Mixing time for a random walk on a ring Stephen Connor Joint work - PowerPoint PPT Presentation

RW on a group Random walk on a ring Problems Mixing time for a random walk on a ring Stephen Connor Joint work with Michael Bate Aspects of Random Walks Durham, April 2014

RW on a group Random walk on a ring Problems Random walks on groups Let G be a finite group and let P be a probability distribution on G ; that is, a function P : G → [0 , 1] such that � g ∈ G P ( g ) = 1.

RW on a group Random walk on a ring Problems Random walks on groups Let G be a finite group and let P be a probability distribution on G ; that is, a function P : G → [0 , 1] such that � g ∈ G P ( g ) = 1. For example, we could have G = S n , the symmetric group on { 1 , 2 , . . . , n } , and we could set  1 if g = 1 is the identity n   2 P ( g ) = if g = ( i , j ) is a transposition n 2  0 otherwise 

RW on a group Random walk on a ring Problems Random walks on groups Let G be a finite group and let P be a probability distribution on G ; that is, a function P : G → [0 , 1] such that � g ∈ G P ( g ) = 1. For example, we could have G = S n , the symmetric group on { 1 , 2 , . . . , n } , and we could set  1 if g = 1 is the identity n   2 P ( g ) = if g = ( i , j ) is a transposition n 2  0 otherwise  A random walk on G is then a Markov chain X with transitions governed by the distribution P . So we fix a starting point X 0 , and then set P ( X t +1 = hg | X t = g ) = P ( h )

RW on a group Random walk on a ring Problems Distribution after repeated steps is given by convolution : � P ( gh − 1 ) P ( h ) P ( X 2 = g | X 0 = 1) = P ∗ P ( g ) = h As long as the probability distribution P isn’t concentrated on a subgroup, the stationary distribution π for X is the uniform distribution; π ( g ) = 1 / | G | for all g ∈ G . When X is ergodic, we’re interested in how long it takes for it to converge to equilibrium.

RW on a group Random walk on a ring Problems Distribution after repeated steps is given by convolution : � P ( gh − 1 ) P ( h ) P ( X 2 = g | X 0 = 1) = P ∗ P ( g ) = h As long as the probability distribution P isn’t concentrated on a subgroup, the stationary distribution π for X is the uniform distribution; π ( g ) = 1 / | G | for all g ∈ G . When X is ergodic, we’re interested in how long it takes for it to converge to equilibrium. Definition The mixing time of X is τ ( ε ) = min { t : � P ( X t ∈ · ) − π ( · ) � TV ≤ ε } .

RW on a group Random walk on a ring Problems Cutoff phenomenon We’re often interested in a natural sequence of processes X ( n ) on groups G ( n ) of increasing size: how does the mixing time τ ( n ) scale with n ?

RW on a group Random walk on a ring Problems Cutoff phenomenon We’re often interested in a natural sequence of processes X ( n ) on groups G ( n ) of increasing size: how does the mixing time τ ( n ) scale with n ? In lots of nice examples a cutoff phenomenon is exhibited: τ ( n ) ( ε ) for all ε > 0 , lim τ ( n ) (1 − ε ) = 1 n →∞ 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0

RW on a group Random walk on a ring Problems Random walk on a ring What if we add some additional structure, and try to move from a walk on a group to a walk on a ring ?

RW on a group Random walk on a ring Problems Random walk on a ring What if we add some additional structure, and try to move from a walk on a group to a walk on a ring ? For example: random walk on Z n ( n odd) with � x + 1 w.p. 1 − p n x → 2 x w.p. p n

RW on a group Random walk on a ring Problems Random walk on a ring What if we add some additional structure, and try to move from a walk on a group to a walk on a ring ? For example: random walk on Z n ( n odd) with p � x + 1 w.p. 1 − p n x → 1 − p 1 − p 2 x w.p. p n 0 6 1 1 − p 1 − p p p p p 5 2 p p 1 − p 1 − p 4 3 1 − p

RW on a group Random walk on a ring Problems Related results Chung, Diaconis and Graham (1987) study the process (used in random number generation)  w.p. 1 ax − 1 3   w.p. 1 x → ax 3  w.p. 1 ax + 1  3

RW on a group Random walk on a ring Problems Related results Chung, Diaconis and Graham (1987) study the process (used in random number generation)  w.p. 1 ax − 1 3   w.p. 1 x → ax 3  w.p. 1 ax + 1  3 When a = 1 there exist constants C and C ′ such that: e − Ct / n 2 < � − π ( n ) � � P ( n ) TV < e − C ′ t / n 2 � � t �

RW on a group Random walk on a ring Problems Related results Chung, Diaconis and Graham (1987) study the process (used in random number generation)  w.p. 1 ax − 1 3   w.p. 1 x → ax 3  w.p. 1 ax + 1  3 When a = 1 there exist constants C and C ′ such that: e − Ct / n 2 < � − π ( n ) � � P ( n ) TV < e − C ′ t / n 2 � � t � When a = 2 and n = 2 m − 1 there exist constants c and c ′ such that: � t n − π ( n ) � � P ( n ) for t n ≥ c log n log log n , TV → 0 as n → ∞ � � � for t n ≤ c ′ log n log log n , � t n − π ( n ) � � P ( n ) TV > ε as n → ∞ � � �

RW on a group Random walk on a ring Problems Back to our process... General problem The distribution of X t isn’t given by convolution. X t = 2 I t X t − 1 + (1 − I t )( X t − 1 + 1) where I t ∼ Bern( p n ).

