Properties of an infinite dimensional EDS system : the Muller s ratchet

Transcription

1 Properties of an infinite dimensional EDS system : the Muller s ratchet LATP June 5, 2011

2 A ratchet source : wikipedia

3 Plan 1 Introduction : The model of Haigh 2 3

4 Hypothesis (Biological) : The population has a fixed sized N, is haploid and asexual. Only the deleterious mutations happen.

5 Hypothesis (Biological) : The population has a fixed sized N, is haploid and asexual. Only the deleterious mutations happen. Purpose : to show that without recombination, deleterious mutations accumulate.

6 Hypothesis (Biological) : The population has a fixed sized N, is haploid and asexual. Only the deleterious mutations happen. Purpose : to show that without recombination, deleterious mutations accumulate.

7 What is the Muller s ratchet? We call the best class the group of individuals with the best fitness ( let say 0 here).

8 What is the Muller s ratchet? We call the best class the group of individuals with the best fitness ( let say 0 here). Note that deleterious mutations can t be lost.

9 What is the Muller s ratchet? We call the best class the group of individuals with the best fitness ( let say 0 here). Note that deleterious mutations can t be lost. If at any given time, the best class is empty, it shall remain empty forever. That is to say, the minimal number of deleterious in this population has increased. It can t go back ( like a ratchet).

10 Haigh s model : This model is in discrete time. 1 α 0 and λ 0 are parameters. We note Y k the number of deleterious mutations of the individual k.at each generation :

11 Haigh s model : This model is in discrete time. 1 α 0 and λ 0 are parameters. We note Y k the number of deleterious mutations of the individual k.at each generation : Each indivual choses a parent from the previous generation with probability to chose the individual k (1 α) Y k N l=0 (1 α)y l

12 Haigh s model : This model is in discrete time. 1 α 0 and λ 0 are parameters. We note Y k the number of deleterious mutations of the individual k.at each generation : Each indivual choses a parent from the previous generation with probability to chose the individual k (1 α) Y k N l=0 (1 α)y l Each individual has a number of mutations equals to the number of his parent + P(λ).

13 In this model, at each generation the ratchet happens with a probability ( λe λ) N. ( everyone mutates)

14 In this model, at each generation the ratchet happens with a probability ( λe λ) N. ( everyone mutates) So the ratchet will happens infinitely many times a.s.

15 In this model, at each generation the ratchet happens with a probability ( λe λ) N. ( everyone mutates) So the ratchet will happens infinitely many times a.s. This means that the fitness of the population.

16 Now the model studied here is the following one, with X k (t) the proportion of individual with k deleterious mutations at time t. k 0 [ ] dx k = α( lx l k)x k + λ(x k 1 X k ) dt + l=0 l N X k (0) = Xk 0 Xk X l N db k,l

17 Now the model studied here is the following one, with X k (t) the proportion of individual with k deleterious mutations at time t. k 0 [ ] dx k = α( lx l k)x k + λ(x k 1 X k ) dt + l=0 l N X k (0) = Xk 0 where B k,l are independent Brownian motions, except for B k,l = B l,k, Xk X l N db k,l

18 Now the model studied here is the following one, with X k (t) the proportion of individual with k deleterious mutations at time t. k 0 [ ] dx k = α( lx l k)x k + λ(x k 1 X k ) dt + l=0 l N X k (0) = Xk 0 where B k,l are independent Brownian motions, except for B k,l = B l,k, and k X k(0) = 1, X k (0) 0 k 0 Xk X l N db k,l

19 Note that these equations are equivalent to Xk (1 X k ) dx k = [α(m 1 k)x k + λ(x k 1 X k )] dt + db k, N X k (0) = Xk 0 with M 1 (t) = k kx k(t) the median number of mutations in the population at time t, and db k standard brownian motions, with k l dx k, dx l (t) = X k X l dt

20 Why this model? Because it is some kind of limit for the Haigh s one, with an infinite population which stay finite at the same time (N). If you look at the equation, you can see : dx k = [ α(m 1 k)x }{{} k +λ(x k 1 X k ) Is the term for selection. ] dt + X k (1 X k ) N db k,

21 Why this model? Because it is some kind of limit for the Haigh s one, with an infinite population which stay finite at the same time (N). If you look at the equation, you can see : dx k = [ α(m 1 k)x k + λ(x k 1 X k ) }{{} Is the term for mutations. ] dt + X k (1 X k ) N db k,

22 Why this model? Because it is some kind of limit for the Haigh s one, with an infinite population which stay finite at the same time (N). If you look at the equation, you can see : Xk (1 X k ) dx k = [α(m 1 k)x k + λ(x k 1 X k )] dt + db k, } N {{} Is the term for resampling.

