)=} {D) B)} g 1) 1 Q {X)=} = g + 1) +1 {B) D)} + g 1) 1 {D) B)} = g + 1) +1 b b + d + g 1) 1 d b + d For =, the reasoning is the same We concude that under Q, {Y n } n is a birth and death Markov chain with discrete parameter and the transition probabiities: p = +1 b b + d, q = 1 d b + d, for 1, and p = 1 Then if we set γ = 1 and γ = i=1 q i p i, for 1, we have: γ = i=1 d i R i 1 b i R i+1 = i=1 q i R i 1 p i R i+1 According to 36) and Proposition 21, {Y n } n is recurrent Furthermore, according to the second point of the Proposition 23), {Xt)} t is recurrent under Q If we denote b and d the birth and death rates of Xt), t ) under Q,we have: b + d = b + d and p = b b + d = +1 b b + d Consequenty, and 222) is proved b = b +1 and d = d 1, 16

18 33 Proofs of Propositions 24 and 25 Proof of Proposition 24 To begin with, we need to find an equivaent, as κ, for T κ), when 1 is an integer For the sake of competeness we state this in the foowing: Lemma 35 Let 1 an integer As κ, T κ) 4 ) 1/2 ) /2 2 q κ 3 /2 4pq) κ /2 31) π p As previousy, we postpone the proof of this emma and we finish the proof of 224)-225) As κ, we obtain the foowing equivaent {A n, T κ} {T κ} = {Λ n, T n} κ Z n T e κ n {T κ} and we deduce the epression of M n, 225) martingae Firsty, {A n, T n}4 ) 2 1/2 ) Zn/2 π Z q n p κ n) 3/2 4pq) /2 κ n) 4 2 π ) 1/2 q p ) /2 κ 3/2 4pq) κ /2 As usuay, we verify the fact that it is a ) Zn )/2 q ) ξn+1/2 M Z n q n+1 {Z n 2} F n = {T n, Z n 2} 4pq) n+1) /2 F n p p ) Zn )/2 ) ξn+1/2 1 q + {T n, Z n 2} 4pq) n+1) /2 q ξ n+1 F n p p ) Zn )/2 ) 1/2 ) 1/2 Z n q = {T n, Z n 2} 4pq) n+1) /2 q q p + q p p p ) Zn )/2 ) 1/2 ) 1/2 1 q + {T n, Z n 2} 4pq) n+1) /2 q q p q = {Z p p p n 2}M n, and, secondy, M n+1 {Z n=1} F n = 2 {T n Z n+1 =2, Z n=1} ) 2 )/2 q 4pq) /2 n+1) p 2 = {T n,z n=1} ) 1 )/2 q 4pq) n /2 p q p ) 1/2 4pq) 1 /2 p = {Zn=1}M n 17

19 Finay, to verify 227), we denote again A n 1 := {Z = a,, Z n 1 = a n 1 } If 1: Q Z n+1 = + 1 Z n =, A n 1 ) = = = + 1 {Z n+1 =+1,Z n=} A n 1 M n+1 {Z n=} A n 1 M n +1 q {T n+1,z n+1 =+1,Z n=,} A n 1 p {T n,z n=,} A n 1 q p ) 1/2 q 4pq) 1 /2 p T n, Z n+1 = + 1 Z n =, A n 1 ) = + 1 2p where we used the Markovian feature of {Z n } under ) +1 )/2 4pq) n+1)/2 ) )/2 4pq) n/2 { T n,z n+1 =+1,Z n=} A n 1 Q Z n+1 = 1 Z n = ) = 1 2 {T n,z n=} A n 1 = + 1 2p Z n+1 = + 1 Z n = ) = + 1 2, Simiary, we show that Aso, it is cear that Q Z n+1 = Z n = 1) = and 227) is proved Proof of Lemma 35 To begin with, reca that see for instance 2, p 351), for a s, 1, s T ) 1 ) 1 4pqs 2 = = T = r)s r 2ps It is a direct computation to get: 1 ) 1 4pqs 2 = 1 4pq) 2ps 2p)! Therefore, if and r are of same evenness we find P T = r) = 4pq)n+r)/2 2p) +r) /2)! 1) k k k= s 2 ) +r)/2 1 i= = 4pq)+r)/2 ) 2p) +r) 2! After some agebraic computations we obtain T = r) = 4pq)+r)/2 π2p) +r)/2)! k = r 1) k k k= ) k 2 i 1 2 k = 2k + 1 ) 1 i= ) +r) 1 2 i= ) k 2 i k i + 1 ) ) 1) k 2k Γ k + 3 ) + r Γ k 1 ) As r, the ony significant term in the atter sum corresponds to k =, hence T = r) r Γ3/2) q π ) +r 2! p ) /2 4pq) r + r 2 Γ 1 ) 2 q 2 2 π p 18 ) /2 r ) 3/2 4pq) r/2 2