RW on a group Random walk on a ring Problems Back to our process... General problem The distribution of X t isn’t given by convolution. X t = 2 I t X t − 1 + (1 − I t )( X t − 1 + 1) where I t ∼ Bern( p n ). But for this relatively simple walk, we can get around this by looking at the process subsampled at jump times . (Here we call a +1 move a ‘step’ and a × 2 move a ‘jump’ .)

RW on a group Random walk on a ring Problems Back to our process... General problem The distribution of X t isn’t given by convolution. X t = 2 I t X t − 1 + (1 − I t )( X t − 1 + 1) where I t ∼ Bern( p n ). But for this relatively simple walk, we can get around this by looking at the process subsampled at jump times . (Here we call a +1 move a ‘step’ and a × 2 move a ‘jump’ .) So consider (with X 0 = Y 0 = 0) k i.i.d. � 2 k +1 − j S j (mod n ) , Y k = S j ∼ Geom( p n ) (mod n ) j =1 (i.e. Y k is the position of X immediately following the k th jump.)

RW on a group Random walk on a ring Problems Plan 1 Find lower bound for mixing time of Y ; 2 Find upper bound for mixing time of Y ; 3 Try to relate these back to the mixing time for X .

RW on a group Random walk on a ring Problems Plan 1 Find lower bound for mixing time of Y ; 2 Find upper bound for mixing time of Y ; 3 Try to relate these back to the mixing time for X . 1 Restrict attention to p n = 2 n α , α ∈ (0 , 1). We expect things to happen (for Y ) sometime around T n := log 2 ( np n ) ∼ (1 − α ) log 2 n

RW on a group Random walk on a ring Problems In fact: Theorem Y exhibits a cutoff at time T n , with window size O (1) .

RW on a group Random walk on a ring Problems In fact: Theorem Y exhibits a cutoff at time T n , with window size O (1) . To prove this, we need to show that (for n odd) � � − 1 � � X ( n ) � � lim inf � P T n − θ ∈ · ≥ 1 − ε ( θ ) � � n n →∞ � TV and � � − 1 � � X ( n ) � � � P lim sup T n + θ ∈ · ≤ ε ( θ ) , � � n n →∞ � TV where ε ( θ ) → 0 as θ → ∞ .

RW on a group Random walk on a ring Problems Lower bound For a lower bound, we simply find a set A n ( θ ) (of considerable size) that Y has very little chance of hitting before time T n − θ , for large θ ∈ N . (Recall the definition of total variation distance!) Y 0 A n ( θ ) E [ Y T n − θ ] (3 / 8 − β ) n

RW on a group Random walk on a ring Problems If A n ( θ ) is chosen as above then π ( A n ( θ )) = 1 / 4 + 2 β .

RW on a group Random walk on a ring Problems If A n ( θ ) is chosen as above then π ( A n ( θ )) = 1 / 4 + 2 β . Using Chebychev’s inequality: 4 1 − θ P ( Y T n − θ ∈ A n ( θ )) ≤ 3(3 / 8 − β ) 2 .

RW on a group Random walk on a ring Problems If A n ( θ ) is chosen as above then π ( A n ( θ )) = 1 / 4 + 2 β . Using Chebychev’s inequality: 4 1 − θ P ( Y T n − θ ∈ A n ( θ )) ≤ 3(3 / 8 − β ) 2 . Now choose β = β ( θ ) to make the difference between these large. . .

RW on a group Random walk on a ring Problems If A n ( θ ) is chosen as above then π ( A n ( θ )) = 1 / 4 + 2 β . Using Chebychev’s inequality: 4 1 − θ P ( Y T n − θ ∈ A n ( θ )) ≤ 3(3 / 8 − β ) 2 . Now choose β = β ( θ ) to make the difference between these large. . . Lemma (Lower bound for Y ) For θ ≥ 3 , � P ( Y T n − θ ∈ · ) − π n ( · ) � TV ≥ 1 − 4 1 − θ/ 3 .

RW on a group Random walk on a ring Problems Upper bound Let P be a probability on a group G . A (complex) representation ρ is a group homomorphism ρ : G → GL d ( C ), where GL d ( C ) denotes the group of d × d invertible complex matrices. We write ˆ � P ( ρ ) = P ( g ) ρ ( g ) g ∈ G for the Fourier transform of P at ρ .