23 Difficulties of this model : There is no known explicit solution.

24 Difficulties of this model : There is no known explicit solution. Each equations use all the X k both in M 1 and in their stochastic term, they can t be separated.

25 Difficulties of this model : There is no known explicit solution. Each equations use all the X k both in M 1 and in their stochastic term, they can t be separated. The diffusion coefficient isn t lipschiztian around 0 and 1.

26 We can calculate the equation of M 1 : dm 1 (t) = (λ αm 2 (t))dt + where M 2 is the variance. M2 (t) N db t,

27 We can calculate the equation of M 1 : dm 1 (t) = (λ αm 2 (t))dt + where M 2 is the variance. M2 (t) N db t, But if we calculate the equation of M 2, M 3 and M 4 appears, and so on. The equation of M k use up to M 2k. This system can t be solved. ( M k is the k-th centered moment).

28 We can calculate the equation of M 1 : dm 1 (t) = (λ αm 2 (t))dt + where M 2 is the variance. M2 (t) N db t, But if we calculate the equation of M 2, M 3 and M 4 appears, and so on. The equation of M k use up to M 2k. This system can t be solved. ( M k is the k-th centered moment). Moreover, we have dx k, dm 1 = X k M 1 dt, they aren t independant.

29 Now the good news :

30 Now the good news : F.Yu, A. Etheridge and C.Cuthbertson have shown that our problem is well posed.

31 Now the good news : F.Yu, A. Etheridge and C.Cuthbertson have shown that our problem is well posed. From this we deduce that our problem has the Markov property.

32 Now the good news : F.Yu, A. Etheridge and C.Cuthbertson have shown that our problem is well posed. From this we deduce that our problem has the Markov property. And.. That s all.

33 What we want to show : Theorem Let T 0 = {inf t 0, X 0 (t) = 0} Then for any initial condition (X k (0)) k N, P(T 0 < + ) = 1. In other words, the ratchet will click a.s.

34 What has been already proved?

35 What has been already proved? A.Etheridge, A. Wakolbinger, P.Pfaffelhuber have shown that in the determinist case (N = + ), the system can be explicitely solved using cumulants on the M k. They obtain that the system will converge at exponential rate to the Poisson distribution with parameter λ α

36 What has been already proved? A.Etheridge, A. Wakolbinger, P.Pfaffelhuber have shown that in the determinist case (N = + ), the system can be explicitely solved using cumulants on the M k. They obtain that the system will converge at exponential rate to the Poisson distribution with parameter λ α But then it never clicks. The ratchet only happens due to the randomness. Moreover the cumulant system in the stochastic case has no known solution.

37 Plan Introduction : The model of Haigh 1 Introduction : The model of Haigh 2 3

38 The proof will use quite a few intermediate results. We will first present a comparison theorem we will use a lot : Our processes are defined on a probability space (Ω, F, P), equipped with a filtration (F t, t 0) which is such that for each k, l 0 {B k,l (t), t 0} is a F t Brownian motion. We denote by P the corresponding σ-algebra of predictable subsets of R + Ω.

39 Lemma Let B t be a standard F t Brownian motion, T a stopping time, σ be a 1/2 Hölder function, b 1 : R R a Lipschitz function and b 2 : Ω R + R R is P B(R) measurable function. Consider the two SDEs { dy1 (t) = b 1 (Y 1 (t))dt + σ(y 1 (t))db t, (3.1) Y 1 (0) = y 1 ; { dy2 (t) = b 2 (t, Y 2 (t))dt + σ(y 2 (t))db t, Y 2 (0) = y 2. (3.2)

40 Lemma Let Y 1 (resp Y 2 ) be a solution of (3.1) ( resp (3.2)). If y 1 y 2 (resp y 2 y 1 ) and outside a measurable subset of Ω of probability zero, t [0, T ], x R, b 1 (x) b 2 (t, x) (resp b 1 (x) b 2 (t, x)), then a. s. t [0, T ], Y 1 (t) Y 2 (t) (resp Y 1 (t) Y 2 (t)).

