Inversions in randomly labelled trees Xing Shi Cai, Cecilia - PowerPoint PPT Presentation

Introduction Random Labels Random Trees Key Lemma Inversions in randomly labelled trees Xing Shi Cai, Cecilia Holmgren, Svante Janson, Tony Johansson and Fiona Skerman

Introduction Random Labels Random Trees Key Lemma Outline Inversions in labelled rooted trees; definitions, pictures Random labelling basic properties tree parameters, total path length results: cumulants, Bernoulli numbers Random labelling of random trees

Introduction Random Labels Random Trees Key Lemma Inversions in a Permutation/Labelled Tree inv (352146)

Introduction Random Labels Random Trees Key Lemma Inversions in a Permutation/Labelled Tree inv (352146)

Introduction Random Labels Random Trees Key Lemma Inversions in a Permutation/Labelled Tree inv (352146) = 6 ρ 3 u 2 5 a π ∈ S n u 3 2 inv ( π ) = #inversions in π u 4 1 a u 5 4 u 6 6

Introduction Random Labels Random Trees Key Lemma Inversions in a Permutation/Labelled Tree inv (352146) = 6 ρ 3 u 2 5 π ∈ S n u 3 2 inv ( π ) = #inversions in π u 4 1 u 5 4 u 6 6

Introduction Random Labels Random Trees Key Lemma Inversions in a Permutation/Labelled Tree inv (352146) = 6 ρ 3 u 2 5 5 > 1 u 2 an ancestor of u 4 u 3 2 u 4 1 u 5 4 u 6 6

Introduction Random Labels Random Trees Key Lemma Inversions in a Permutation/Labelled Tree inv (352146) = 6 ρ 3 T rooted tree with n nodes, u 2 5 5 > 1 write u < v if u is an ancestor of v u 2 an ancestor of u 4 π : V ( T ) → [ n ] u 3 2 u 4 1 u 5 4 u 6 6

Introduction Random Labels Random Trees Key Lemma Inversions in a Permutation/Labelled Tree inv (352146) = 6 ρ 3 T rooted tree with n nodes, u 2 5 5 > 1 write u < v if u is an ancestor of v u 2 an ancestor of u 4 π : V ( T ) → [ n ] u 3 2 � inv ( T , π ) = 1 [ π ( u ) > π ( v )] u 4 1 u < v u 5 4 u 6 6

Introduction Random Labels Random Trees Key Lemma Inversions in a Permutation/Labelled Tree inv (352146) = 6 ρ 3 T rooted tree with n nodes, u 2 5 a write u < v if u is an ancestor of v π : V ( T ) → [ n ] u 3 2 � inv ( T , π ) = 1 [ π ( u ) > π ( v )] u 4 1 a u < v P n path with n vetices, u 5 4 inv ( P n , π ) = inv ( π ) u 6 6

Introduction Random Labels Random Trees Key Lemma Inversions in a Permutation/Labelled Tree inv (352146) = 6 ρ T rooted tree with n nodes, write u < v if u is an ancestor of v π : V ( T ) → [ n ] � inv ( T , π ) = 1 [ π ( u ) > π ( v )] u < v P n path with n vetices, inv ( P n , π ) = inv ( π )

Introduction Random Labels Random Trees Key Lemma Inversions in a Permutation/Labelled Tree inv (352146) = 6 ρ 2 T rooted tree with n nodes, write u < v if u is an ancestor of v 9 11 π : V ( T ) → [ n ] � inv ( T , π ) = 1 [ π ( u ) > π ( v )] 8 12 3 6 5 u < v P n path with n vetices, inv ( P n , π ) = inv ( π ) 10 4 1 7

Introduction Random Labels Random Trees Key Lemma Inversions in a Permutation/Labelled Tree inv (352146) = 6 ρ 2 T rooted tree with n nodes, write u < v if u is an ancestor of v 9 11 π : V ( T ) → [ n ] � inv ( T , π ) = 1 [ π ( u ) > π ( v )] 8 12 3 6 5 u < v P n path with n vetices, inv ( P n , π ) = inv ( π ) 10 4 1 7 # inversions = 8

Introduction Random Labels Random Trees Key Lemma Random Node Labels: Expectation Start with fixed tree T , | T | = n ρ Choose π : V ( T ) → [ n ] uniformly

Introduction Random Labels Random Trees Key Lemma Random Node Labels: Expectation Start with fixed tree T , | T | = n ρ 2 Choose π : V ( T ) → [ n ] uniformly 9 11 8 12 3 6 5 10 4 1 7

