Critical branching Brownian motion with absorption. by Jason Schweinsberg University of California at San Diego

Transcription

1 Critical branching Brownian motion with absorption by Jason Schweinsberg University of California at San Diego (joint work with Julien Berestycki and Nathanaël Berestycki) Outline 1. Definition and motivation 2. Tools 3. Main results 4. Future work and open problems

2 Branching Brownian motion with absorption Begin with one particle at x > 0. Each particle independently moves according to standard onedimensional Brownian motion with drift µ. Each particle splits into two at rate 1. Particles are killed if they reach the origin. t = 0 t = T

3 Motivation #1: population models with selection Can be used to model populations subject to natural selection, as proposed by Brunet, Derrida, Mueller, and Munier (2006). particles positions of particles branching events absorption at 0 individuals in the population fitness of individuals births deaths of unfit individuals movement of particles changes in fitness over generations

4 Motivation #2: connections with PDEs Consider BBM without drift (µ = 0) and no killing at 0. Let M(t) be the position of the right-most particle at time t. Let u(t,x) = P(M(t) x). Then (McKean, 1975): u t = 1 2 u 2 x 2 +u2 u, with u(0,x) = 1 {x 0}. This is the FKPP equation, introduced by Fisher (1937) and by Kolmogorov, Petrovskii, and Piscounov (1937). FKPP equation has traveling wave solutions u(t,x) = w(x vt), 1 2 w +vw +w 2 w = 0. McKean (1975), Bramson (1978, 1983), and Neveu (1988) studied this equation with w( ) = 0, w( ) = 1 using BBM. Harris, Harris, and Kyprianou (2006) studied this equation on R + with w(0+) = 0, w( ) = 1 using BBM with absorption.

5 Motivation #3: correlated Gaussian random variables Consider BBM without drift (µ = 0) and no killing at 0. Let X 1 (t),x 2 (t),...,x N(t) (t) be the particle positions at time t. Let T(i,j) be time when trajectories of particles i and j separate. Conditional on the T(i,j), the random variables X 1 (t),...,x N(t) (t) are centered Gaussian with Var(X i ) = t, Cov(X i,x j ) = T(i,j). Resembles correlation structure in some spin glass type models. Ding and Zeitouni (2013) obtained results about the behavior of points near the maximum of the two-dimensional discrete Gaussian free field by using results about branching random walks.

6 Condition for extinction Theorem (Kesten, 1978): Suppose that, at time zero, there is one particle at x > 0. Branching Brownian motion with absorption dies out almost surely if µ 2. If µ < 2, the process survives forever with positive probability. Hereafter, we always assume µ = 2 (critical case). Questions As x, What is the typical extinction time? How does the number of particles evolve up to extinction? What is the configuration of particles before the extinction? For fixed x > 0, what is the probability of survival until time t?

7 First moment calculations Consider first a single Brownian particle started at x, with drift of 2 and absorption at 0. The density of the position of the particle at time t is p t (x,y) = 1 2πt (e (x y)2 /2t e (x+y)2 /2t ) e 2(x y) t. Returning to BBM with absorption, let N(t) be the number of particles at time t. Let X 1 (t) X 2 (t) X N(t) (t) be the positions of particles at time t. Theorem (Many-to-One Lemma): If f : (0, ) R, then E [ N(t) i=1 f(x i (t)) ] = e t 0 f(y)p t(x,y)dy. Take f = 1 A to get expected number of particles in a set A.

8 Second moment calculations Theorem (Sawyer, 1976; Kesten, 1978): If f : (0, ) R, then E [( N(t) f(x i (t)) ) 2 ] i=1 2e 2t s t = e t 0 f(y)2 p t (x,y)dy + 0 f(y 1)f(y 2 )p s (x,z)p t s (z,y 1 )p t s (z,y 2 )dy 1 dy 2 dzds. Moments are dominated by rare events in which one particle drifts unusually far to the right and has many surviving offspring. We use truncation: kill particles that get too far to the right. Moments can be calculated the same way, after adjusting p t (x,y). Must show that the probability that a particle gets killed is low.