41 And, last but not least, a remark : Remark let A, B Ω. Then P(A B) P(A) + P(B) 1.

42 Plan Introduction : The model of Haigh 1 Introduction : The model of Haigh 2 3

43 Now we will compare the equation of M 1 to a more friendly one. M2 (t) (λ αm 2 (t))dt + N db t = (λ 1 2 αm M2 (t) 2(t))dt + N db t 1 2 αm 2(t)dt (λ 1 2 αx 0M1 2 M2 (t) )dt + N db t 1 2 αm 2(t)dt

44 We will show that M 1 can t grow too fast : Lemma c > 0, t > 0, P( sup 0 r t 2 M 1 (r +t) M 1 (t) λt 2 +c) 1 exp( 2αNc) > 0 : We use Z s,t = t+s t M 2 (r) N db r α s+t t We note that, at set t, exp(2αnz u,t) is both a local martingale and a surmartingale. We also have sup M 1 (s + t) M 1 (t) sup Z s,t + λt, 0 s t 0 s t using the comparison theorem and the previous remark. M 2 (r)dr,.

45 And c > 0 P( sup 0 u t Z u,t c) P( sup 0 u t exp(2αnz u,t) exp(2αnc)) exp( 2αNc) < 1 Where we have taken advantage of that exp(2αnz u,t) is a local martingale and Doob inequality, hence P(sup u 0 Z u,t c) exp( 2αNc) < 1 Then P( sup 0 r t M 1 (r + t) M 1 (t) λt + c) 1 exp( 2αNc) > 0

46 Plan Introduction : The model of Haigh 1 Introduction : The model of Haigh 2 3

47

48 What is Ω 1? For this we use a simpler model. Let X be the solution of the following system : { X (0) =δ X (t) =dt + 2 X (t)db 0

49 If X 0 (0) δ, M 1 (0) M, as long as X 0 M 1 < 2ε we can compare X 0 and X. Lemma Let T min = inf{t > 0, X 0 (t)m 1 (t) 2ε or X 0 (t) δ + µ}. Then t [0, T min ], we have X 0 (t) X (A(t)), where A(t) = 1 1 X 0 (s) ds and σ(t) = inf{u > 0, A(u) t}. 4 N t 0 t Note that A(t) t because 4 1 X 5N 4N ( thanks to the choices of µ and δ), and T was chosen such as A(T ) ε. 12λ

50 : We note X 0 (t) = X 0 (σ(t)) ( resp M 1 (t) = M 1 (σ(t))) d X 0 (t) = (α M 1 (t) λ) X 4N 0 (t) 1 X 0 (t) dt + 2 X 0 (t)dw t Since M 1 (t) X 0 (t) 2ε and 1 δ 1, 2 (α M 1 (t) λ) X 4N 0 (t) 1 X 0 (t) 1 Then, using the comparaison theorem, we obtain the conclusion.

51 Lemma Let X(t) be the solution of the previous system. and T 0 = inf{t > 0, X (t) = 0}. Then T > 0, p 2 < 1 δ > 0 such as P(T 0 T ) > p 2

52 : Let We have Then T > 0, P( 0 Introduction : The model of Haigh Y (t) = δexp ( t + 2B 0 (t)) D(t) = t 0 Y (s)ds σ(t) = inf {s > 0, D(s) > t} Y (t)dt T ) = P( X t = Y σ(t) D( ) < 0 exp ( t + 2B 0 (t)) dt T δ ) But T > 0, lim δ 0 T δ = a. s.

53 Corollary Let X(t) be the solution of (3.1), T 0 = inf{t > 0, X (t) = 0}, T µ = inf{t > 0, X (t) = δ + µ}. Then T > 0, p 2 < 1, µ > 0, δ > 0 such as P(T 0 T T µ) > p 2 : We use the same proof as before, noticing that P(T 0 T T µ) P({ {sup t 0 0 δ 0 1 exp( t + 2B 0 (t))dt T δ } exp( t + 2B 0 (t)) δ + µ }) δ

54 Now we chose a value M > 0 for M 1 (0), and we set where A is defined below, T = εn 3λ, M max = M + λa(t ) + ε, p 2 = exp( αn ε 3 ), µ = ε 6M max ε Now δ is given in terms T, M max, p 2 and µ by the previous corollary, and we let. δ = δ 1 10 ε M

55 Now we need to control T m in. We need to prove is X 0 (t)m 1 (t) 2ε 0 t A(T ) A(T µ) with probability p 3 > 1 p 2. Lemma p 3 > 1 p 2 such as P(sup 0 t A(T ) A(T µ) X 0 (t)m 1 (t) 2ε) p 3

