Seminar on Markov Chains and Mixing Times, Fall 16/17 Jay Tenenbaum Coupling Seminar on Markov Chains and Mixing Times, Fall 16/17 Jay Tenenbaum

Reminders In case you forgot…

Definition - Total variation distance Let 𝜇 and 𝜈 be two probability distributions on Ω. The total variation distance between µ and 𝜈 is given by: ||𝜇−𝜈|| 𝑇𝑉 = max 𝐴⊂Ω |𝜇 𝐴 −𝜈(𝐴)| 1 − γ = α = β = ||µ − ν||TV

Definition - Total variation distance Proposition: ||𝜇−𝜈|| 𝑇𝑉 = max 𝐴⊂Ω |𝜇 𝐴 −𝜈(𝐴)| = 1 2 Σ 𝑥∈Ω |𝜇 𝑥 −𝜈(𝑥)| 1 − γ = α = β = ||µ − ν||TV

Definition - Coupling A coupling of two probability distributions µ and 𝜈 (both over Ω) is a pair of random variables (𝑋, 𝑌) defined over Ω×Ω) such that: the marginal distribution of 𝑋 is µ [𝑃 𝑋 = 𝑥 = Σ 𝑦∈Ω 𝑃 𝑥,𝑦 = µ(𝑥) ] the marginal distribution of 𝑌 is 𝜈 [𝑃 𝑌 =𝑦 = Σ 𝑥∈Ω 𝑃 𝑥,𝑦 =𝜈(𝑦) ]

Proposition: Let 𝜇 and 𝜈 be two probability distributions on Ω. Then: ||𝜇−𝜈|| 𝑇𝑉 =inf P X≠𝑌 X,Y coupling of 𝜇 𝑎𝑛𝑑 𝜈}

Definition – Distance to Stationary From now on we assume That P is an ergodic (aperiodic, irreducible) Markov process with stationary distribution 𝜋. Bounding the maximal distance (over 𝑥 0 ∈Ω) between 𝑃 𝑡 ( 𝑥 0 ,⋅) and 𝜋 is among our primary objective. It is therefore convenient to define: 𝑑 𝑡 ≔ max 𝑥∈Ω || 𝑃 𝑡 𝑥,⋅ −𝜋|| 𝑇𝑉

Definition – Mixing Time It is useful to introduce a parameter which measures the time required by a Markov chain for the distance to stationarity to be small. The mixing time is defined by: 𝑡 𝑚𝑖𝑥 𝜖 ≔ min 𝑡 𝑑 𝑡 <𝜖} We standardize: 𝑡 𝑚𝑖𝑥 ≔ 𝑡 𝑚𝑖𝑥 1 4

Random Walk Coupling example

Simple random walk on {0, 1 . . . , 𝑛} (Markov Chain Coupling Motivation) Move up or down with probability 1 2 if possible. Do nothing if attempt to move outside interval.

Simple random walk on {0, 1 . . . , 𝑛} Claim 4. If 0 ≤ 𝑥 ≤ 𝑦 ≤ 𝑛, 𝑡≥0 𝑃 𝑡 (𝑦,0)≤ 𝑃 𝑡 (𝑥,0)

Simple random walk on {0, 1 . . . , 𝑛} Claim 4. If 0 ≤ 𝑥 ≤ 𝑦 ≤ 𝑛, 𝑡≥0 𝑃 𝑡 (𝑦,0)≤ 𝑃 𝑡 (𝑥,0) Proof: Define a coupling (Xt, Yt ) of 𝑃 𝑡 𝑥,⋅ and 𝑃 𝑡 (𝑦,⋅): 𝑋0 = 𝑥 , 𝑌0 = 𝑦 . Let 𝑏1, 𝑏2 . . . be i.i.d. {±1}-valued Bernoulli (1/2). At the 𝑖′th step, attempt to add 𝑏𝑖 to both 𝑋𝑖−1 and 𝑌𝑖−1. Stop after t steps.

