Competing Brownian Particles

Transcription

1 Competing Brownian Particles Andrey Sarantsev University of California, Santa Barbara February 9, 2018 Andrey Sarantsev California State University, Los Angeles 1 / 16

2 Brownian Motion A Brownian motion is a stochastic process W = (W (t), t 0) which starts from W (0) = 0, has continuous trajectories, and independent Gaussian increments: W (t) W (s) is independent of W (u), u s, and is distributed according to the normal distribution N (0, t s) Recall: a random variable Z N (0, σ 2 ) satisfies P(a Z b) = 1 2πσ b a e z2 /2σ 2 dz. Andrey Sarantsev California State University, Los Angeles 2 / 16

3 Atlas Model Stochastic processes X 1 (t),..., X N (t) t X i (t) = X i (0) + 1 ( X i (s) = min X j(s) ) ds + W i (t), 0 j=1,...,n for i = 1,..., N, t 0 W 1 (t),..., W N (t) are independent Brownian motions The bottom particle has upward drift 1, all other particles move as driftless Brownian motions Called the Atlas model because the bottom particle, called the Atlas particle, supports the system, similarly to the Greek Atlas hero supporting the sky on his shoulders. Andrey Sarantsev California State University, Los Angeles 3 / 16

4 Motivation Originally introduced in (Banner, Fernholz, Karatzas, 2005) Financial modeling: Stocks with smaller capitalizations have larger growth rates and larger volatilities Subsequently studied in (Pal, Pitman, 2008; Ichiba et al 2011) Similar models have other applications in physics Andrey Sarantsev California State University, Los Angeles 4 / 16

5 Gap Process Z(t) = (Z 1 (t),..., Z N 1 (t)), with Z k (t) = X (k+1) (t) X (k) (t), k = 1,..., N 1, t 0. Takes values in [0, ) N 1. Z k is the distance between the kth and k + 1st ranked particles. A stationary gap distribution is a probability measure π on [0, ) N 1 such that if Z(0) π, then Z(t) π for all t 0. Andrey Sarantsev California State University, Los Angeles 5 / 16

6 Stationary Gap Distributions Theorem (Pal, Pitman, 2008) There exists a unique stationary gap distribution, given by product measure N 1 ( π = Exp 2 N k ). N k=1 Recall: Z Exp(α) is the exponential random variable with values on [0, ): P(a Z b) = b a αe αz dz. Andrey Sarantsev California State University, Los Angeles 6 / 16

7 Infinite Atlas Model Infinite system of Brownian particles (X i (t)) i 1. Ranked from bottom to top: X (1) (t) X (2) (t)... Bottom particle moves as a Brownian motion with drift 1. All other particles moves as independent Brownian motions. t X i (t) = X i (0) + 1 ( X i (s) = min X j(s) ) ds + W i (t), 0 j=1,...,n W 1, W 2,... i.i.d. Brownian motions Andrey Sarantsev California State University, Los Angeles 7 / 16

8 Infinite Atlas Model Called the infinite Atlas model because the bottom particle, called the Atlas particle, supports the system, similarly to the Greek Atlas hero supporting the sky on his shoulders. Introduced in (Pal, Pitman, 2008). Studied in Shkolnikov, 2011; Ichiba, Karatzas, Shkolnikov, 2013; S, 2016; Dembo, Tsai, 2015; S, Tsai, 2017; Tsai, 2017; Cabezas, Dembo, S, Sidoravicius, 2017; Dembo, Jara, Olla, Andrey Sarantsev California State University, Los Angeles 8 / 16

9 Stationary Gap Distribution: Heuristics Find stationary gap distribution π for infinite Atlas model. Stationary gap distribution for the Atlas model with N particles is N 1 ( π N = Exp 2 N k ). N k=1 For each k, letting N, we get: 2 N k N 2. It is reasonable to conjecture that π = k=1 Exp(2) Andrey Sarantsev California State University, Los Angeles 9 / 16

10 Conjecture A stationary gap distribution for the infinite Atlas model is π = Exp(2). k=1 This was formally proved in (Pal, Pitman, 2008) Conjecture: For the infinite Atlas model, this is the unique stationary gap distribution (Pal, Pitman, 2008) Andrey Sarantsev California State University, Los Angeles 10 / 16

11 Conjecture is Resolved in the Negative For every a 0, the following is a stationary gap distribution: π a = Exp(2 + ka), a 0. k=1 This includes the result from (Pal, Pitman, 2008): π 0 = Exp(2). k=1 Open Question: Are there othes, except π a and their mixtures? (S, Tsai, 2017) Andrey Sarantsev California State University, Los Angeles 11 / 16

12 Exponentially Many Particles Stationary gap distribution: Z(t) π a = Exp(2 + ka), a > 0. k=1 There are O(exp(aL)) particles on [X (1) (t), X (1) (t) + L] for large L. The quantity of particles is growing exponentially fast when you go to infinity. This creates a negative drift: E [ X (k) (t) X (k) (0) ] = a 2 t. The sheer density of particles at the top pushes the bottom particles down with (on average) linear speed. (S, Tsai, 2017) Andrey Sarantsev California State University, Los Angeles 12 / 16

13 Equilibrium Distribution Stationary gap distribution: π 0 = Exp(2). k=1 If Z(0) π 0 then Z(t) π 0 for all t 0. Under this distribution, E [ X (k) (t) X (k) (0) ] = 0 for all k. There are 2L particles on [X (1) (t), X (1) (t) + 2L] for large L. Pressure from above by Brownian crowd and pressure from below by unit drift form an equilibrium. (Pal, Pitman, 2008; S, 2016) Andrey Sarantsev California State University, Los Angeles 13 / 16

14 Long-Term Convergence to Equilibrium If we start not from the stationary gap distribution, as t, does the gap process converge to this stationary gap distribution? We understand convergence in the weak sense: P(Z(t) A) π(a) for every subset A with nice boundary : π( A) = 0, as t For the finite Atlas model, the answer is yes (Ichiba et al 2011). For the infinite Atlas model, this cannot be true, since we have multiple stationary gap distributions. Convergence domains? We have a partial description of convergence domain for π 0 = k=1 Exp(2). Open Question: Other convergence domains? Andrey Sarantsev California State University, Los Angeles 14 / 16

15 Convergence Domains Partial description of convergence domains for π 0 = k=1 Exp(2). Theorem (S, 2016) If the initial gap Z(0) is larger than π 0. Andrey Sarantsev California State University, Los Angeles 15 / 16

16 Convergence Domains Partial description of convergence domains for π 0 = k=1 Exp(2). Theorem (Dembo, Jara, Olla, 2017) (1) z k k(ln k) 1 almost surely (2) z k Exp(λ k ) independent with λ k but slow enough: lim m 1 m m ln m j=1 λ 1 j =. (3) Other cases, when z k 0 not too fast... Andrey Sarantsev California State University, Los Angeles 16 / 16