Branching Brownian motion seen from the tip

Transcription

1 Branching Brownian motion seen from the tip J. Berestycki 1 1 Laboratoire de Probabilité et Modèles Aléatoires, UPMC, Paris 09/02/2011 Joint work with Elie Aidekon, Eric Brunet and Zhan Shi J Berestycki (Paris) ANR MANEGE 09/02/ / 35

2 Outline 1 BBM and FKPP J Berestycki (Paris) ANR MANEGE 09/02/ / 35

3 Outline 1 BBM and FKPP 2 BBM seen from the tip J Berestycki (Paris) ANR MANEGE 09/02/ / 35

4 Outline 1 BBM and FKPP 2 BBM seen from the tip J Berestycki (Paris) ANR MANEGE 09/02/ / 35

5 BBM The model = Particles X 1 (t),..., X N(t) (t) on R Start with one particle at 0. Movement = independent Brownian motions Branching = at rate 1 into two new particles (more general possible). J Berestycki (Paris) ANR MANEGE 09/02/ / 35

6 Rightmost particle Define M(t) = max i=1,...,n(t) X i (t). Theorem (Rightmost particle M(t)) c = 2. lim t M(t) t = c. c t M(t) a.s. Proof by martingale techniques J Berestycki (Paris) ANR MANEGE 09/02/ / 35

7 Rightmost particle Define M(t) = max i=1,...,n(t) X i (t). Define u(t, x) := P(M(t) < x). Theorem (Rightmost particle M(t)) c = 2. lim t M(t) t = c. c t M(t) a.s. Proof by martingale techniques J Berestycki (Paris) ANR MANEGE 09/02/ / 35

8 Rightmost particle Define M(t) = max i=1,...,n(t) X i (t). Theorem (Rightmost particle M(t)) c = 2. lim t M(t) t = c. c t M(t) a.s. Proof by martingale techniques Define u(t, x) := P(M(t) < x). The map u(t, x) = P(M(t) < x) solves t u = x u + u(u 1) ( avec 1 u(x, 0) = ) J Berestycki (Paris) ANR MANEGE 09/02/ / 35

9 Rightmost particle Define M(t) = max i=1,...,n(t) X i (t). Theorem (Rightmost particle M(t)) c = 2. lim t M(t) t = c. c t M(t) a.s. Proof by martingale techniques Define u(t, x) := P(M(t) < x). The map u(t, x) = P(M(t) < x) solves t u = x u + u(u 1) ( avec 1 u(x, 0) = ) Idea : Initial particle in 0. After dt - with proba (1 dt) no split but diffuse : u(t + dt, x) u(t, x ξ t ) - with proba dt branch u(t + dt, x) u(t, x) 2 J Berestycki (Paris) ANR MANEGE 09/02/ / 35

10 Bramson s result Define m t by u(t, m t ) = 1/2. i.e. m t is the median of M(t) (results still valid if we take the expectation). KPP 37 u(t, m t + x) w(x) unif. in x as t where m t = 2t + a(t) and a(t). (McKean shows that a(t) 2 3/2 log t) and w solution to 1 2 w + 2w + (w 2 w) = 0. Purely analytical method. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

11 Bramson s result Define m t by u(t, m t ) = 1/2. i.e. m t is the median of M(t) (results still valid if we take the expectation). KPP 37 u(t, m t + x) w(x) unif. in x as t where m t = 2t + a(t) and a(t). (McKean shows that a(t) 2 3/2 log t) and w solution to 1 2 w + 2w + (w 2 w) = 0. Purely analytical method. Bramson 83 m t = 2t 3 2 3/2 log t + c + o t (1) J Berestycki (Paris) ANR MANEGE 09/02/ / 35