9 Branching Brownian motion in a strip Consider Brownian motion killed at 0 and L. If there is initially one particle at x, the density of the position at time t is: qt L (x,y) = 2 ( ) ( ) e π2 n 2 t/2l 2 nπx nπy sin sin. L L L n=1 Add branching and drift of 2, density becomes: p L t (x,y) = q L t (x,y) e 2(x y) t e t, meaning that if B (0,L), the expected number of particles in B at time t is B pl t (x,y)dy. For t L 2, p L t (x,y) 2 L e π2 t/2l 2 e 2x sin ( πx L ) e 2y sin ( πy L ).

10 Because N(t) decreases over time, need to move position of the right boundary closer to 0. As in Kesten (1978), consider L(t) = ( 1 t ) 1/3 τx 3 x, τ = 2 2 3π 2. Use methods of Harris and Roberts (2012) for calculations. Main results (Rough Version): Right-most particle at time t is close to L(t). Extinction occurs near time τx 3. Density of particles near y at time t is proportional to Number of particles satisfies d dt e 2y sin ( πy L(t) N(t) π2 2L(t) 2N(t), N(0) C x 3e 2x. ).

11 Number of particles Let N(t) be the number of particles at time t. Let ζ = inf{t : N(t) = 0} be the extinction time. Theorem (BBS, 2012): Let ε > 0. Then there exists β > 0 such that for sufficiently large x, P ( τx 3 βx 2 < ζ < τx 3 +βx 2) > 1 ε. Theorem (BBS, 2012): Let ε > 0 and δ > 0. Then there are positive constants B, C 1, C 2 such that for sufficiently large x, ( C1 P 2L(t) C ) x 3e N(t) 2 2L(t) x 3e > 1 ε for all t [Bx 2,(1 δ)τx 3 ].

12 Configuration of particles Let t = sx 3, where 0 < s < τ. Let X 1 (t) X 2 (t) X N(t) (t) denote the positions of the particles at time t. Idea: density of particles at y (0,L(t)) is proportional to Theorem (BBS, 2012): Let e 2y sin ( πy L(t) χ(t) = 1 N(t) N(t) i=1 ). δ Xi (t). Then χ(t) µ as x, where µ is the probability measure on (0, ) whose density is g(y) = 2ye 2y. Theorem (BBS, 2012): Define the probability measure η(t) = 1 Y(t) N(t) i=1 N(t) e 2Xi (t) δxi (t)/l(t), Y(t) = 2Xi (t) e. i=1 Then η(t) ν as x where ν is the probability measure on (0,1) whose density is h(y) = π 2 sin(πy).

13 Location of right-most particle (without absorption) Consider BBM without drift (µ = 0) and no killing at 0. Bramson (1983) showed that the median position m(t) of the right-most particle is: m(t) = 2t logt+o(1). Theorem (Lalley and Sellke, 1987): Let M(t) be the position of the right-most particle at time t. Then lim P(M(t) m(t)+x) = E[exp( CZe 2x)], t where C > 0 is a constant and N(t) Z = lim ( 2t X 2Xi (t) 2t i (t))e. t i=1 A full description of the configuration near the right edge was discovered independently by Arguin, Bovier, and Kistler (2011), and by Aïdékon, Berestycki, Brunet, and Shi (2011).

14 Location of right-most particle (with absorption) Theorem (BBS, 2012): Recall that M(t) is the position of the right-most particle at time t. Let ε > 0. Suppose t = sx 3, where 0 < s < τ. Then there exist positive constants d 1 and d 2 such that for sufficiently large x, P ( L(t) 3 2 logx d 1 M(t) L(t) 3 2 logx+d 2 ) > 1 ε. Proof strategy: Consider the process at time t γx 2, where γ > 0 is small. By the previous result, we have a rather precise description of the particle configuration at this time. Turn off absorption at 0 between times t γx 2 and time t. Results of Bramson (1983) allow us to estimate the probability that a particle at z at time t γx 2 has a descendant to the right of y at time t. We can show that for γ sufficiently small, the trajectory of such a particle is unlikely to cross 0.