Simple random walk on {0, 1 . . . , 𝑛} For all t, Xt ≤ Yt . Therefore, 𝑃𝑡 𝑦 , 0 = 𝑃 𝑌𝑡 = 0 ≤ 𝑃 𝑋𝑡 = 0 =𝑃𝑡 𝑥 , 0 Note: In this case, when the coupling meets, it sticks together

Markov Chain Coupling

Markov Chain Coupling

Coupling Example

Coupling Example

Coupling Example

Coupling Example

Coupling Example

Markov Chain Coupling Definition Def: A coupling of a Markov Chain with transition matrix P is a Markovian Process 𝑋 𝑡 , 𝑌 𝑡 𝑡=0 ∞ such that: Each process {Xt}, {Yt} is a Markov Chain with transition matrix P. (∀𝑎,𝑏∈Ω Pr 𝑋 𝑡+1 =𝑏 𝑋 𝑡 =𝑎 =𝑃 𝑎,𝑏 =Pr⁡[ 𝑌 𝑡+1 =𝑏| 𝑌 𝑡 =𝑎]) Note that 𝑋 𝑡 𝑡=0 ∞ , 𝑌 𝑡 𝑡=0 ∞ may have different starting distributions 𝑋 0 , 𝑌 0 .

Markov Chain Coupling - Stickyness We can modify the Markovian Chain coupling so that chains stay together after meeting: 𝑖𝑓 𝑋 𝑠 = 𝑌 𝑠 , 𝑡ℎ𝑒𝑛 ∀𝑡≥𝑠 𝑋 𝑡 = 𝑌 𝑡 [Simply run according to the Markovian chain coupling, until they meet. Then run them together based on original Markov Chain] We denote such Couplings as Sticky Couplings

Markov Chain Coupling - Notation Given a Markovian Chain Coupling 𝑋 𝑡 , 𝑌 𝑡 𝑡=0 ∞ of a Markov Chain with transition P, we denote 𝑃 𝑥,𝑦 to be: 𝑃 𝑥,𝑦 𝐴 =𝑃(𝐴| 𝑋 0 =𝑥, 𝑌 0 =𝑦) Which is the probability assuming 𝑋 0 is at x, and 𝑌 0 is at y [recall that 𝑋 𝑡 𝑡=0 ∞ , 𝑌 𝑡 𝑡=0 ∞ may have different starting distributions]

Markov Chain Coupling - Observation Let 𝑋 𝑡 , 𝑌 𝑡 𝑡=0 ∞ be a coupling of P where X 0 = 𝑥, 𝑌 0 =𝑦. Observation: ∀𝑡≥0, ( 𝑋 𝑡 , 𝑌 𝑡 ) is a coupling of 𝑃 𝑡 (𝑥,⋅) with 𝑃 𝑡 𝑦,⋅ Proof. 𝑃 𝑡 (𝑥, 𝑧) = 𝑃 𝑥,𝑦 { 𝑋 𝑡 = 𝑧} and 𝑃 𝑡 (𝑦, 𝑧) = 𝑃 𝑥,𝑦 { 𝑌 𝑡 = 𝑧}

Bounding Distance to Stationary Using Couplings

Bounding Total Variation Distance As usual, P is an ergodic MC over Ω with stationary distribution 𝜋. Theorem: Let 𝑋 𝑡 , 𝑌 𝑡 𝑡=0 ∞ be a sticky coupling of P where X 0 = 𝑥, 𝑌 0 =𝑦. Let 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 be the first time the chains meet ( 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 := min 𝑡 𝑋 𝑡 = 𝑌 𝑡 }). Then: || 𝑃 𝑡 𝑥,⋅ − 𝑃 𝑡 𝑦,⋅ || 𝑇𝑉 ≤ 𝑃 𝑥,𝑦 { 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑡} Proof. || 𝑃 𝑡 𝑥,⋅ − 𝑃 𝑡 𝑦,⋅ || 𝑇𝑉 ≤ 𝑃 𝑥,𝑦 𝑋 𝑡 ≠ 𝑌 𝑡 = 𝑃 𝑥,𝑦 { 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑡} ( ||𝜇−𝜈|| 𝑇𝑉 =inf P X≠𝑌 X,Y coupling of 𝜇 𝑎𝑛𝑑 𝜈}) (definition of 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 + stickyness)