12 Bramson s result Define m t by u(t, m t ) = 1/2. i.e. m t is the median of M(t) (results still valid if we take the expectation). KPP 37 u(t, m t + x) w(x) unif. in x as t where m t = 2t + a(t) and a(t). (McKean shows that a(t) 2 3/2 log t) and w solution to 1 2 w + 2w + (w 2 w) = 0. Purely analytical method. Bramson 83 m t = 2t 3 log t + c + o 2 3/2 t (1) ( ) ( ) 1 1 If change initial condition from, to Then only the constant c changes J Berestycki (Paris) ANR MANEGE 09/02/ / 35

13 Bramson s result Define m t by u(t, m t ) = 1/2. i.e. m t is the median of M(t) (results still valid if we take the expectation). KPP 37 u(t, m t + x) w(x) unif. in x as t where m t = 2t + a(t) and a(t). (McKean shows that a(t) 2 3/2 log t) and w solution to 1 2 w + 2w + (w 2 w) = 0. Purely analytical method. Bramson 83 m t = 2t 3 log t + c + o 2 3/2 t (1) ( ) ( ) 1 1 If change initial condition from, to Then only the constant c changes Hence C B R such that with P(W x) = w(x). M t ( 2t log t + C B) dist W 0 J Berestycki (Paris) ANR MANEGE 09/02/ /

14 Lalley-Sellke KPP, Bramson P(M t m t < x) w(x) so M t m t converges in law to a variable with dist. w. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

15 Lalley-Sellke KPP, Bramson P(M t m t < x) w(x) so M t m t converges in law to a variable with dist. w. Can t be ergodic i.e. 1 t 1 Ms<m t s+xds w(x) a.s.. 0 J Berestycki (Paris) ANR MANEGE 09/02/ / 35

16 Lalley-Sellke KPP, Bramson P(M t m t < x) w(x) so M t m t converges in law to a variable with dist. w. Can t be ergodic i.e. 1 t 1 Ms<m t s+xds w(x) a.s.. 0 Suppose it holds. Then t 1 t 0 1 M x s <m s+xds w(0) a.s.. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

17 Lalley-Sellke KPP, Bramson P(M t m t < x) w(x) so M t m t converges in law to a variable with dist. w. Can t be ergodic i.e. 1 t 1 Ms<m t s+xds w(x) a.s.. 0 Suppose it holds. Then t 1 t 0 1 M x s <m s+xds w(0) a.s.. Start two independent BBM, one at x and one at 0. Positive probability that they meet before any branching successfull coupling so w(0) = w(x). Contradiction. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

18 The derivative martingale The derivative martingale Z(t) = u N(t) ( 2t X u (t))e 2Xu(t) 2t (additive martingale W 2 (t) = u N(t) e 2X u(t) 2t 0) Z := lim Z(t) exists, is finite and positive with proba. 1. Not UI. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

19 The derivative martingale The derivative martingale Z(t) = u N(t) ( 2t X u (t))e 2Xu(t) 2t (additive martingale W 2 (t) = u N(t) e 2X u(t) 2t 0) Z := lim Z(t) exists, is finite and positive with proba. 1. Not UI. Theorem C > 0 s.t. x R lim lim P (M(t + s) m(t + s) x F s) = exp{ CZe 2x }, a.s. s t J Berestycki (Paris) ANR MANEGE 09/02/ / 35

20 The derivative martingale The derivative martingale Z(t) = u N(t) ( 2t X u (t))e 2Xu(t) 2t (additive martingale W 2 (t) = u N(t) e 2X u(t) 2t 0) Z := lim Z(t) exists, is finite and positive with proba. 1. Not UI. Theorem C > 0 s.t. x R lim lim P (M(t + s) m(t + s) x F s) = exp{ CZe 2x }, a.s. s t Thus the TW has representation [ ] w(x) = E exp{ CZe 2x } J Berestycki (Paris) ANR MANEGE 09/02/ / 35

