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Biased card shuffling and the asymmetric

exclusion process

Elchanan Mossel, Microsoft Research

Joint work with

Itai Benjamini, Microsoft Research

Noam Berger, U.C. Berkeley

Chris Hoffman, University of Washington

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Card Shuffling

Consider the following Markov chain on the space of permutations on N elements:

• Choose uniformly at random two adjacent cards.

• With probability p order them in increasing order.

• With probability q = 1-p order them in decreasing order.

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4 2 7 3 5 6

4 2 3 7 5 6

4 2 3 7 5 6

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•If p = q = 0.5, we call the card shuffling unbiased.

•Otherwise, we say that the system is biased. In this case we assume W.L.O.G that p>q.

Terminology

Motivation

• Analytic methods don’t give the mixing time (more later).

• Are biased system mixing faster than non-biased?

• Robustness analysis of bubble-sort.

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Mixing times

• The “total-variation” distance between μ and ν is:

• Let tσ be the distribution on the permutations after t steps when starting at the permutation σ. The “mixing time” of the dynamics is defined by:

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We prove the following conjecture of Diaconis and Ram (2000):

For all p > ½, the mixing time for the biased card shuffling is (N2).

Our Main Result

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Related Card Shuffling Results

• The mixing time for the unbiased card shuffling is Θ(N3 log N) (Wilson). Sharp results – using height functions and approximate eigen-functions.

• The mixing time for the “deterministic biased” card shuffling is (N2) (Diaconis, Ram) – uses representation theory.

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Methods for bounding Mixing Time

• Coupling• Spectral gap• Log Sobolev constant• Representation theory.

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Spectral gap and mixing time

• The card shuffling defines a stochastic matrix with spectrum 1 > γ1 > … >.

• The “spectral gap” of the dynamics is 1-γ1.

• In general:

Problem: For the biased card shuffling,

• 1-γ1 = O(1/n), and

• log(1/(min π(σ)) = Ω(n2), • so we get a bound of order n3.

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Log Sobolev and Mixing Time

The Log Sobolev constant (won’t define) gives a bound on the mixing time:

Problem. For the biased card shuffling: 1/ = Ω(n3).

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Our proof – coupling:

Let x and y be permutations. We choose simultaneously the location and the direction for updating x and y. This defines a coupling .

1 2 3 4 5 6

6 5 4 3 2 1

1 2 3 4 5 6

6 5 3 4 2 1

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The Exclusion Process

The state space for the exclusion process is{0,1}N

where ones represent particles and zeroesrepresent their absence.

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If there are zero or two particles we do nothing.

First we pick a pair of adjacent positions.

Dynamics of the Exclusion process

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with probability 1-p we put the particle on the right.

with probability p we put the particle on the left

If there is one particle then

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Projections

For any J<N consider the following height functions hJ:SN{0,1}N

The transition probabilities of biased card shufflingproject to the probabilities of the exclusion process.(Used by Wilson for the unbiased case).

1 5 3 2 4 6

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• The coupling on the card shuffling generates a coupling J on the exclusion process with J particles.

• The projections determine the permutation. Thus

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A Partial OrderWe define a partial order on states of the exclusion process. For x and y with Σyi=Σxi, we write

y x

if, for all i, the i-th particle of y is to the left of the i-th particle of x.

y

x

NOTE: The coupling preserves the partial ordering.

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For any N and J < N, let HJ,N be the hitting time of

Starting at

before time T. Since the coupling preserves the order

J

The partial Order and Coupling

J

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If there exists C such that for all N and j<N

Then

CN2

The partial Order and Coupling

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Reduction:

It is sufficient to prove that there exists a constant C, such that for all N, the discrete time exclusion process starting at

will hit

before time CN2 with probability at least 1-1/Ne.

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Equivalent Formulation:

There exists a constant C, such that the continuous time exclusion process starting at

will hit

before time CN with probability at least 1-1/eN.

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To infinite systems

We can couple with the following process on .Starting at

How much time will it take until we hit

JN-J

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The motionless process• The product measure with probabilities

Is a stationary measure.• It’s not ergodic. Take the ergodic

component

• By Poincaré, the ground configuration is recurrent.

• We prove that it’s hitting time from the stationary measure has tail exp(-Ω(n½)) (Not easy).

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Kipnis results for product measures

Kipnis proved that starting with i.i.d. measure on Z with density ρ, the location of a tagged particle x(t) satisfies the following.

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Half-line results We need a similar result: starting with all

particles on the left half-line, and a product measure on the right half-line, the particles pile up with a linear speed.

By duality, and reflection, suffices to prove that for the one sided process Kipnis results still holds.

Note that here the tagged particle moves slower than in the two

sided process.

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Second class particles

In order to prove the result we couple the one sided process, two sided process and a third process with “second” class particles with the following drift rule:

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Second class particles Consider the following coupling of the 3 systems

1:

2:

3:

Let x1(t) be the location of the tagged particle in system 1. Similarly, let x2(t), x3(t) and y3(t).

Then for all t, 0 ≤ x1(t) - x2(t) = x3(t) - x2(t) ≤

max{0, x3(t) - y3(t)}.

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Second class particles

In order to analyze x3(t) - y3(t), we note that

• The distance between consecutive particles is geometric.

3:

• By deleting all non-occupied sites, we obtain the motionless process, in which the distance between the tagged particles has an exponential tail.

• Therefore distance has exp. tail as needed

Actual argument goes via coupling of system 3 with a stationary system of two particles which projects to the stationary motionless measure

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JN - J

We couple the following 3 processes:

Main steps of main result

Is dominated by a process with geometric gaps

Which behaves similarly to a process with geometric gaps and infinite number of particles to the right.