Infinite Atlas Model One Year Later nite Atlas Model One Year Later Andrey Sarantsev ... Seattle 12/24. ... fx 0g)c 1 + c 2 = : At time t = 0,

Infinite Atlas ModelOne Year Later

Andrey Sarantsev

University of California, Santa Barbara

October 2, 2017

Andrey Sarantsev University of Washington, Seattle 1 / 24

Infinite Atlas Model

Infinite system of Brownian particles (Xi (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.

dXi (t) = 1(Xi (t) = X(1)(t))dt + dWi (t),

W1,W2, . . . i.i.d. Brownian motions


Infinite Atlas Model

Called the infinite Atlas model because the bottom particle, calledthe Atlas particle, “supports the system”, similarly to the GreekAtlas 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, 2017.


Motivation

Stochastic Portfolio TheoryBanner, Fernholz, Karatzas, 2005Pal, Chatterjee, 2010Jourdain, Reygner, 2013

Similar systemsFerrari, Spohn, Weiss, 2015Aizenman, Ruzmaikina, 2005Aizenman, Arguin, 2009

Scaling limits of asymmetrically colliding random walksKaratzas, Pal, Shkolnikov, 2016

Discretized version of McKean-Vlasov nonlinear diffusionJourdain, Reygner, 2013Dembo, Shkonlikov, Varadhan, Zeitouni, 2016Kolli, Shkolnikov, 2016


Weak Existence and Uniqueness

Unlike the finite system, the infinite system might not always exist,even in the weak sense.

Why? Because it might not be possible to rank these particles.

Assume for simplicity that there is no bottom drift. Start infinitelymany i.i.d. Brownian motions from the same initial point. Then itis impossible to rank them from bottom to top.

We need a condition on initial values xn = Xn(0), n = 1, 2, . . .They should be far enough apart, or, alternatively, tend to infinityxn →∞ fast enough.


Existence and Uniqueness

An assumption on initial conditions xn = Xn(0), n = 1, 2, . . .

Tending to infinity fast enough?

∞∑n=1

exp(−αx2n ) <∞ for all α > 0.

xn = cnα for α > 0 worksxn = c log n worksxn = c(log n)1/2 does not work

Theorem (Ichiba, Karatzas, Shkolnikov, 2013; S, 2016)

Suppose Xn(0) = xn →∞ fast enough. Then the system exists inthe strong sense and is pathwise unique.


Questions of Interest

Fluctuations of the Atlas particle X(1)(t) as t →∞Scaling limit of empirical measure

∑i≥1 δXi (t)

Stationary gap distributions for the gap process

Z (t) = (Zk(t) = X(k+1)(t)− X(k)(t))k≥1

Long-term weak convergence of Z (t) as t →∞


Equilibrium Distribution

Exponential distribution Exp(λ): mean λ−1, density λe−λx dx

π0 =∞⊗n=1

Exp(2).

If Z (0) ∼ π0 then Z (t) ∼ π0 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 frombelow by unit drift form an equilibrium.

(Pal, Pitman, 2008; S, 2016)


Equilibruim Fluctuations

In the equilibrium distribution π0, we have:(ε1/4X(1)(t/ε), t ≥ 0

)⇒ cB1/4, ε ↓ 0.

B1/4 is the fractional Brownian motion: a Gaussian process with

EB1/4(t) = 0, E[B1/4(s)B1/4(t)

]=

1

2

(t1/2 + s1/2 − |t − s|1/2

).

In particular, EB21/4(t) = t1/2. And c = (2/π)1/4.

As t →∞, X(1)(t) behaves as B1/4, not a Brownian motion.

Not a Brownian scaling. (Brownian scaling would be ε1/2.)

(Dembo, Tsai, 2015)


Equilibrium Fluctuations

Let iε(x) := bxε−1/2/2c. As ε ↓ 0,

ε1/4[iε(x)− 2Xiε(x)(ε

−1t)]⇒ X (t, x).

X (t, x) is the solution to additive stochastic heat equation:

∂X∂t− 1

2

∂2X∂x2

=√

2W.

W(t, x) is space-time white noise: a field Wf for f ∈ L2(R2)

EWf = 0, E [WfWg ] = (f , g)L2(R2)

(Dembo, Tsai, 2015)


Long-Term Convergence to Equilibrium

Recall that the gap process is defined as

Z (t) = (Zk(t) = X(k+1)(t)− X(k)(t))k≥1

(a) The family (Z (t), t ≥ 0) of random variables is tight in R∞.

(b) Every weak limit point of Z (t) as t →∞ is stochasticallysmaller than the stationary distribution π0.