21 The derivative martingale The derivative martingale Z(t) = u N(t) ( 2t X u (t))e 2Xu(t) 2t (additive martingale W 2 (t) = u N(t) e 2X u(t) 2t 0) Z := lim Z(t) exists, is finite and positive with proba. 1. Not UI. Theorem C > 0 s.t. x R lim lim P (M(t + s) m(t + s) x F s) = exp{ CZe 2x }, a.s. s t Thus the TW has representation 1 w(x) Cxe 2x. [ ] w(x) = E exp{ CZe 2x } J Berestycki (Paris) ANR MANEGE 09/02/ / 35

22 Structure of W Suggests that, once we "know" Z, P (M(t) m(t) x) exp{ e 2x+log CZ }, a.s. so that P( 2(M(t) m(t)) log(cz) x) exp( e x ) and hence M(t) m(t) 2 1/2 log(cz) dist 2 1/2 G where G is a Gumbel variable. W = dist (G + log(cz))/ 2 J Berestycki (Paris) ANR MANEGE 09/02/ / 35

23 Structure of W Suggests that, once we "know" Z, P (M(t) m(t) x) exp{ e 2x+log CZ }, a.s. so that P( 2(M(t) m(t)) log(cz) x) exp( e x ) and hence M(t) m(t) 2 1/2 log(cz) dist 2 1/2 G where G is a Gumbel variable. W = dist (G + log(cz))/ 2 Purple line is m(t) J Berestycki (Paris) ANR MANEGE 09/02/ / 35

24 Structure of W Suggests that, once we "know" Z, P (M(t) m(t) x) exp{ e 2x+log CZ }, a.s. so that P( 2(M(t) m(t)) log(cz) x) exp( e x ) and hence M(t) m(t) 2 1/2 log(cz) dist 2 1/2 G where G is a Gumbel variable. W = dist (G + log(cz))/ 2 Purple line is m(t) Because of early fluctuation a random delay is created (2 1/2 log(cz)) J Berestycki (Paris) ANR MANEGE 09/02/ / 35

25 Structure of W Suggests that, once we "know" Z, P (M(t) m(t) x) exp{ e 2x+log CZ }, a.s. so that P( 2(M(t) m(t)) log(cz) x) exp( e x ) and hence M(t) m(t) 2 1/2 log(cz) dist 2 1/2 G where G is a Gumbel variable. W = dist (G + log(cz))/ 2 Purple line is m(t) Because of early fluctuation a random delay is created (2 1/2 log(cz)) Fluctuations around m(t) + 2 1/2 log(cz) are Gumbel. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

26 Outline 1 BBM and FKPP 2 BBM seen from the tip J Berestycki (Paris) ANR MANEGE 09/02/ / 35

27 Conjecture Let X 1 (t) > X 2 (t) >... be the particles ordered by position at time t. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

28 Conjecture Let X 1 (t) > X 2 (t) >... be the particles ordered by position at time t. Lalley and Sellke conjecture that the point process of particles converges to an equilibrium state. X i (t) m(t) + c log(cz(t))/ 2 J Berestycki (Paris) ANR MANEGE 09/02/ / 35

29 Conjecture Let X 1 (t) > X 2 (t) >... be the particles ordered by position at time t. Lalley and Sellke conjecture that the point process of particles X i (t) m(t) + c log(cz(t))/ 2 converges to an equilibrium state. First we show a simple argument due to Brunet Derrida to show that the PP X i (t) m(t) converges. Uses Bramson and McKean representation. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

30 McKean Representation Recall : u(t, x) := P(M(t) < x) solves t u = x u + u(u 1) with ( ) 1 1 u(x, 0) =. More generally, let g : R [0, 1] then Theorem (McKean, 1975) 0 If u : R R + [0, 1] solves the FKPP equation 0 t u = x u + u(u 1) with initial condition u(0, x) = g(x), then u(t, x) = E[ u N(t) g(x u (t) + x)]. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

31 McKean Representation Recall : u(t, x) := P(M(t) < x) solves t u = x u + u(u 1) with ( ) 1 1 u(x, 0) =. More generally, let g : R [0, 1] then Theorem (McKean, 1975) 0 If u : R R + [0, 1] solves the FKPP equation 0 t u = x u + u(u 1) with initial condition u(0, x) = g(x), then u(t, x) = E[ u N(t) In general t u = x u + β(f (u) u). g(x u (t) + x)]. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