20 Finay, since T κ) = r κ T = r), it is no difficut to prove that, as κ, T κ) T = κ), the first term again The proof of 31) is done Proof of Propositions 25 It is a cassica resut, using the Girsanov theorem see, for instance 4, p 197), that, for > : ) T B,µ) > κ) = T B,µ) > κ) = ep µs)2 ds, κ 2πs 3 2s and we easiy deduce that: T B,µ) > κ) κ 2πκ 3 e µ κµ 2 /2 Then, as in the proof of Proposition 24: A t {T B,µ) >κ} {T B,µ) >κ} κ κ A t {T B,µ) >κ} A t {T B,µ) Bt) {T B,µ) >κ} { T eb,µ) >κ t} >κ} Bt) / 2πκ t) 3 ) ep µbt) κ t)µ2 /2) / 2πκ 3 ) ep µ κµ 2 /2) and we deduce the epression of M B,µ) We prove the fact that it is a F B,µ) -martingae by using the cassica Itô s formua We write Ceary, M B,µ) t) = eµ Bt)Y t), where Y t) = e µbt) µ 2 t/2 311) Y t) = e µ µ and then, by the integration by parts formua, M B,µ) t) = eµ eµ Bt)Y t) = 1 Y s)dbs) ) Y s)1 µbs))dbs) 312) Let us now study the process {Bt)} t under Q B,µ) We consider an arbitrary function f C 2 ; ) Then, using Itô s formua and 312) we can write { } fbt))m B,µ) t) = f) + fbs))dm B,µ) s) + M B,µ) s)dfbs)) { e + fb), M B,µ) µ } t) = f) + fbs))y s)1 µbs)) + f Bs))M B,µ) s) dbs) + {µf Bs))M B,µ) s) + 12 } f Bs))M B,µ) s) + eµ Y s)1 µbs))f Bs)) ds 19

21 Taking the epectation and using 311) we get B,µ) ) f Bt) = f Bt) ) M B,µ) t) { = f) f Bs))M B,µ) s) + eµ = f) + LfBs))M B,µ) s) ds = } Y s)f Bs)) B,µ) f) + ds LfBs))ds, where L = 1 /2 d2 /d / d /d We deduce that under Q B,µ), the infinitesima generator of {Bt)} t is the infinitesima generator of a Besse process of dimension 3 4 Besse chains and other eampes 41 Besse chains We sha iustrate our resuts with two eampes In the discrete situation we consider the so-caed Besse random wak with inde α >, whie in the continuous parameter setting we adapt this eampe According to Theorem 21 one can obtain: Coroary 41 Let {X n } n the Besse random wak with inde α > 1 For any integer n and any event A n F n, 27) is verified with the positive no uniformy integrabe martingae M n := Γ2α + 1)X n! ΓX n + 2α + 1)2α + 1) Un+ 41) {Xn=} 2 Let Q be the probabiity measure on Ω, F ) given by 29) Then, under Q, {X n } n is a recurrent birth and death chain with transition probabiities given by: p = 1 and p := α + 1, q := + 2α, for 1 42) 2 + 2α + 1 Proof We compute R n, n First, by 23): γ = i=1 q i p i = i=1 Then, by 32), for n 1: R n = 2α + 1) n i i + 2α + 1 = Γ 2α + 2) Γ + 1) = 2α + 1)β2α + 1, + 1) Γ + 2α + 2) β2α + 1, + 1) = 2α + 1) β2α + 1, + n + 1) = 2α + 1)β2α, n + 1), where we used the foowing equaity k β, y + k) = β 1, y) see for instance, 3, p 95) Hence, R n = 2α + 1)Γ2α)n! Γn + 2α + 1), for n 1, and R = S = 1 + R 1 = 2α + 1 2α 43) 2