56 : P( sup 0 t A(T ) A(T µ ) M 1 (t) M 1 (0) + λa(t ) + ε 6 ) P( sup 0 t A(T ) 1 exp( αn ε 6 ) M 1 (t) M 1 (0) + λa(t ) + ε 6 )

57 Then sup X 0 (t)m 1 (t) (δ + µ)(m 1 (0) + λa(t ) + ε 0 t A(T ) A(T µ ) 6 ) δm 1 (0) + µm 1 (0) + λa(t ) + ε 6 ε + ε 6 + ε 12 + ε 6 2ε

58 So, to sum up, with probability p fin = p 2 + p 3 1, we have : T min A(T µ) A(T ), T µ T, hence on the interval [ 0, A(T ) A(T µ) ] = [0, A(T )], we have both X 0 (t) X (A(t)) and X (A(t)) reaches 0.Hence the result.

59

60 Now for the set Ω 2 = {(X, M 1 ), X δ and XM 1 δm( ε)} We will show that X 0(0), M 1(0) has a better probability to reach 0 before time T. Let C = M 1 1 (because of the definition of X, M). Then we M have X 0 ε. C The probability to reach zero for X (t) is decreasing in δ, we increase this probability by starting from X (0) = X 0(0) δ. C Moreover, we start with X 0(0)M 1(0) ε. The only thing which is worse than the original case is M 1(0) which is greater than M 1 (0), hence a greater M max.

61 But this only appears in one place : when we defined µ. Note that if we define M 1,max = M 1(0) + λt + ε 6, the maximum reached by M 1, we have : M 1,max CM max P(M 1,max sup M 1(t)) 1 exp( αn ε 0 t T 6 ) By definition of µ, if we define µ with M 1,max we have µ µ C.

62 But if we look at the demonstration, we have, since T CT T and δ +µ = 1 + µ 1 + µ δ δ δ δ δ δ P(T 0 T T µ ( ) P { exp( t + 2B 1 (t))dt T 0 δ } ) {sup exp( t + 2B 1 (t)) δ + µ } t 0 δ ( P { exp( t + 2B 1 (t))dt T 0 δ } {sup exp( t + 2B 1 (t)) δ + µ ) } t 0 δ

63 We sum up the result in the following proposition, where ε = XM ( it s a different ε, smaller than the previous one, but we keep the notation). Proposition Let (X k (t)) k N be the solution of the initial model, and M 1 its mean as defined in the beginning. Then p p fin > 0 such as if t > 0 such as X 0 (t) δ and X 0 (t)m 1 (t) ε, P(T 0 < t + T ) p > 0

64 Plan Introduction : The model of Haigh 1 Introduction : The model of Haigh 2 3

65 First an inequality Lemma Let µ be a probability on N, x k = µ(k) k 0, M 1 = k N kx k, and M 2 = k N (k M 1) 2 x k Then M 2 1 x 0 M 2 x 0 M 2 1.

66 : By Jensen we have

67 : By Jensen we have ( k 1 X k 1 X 0 k ) 2 k 1 X k 1 X 0 k 2 with equality if and only if there exists only one k 1 such as X k > 0. Then :

68 : By Jensen we have ( k 1 X k 1 X 0 k ) 2 k 1 X k 1 X 0 k 2 with equality if and only if there exists only one k 1 such as X k > 0. Then : M 2 1 (1 X 0 ) k 1 X k k 2, hence

69 : By Jensen we have ( k 1 X k 1 X 0 k ) 2 k 1 X k 1 X 0 k 2 with equality if and only if there exists only one k 1 such as X k > 0. Then : M 2 1 (1 X 0 ) k 1 X k k 2, hence X 0 M 2 1 (1 X 0 ) k 1 X kk 2 (1 X 0 )M 2 1

70 : By Jensen we have ( k 1 X k 1 X 0 k ) 2 k 1 X k 1 X 0 k 2 with equality if and only if there exists only one k 1 such as X k > 0. Then : M 2 1 (1 X 0 ) k 1 X k k 2, hence X 0 M1 2 (1 X 0 ) k 1 X kk 2 (1 X 0 )M1 2 X 0 M1 2 (1 X 0 ) ( k 1 X ) kk 2 M1 2

71 Now we can obtain some kind of recurrence for M 1. Lemma Let Sλ T := inf{t T, X 0(t)M 1 (t) 2 2 λ+1 }. Then for any α T > 0, we have Sλ T < +

72 :

73 : On the interval [0, S λ ],we have α 2 M 2 α 2 X 0M 2 1 (λ + 1),

74 : On the interval [0, S λ ],we have α 2 M 2 α 2 X 0M 2 1 (λ + 1), and then the process M 1 is bounded from above by the process Y, solution of the SDE dy t = dt α 2 M M2 (t) 2(t)dt + N db t, Y 0 = M 1 (0).