32 Brunet Derrida argument (seen from m t ) 1 0 exp( k) H φ (x, t) = E[ φ(x X u (t))], H φ solves KPP with H φ (x, 0) = φ(x). If φ H φ (x, t) = P(M(t) < x) Bramson : H φ (m(t) + z, t) w(z + δ(φ)). If φ H φ (x, t) = E[e kn(x,t) ] with N(x, t) = #u : X u (t) > x. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

33 Brunet Derrida argument (seen from m t ) 1 0 exp( k) H φ (x, t) = E[ φ(x X u (t))], H φ solves KPP with H φ (x, 0) = φ(x). If φ H φ (x, t) = P(M(t) < x) If φ H φ (x, t) = E[e kn(x,t) ] with N(x, t) = #u : X u (t) > x. Bramson : H φ (m(t) + z, t) w(z + δ(φ)). For any Borel set A R, the Laplace transform of #{u : X u (t) m(t) + A} converges. Theorem (Brunet Derrida 2010) The point process of the particles seen from m(t) converges in distribution as t. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

34 Not too hard to show : the limit point process (X i, i = 1, 2,...) has the superposition property, i.e., α, β s.t. e α + e β = 1 : X α + X β = d X. Clearly the PPP (e x ) has this property. Who else? J Berestycki (Paris) ANR MANEGE 09/02/ / 35

35 Not too hard to show : the limit point process (X i, i = 1, 2,...) has the superposition property, i.e., α, β s.t. e α + e β = 1 : X α + X β = d X. Clearly the PPP (e x ) has this property. Who else? each atom is replaced by an iid copy of a point process Maillard (2011) show those are the only superposable PP. What is the decoration of BBM? J Berestycki (Paris) ANR MANEGE 09/02/ / 35

36 Not too hard to show : the limit point process (X i, i = 1, 2,...) has the superposition property, i.e., α, β s.t. e α + e β = 1 : X α + X β = d X. Clearly the PPP (e x ) has this property. Who else? each atom is replaced by an iid copy of a point process Maillard (2011) show those are the only superposable PP. What is the decoration of BBM? Independently, Bovier, Arguin and Kestler 2010 obtain results that are similar to part of what follows. Seem to use techniques. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

37 Normalization We do the following change of coordinates : we suppose that the Brownian motions have diffusion σ 2 = 2 and drift ρ = 2. Instead of rightmost focus on leftmost. Tilts the cone in which the BBM lives. Left-most particle m(t) = 3 2 log t + C B. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

38 Normalization We do the following change of coordinates : we suppose that the Brownian motions have diffusion σ 2 = 2 and drift ρ = 2. Instead of rightmost focus on leftmost. Tilts the cone in which the BBM lives. Left-most particle m(t) = 3 2 log t + C B. In this framework Z(t) := u N(t) X u (t)e Xu(t) is the derivative martingale. Recall its limit exists and is positive a.s. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

39 BBM seen from the tip Y i (t) := X i (t) m t + log(cz), 1 i N(t). Theorem (Aidekon, B., Brunet, Shi 11) As t the point process (Y i (t), 1 i N(t)) converges in distribution to the point process L obtained as follows (i) Define P a Poisson point process on R +, with intensity measure 2 σ e 2x/σ dx where a is a constant. (ii) For each atom x of P, we attach a point process x + Q (x) where Q (x) are i.i.d. copies of a certain point process Q. (iii) L is then the superposition of all the point processes x + Q (x) : L := {x + y : x P {0}, y Q (x) } Bovier Arguin and Kistler (2010) obtain a very similar result (and much more). J Berestycki (Paris) ANR MANEGE 09/02/ / 35

