1. 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

2. Reminders
In case you forgot…

3. 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

4. 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

5. 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) ]

6. 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}

7. 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

8. 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

9. Random Walk
Coupling example

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

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

12. 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, 𝑏 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.

13. 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

14. Markov Chain Coupling

15. Markov Chain Coupling

16. Coupling Example

17. Coupling Example

18. Coupling Example

19. Coupling Example

20. Coupling Example

21. 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 .

22. 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

23. 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]

24. 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}

25. Bounding Distance to Stationary Using Couplings

26. 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)

27. 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}

28. 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

29. 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)

30. 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!

31. Lazy Random Walk On The Cycle
Bounding mixing time

32. 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

33. 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…

34. Random Lazy Walk On The Cycle
Coupling summary: choose 𝑋 𝑡 , 𝑌 𝑡 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.

35. Random Lazy Walk On The Cycle – Bounding 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒
Coupling summary: choose 𝑋 𝑡 , 𝑌 𝑡 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

36. Random Lazy Walk On The Cycle
Coupling summary: choose 𝑋 𝑡 , 𝑌 𝑡 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 ≤ 1 𝑛 2 ∗ 𝑛 2 4 ≤ 1 4 , therefore:
𝑡 𝑚𝑖𝑥 = min 𝑡 𝑑 𝑡 ≤ ≤ 𝑛 2

37. Lazy Random Walk On The Hypercube
Bounding mixing time

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

39. 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

40. 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)

41. 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))

42. 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

43. Card Shuffling – Random Transposition
Bounding mixing time

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

45. 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!!!

46. 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

47. 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

48. 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

49. 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

50. 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 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− 𝑛−𝑖 2 < 𝑛 2 Σ 𝑖=1 ∞ 1 𝑖 2 = 𝜋 𝑛 2

51. Random Transposition – Bounding Mixing Time
We’ve just seen that 𝐸 𝜏 𝑐𝑜𝑢𝑝𝑙𝑒 < 𝜋 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 𝑛 2
Therefore: For 𝑡≥ 𝜋 𝑛 2 we have 𝑑 𝑡 ≤ 1 𝑡 𝜋 𝑛 2 ≤ 1.5 𝑛 2 𝜋 2 𝜋 𝑛 2 ≤ 1 4 , therefore:
𝑡 𝑚𝑖𝑥 = min 𝑡 𝑑 𝑡 ≤ ≤ 𝜋 𝑛 2 =𝑂( 𝑛 2 )

52. Lazy Random Walk On a Binary Tree
Bounding mixing time

53. Finite Binary Tree

54. 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

55. 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

56. 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.

57. 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!!!

58. 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.

59. 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 ( 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]

60. 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 ≤ ≤16𝑛=𝑂(𝑛)

61. Thanks!!!
Jay tenenbaum