22 The epression 41) is then a consequence of 43) and 28) 21), p k = p k R k+1/r k, and using again 43) we obtain 42) Furthermore, according to Remark 41 In 11 one considers the penaization of the transient continuous Besse process {R t } t with dimension d > 2 or of inde n = d /2 1 > ) by its sojourn time in an interva Set, for r >, U r) t =,rr s )ds t ) One studies the imit im t A s One proves in 11 that : {U r) >t} {U r) >t} the positive martingae M r) s = A s M r) s is given by M r) t =: Q r) A s), A s F s s fied) := h r) R t ) ep z 1U r) t /2r 2 ), where h r) z) := Γn + 1) 2r /z 1 z) n J n z 1z/r) z r + 2r2 /z 1 z 2 ) n J n z 1 ) z r, and z 1 is the first positive zero of the Besse function J n 1 for r, under Q r), the process {R t } t has the infinitesima generator L := 1 /2 d2 /d n)/ d /d This process is recurrent and behaves, in a certain sense, as r, as a Besse process of dimension 4 d or of inde n) We return to the case of Besse random wak with inde α > 1 and transition probabiities given by 212) and we iustrate now a simiar phenomenon as for the continuous Besse process Let us remak that the Besse random wak coud aso be characterized by the foowing transition probabiities: X n+1 = 1 X n = ) = α + 1 /2 + α + 1 /2 X n+1 = + 1 X n = ) = 1 2 ), 1 + α + 1 /2 + α + 1/2 Ceary, if α, 1) \ { 1 /2} or of dimension δ 2, 4) \ {3}), with As, we can write QX n+1 = 1 X n = ) = 1 2 H) = QX n+1 = 1 X n = ) = α α α/2 α 2 α + 1 /2) + α + 1 /2) 1 α α ) ) ), for, n 1 integers + H) 2, + o 1 /) In other words, if α 1/2, the behaviour of the chain {X n } n under Q, is that of a Besse random wak with inde α = α or with dimension δ = 4 δ, 2) \ {1}), at east for arge states If α = 1/2, the chain {X n } n under Q is a standard symmetric random wak 21

23 Let us turn now to the continuous parameter eampe which is buit using the previous eampe Let us ca it Besse jump process with inde α > It is a pure jump process {Xt)} t with birth and death rates: b = + 2α + 1, d =, for integer We can state a consequence of Theorem 22: Coroary 42 1 For any t and any event A t F t, 219) is verified with the positive no uniformy integrabe martingae Mt) := 2α 2α + 1 e2αv t) Γ2α + 1) Xt)! Γ2α + Xt) + 1) 44) 2 Let Q be the probabiity measure on Ω, F ) given by 221) Then, under Q, {X t } t is a recurrent birth and death chain with transition probabiities given by: b := + 1,, d := + 2α, 1 45) The proof of this coroary is based on the foowing fact : the underying jump chain associated to {Xt)} t is a Besse random wak with inde α The remaining detais are eft to the reader 42 Other eampes Let us state other natura questions: assuming the hypothesis of Theorem 22, the process {Xt)} t, under Q, is it recurrent positive? Does there eist an invariant probabiity distribution? The answer at these questions is no Here are two simpe eampes Set, for, b := a, d := a 2, where < a < 1 Therefore, the associated birth and death process is transient since γ n = n =1 a = a nn+1)/2, S = n a nn+1) /2 < Again, the process is recurrent under Q, by Theorem 22 Using 222) we get We compute b = b +1 = a +1, d = d 1 = a 2 1 j j 1 =1 b 1 d = j 1 ) 2 Rj j 1 S a +1 = 1 S 2 =1 j 1 R 2 j a j2 +3j)/2 = 1 S 2 U j, j 1 22