40 Structure of Q τ X 1,t 2 τ X 1,t 1 t X 1,t N τ 1 t,x 1,t N τ 2 t,x 1,t J Berestycki (Paris) ANR MANEGE 09/02/ / 35

41 Structure of Q τ X 1,t 2 τ X 1,t 1 Y (s) := X 1,t(t s) X 1,t(t) is the red path. Q is the superposition of the BBM that branch on this path. t X 1,t N τ 1 t,x 1,t N τ 2 t,x 1,t J Berestycki (Paris) ANR MANEGE 09/02/ / 35

42 Structure of Q τ X 1,t 2 τ X 1,t 1 Y (s) := X 1,t(t s) X 1,t(t) is the red path. Q is the superposition of the BBM that branch on this path. t X 1,t N τ 1 t,x 1,t N τ 2 t,x 1,t N x (t) = BBM at time t started from one ptc at x. Nx (t) = BBM at time t conditioned to min Nx (t) > 0 started from one ptc at x. G t (x) := P{min N 0 (t) x}, so that G t (x + m t ) P(σW x) by Bramson. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

43 Law of Y. U (b) v := B v, if v [0, T b ], b R v Tb, if v T b. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

44 Law of Y. U (b) v := B v, if v [0, T b ], b R v Tb, if v T b. Take b random : P(b dx) = f (x)/c 1 where f (x) := E [e 2λ 0 ] G v (σu v (x) )dv (1) and c 1 = constant. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

45 Law of Y. U (b) v := B v, if v [0, T b ], b R v Tb, if v T b. Take b random : P(b dx) = f (x)/c 1 where f (x) := E [e 2λ 0 ] G v (σu v (x) )dv (1) and c 1 = constant. Conditionally on b, Y has a density /U (b) given by 1 f (b) e 2λ 0 G v (σu v (b) )dv (2) J Berestycki (Paris) ANR MANEGE 09/02/ / 35

46 Structure of Q Theorem (Aidekon, B., Brunet, Shi 11) Let (Y (t), t 0) be as above. Start independent BBMs N Y (t) (t) conditioned to finish to the right of X 1,t along the path Y at rate 2λ(1 P(min N Y (t) (t) 0))dt. Then t π N Y (t) (t) is distributed as Q. t X 1,t τ X 1,t 1 τ X 1,t 2 N τ 1 t,x 1,t N τ 2 t,x 1,t D(ζ, t) := τ X N (i) 1,t i t ζ t,x 1,t particles born off X 1,t less than ζ unit of time ago, then we have the following joint convergence in distribution lim lim {(X 1,t(t s) X 1,t (t), s 0), D(ζ, t)} = {(Y (s), s 0), Q}. ζ t J Berestycki (Paris) ANR MANEGE 09/02/ / 35

47 Structure of Q { ( I ζ (t) = E exp i 1 )} nj=1 t τ i <ζ α j #[N τ i t,x 1,t (X 1 (t) + A j )] J Berestycki (Paris) ANR MANEGE 09/02/ / 35

48 Structure of Q { ( I ζ (t) = E exp i 1 )} nj=1 t τ i <ζ α j #[N τ i t,x 1,t (X 1 (t) + A j )] Theorem (Aidekon, B., Brunet, Shi 11) α j 0 (for 1 j n), E {e n j=1 α jq(a j ) } = lim lim I 0 ζ(t) = ζ t 0 E(e 2λ 0 E(e 2λ 0 G v (σu(b) v )dv )db G v (σu v (b) )dv )db, where G v (x) := P{min N (v) x}, Gv (x) := 1 E [e n j=1 α j#[n (v) (x+a j )] 1{min N (v) x} ]. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

49 Define { I(t) := E F (X 1,t (s), s [0, t]) ( exp i f (t τ X 1,t i ) n j=1 )} α j #[N τ i t,x 1,t (X 1 (t) + A j )], τ X 1,t 2 τ X 1,t 1 t X 1,t N τ 1 t,x 1,t N τ 2 t,x 1,t J Berestycki (Paris) ANR MANEGE 09/02/ / 35