Bounding Total Variation Distance [Theorem: || 𝑃 𝑡 𝑥,⋅ − 𝑃 𝑡 𝑦,⋅ || 𝑇𝑉 ≤ 𝑃 𝑥,𝑦 { 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑡}] Corollary: 𝑑 𝑡 ≤ max 𝑥,𝑦∈Ω 𝑃 𝑥,𝑦 { 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑡}

Reminder Given P an ergodic Markov process with stationary distribution 𝜋, We defined: 𝑑 𝑡 ≔ max 𝑥∈Ω || 𝑃 𝑡 𝑥,⋅ −𝜋|| 𝑇𝑉 However, we also defined: 𝑑 𝑡 ≔ max 𝑥,𝑦∈Ω || 𝑃 𝑡 𝑥,⋅ − 𝑃 𝑡 𝑦,⋅ || 𝑇𝑉 And showed that: 𝑑 𝑡 ≤ 𝑑 𝑡 ≤2𝑑 𝑡 We will use the fact that 𝑑 𝑡 ≤ 𝑑 𝑡

Bounding Distance to Stationary [Theorem: || 𝑃 𝑡 𝑥,⋅ − 𝑃 𝑡 𝑦,⋅ || 𝑇𝑉 ≤ 𝑃 𝑥,𝑦 { 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑡}] Corollary: 𝑑 𝑡 ≤ max 𝑥,𝑦∈Ω 𝑃 𝑥,𝑦 { 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑡} Proof. 𝑑 𝑡 ≤ 𝑑 𝑡 = max 𝑥,𝑦∈Ω 𝑃 𝑡 𝑥,⋅ − 𝑃 𝑡 𝑦,⋅ 𝑇𝑉 ≤ max 𝑥,𝑦∈Ω 𝑃 𝑥,𝑦 { 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑡} Using Theorem for each pair of 𝑥,𝑦∈Ω Design a coupling that brings X and Y together fast (for each 𝑥,𝑦∈Ω)

Examples – Bounding Mixing Time In all the following examples we consider Markov Chains. For each such example we: Define a suitable coupling 𝑋 𝑡 , 𝑌 𝑡 𝑡=0 ∞ . Bound the value 𝑃 𝑥,𝑦 { 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑡} for each pair of 𝑥,𝑦∈Ω Use the Corollary: 𝑑 𝑡 ≤ max 𝑥,𝑦∈Ω 𝑃 𝑥,𝑦 { 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑡} to bound 𝑑(𝑡) Find the minimal t that ensures that 𝑑 𝑡 < 1 4 That t is an upper bound for the Mixing Time!

Lazy Random Walk On The Cycle Bounding mixing time

Random Lazy Walk On The Cycle 𝛀= ℤ 𝒏 = 𝟏,…,𝒏 𝑷 𝒋,𝒌 = 𝟏 𝟒 𝒊𝒇 𝒌≡𝒋+𝟏 𝟏 𝟒 𝒊𝒇 𝒌≡𝒋−𝟏 𝟏 𝟐 𝒊𝒇 𝒌≡𝒋 𝟎 𝑶.𝑾. 1/2 1 2 3 4 n 1/2 1/4 1/4 1/2 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/2 1/2

Random Lazy Walk On The Cycle We construct a coupling 𝑋 𝑡 , 𝑌 𝑡 𝑡=0 ∞ of two particles walking lazily on the circle, one starting from 𝑥, the other from 𝑦. At each move: Flip a coin If heads, ( 𝑋 𝑡 ) moves CW or CCW based on additional coin flip. If tails, ( 𝑌 𝑡 ) moves CW or CCW based on additional coin flip. Assume stickiness…

Random Lazy Walk On The Cycle Coupling summary: 50-50 choose 𝑋 𝑡 , 𝑌 𝑡 . 50-50 choose CW,CCW. Argument: This is indeed a Markov Chain Coupling Proof: From each unique particle’s point of view we have 50% chance to move it and if we move it, 50% chance cw or ccw.