24 where U j := 1 a j2 +3j)/2 j a +1) /2 2 Since U j+1 U j 2j 4)/2 aj+1)j+2) = a a jj+1) a+2j+3) /2 a+2j+1) /2 ) 2 ) = a 3j a+2j+3) /2 2 a+2j+1) /2 and, by dominated convergence theorem, im j a+2j+3) /2 a+2j+1) /2 = 1, then im j U j+1/u j = By D Aembert criterion, the series 1 /S 2 ) j 1 U j converges and therefore the necessary condition for the eistence of an invariant distribution see, for instance 6, p 78 or 1, p 412) is satisfied Hence, in this case there eists an invariant distribution for the process {Xt)} t under Q Set, for, b := a, d := a +1, where < a < 1 Again, the associated birth and death process is transient since Using 222) we obtain γ n = n =1 a = a n, S = n γ n = 1 1 a < b = b +1 = b a +1 a = a +1, d = d 1 = d a 1 a = a On the one hand, by Theorem 22 one knows that the process is recurrent under Q On the other hand: j b 1 = 1 = d j 1 j 1 =1 Hence in this case there is not invariant distribution for {Xt)} t under Q Let us note that in previous eampes there is no eposion A natura question is then the foowing : if the process epodes is it possibe to prove the eistence of an invariant distribution? Again the answer is negative as we can see beow Let us reca, according to 9 p 9, that {Xt)} t can epode if sup b + d = Set, for, b = 1 + ) a and d = a, with a > 1 Then, obviousy sup b + d =, and γ n = n =1 a 1 + ) a, S = ) a < As previousy, we have: j j 1 =1 b 1 d = j 1 ) 2 Rj j b 1 S d =1 = 1 S 2 j 1 j ) a 2 23

25 Since j ) a j j 1 + y) a = 1 a 1)j + 1) a 1, the series j 1 j =1 b 1/ d behaves as the series j 1 j+1)2 2a, that is convergent if a > 3 /2 and divergent if a 3 /2 Finay, the eposion is not a criterion for the eistence of an invariant distribution Acknowedgement: We thank Bernard Roynette for many vauabe discussions on the subject and especiay about Proposition 25 and Remark 41 References 1 Debs, P 27) Pénaisation de a marche aéatoire standard par une fonction du maimum uniatère, du temps oca en zéro et de a ongueur des ecursions Preprint IECN 2 Feer, W 197) An introduction to probabiity theory and its appications vo 1, 3rd edition, Wiey, New York 3 Gradstein, IS, Ryzhik, IM 198) Tabe of integras, Series and Products, Academic Press New York 4 Karatzas, I, Shreve, SE 1991) Brownian motion and stochastic cacuus, 2nd edition, Springer New York 5 Kebaner, FC 25) Introduction to Stochastic Cacuus with Appications 2nd edition, Imperia Coege Press, London 6 Lawer, G 26) Introduction to Stochastic Processes 2nd edition, Chapman & Ha/CRC, Boca Raton 7 Le Ga, J-F 1986) Une approche éémentaire des théorémes de décomposition de Wiiams Séminaire de Probabiités, XX, 1984/85, , Lecture Notes in Math 124, Springer, Berin 8 Mishchenko, AS 26) A discrete Besse process and its properties Theory Probab App 5, Norris, J 1998) Markov chains 2nd printing, Cambridge University Press, Cambridge 1 Resnick, SI 25) Adventures in Stochastic Processes, 4th printing, Birkhäuser, Boston, Roynette, B, Yor, M 27) Brownian penaizations: metatheorem and correct resuts Book in progress 12 Roynette, B, Vaois, P, Yor, M 26) Pénaisations et queques etensions du théorème de Pitman, reatives au mouvement brownien et à son maimum uniatère, Séminaire de Probabiités XXXVIII : In memoriam Pau-André Meyer, , Lecture Notes in Math 1874, Springer, Berin 24