Branching Brownian motion seen from the tip

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 09/02/2011 1 / 35

Outline 1 BBM and FKPP J Berestycki (Paris) ANR MANEGE 09/02/2011 2 / 35

Outline 1 BBM and FKPP 2 BBM seen from the tip J Berestycki (Paris) ANR MANEGE 09/02/2011 2 / 35

Outline 1 BBM and FKPP 2 BBM seen from the tip J Berestycki (Paris) ANR MANEGE 09/02/2011 3 / 35

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 09/02/2011 4 / 35

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 09/02/2011 5 / 35

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 09/02/2011 5 / 35

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 = 1 2 2 x u + u(u 1) ( avec 1 u(x, 0) = 1 0 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 09/02/2011 5 / 35

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 09/02/2011 6 / 35

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 0 0 0 0 J Berestycki (Paris) ANR MANEGE 09/02/2011 6 / 35

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 3 2 2 log t + C B) dist W 0 J Berestycki (Paris) ANR MANEGE 09/02/2011 6 / 35 0 0 0

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 09/02/2011 7 / 35

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 09/02/2011 7 / 35

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 09/02/2011 7 / 35

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 09/02/2011 8 / 35

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 09/02/2011 9 / 35

Outline 1 BBM and FKPP 2 BBM seen from the tip J Berestycki (Paris) ANR MANEGE 09/02/2011 10 / 35

Conjecture Let X 1 (t) > X 2 (t) >... be the particles ordered by position at time t. J Berestycki (Paris) ANR MANEGE 09/02/2011 11 / 35

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 09/02/2011 11 / 35

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 09/02/2011 11 / 35

McKean Representation Recall : u(t, x) := P(M(t) < x) solves t u = 1 2 2 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 = 1 2 2 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 09/02/2011 12 / 35

McKean Representation Recall : u(t, x) := P(M(t) < x) solves t u = 1 2 2 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 = 1 2 2 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 = 1 2 2 x u + β(f (u) u). g(x u (t) + x)]. J Berestycki (Paris) ANR MANEGE 09/02/2011 12 / 35

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 09/02/2011 13 / 35

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 09/02/2011 13 / 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? J Berestycki (Paris) ANR MANEGE 09/02/2011 14 / 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 09/02/2011 14 / 35

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 09/02/2011 15 / 35

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 09/02/2011 16 / 35

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 09/02/2011 17 / 35

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 09/02/2011 17 / 35

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 09/02/2011 17 / 35

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 09/02/2011 18 / 35

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 09/02/2011 18 / 35

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 09/02/2011 18 / 35

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 09/02/2011 19 / 35

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 09/02/2011 20 / 35

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 09/02/2011 20 / 35

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 09/02/2011 21 / 35

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 09/02/2011 21 / 35

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 09/02/2011 22 / 35

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 09/02/2011 23 / 35

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 09/02/2011 23 / 35

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 09/02/2011 24 / 35

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 09/02/2011 25 / 35

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 09/02/2011 25 / 35

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 09/02/2011 26 / 35

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 09/02/2011 27 / 35

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 09/02/2011 28 / 35

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 09/02/2011 28 / 35

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 09/02/2011 29 / 35

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 09/02/2011 30 / 35

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 09/02/2011 31 / 35

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 09/02/2011 32 / 35

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 09/02/2011 33 / 35

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 09/02/2011 34 / 35

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 09/02/2011 34 / 35

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 09/02/2011 34 / 35

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 09/02/2011 35 / 35

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 09/02/2011 35 / 35

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 09/02/2011 35 / 35