Random Lazy Walk On The Cycle – Bounding 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 Coupling summary: 50-50 choose 𝑋 𝑡 , 𝑌 𝑡 . 50-50 choose CW,CCW. Bounding: let 𝐷 𝑡 the clockwise distance between 𝑋 𝑡 to 𝑌 𝑡 . Note that 𝐷 𝑡 is a simple random walk on intertior of {0,…,𝑛} and gets absorbed on either zero or n. We’ve seen in first lecture that if 𝜏 is the time required to get absorbed, and 𝐷 0 =𝑘 then: 𝐸 𝑘 𝜏 =𝑘 𝑛−𝑘 Notice that 𝜏= 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 , 𝐷 0 is the clockwise distance between 𝑥,𝑦. Therefore: ∀𝑥,𝑦 𝐸 𝑥,𝑦 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 ≤ max 𝑘∈{0,…,𝑛−1} 𝑘 𝑛−𝑘 ≤ 𝑛 2 4

Random Lazy Walk On The Cycle Coupling summary: 50-50 choose 𝑋 𝑡 , 𝑌 𝑡 . 50-50 choose CW,CCW. Bounding: ∀𝑥,𝑦 𝐸 𝑥,𝑦 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 ≤ 𝑛 2 4 Therefore: 𝑑 𝑡 ≤ max 𝑥,𝑦∈Ω 𝑃 𝑥,𝑦 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑡 ≤ 1 𝑡 max 𝑥,𝑦∈Ω 𝐸 𝑥,𝑦 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 ≤ 𝑛 2 4𝑡 Therefore: For 𝑡≥ 𝑛 2 we have 𝑑 𝑡 ≤ 1 𝑡 𝑛 2 2 ≤ 1 𝑛 2 ∗ 𝑛 2 4 ≤ 1 4 , therefore: 𝑡 𝑚𝑖𝑥 = min 𝑡 𝑑 𝑡 ≤ 1 4 ≤ 𝑛 2

Lazy Random Walk On The Hypercube Bounding mixing time

Sampling a Hypercube Toy problem: Randomly sample from the vertices of a k-dimensional hypercube. k = 3

Lazy Random Walk On Hypercube Markov Chain: [equivalent to] pick coordinate uniformly i {1,…,n} pick value uniformly b {0,1} set x(i)=b Markov Chain is ergodic & symmetric => Stationary distribution  is uniform ={0,1}n 1/2 1/6 1/6 1/6

Coupling For Lazy Random Walk On Hypercube Assuming X 0 =𝑥, 𝑌 0 =𝑦, define transition 𝑋 𝑡 , 𝑌 𝑡 → 𝑋 𝑡+1 , 𝑌 𝑡+1 : pick coordinate uniformly i {1,…,n} pick value uniformly b {0,1} set 𝑋 𝑡 𝑖 =𝑏 , 𝑌 𝑡 (𝑖)=𝑏 This is indeed a coupling. ( 0 , 0 , 1 , 0 , 1) t=0 ( 1 , 1 , 0 , 0 , 1) i=3, b=0 ( 0 , 0 , 0 , 0 , 1) t=1 ( 1 , 1 , 0 , 0 , 1) i=5, b=1 ( 0 , 0 , 0 , 0 , 1) t=2 ( 1 , 1 , 0 , 0 , 1) i=3, b=1 ( 0 , 0 , 1 , 0 , 1) t=3 ( 1 , 1 , 1 , 0 , 1) i=1, b=0 (0 , 0 , 1 , 0 , 1) t=4 (0 , 1 , 1 , 0 , 1) i=2, b=1 (0 , 1 , 1 , 0 , 1) t=5 (0 , 1 , 1 , 0 , 1)

Coupling For Lazy Random Walk On Hypercube Denote by 𝜏 the first time where all the coordinates have been selected at least once. The two walkers agree by time 𝜏 ( 𝑡 𝑐𝑜𝑢𝑝𝑙𝑒 ≤𝜏). (For each initial 𝑥,𝑦!!!) 𝜏 distributes like the coupon collector random variable (with n coupons). Therefore: (Assumption) 𝑃 𝜏>𝑛𝑙𝑛(𝑛)+𝑐𝑛 ≤ 𝑒 −𝑐 Therefore: 𝑑 𝑛𝑙𝑛(𝑛)+ l𝑛 4 ∗𝑛 ≤ max 𝑥,𝑦∈Ω 𝑃 𝑥,𝑦 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑛𝑙𝑛(𝑛)+ln⁡(4)∗𝑛 ≤ max 𝑥,𝑦∈Ω 𝑃 𝑥,𝑦 𝜏>𝑛𝑙𝑛(𝑛)+ln⁡(4)∗𝑛 ≤ e − ln 4 = 1 4 Therefore: 𝑡 𝑚𝑖𝑥 ≤𝑛𝑙𝑛(𝑛)+ l𝑛 4 ∗𝑛=𝑂(𝑛𝑙𝑜𝑔(𝑛))

