Biased card shuffling and the asymmetric exclusion process

1. 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

2. Card Shuffling Consider the following Markov chain on the space ofpermutationson 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.

3. 4 2 7 3 5 6 4 2 3 7 5 6 4 2 3 7 5 6

4. Terminology • If p = q = 0.5, we call the card shufflingunbiased. • Otherwise, we say that the system isbiased. In this case we assume W.L.O.G that p>q. Motivation • Analytic methods don’t give the mixing time (more later). • Are biased system mixing faster than non-biased? • Robustness analysis of bubble-sort.

5. 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 permutationσ. The “mixing time” of the dynamics is defined by:

6. Our Main Result We prove the following conjecture of Diaconis and Ram (2000): For all p > ½, the mixing time for the biased card shuffling isO(N2).

7. 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 O(N2) (Diaconis, Ram) – uses representation theory.

8. Methods for bounding Mixing Time Coupling Spectral gap Log Sobolev constant Representation theory.

9. 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.

10. 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).

11. 2 3 1 2 4 5 6 6 1 2 5 1 2 4 5 6 6 5 4 1 4 3 3 3 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 .

12. The Exclusion Process The state space for the exclusion process is {0,1}N where ones represent particles and zeroes represent their absence.

13. Dynamics of the Exclusion process First we pick a pair of adjacent positions. If there are zero or two particles we do nothing.

14. 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.

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

16. The coupling on the card shuffling generates a coupling J on the exclusion process with J particles. • The projections determine the permutation. Thus

17. A Partial Order We 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.

18. J J The partial Order and Coupling 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

19. The partial Order and Coupling If there exists C such that for all N and j<N Then CN2

20. 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.

21. 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.

22. We can couple with the following process on . Starting at J N-J To infinite systems How much time will it take until we hit

23. 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).

24. Kipnis results for product measures Kipnis proved that starting with i.i.d. measure on Z with density ρ, the location of a tagged particlex(t) satisfies the following.

25. 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.

26. 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:

27. Second class particles 1: Consider the following coupling of the 3 systems 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)}.

28. 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

29. N - J J 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.