50 Define { I(t) := E F (X 1,t (s), s [0, t]) ( exp i f (t τ X 1,t i ) n j=1 )} α j #[N τ i t,x 1,t (X 1 (t) + A j )], τ X 1,t 2 r 0, x R define τ X 1,t 1 G r (x) := P{min N (r) x}, n ] G (f ) r (x) := E [e f (r) α j #[N (r) (x+a j )] j=1 1{min N (r) x}. Hence, when f 0 we have G (f ) r (x) = 1 G r (x)). t X 1,t N τ 1 t,x 1,t N τ 2 t,x 1,t J Berestycki (Paris) ANR MANEGE 09/02/ / 35

51 Define { I(t) := E F (X 1,t (s), s [0, t]) ( exp i f (t τ X 1,t i ) n j=1 )} α j #[N τ i t,x 1,t (X 1 (t) + A j )], τ X 1,t 2 r 0, x R define τ X 1,t 1 G r (x) := P{min N (r) x}, n ] G (f ) r (x) := E [e f (r) α j #[N (r) (x+a j )] j=1 1{min N (r) x}. Hence, when f 0 we have G (f ) r (x) = 1 G r (x)). t X 1,t N τ 1 t,x 1,t N τ 2 t,x 1,t Theorem (Aidekon, B., Brunet, Shi 11) We have I(t) = E [e σbt F (σb s, s [0, t]) e 2λ t where B is BM. 0 [1 G(f t s(σb ) t σb s)]ṣ ], J Berestycki (Paris) ANR MANEGE 09/02/ / 35

52 Theorem (Aidekon, B., Brunet, Shi 11) In particular, the path (s X 1,t (s), 0 s t) is a standard Brownian motion in a potential : [ ] E F (X 1,t (s), s [0, t]) = E [e σbt F (σb s, s [0, t]) e 2λ t ] G 0 t s(σb t σb s)ds. (3) J Berestycki (Paris) ANR MANEGE 09/02/ / 35

53 path of the leftmost particle Theorem (Aidekon, B., Brunet, Shi 11) ɛ > 0 x > 0 s.t. P(A t (x)) > 1 ɛ for all t large enough. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

54 path of the leftmost particle Theorem (Aidekon, B., Brunet, Shi 11) ɛ > 0 x > 0 s.t. P(A t (x)) > 1 ɛ for all t large enough. Again see also Arguin Bovier Kistler. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

55 Theorem (Aidekon, B., Brunet, Shi 11) As t the point process (Y i (t), 1 i N(t)) converges in distribution to the point process L obtained as follows (i) Define P a Poisson point process on R +, with intensity measure 2 σ e 2x/σ dx where a is a constant. (ii) For each atom x of P {0}, we attach a point process x + Q (x) where Q (x) are i.i.d. copies of a certain point process Q. (iii) L is then the superposition of all the point processes x + Q (x) : L := {x + y : x P {0}, y Q (x) } With extreme value theory. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

56 Idea : Fix K > 0 large and stop particles when they hit k for the first time. (no escape). H k = # particles stoped at k k Z = lim k 2 1/2 ke k H k, (4) exists a.s., is in (0, ). ξ k,u u J Berestycki (Paris) ANR MANEGE 09/02/ / 35

57 Idea : Fix K > 0 large and stop particles when they hit k for the first time. (no escape). H k = # particles stoped at k k Z = lim k 2 1/2 ke k H k, (4) ξ k,u u exists a.s., is in (0, ). Start iid BBM so that u H k, X u 1 (t) = d k + X 1 (t ξ k,u ), where ξ k,u = time when u reaches k. By Bramson k 1, u H k, X u 1 (t) m t law k + Wu, t. (see also Bovier Arguin and Kistler (2010) : extremal particles in BBM either branch at the very beginning or at the end) J Berestycki (Paris) ANR MANEGE 09/02/ / 35

58 Define Pk, := δ k+w (u)+log Z. u H k Recall that P = PPP with intensity ae x dx. Proposition Pk, P In the sense of convergence in distribution. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