(c) If Z (0) is stochastically larger than π0, then Z (t)⇒ π0.

(S, 2016)


Long-Term Convergence to Equilibrium

For which initial conditions Zk(0) = zk do we have Z (t)⇒ π0?

Case 1. zk√k(ln k)−1 →∞ a.s.

Case 2. zk ∼ Exp(λk) independent with λk →∞ but slow enough:

limm→∞

1√m lnm

m∑j=1

λ−1j =∞.

Other cases, when zk → 0 not too fast...

(Dembo, Jara, Olla, 2017)


Stationary Gap Distributions

πa =∞⊗k=1

Exp(2 + ka), a ≥ 0.

This includes the result from (Pal, Pitman, 2008):

π0 =∞⊗k=1

Exp(2).

Open Question: Are there othes, except πa and their mixtures?

(S, Tsai, 2017)


Exponentially Many Particles

Z (t) ∼ πa =∞⊗k=1

Exp(2 + ka), a > 0.

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 goto infinity. This creates a negative drift:

E[X(k)(t)− X(k)(0)

]= −a

2t.

The sheer density of particles at the top pushes the bottomparticles down with (on average) linear speed.

(S, Tsai, 2017)


Damped Fluctuations

Z (t) ∼ πa =∞⊗k=1

Exp(2 + ka), a > 0.

The Atlas particle does not fluctuate:

(X(1)(t) +at

2, t ≥ 0) is tight.

The density of particles dampens the Brownian fluctuation.

(Tsai, 2017)


Out of Equilibrium Behavior

Assume we start from

Z (0) ∼∞⊗k=1

Exp(λ), λ 6= 2.

Then as ε ↓ 0, measure-valued processes converge:

ε

∞∑k=1

δεXi (ε−2t) ⇒ p(t, x) dx ,

p(t, x) = c1 + c2Φ(t−1/2x

), x ≥ k

√t.

Φ(x) is the standard normal CDF, k ∈ R(Cabezas, Dembo, S, Sidoravicius, 2017)



The density p(t, x) satisfies Stefan problem for the heat equation:

∂u

∂t=

1

2

∂2u

∂x2, x > y(t),

and u = 0 for x < y(t), where y(t) is the free boundary.

Stefan problem is a particular case of a free-boundary problem forPDE.

Boundary evolves with time.

An example is melting ice.


Out of Equilibrium Behavior: Initial Condition

Limiting density:

p(t, x) = c1 + c2Φ(t−1/2x

), x ≥ k

√t.

u(0, x) = λ1{x≥0} ⇒ c1 + c2 = λ.

At time t = 0, particles are distributed evenly with density λ.


Out of Euilibrium Behavior: Boundary Condition 1

Limiting density:

p(t, x) = c1 + c2Φ(t−1/2x

), x ≥ k

√t.

u(t, y(t)+) = 2 ⇒ c1 + c2Φ(k) = 2.

The particle density near the Atlas particle quickly reaches itsequilibrium value 2.


Out of Euilibrium Behavior: Boundary Condition 2

Limiting density:

p(t, x) = c1 + c2Φ(t−1/2x

), x ≥ k

√t.

u(t, y(t)+) y ′(t) +1

2

∂u

∂x(t, y(t)+) = 0.

The left boundary of our particle density (the rescaled Atlasparticle) moves as the density flux. This gives us

c1 +c22

Φ′(k) = 0.


Out of Equilibrium Behavior: Boundary Conditions

From previous slides, we got:

c1 + c2 = λ;

c1 + c2Φ(k) = 2;

c1 +c22

Φ′(k) = 0.

Solving this system of three equations, we get c1, c2, k .



The constant k is the unique solution of

k(1− Φ(k))

Φ′(k)= 1− λ

2.

λ < 2 ⇒ k > 0 (low pressure from Brownian crowd above)

λ = 2 ⇒ k = 0 (back in equilibrium case, scaling ε1/4);

λ > 2 ⇒ k < 0 (high pressure from Brownian crowd above)



Assume we start from

Z (0) ∼∞⊗k=1

Exp(λ), λ 6= 2.

Then as ε ↓ 0,εX(1)(tε

−2)⇒ k√t.

This is Brownian scaling: Pressure from particles above is notenough to change the character of Brownian fluctuations.

(Cabezas, Dembo, S, Sidoravicius, 2017)


Thank You For Your Attention!


Documents

Infinite Atlas Model One Year Later nite Atlas Model One Year Later Andrey Sarantsev ... Seattle 12/24. ... fx 0g)c 1 + c 2 = : At time t = 0,