75 Since M 1 cannot become negative, it now suffice to show that t M2 (r) Z t := N db r α t M 2 (r)dr is bounded from above a.s. If we define C(t) = 1 M N 0 2(s)ds, we have Z t = W (C(t)) N αc(t) where W is a standard 2 Brownian motion. Now, if C( ) = then lim t Z t =, hence Z t is bounded from above. Or else C( ) <, and we have sup t>0 Z t = sup 0<s<C( ) W (s) N αs < a.s. 2 t

76

77 As a consequence of the previous results : Lemma Let (X k (t)) k N be the solution of the Muller s ratchet. Then we have { lim t + X 0(t) = 0} {T 0 < + }

78 : If lim t + X 0 (t) = 0 then T 0 > 0 such as t > T 0, we have X 0 (t) δ ε2 α. 4(λ+1) Then we have a T 1 > T 0 such as X 0 (t)m 2 1 (t) 2 λ + 1 α. Let us suppose that X 0 M 1 > ε. Then M 1 = X 0M 1 > 4 λ + 1 X 0 αε X 0 M1 2 = X 0 M 1 M 1 4 λ + 1 α which is absurd. Then we have X 0 M 1 ε.

79 Then we have P(T 0 < T 1 + T ) > p. This situation presents itself infinitely many time, as long as the ratchet doesn t click, we obtain the conclusion, and since the system (X k ) k N has the markov property.

80 Now the recurrence on M 1 Let S β t = inf {t > t 0, M 1 (t ) β}. Then we will prove the following lemma : Lemma t > 0, we have P(T 0 S β t < ) = 1

81

82 : First, we note δ inf = δ ε2 α. 4(λ+1) Now we introduce the process Yt t, defined t 0, t t which is the solution of the following system : dy t t = Y t t = δ inf Y t t (1 Y t N t ) db 0 (3.3) We note that Yt t is an instance of the processes studied in the addendum, because here we have 2 < f (t) = 0 < 2. Then using Ru t = inf {t t, Yt t = u},

83 We have E(R0 t R1 t ) < + and P(R1 t < R0 t ) > 0. From this we deduce that K > 0, p > 0 such as P(R1 t K R0 t ) p > 0). In particular P(R1 t K) p > 0. We use L = K T. ( T from the step 3). Now we construct the following sequence : U 0 = U n = the first time where x 0 M1 2 2 λ+1, and n 1 U α n = the first time after U n 1 + L where x 0 M1 2 2 λ+1. This time exists a.s. and is < + α thanks to prop 1.2.

84 Now, at U 0 : Either X 0 (U 0 ) δ i nf, then using the same reasoning as 4.1, we have P(T 0 U 0 + T ) = p fin > 0. Or X 0 (U 0 ) > δ i nf. And in that case there is 2 case : Either inf U0 t U 0 +K M 1 (t) β. In that case we have (αm 1 λ)x 0 0, and then we can use the comparaison theorem and we get X 0 (t) Y t U 0. Then P(T 1 U 0 + K) p > 0. But if X 0 (t) = 1, then M 1 (t) = 0. Hence P(S β U 0 + K) p > 0. In the other case inf U0 t U 0 +K M 1 (t) < β, hence S β U 0 + K.

85 To conclude, if we use q = p p fin, we have P(T 0 S β t = + ) P(T 0 S β t U 0 + L) 1 q. Now since the processes X 0,M 1 are markovian, by iterating we obtain P(T 0 S β t = + ) = 0

86

87 Plan Introduction : The model of Haigh 1 Introduction : The model of Haigh 2 3

88 Now, starting from ((X k ) k N, M 1 ) withm 1 β we want to study a path that will reach zero. To simplify notations we reset the time. One of the main problem here is that the quadratic variation of X 0 is X 0(1 X 0 ), which is a difficulty around 1 and 0. We N need to study three separate cases : X 0 [δ 1 ; 1] X 0 δ or X 0 M 1 > ε X 0 δandx 0 M 1 ε