Proof of Assumption Theorem: Let τ be a coupon collector random variable, as seen before. For any c > 0 we have: 𝑃 𝜏>𝑛𝑙𝑛(𝑛)+𝑐𝑛 ≤ 𝑒 −𝑐 Proof: Let 𝐴 𝑖 be the event that the I’th type does not appear among the first 𝑛𝑙𝑛(𝑛)+𝑐𝑛 coupons drawn. Then, from independence of trials 𝑃 𝐴 𝑖 = 1− 1 𝑛 𝑛𝑙𝑛(𝑛)+𝑐𝑛 Now, 𝑃 𝜏>𝑛𝑙𝑛(𝑛)+𝑐𝑛 =P ∪ 𝑖=1 𝑛 𝐴 𝑖 ≤ Σ 𝑖=1 𝑛 𝑃 𝐴 𝑖 = Σ 𝑖=1 𝑛 1− 1 𝑛 𝑛𝑙𝑛(𝑛)+𝑐𝑛 = =𝑛 1− 1 𝑛 −𝑛(−ln⁡(𝑛)−𝑐) ≤𝑛 𝑒 −ln⁡(𝑛) 𝑒 −𝑐 ≤ 𝑒 −𝑐

Card Shuffling – Random Transposition Bounding mixing time

Card Shuffling – Random Transposition Irreducible Aperiodic (Transpose card with itself) P is symmetric: x,y P(x,y)=P(y,x)   is uniform

Random Transposition – Coupling Definition We construct a coupling 𝜎 𝑡 , 𝜎 𝑡 ′ 𝑡=0 ∞ of the Random Transposition MC: Ω= 𝑆 𝑛 At each move: 𝜎 𝑡 , 𝜎 𝑡 ′ →( 𝜎 𝑡+1 , 𝜎 𝑡+1 ′ ) Choose card 𝑋 𝑡 and an independent position 𝑌 𝑡 uniformly. Switch the card 𝑋 𝑡 with the card at position 𝑌 𝑡 in both 𝜎 𝑡 , 𝜎 𝑡 ′ . Let 𝑀𝑡 = # of cards at the same position in 𝜎 𝑡 and 𝜎 𝑡 ′ Indeed a coupling!!!

Random Transposition – Case 1 At each move: 𝜎 𝑡 , 𝜎 𝑡 ′ →( 𝜎 𝑡+1 , 𝜎 𝑡+1 ′ ) Choose card 𝑋 𝑡 and an independent position 𝑌 𝑡 uniformly. Switch the card 𝑋 𝑡 with the card at position 𝑌 𝑡 in both 𝜎 𝑡 , 𝜎 𝑡 ′ . 𝑀 𝑡 = # of cards at the same position in 𝜎 𝑡 and 𝜎 𝑡 ′ Case 1: 𝑋 𝑡 in same position => 𝑀 𝑡+1 = 𝑀 𝑡

Random Transposition – Case 2 At each move: 𝜎 𝑡 , 𝜎 𝑡 ′ →( 𝜎 𝑡+1 , 𝜎 𝑡+1 ′ ) Choose card 𝑋 𝑡 and an independent position 𝑌 𝑡 uniformly. Switch the card 𝑋 𝑡 with the card at position 𝑌 𝑡 in both 𝜎 𝑡 , 𝜎 𝑡 ′ . 𝑀 𝑡 = # of cards at the same position in 𝜎 𝑡 and 𝜎 𝑡 ′ Case 2: 𝑋 𝑡 in different positions 𝜎 𝑡 𝑌 𝑡 = 𝜎 𝑡 ′ ( 𝑌 𝑡 ) => 𝑀 𝑡+1 = 𝑀 𝑡

