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Basics Application: Erd¨ os-R´ enyi degree sequence Application: Harmonic functions on lattices and trees Modern Discrete Probability IV - Coupling Review S´ ebastien Roch UW–Madison Mathematics November 29, 2014 S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling

Basics Definitions and examples Application: Erd¨ os-R´ enyi degree sequence Coupling inequality Application: Harmonic functions on lattices and trees Basics 1 Definitions and examples Coupling inequality Application: Erd¨ os-R´ enyi degree sequence 2 Application: Harmonic functions on lattices and trees 3 S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling

Basics Definitions and examples Application: Erd¨ os-R´ enyi degree sequence Coupling inequality Application: Harmonic functions on lattices and trees Basic definitions Definition (Coupling) Let µ and ν be probability measures on the same measurable space ( S , S ) . A coupling of µ and ν is a probability measure γ on the product space ( S × S , S × S ) such that the marginals of γ coincide with µ and ν , i.e., γ ( A × S ) = µ ( A ) and γ ( S × A ) = ν ( A ) , ∀ A ∈ S . Similarly, for two random variables X and Y taking values in ( S , S ) , a coupling of X and Y is a joint variable ( X ′ , Y ′ ) taking values in ( S × S , S × S ) whose law is a coupling of the laws of X and Y . Note that X and Y need not be defined on the same probability space—but X ′ and Y ′ do need to. S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling

Basics Definitions and examples Application: Erd¨ os-R´ enyi degree sequence Coupling inequality Application: Harmonic functions on lattices and trees Examples I Example (Bernoulli variables) Let X and Y be Bernoulli random variables with parameters 0 ≤ q < r ≤ 1 respectively. That is, P [ X = 0 ] = 1 − q and P [ X = 1 ] = q , and similarly for Y . Here S = { 0 , 1 } and S = 2 S . - (Independent coupling) One coupling of X and Y is ( X ′ , Y ′ ) where d d X ′ = X and Y ′ = Y are independent . Its law is � ( 1 − q )( 1 − r ) � ( 1 − q ) r � � P [( X ′ , Y ′ ) = ( i , j )] i , j ∈{ 0 , 1 } = . q ( 1 − r ) qr - (Monotone coupling) Another possibility is to pick U uniformly at random in [ 0 , 1 ] , and set X ′′ = ✶ { U ≤ q } and Y ′′ = ✶ { U ≤ r } . The law of coupling ( X ′′ , Y ′′ ) is � 1 − r � r − q � � P [( X ′′ , Y ′′ ) = ( i , j )] i , j ∈{ 0 , 1 } = . 0 q S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling

Basics Definitions and examples Application: Erd¨ os-R´ enyi degree sequence Coupling inequality Application: Harmonic functions on lattices and trees Examples II Example (Bond percolation: monotonicity) Let G = ( V , E ) be a countable graph. Denote by P p the law of bond percolation on G with density p . Let x ∈ V and assume 0 ≤ q < r ≤ 1. - Let { U e } e ∈ E be independent uniforms on [ 0 , 1 ] . - For p ∈ [ 0 , 1 ] , let W p be the set of edges e such that U e ≤ p . Thinking of W p as specifying the open edges in the percolation process on G under P p , we see that ( W q , W r ) is a coupling of P q and P r with the property that P [ W q ⊆ W r ] = 1 . Let C ( q ) and C ( r ) be the open clusters of x under W q x x and W r respectively. Because C ( q ) ⊆ C ( r ) x , x θ ( q ) := P q [ |C x | = + ∞ ] = P [ |C ( q ) x | = + ∞ ] ≤ P [ |C ( r ) x | = + ∞ ] = θ ( r ) . S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling

Basics Definitions and examples Application: Erd¨ os-R´ enyi degree sequence Coupling inequality Application: Harmonic functions on lattices and trees Examples III Example (Biased random walks on Z ) For p ∈ [ 0 , 1 ] , let ( S ( p ) ) be nearest-neighbor random walk on Z t started at 0 with probability p of jumping to the right and probability 1 − p of jumping to the left. Assume 0 ≤ q < r ≤ 1. - Let ( X ′′ i , Y ′′ i ) i be an infinite sequence of i.i.d. monotone Bernoulli couplings with parameters q and r respectively. - Define ( Z ( q ) , Z ( r ) ) := ( 2 X ′′ i − 1 , 2 Y ′′ i − 1 ) . i i - Let ˆ S ( q ) i ≤ t Z ( q ) and ˆ S ( r ) i ≤ t Z ( r ) = � = � . t i t i Then (ˆ S ( q ) , ˆ S ( r ) ) is a coupling of ( S ( q ) , S ( r ) ) such that t t t t S ( q ) ˆ ≤ ˆ S ( r ) for all n almost surely. So for all y and all t t t P [ S ( q ) ≤ y ] = P [ˆ S ( q ) ≤ y ] ≥ P [ˆ S ( r ) ≤ y ] = P [ S ( r ) ≤ y ] . t t t t S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling

Basics Definitions and examples Application: Erd¨ os-R´ enyi degree sequence Coupling inequality Application: Harmonic functions on lattices and trees Coupling inequality I Let µ and ν be probability measures on ( S , S ) . Recall the definition of total variation distance: � µ − ν � TV := sup | µ ( A ) − ν ( A ) | . A ∈S Lemma Let µ and ν be probability measures on ( S , S ) . For any coupling ( X , Y ) of µ and ν , � µ − ν � TV ≤ P [ X � = Y ] . S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling

Basics Definitions and examples Application: Erd¨ os-R´ enyi degree sequence Coupling inequality Application: Harmonic functions on lattices and trees Coupling inequality II Proof: µ ( A ) − ν ( A ) = P [ X ∈ A ] − P [ Y ∈ A ] = P [ X ∈ A , X = Y ] + P [ X ∈ A , X � = Y ] − P [ Y ∈ A , X = Y ] − P [ Y ∈ A , X � = Y ] = P [ X ∈ A , X � = Y ] − P [ Y ∈ A , X � = Y ] ≤ P [ X � = Y ] , and, similarly, ν ( A ) − µ ( A ) ≤ P [ X � = Y ] . Hence | µ ( A ) − ν ( A ) | ≤ P [ X � = Y ] . S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling

Basics Definitions and examples Application: Erd¨ os-R´ enyi degree sequence Coupling inequality Application: Harmonic functions on lattices and trees Example: Poisson distributions Let X ∼ Poi ( λ ) and Y ∼ Poi ( ν ) with λ > ν . Recall that a sum of independent Poisson is Poisson. This fact leads to a natural coupling: let ˆ Y ∼ Poi ( ν ) , ˆ Z ∼ Poi ( λ − ν ) independently of Y , and ˆ X = ˆ Y + ˆ Z . Then (ˆ X , ˆ Y ) is a coupling and Z > 0 ] = 1 − e − ( λ − ν ) ≤ λ − ν. � µ X − µ Y � TV ≤ P [ˆ X � = ˆ Y ] = P [ˆ S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling

Basics Definitions and examples Application: Erd¨ os-R´ enyi degree sequence Coupling inequality Application: Harmonic functions on lattices and trees Maximal coupling I In fact, the inequality is tight. For simplicity, we prove this in the finite case only. Lemma Assume S is finite and let S = 2 S . Let µ and ν be probability measures on ( S , S ) . Then, � µ − ν � TV = inf { P [ X � = Y ] : coupling ( X , Y ) of µ and ν } . Let A = { x ∈ S : µ ( x ) > ν ( x ) } , B = { x ∈ S : µ ( x ) ≤ ν ( x ) } and � � � p := µ ( x ) ∧ ν ( x ) , α := [ µ ( x ) − ν ( x )] , β := [ ν ( x ) − µ ( x )] . x ∈ S x ∈ A x ∈ B S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling

Basics Definitions and examples Application: Erd¨ os-R´ enyi degree sequence Coupling inequality Application: Harmonic functions on lattices and trees Maximal coupling II Figure : Proof by picture that: 1 − p = α = β = � µ − ν � TV . S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling

Basics Definitions and examples Application: Erd¨ os-R´ enyi degree sequence Coupling inequality Application: Harmonic functions on lattices and trees Maximal coupling III Proof: Lemma: � x ∈ S µ ( x ) ∧ ν ( x ) = 1 − � µ − ν � TV . Proof of lemma: � 2 � µ − ν � TV = | µ ( x ) − ν ( x ) | x ∈ S � � = [ µ ( x ) − ν ( x )] + [ ν ( x ) − µ ( x )] x ∈ A x ∈ B � � � = µ ( x ) + ν ( x ) − µ ( x ) ∧ ν ( x ) x ∈ A x ∈ B x ∈ S � � � = 2 − µ ( x ) − ν ( x ) − µ ( x ) ∧ ν ( x ) x ∈ B x ∈ A x ∈ S � = 2 − 2 µ ( x ) ∧ ν ( x ) . x ∈ S Lemma: � x ∈ A [ µ ( x ) − ν ( x )] = � x ∈ B [ ν ( x ) − µ ( x )] = � µ − ν � TV = 1 − p . Proof: First equality is immediate. Second equality follows from second line in previous lemma. S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling

Basics Definitions and examples Application: Erd¨ os-R´ enyi degree sequence Coupling inequality Application: Harmonic functions on lattices and trees Maximal coupling IV The maximal coupling is defined as follows: - With probability p , pick X = Y from γ min where γ min ( x ) := 1 p µ ( x ) ∧ ν ( x ) , x ∈ S . - Otherwise, pick X from γ A where γ A ( x ) := µ ( x ) − ν ( x ) , x ∈ A , and, 1 − p independently, pick Y from γ B ( x ) := ν ( x ) − µ ( x ) , x ∈ B . Note that X � = Y 1 − p in that case because A and B are disjoint. The marginal law of X at x ∈ S is p γ min ( x ) + ( 1 − p ) γ A ( x ) = µ ( x ) , and similarly for Y . Finally P [ X � = Y ] = 1 − p = � µ − ν � TV . S´ ebastien Roch, UW–Madison Modern Discrete Probability – Coupling