89 The following lemma will show that if X 0 starts too close from 1, it will quickly goes under δ 1 : Lemma Let t 1 = 8 λ 2 and δ 1 = max{ 9 3λ + 5α, 10 5(λ + α), 1 2 λ }. Then, t > 0, if X 0 (t) > δ 1, then q > 0 such as P(inf{s > t, X 0 (s) δ 1 } < t + t 1) 1 exp( N) > 0

90 : During the interval [t, T δ1 (t)], we have, since X 1 1 X 0, 2λ X 1 (t) 5(λ + α). 2λ αm 1 X 1 + λx 0 (λ + α)x 1 λx 0 (λ + α) 5(λ + α) λx 0 2λ 5 since X 0 (s) > 0, 9 as well. λ 2

91 Hence X 1 Y 1 when s [t, T δ1 (t)], where Y 1 is the solution of the following system Y 1 (t) = X 1 (t) dy 1 (s) = λ 2 ds + Y1 (1 Y 1 ) (3.4) db 1 N

92 Noticing that Y 1 (1 Y 1 ) 1 4, we have ( with C = 2 λ ). ( P { t+t 1 t ( = P ) Y1 (1 Y 1 ) db 1 < C} {T δ1 > t + t N 1} ) Y1 (1 Y 1 ) db 1 > C N t+t 1 t

93 ( ( t+t 1 Y1 (1 Y 1 ) = P exp p db 1 t N > exp (pc )) p2 t+t 1 Y t 1 (1 Y 1 ) ds 2N ( ( t+t 1 Y1 (1 Y 1 ) P exp p db 1 t N )) > exp (pc p2 8N t 1 ) exp ( pc + p2 8N t 1 ) Y t 1 (1 Y 1 ) ds 2N p2 t+t 1 t+t 1 t ) p 2 Y 1 (1 Y 1 ) ds 2N

94 Where p = 4CN t minimizes the quantity. ( ) t+t 1 Y1 (1 Y 1 ) P db 1 C 1 exp ( N) > 0 N t Now, since t+t 1 t λ 2 ds = 4 λ = 2C

95 We have t+t 1 Y1 (1 Y 1 ) { db 1 C} {T δ1 < t + t t N 1} Which implies that P (T δ1 < t + t 1) 1 exp( N)

96 Now we add a control on M 1 Lemma Let δ 1 = max{ 9, 3λ+5α, 1 2 }, 10 5(λ+α) λ t 1 = 8, ( ) λ 2 ε 0 = 1 ln 2 and β = β + λt αn 1 exp( N) 1 + ε 0. Then, t > 0, if X 0 (t) > δ 1 and M 1 (t) < β, then P ({T δ1 t + t 1} {M 1 (T δ1 ) β }) = p init > 0

97 : Here, Then, P(T δ1 t 1) q P(M 1 (T δ1 ) β ) 1 exp( snε 0 ) P ({T δ1 t 1} {M 1 (T δ1 ) β }) = p init 1 exp( N) exp( αnε 0 ) 1 exp( N) 2 > 0

98

99 Now the second part : X 0 δ 1 but either X 0 > δ or X 0 M 1 > ε. First some inequalities : Lemma Let {V t, t 0} be a standard Brownian motion, and c > 0 a constant. The for any t > 0, δ > 0, ( ) P inf {cs + B s} δ 0 s t 1 ( ) 2 δ π t + c t.

100 : We have, with Z denoting a N(0, 1) random variable, ( ) ( ) P inf s} δ P inf s δ ct 0 s t 0 s t ( ) P sup 0 s t V s δ + ct 2P(V t δ + ct) 1 P( Z δ t + c t), from which the result clearly follows.

101 From this we deduce : Lemma Let {V t, t 0} be a standard Brownian motion, and c > 0 a constant. The for any t > 0, δ > 0, µ > 0, ( ) P inf {cs + V s} δ, sup {cs + V s } µ 0 s t 0 s t ( ) [ 2 1 δ π t + c t 2 exp 1 ( ) ] 2 µ t 2 c t.

102 : ( ) P sup (cs + V s ) µ 0 s t ( ) P sup 0 s t V s µ ct = 1 2P(V t µ ct). Now, Z denoting a N(0, 1) random variable, for all p > 0, ( P(B t µ ct) = P Z µ c ) t t ( ( [ ] ) µ t = P exp(pz p 2 /2) exp p c t p2 2 ( [ ] ) µ t exp p c t + p2 2

103 Choosing p = µ/ t c t, we conclude from the above computations that ( ) [ P sup (cs + V s ) µ 1 2 exp 1 ( ) ] 2 µ t 0 s t 2 c t.