15 Long-run survival probability Recall that ζ = inf{t : N(t) = 0} is the extinction time. Theorem (Kesten, 1978): There exists K > 0 such that for each x > 0, we have for sufficiently large t: xe 2x K(logt) 2 (3π 2 t) 1/3 P(ζ > t) (1+x)e 2x+K(logt) 2 (3π 2 t) 1/3. Theorem (BBS, 2012): There exist C 1 > 0 and C 2 > 0 such that for each x > 0, we have for sufficiently large t: C 1 xe 2xe (3π 2 t) 1/3 P(ζ > t) C 2 xe 2xe (3π 2 t) 1/3. Note: result obtained nonrigorously by Derrida and Simon (2007).

16 Proof Technique Let L(s) = c(t s) 1/3, where c = τ 1/3. time 0 s L(s) time t Step 1: Show there exists C > 0 such that if there is a particle at L(s) at time s [0,t 1], then the probability that a descendant of this particle survives until time t is at least C. Step 2: Use moment calculations to bound the probability that a particle hits the curve s L(s). Idea of counting particles that hit this curve in Kesten (1978).

17 A branching process approximation Consider the curve s L(s), where L(s) = c(t s) 1/3. Start with one generation 0 particle at L(s) at time s. Fix a large constant y > 0. Particles that return to the curve after hitting L(s) y are called generation 1 particles. If a generation 1 particle is at L(u) at time u, then descendants of that particle that return to the curve after hitting L(u) y are called generation 2 particles, and so on. Leads to comparison with a branching process. If supercritical, there is a positive probability that many particles will hit the curve near time t, ensuring one will survive until time t.

18 Mean of the offspring distribution Theorem (Neveu, 1987): Suppose there is a particle at L(s). Let K(y) be the number of descendants that reach L(s) y, if particles are killed upon reaching L(s) y. There exists a random variable W such that Also, E[W] =. lim y ye 2y K(y) = W a.s. A particle at L(s) y has, on average, Cye 2y descendants that hit the curve after time s. When a particle hits the curve at L(s), it will have CK(y)ye 2y CW descendants that return to the curve after hitting L(s) y. Because E[W] =, the process is supercritical.

19 Future work and open problems Conjecture: Suppose at time zero there is one particle at x > 0. There exists a positive constant C such that as t, P(ζ > t) Cxe 2xe (3π 2 t) 1/3. One can also consider the question of proving a Yaglom-type limit theorem: conditional on survival until a large time t, what does the configuration of particles look like at time t? Theorem (Kesten, 1978): Suppose, at time zero, there is one particle at x > 0. There are positive constants K 1 and K 2 such that and lim t P(N(t) > ek 1t 2/9 (logt) 2/3 ζ > t) = 0 lim t P(M(t) > K 2t 2/9 (logt) 2/3 ζ > t) = 0.

20 A Yaglom-type result Conjecture: Suppose at time zero there is one particle at x > 0. Let Y have an Exponential(1) distribution. Then, conditional on survival until time t, the following hold as t : Proof Idea: t 2/9 logn(t) (3π 2 ) 1/3 Y 1/3 t 2/9 M(t) cy 1/3 If the previous conjecture holds, then P(ζ > t+yt 2/3 ζ > t) = P(ζ > t+yt2/3 ) P(ζ > t) e (3π2 ) 1/3 y/3. If M(t) c(yt 2/3 ) 1/3, then the process should survive until t+yt 2/3. Because we know what the configuration of particles should look like, we can find the distribution of N(t) as well.