Biased card shuffling and the asymmetric exclusion process

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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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Terminology

• 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 pgtq.

Motivation

• Analytic methods dont 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 ?
  is
• Let ?ts be the distribution on the permutations
  after t steps when starting at the permutation s.
  
• The mixing time of the dynamics is defined by

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Our Main Result
We prove the following conjecture of Diaconis
and Ram (2000) For all p gt ½, the mixing time
for the biased card shuffling is O(N2).
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Related Card Shuffling Results

• The mixing time for the unbiased card shuffling
  is T(N3 log N) (Wilson). Sharp results using
  height functions and approximate eigen-functions.
• The mixing time for the deterministic biased
  card shuffling is O(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 gt ?1 gt gt.
• The spectral gap of the dynamics is 1-?1.
• In general

• Problem For the biased card shuffling,
• 1-?1 O(1/n), and
• log(1/(min p(s)) O(n2),
• so we get a bound of order n3.

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Log Sobolev and Mixing Time
The Log Sobolev constant ? (wont define) gives
a bound on the mixing time
Problem. For the biased card shuffling 1/?
O(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 ?.

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The Exclusion Process
The state space for the exclusion process
is 0,1N where ones represent particles and
zeroes represent their absence.
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Dynamics of the Exclusion process
First we pick a pair of adjacent positions.
If there are zero or two particles we do nothing.
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If there is one particle then
with probability p we put the particle on the
left
with probability 1-p we put the particle on the
right.
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Projections
For any JltN consider the following height
functions hJSN?0,1N The transition
probabilities of biased card shuffling project
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 Order
We define a partial order on states of the
exclusion process. For x and y with SyiSxi, 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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The partial Order and Coupling
For any N and J lt N, let HJ,N be the hitting time
of
Starting at
before time T. Since the coupling preserves the
order
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The partial Order and Coupling
If there exists C such that for all N and jltN
Then
CN2
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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
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The motionless process

• The product measure with probabilities
• Is a stationary measure.
• Its not ergodic. Take the ergodic component
• By Poincaré, the ground configuration is
  recurrent.
• We prove that its hitting time from the
  stationary measure has tail exp(-O(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 ?, 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) max0, 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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Main steps of main result
We couple the following 3 processes
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.