104 We set : ε = log(4) αn, so that e Nα ε = 1 4 µ = 1 δ 1 2 We will use another time change, and we use A and σ again for the time change (but they are not the same) and aim at proving that X 0 will go down to δ in a finite number of steps, while staying below X + µ (so that 1 X 0 (t) a := 1 (X + µ)), and while M 1 does not go too far on the right, all that with positive probability.

105 We have : X0 (t)[1 X 0 (t)] dx 0 (t) = (αm 1 (t) λ)x 0 (t)dt + db 0. N Let t X 0 (s)[1 X 0 (s)] A t := ds, 0 N σ t := inf{s > 0, A s > t}, X 0 (t) := X 0 (σ t ), M 1 (t) := M 1 (σ t ), and

106 We deduce σ t = t 0 X 0 (t) = X + N N X 0 (s)(1 X 0 (s)) ds, t 0 (α M 1 (s) λ) ds + B t, 1 X 0 (s) where B t is a new standard Brownian motion. At the k th step of our iterative procedure, we let X 0 start from X k 1 j=1 δ j, and we stop the process X 0 at the first time that it reaches the level X k j=1 δ j.

107 We will choose δ k and t k such as for each 1 k K (K to be defined below), ( ) P inf {A k s + B s } δ k, sup {A k s + B s } µ > 2 0 s t k 0 s t k 3, where, with T k := σ tk, A k = Nα a ( β + k ε + λ ) k T j, j=1 so that we have from Lemma 4 and our choice of ε that P( sup 0 s T k M 1 (s) A k M1 (0) A k 1 ) 1/3.

108 We show that we can choose the two sequences δ k and t k for k 1 in such a way that not only the two previous inequalities hold, but also that there exists K < such that X K δ k δ. k=1 Since during the k th step we are considering the event that 1 X 0 (t) a, and also X 0 (t) X k j=1 δ j, we have that T k N a(x k j=1 δ j) t k,

109 Then we choose Introduction : The model of Haigh A k := N α a [ First we want to insure β + k ε + N λ a k j=1 δ k tk + A k tk 0.4, which we achieve by requesting both that t j X j i=1 δ i δ k = 0.2 t k (3.5) ]. and ( ) A k tk 0.2 t k. (3.6) A k

110 On the other hand, we shall also request that for each k 1, δ k X k 1 δ j It follows from A k N αβ a C N = N αβ a A k C N + 25 N 2 λα a 2 C N + D N k 1 δ k 1 k 1 2 (X δ j ). that with and D N = N α a ( ( ε + j=1 λ ), αβ ) sup δ j k + kε Nα 1 j k a

111 Since we have j 0 δ j 0.2 t j (0.2)2 1 a 25 Nαβ Finally this leads to choosing, with κ 1 to be chosen below 25 ( ) κ δ k = inf (C N + D N k), 1 k 1 2 (x δ j ) t k = 25δ 2 k. A j j=1

112 Then from a certain rank on we have Lemma K > 0, k > K, δ k = 1 k 1 2 (X δ j ). j=1

113 : Since k N 1 = +, K 25(C N +D N > 0 such as k) K 1 k=2 > 1. Then 2 K K 25(C N +D N such as k) ( 1 inf, 1(X ) K 1 25(C N +D N K) 2 j=1 δ j) = 1(X K 1 2 j=1 δ j). Then by recurrence, if we have the previous equality at rank k, for the rank k + 1 we have 1 25(C N +D N (k+1)) δ k 1 25(C N +D N (k+1)) 1 25(C N +D N k) 1 2 since k 2

114 That is to say Introduction : The model of Haigh Hence δ k+1 = inf ( 1 25(C N + D N (k + 1)) 1 2 δ k 1 k 1 4 (X δ j ) j=1 1 k 2 (X δ j ) 1, 1(X ) k 25(C N +D N (k+1)) 2 j=1 δ j) j=1 = 1 2 (X k j=1 δ j)

115 Then for each k > K, X 0 progresses by a step equal to half the remaining distance to zero. Consequently c > 0 X k = X k j=1 δ j c2 k. We are looking for the smallest integer k such that c2 k δ, which implies that [ ] log(c) log(δ ) k 1. log(2) Since moreover A k (25δ k ) 1, there exists a constant c such that δ A k c δ log( 1 δ ). Hence there exists a δ δ (which depends only upon C N, D N, c which are constants) such that at the end of the k th step, both X 0 δ and X 0 M 1 ε.