Random Transposition – Cases 3 𝑋 𝑡 in different positions 𝜎 𝑡 𝑌 𝑡 ≠ 𝜎 𝑡 ′ ( 𝑌 𝑡 ) => 𝑀 𝑡+1 > 𝑀 𝑡 𝑀 𝑡+1 = 𝑀 𝑡 +2 𝑀 𝑡+1 = 𝑀 𝑡 +3 𝑀 𝑡+1 = 𝑀 𝑡 +1

Random Transposition – Cases Summary 𝑋 𝑡 in same position 𝑀 𝑡+1 = 𝑀 𝑡 Case 2: 𝑋 𝑡 in different positions 𝜎 𝑡 𝑌 𝑡 = 𝜎 𝑡 ′ ( 𝑌 𝑡 ) Cases 3: 𝜎 𝑡 𝑌 𝑡 ≠ 𝜎 𝑡 ′ ( 𝑌 𝑡 ) 𝑀 𝑡+1 > 𝑀 𝑡 We now calculate: 𝑃( 𝑀 𝑡+1 > 𝑀 𝑡 | 𝑀 𝑡 =𝑖) [0≤𝑖≤𝑛−1] The only case in which 𝑀 𝑡+1 > 𝑀 𝑡 is cases 3, which has a probability of 𝑛−𝑖 𝑛 ∗ 𝑛−𝑖 𝑛 = 𝑛−𝑖 2 𝑛 2 Therefore: 𝑃 𝑀 𝑡+1 > 𝑀 𝑡 𝑀 𝑡 =𝑖 = 𝑛−𝑖 2 𝑛 2

Random Transposition – Bounding MixingTime Theorem: Let 𝜏 be the first time 𝑀 𝑡 =𝑛.(The time it takes the decks to match each other) Then for every pair of initial permutations 𝜎 0 , 𝜎 0 ′ : 𝐸 𝜏 < 𝜋 2 6 𝑛 2 Proof: Let 𝜏 𝑖 be the amount of steps between the first time 𝑀 𝑡 ≥i−1 and the first time 𝑀 𝑡 ≥i. [since 𝑀 𝑡 could increase by 1,2,3, 𝜏 𝑖 could be equal zero] Then 𝜏= 𝜏 1 +…+ 𝜏 𝑛 We’ve seen 𝑃 𝑀 𝑡+1 > 𝑀 𝑡 𝑀 𝑡 =𝑖 = 𝑛−𝑖 2 𝑛 2 , therefore: 𝐸 𝜏 𝑖+1 𝑀 𝑡 =𝑖 = 𝑛 2 𝑛−𝑖 2 When no value of 𝑡 satisfies 𝑎 𝑡 =𝑖, 𝜏 𝑖+1 =0 therefore: 𝐸 𝜏 ≤ Σ 𝑖=0 𝑛−1 𝑛 2 𝑛−𝑖 2 = 𝑛 2 Σ 𝑖=0 𝑛−1 1 𝑛−𝑖 2 < 𝑛 2 Σ 𝑖=1 ∞ 1 𝑖 2 = 𝜋 2 6 𝑛 2

Random Transposition – Bounding Mixing Time We’ve just seen that 𝐸 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 < 𝜋 2 6 𝑛 2 , no matter what the initial permutations 𝜎 0 , 𝜎 0 ′ are. Therefore: 𝑑 𝑡 ≤ max 𝜎 0 , 𝜎 0 ′ ∈Ω 𝑃 𝜎 0 , 𝜎 0 ′ 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑡 ≤ 1 𝑡 max 𝜎 0 , 𝜎 0 ′ ∈Ω 𝐸 𝜎 0 , 𝜎 0 ′ 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 ≤ 1 𝑡 𝜋 2 6 𝑛 2 Therefore: For 𝑡≥ 𝜋 2 1.5 𝑛 2 we have 𝑑 𝑡 ≤ 1 𝑡 𝜋 2 6 𝑛 2 ≤ 1.5 𝑛 2 𝜋 2 𝜋 2 6 𝑛 2 ≤ 1 4 , therefore: 𝑡 𝑚𝑖𝑥 = min 𝑡 𝑑 𝑡 ≤ 1 4 ≤ 𝜋 2 1.5 𝑛 2 =𝑂( 𝑛 2 )