59 Take (X i, i N) a sequence of i.i.d. r.v. such that P(X i x) Cxe x, as x. Call M n = max i=1,...,n X i the record. Define b n = log n + log log n. Then P(M n b n y) = (P(X i y + b n )) n = (1 (1 + o(1))c(y + b n )e (y+bn) ) n ( ) 1 exp nc(y + b n ) n log n e y exp( Ce y ) J Berestycki (Paris) ANR MANEGE 09/02/ / 35

60 Take (X i, i N) a sequence of i.i.d. r.v. such that P(X i x) Cxe x, as x. Call M n = max i=1,...,n X i the record. Define b n = log n + log log n. Then P(M n b n y) = (P(X i y + b n )) n = (1 (1 + o(1))c(y + b n )e (y+bn) ) n ( ) 1 exp nc(y + b n ) n log n e y exp( Ce y ) Therefore P(M n (b n + log C) y) exp( e y ). By classical results the point process n ζ n := δ Xi b n log C i=1 converges in distribution to a Poisson point process on R with intensity e x dx. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

61 Recall that P(W x) Cxe x so apply to u H k δ W (u)+(log Hk +log log H k )+log C which converges in dist. to a PPP on R with intensity e x dx. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

62 Recall that P(W x) Cxe x so apply to u H k δ W (u)+(log Hk +log log H k )+log C which converges in dist. to a PPP on R with intensity e x dx. Use k2 1/2 e k H k Z to obtain H k σk 1 e k Z and therefore and hence log H k + log log H k k + log Z + c P k, = u H k δ k+w (u)+log Z also converges (as k ) towards a PPP on R with intensity 2 σ e 2x/σ dx. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

63 many-to-one W t := u N (t) e X(u), t 0, is the additive martingale. Because critical not UI and 0. Q F t = W t P F t. Under Q spine = BM drift (0), branch at rate 2 into two particles. Q{Ξ t = u F t } = e X(u) W t. For each u N (t), G u = a r.v. in F t. E P [ u N (t) G u ] = E Q 1 W t u N (t) [ ] = E Q e X(Ξt) G Ξ. G u Suppose that we want to check if a path X u (s) with some property A P( u = t : (X u (s), s [0, t]) A) = P(e σbt ; (σb s, s [0, t]) A). J Berestycki (Paris) ANR MANEGE 09/02/ / 35

64 path of the leftmost particle Theorem (Aidekon, B., Brunet, Shi 11) ɛ > 0 x > 0 s.t. P(A t (x)) > 1 ɛ for all t large enough. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

65 A t (x) := E 1 (x) E 2 (x) E 3 (x) E 4 (x) where the events E i are defined by { X1,t E 1 (x) := 3 } 2 log t x, { E 2 (x) := E 3 (x) := E 4 (x) := min X 1,t(s) x, min X 1,t(s) 3 [0,t] [t/2,t] 2 log t x { s [x, t/2], X 1,t (s) s 1/3} { } s [t/2, t x], X 1,t (s) X 1,t [(t s) 1/3, (t s) 2/3 ]. } J Berestycki (Paris) ANR MANEGE 09/02/ / 35

66 the event E 3 Suppose we have E 1(z) and E 2(z) for z large enough. By the many-to-one lemma, we get P(E 3(x) c, E 1(z), E 2(z)) { e z t 3/2 P s [x, t/2] : B s s 1/3, min B s z, [0,t/2] min B s 3 t/2,t 2 log t z, Bt 3 }. 2 log t + z Applying the Markov property at time t/2, it yields that { P s [x, t/2] : B s s 1/3, min B s z, min B s 3 [0,t/2] t/2,t 2 log t z, Bt 3 } 2 log t + z = E [ 1 { s [x,t/2]:bs s 1/3 } 1 {min [0,t/2] B s z} { P Bt/2 min B s 3 2 log t z, B t/2 3 } ] 2 log t + z s [0,t/2] c2zt 3/2 E [1 { s [x,t/2]:bs s 1/3} 1 {min[0,t/2] Bs z}(b t/2 3 ] 2 log t + z) c2zt 3/2 E [ 1 { s [x,t/2]:bs s 1/3 } 1 {min [0,t/2] B s z}(b t/2 + z) ] where { the second inequality comes from } bound on P min s [0,t] B s x, B t x + y. We recognize the h-transform of the Bessel. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