116 . We just need to check that the probability of the previous pathes is = p trans > 0. Given the choice that we have made for ε, it suffices to make sure that ( ) 2 δk + A k tk < 1/3, k 1, π tk which is a consequence of 3 1 π/2 > 0.4, and [ 2 exp 1 ( ) ] µ 2 tk A k tk < 1/3, k 1. 2

117 This is equivalent to ( µ tk A k tk ) 2 > 2 log 6, which follows from κ 2 log 6+4C N µ 2 log 6. We therefore 10 choose κ = 1 2 log CN µ 2 log 6. 10

118 Then P( the k-th step happens ) = 1, hence, ( using the markovian property, we have p trans inf 1 k, k 12) but k < for any initial condition X, and increasing in X, so the worst k, noted k max is reached for X = δ 1, and then p trans inf k ( ) kmax 1 > 0 12 Note that the time ( in the initial scale of time) in which we reach δ is bounded by t 2 = 100Nk max. We ve finally reached third situation, which leads to the conclusion.

119 X 0 δ and X 0 M 1 ε Here we can immediatly apply the step 3 of this document, and we have a probability p fin to reach 0 before T elapsed. So to sump up, using again the markovian properties of the system, Lemma t > 0, if M 1 (t) β, then P(T 0 < t + t 1 + t 2 + T ) p fin p trans p ini > 0

120 Plan Introduction : The model of Haigh 1 Introduction : The model of Haigh 2 3

121 Here we will prove the following stronger theorem Theorem For any choice of initial condition, let (X k (t)) k Z+ the solution of (8). Then E(T 0 ) <. We first note that the reasoning of section 5 can be done with any initial value ρ for M 1, instead of β. That is to say, with S t ρ = inf {s > t, M 1 (s) ρ} (and S ρ = S 0 ρ ), Lemma t ρ 1, tρ 2, tρ 3 <, and pρ ini, pρ trans, p ρ fin > 0 such that P(T 0 < S t ρ + t ρ 1 + tρ 2 + tρ 3 ) pρ ini pρ transp ρ fin

122 Now let us choose ρ = ε 2λ. We have : δ α Lemma Let K = L + t 3 ( L to be defined below). Then p > 0, such that for any initial condition, P(T 0 S ρ K) p : We are going to argue like in the recurrence for M 1. We introduce the process Y s, defined s 0 which is the solution of the following system : dy s = αε 2 ds + Ys (1 Y s ) db 0 (s) N (3.7) Y 0 = 0

123 We define for any 0 u 1 R u = inf {s 0, Y s = u}. Since αε > 0 we deduce that L > 0, p > 0 such as 2 P(R 1 L) p > 0. We use K = L + t 3. ( t 3 from Proposition 3.2 ). Now there are several possibilities : Either inf 0 s L M 1 (s) ρ, then S ρ < L < K. Or else inf 0 s L M 1 (s) ρ. Then either inf 0 s L X 0 (s)m 1 (s) ε, then t < L such as X 0 (t)m 1 (t) ε (which implies X 0 (t) δ, becausem 1 (t) ρ ε ). In that δ case we can use Proposition 3.2, and we have P(T 0 K) = p fin > 0, which implies P(T 0 S ρ K) = p fin > 0,

124 Or else we have both inf 0 s L M 1 (s) ρ and inf 0 s L X 0 (s)m 1 (s) ε. In that last sub case we have (since X 0 ε M 1, and αm 1 λ λ > 0) inf (αm 1(s) λ)x 0 (s) 0 s L inf ε(α 0 s L αε 2, λ M 1 (s) ) and consequently we can use the comparison theorem (Lemma 2), which implies that s [0, L], X 0 (s) Y s. Then P(T 1 L) p > 0. But when X 0 hits 1, M 1 hits 0. Hence P(S ρ L) p > 0.

125 We can now conclude. We deduce from the previous results and the strong Markov property that K, p > 0 such as for all n 0, P(T 0 > nk) (1 p) n. Consequently E(T 0 ) = n=0 (n+1)k nk KP(T > nk) n=0 P(T > t)dt = K p

126 THANK YOU FOR YOUR ATTENTION!