Lazy Random Walk On a Binary Tree Bounding mixing time

Finite Binary Tree

Reminders - Tree A tree is a connected (undirected!) graph with no cycles. The root is a distinguished vertex. The depth of a vertex v is the distance to the root. A level of a tree is a subset of vertices all at same depth Children of a vertex 𝑣∈𝑉 are neighbors of v with depth larger than v. A leaf is a vertex with degree one. 1 2

Reminders – Binary Tree of Depth k A tree where: Root has degree 2 Every vertex of distance≠0,𝑘 from root has degree 3 Vertices distance k from root are leaves

Lazy Random Walk On a Binary Tree We now consider the lazy random walk on the finite binary tree. [we denote by 𝑁(𝑣) the number of neighbors of given 𝑣∈𝑉] 𝑷 𝒖,𝒗 = 𝟏 𝟐𝑵(𝒖) 𝒊𝒇 (𝒖,𝒗)∈𝑬 𝟏 𝟐 𝒊𝒇 𝒖=𝒗 𝟎 𝑶.𝑾.

Coupling Definition We construct a coupling 𝑋 𝑡 , 𝑌 𝑡 𝑡=0 ∞ of two particles walking lazily on the binary tree by 2 phases. Phase 1 (as long as 𝑋 𝑡 , 𝑌 𝑡 not in same depth): Toss a coin to decide with chain of the 2 moves. Make random walk on chosen one. Phase 2 ( 𝑋 𝑡 , 𝑌 𝑡 in same depth): Make a random lazy walk on chain 𝑋 𝑡 and move accordingly on 𝑌 𝑡 [up/down-right/down-left]. Assume to be sticky… Indeed a coupling!!!

Phase 1 (as long as 𝑋 𝑡 , 𝑌 𝑡 not in same depth): Phase 2 ( 𝑋 𝑡 , 𝑌 𝑡 in same depth): Make a random lazy walk on chain 𝑋 𝑡 and move accordingly on 𝑌 𝑡 [up/down-right/down-left]. Phase 1 (as long as 𝑋 𝑡 , 𝑌 𝑡 not in same depth): Toss a coin to decide with chain of the 2 moves. Make random walk on chosen one.

Binary Tree Coupling - Observation Define the time 𝑡 0 of a run of the coupling as the first time 𝑋 𝑡 𝑡=0 ∞ has first visited a leaf and then visited the root. We argue that 𝑋 𝑡 0 = 𝑌 𝑡 0 . ( 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 ≤ 𝑡 0 ) Definition: Define the commute time as the time it takes in a rlw starting from root, visiting a leaf and then returning to the root. Coupling: Phase 1: move 1 or 2 uniformly Phase 2: move both sync. ≤ 𝑛𝑜𝑛−𝑡𝑟𝑖𝑣𝑖𝑎𝑙 𝑙𝑒𝑚𝑚𝑎 4𝑛 (𝑛=|𝑉|) ∀𝑥,𝑦 𝐸 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 ≤ 𝐸 𝑡 0 ≤𝐸[𝑐𝑜𝑚𝑚𝑢𝑡𝑒 𝑡𝑖𝑚𝑒]

Binary Tree Coupling - Observation ∀𝑥,𝑦 𝐸 𝑥,𝑦 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 ≤4𝑛 Therefore: 𝑑 𝑡 ≤ max 𝑥,𝑦∈Ω 𝑃 𝑥,𝑦 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 >𝑡 ≤ 1 𝑡 max 𝑥,𝑦∈Ω 𝐸 𝑥,𝑦 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 ≤ 4𝑛 𝑡 Therefore: For 𝑡≥16𝑛 we have 𝑑 𝑡 ≤ 4𝑛 16𝑛 ≤ 1 4 , therefore: 𝑡 𝑚𝑖𝑥 = min 𝑡 𝑑 𝑡 ≤ 1 4 ≤16𝑛=𝑂(𝑛)

Thanks!!! Jay tenenbaum