67 We end up with P(E 3 (x) c, E 1 (z), E 2 (z)) e z c2zp z ( s [x, t/2] : R s s 1/3 ) e z c2zp z ( s x : R s s 1/3 ) which is less than ε for x large enough. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

68 proof Step 1 : The process V x (t) := u N(t) w( 2t X i (t) + x) is a F t -martingale J Berestycki (Paris) ANR MANEGE 09/02/ / 35

69 proof Step 1 : The process V x (t) := u N(t) w( 2t X i (t) + x) is a F t -martingale Step 2 : u N(t) e 2X u(t) 2t is a positive martingale, converges to a finite value, so min u ( 2t X u (t)) = + a.s. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

70 proof Step 1 : The process V x (t) := u N(t) w( 2t X i (t) + x) is a F t -martingale Step 2 : u N(t) e 2X u(t) 2t is a positive martingale, converges to a finite value, so min u ( 2t Xu x (t)) = + a.s. Step 3 : 1 w(y) = Cye 2y. log V x (t) = u N(t) u N(t) log w( 2t X i (t) + x) C( 2t X i (t) + x)e 2t+ 2X i (t) 2x CZ(t)e 2x CY (t)xe 2x with Y (t) = u N(t) e 2X i (t) 2t. Clearly lim Y /Z = 0. lim Y (t) = Y 0 exists a.s. so Z(t) a.s. on the event Y > 0, this V x = 0. But since E(V x ) = w(x) 1 when x, P(Y > 0) = 0. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

71 proof 2 Step 4 : Thus lim Z(t) = e 2x C 1 log V x. We conclude that lim Z(t) exists and > 0 a.s. [ { }] w(x) = E[V x ( )] = E exp CZe 2x. J Berestycki (Paris) ANR MANEGE 09/02/ / 35

72 proof 2 Step 4 : Thus lim Z(t) = e 2x C 1 log V x. We conclude that lim Z(t) exists and > 0 a.s. [ { }] w(x) = E[V x ( )] = E exp CZe 2x. Step 5 : P(M(t +s) m(t +s)+x F s ) = u N(t) u(t, x +m(t +s) X u(s)). J Berestycki (Paris) ANR MANEGE 09/02/ / 35

73 proof 2 Step 4 : Thus lim Z(t) = e 2x C 1 log V x. We conclude that lim Z(t) exists and > 0 a.s. [ { }] w(x) = E[V x ( )] = E exp CZe 2x. Step 5 : P(M(t +s) m(t +s)+x F s ) = u N(t) u(t, x +m(t +s) X u(s)).recall that u(t, x + m(t)) = P(M(t) m(t) + x) w(x) and that lim t (m(t + s) m(t) 2s) = 0 J Berestycki (Paris) ANR MANEGE 09/02/ / 35

74 proof 2 Step 4 : Thus lim Z(t) = e 2x C 1 log V x. We conclude that lim Z(t) exists and > 0 a.s. [ { }] w(x) = E[V x ( )] = E exp CZe 2x. Step 5 : P(M(t +s) m(t +s)+x F s ) = u N(t) u(t, x +m(t +s) X u(s)).recall that u(t, x + m(t)) = P(M(t) m(t) + x) w(x) and that lim t (m(t + s) m(t) 2s) = 0 so that lim t P(M(t + s) m(t + s) + x F s ) = u N(t) = u N(t) := V x (s) w(x + m(t + s) X u (s) m(t) w(x + 2s X u (s)) J Berestycki (Paris) ANR MANEGE 09/02/ / 35