Introduction Random Labels Random Trees Key Lemma Random Node Labels: Expectation Start with fixed tree T , | T | = n ρ Choose π : V ( T ) → [ n ] uniformly u � Inv ( T , π ) = 1 [ π ( u ) > π ( v )] u < v v

Introduction Random Labels Random Trees Key Lemma Random Node Labels: Expectation Start with fixed tree T , | T | = n ρ Choose π : V ( T ) → [ n ] uniformly � Inv ( T , π ) = 1 [ π ( u ) > π ( v )] u < v 1 2 = 1 � E [ Inv ( T )] = 2Υ( T ) u < v Υ( T ) is the total path length Υ( T ) = � v h ( v ), height of v is distance from ρ

Introduction Random Labels Random Trees Key Lemma Random Node Labels: Expectation Start with fixed tree T , | T | = n ρ Υ = 24 Choose π : V ( T ) → [ n ] uniformly � Inv ( T , π ) = 1 [ π ( u ) > π ( v )] u < v 1 2 = 1 � E [ Inv ( T )] = 2Υ( T ) u < v Υ( T ) is the total path length Υ( T ) = � v h ( v ), height of v is distance from ρ

Introduction Random Labels Random Trees Key Lemma Results on Fixed Trees For a path P n , asymptotic normality. Theorem Feller ’68, Let π be uniformly random permutation. Moment generating function e jt − 1 n E [ e tInv ( π ) ] = � j ( e t − 1) , j =1 and Inv ( π ) − E ( Inv ( π )) → N (0 , 1) . V ( Inv ( π )) Cumulant moments of r.v. X , κ k ( X ): ln E [ e tX ] = � ∞ k =0 κ k t k k ! For node u , let z u denote the number of nodes in the tree rooted at u . B k denotes the k -th Bernoulli number.

Introduction Random Labels Random Trees Key Lemma Results on Fixed Trees Theorem CHJJS ’17+ Let T be a fixed tree. Write X = Inv ( T , π ). Let κ k ( X ) be the k -th cumulant of X . Then E [ X ] = 1 � ( z v − 1) 2 v ∈ V � ( z 2 V [ X ] = v − 1) v ∈ V and more generally, κ k ( X ) = B k k ( − 1) k � ( z k v − 1) v ∈ V

Introduction Random Labels Random Trees Key Lemma Results on Fixed Trees Theorem CHJJS ’17+ Let T be a fixed tree. Write X = Inv ( T , π ). Let κ k ( X ) be the k -th cumulant of X . Then E [ X ] = 1 � ( z v − 1) 2 v ∈ V � ( z 2 V [ X ] = v − 1) v ∈ V and more generally, κ k ( X ) = B k k ( − 1) k � ( z k v − 1) v ∈ V e z v t − 1 � E [ e tX ] = z v ( e t − 1) . v ∈ V

Introduction Random Labels Random Trees Key Lemma Results: Combinatorial Interpretation v z k To express the sum � v . ρ Denote number of common ancestors c ( v 1 , . . . , v k ) = |{ u : u ≤ v i , ∀ i }|

Introduction Random Labels Random Trees Key Lemma Results: Combinatorial Interpretation v z k To express the sum � v . ρ Denote number of common ancestors c ( v 1 , . . . , v k ) = |{ u : u ≤ v i , ∀ i }|

Introduction Random Labels Random Trees Key Lemma Results: Combinatorial Interpretation v z k To express the sum � v . ρ Denote number of common ancestors c ( v 1 , . . . , v k ) = |{ u : u ≤ v i , ∀ i }| � Υ k ( T ) := c ( v 1 , . . . , v k ) v 1 ,..., v k

Introduction Random Labels Random Trees Key Lemma Results: Combinatorial Interpretation v z k To express the sum � v . ρ Denote number of common ancestors c ( v 1 , . . . , v k ) = |{ u : u ≤ v i , ∀ i }| � � z k Υ k ( T ) := c ( v 1 , . . . , v k ) = v v 1 ,..., v k v For one vertex c ( u ) = h ( u ) + 1, So Υ 1 ( T ) = Υ( T ) + n . Υ( T ) is the total path length Υ( T ) = � v h ( v ), height of v is distance from ρ

Introduction Random Labels Random Trees Key Lemma Results: Combinatorial Interpretation v z k To express the sum � v . ρ Denote number of common ancestors c ( v 1 , . . . , v k ) = |{ u : u ≤ v i , ∀ i }| � � z k Υ k ( T ) := c ( v 1 , . . . , v k ) = v v 1 ,..., v k v For one vertex c ( u ) = h ( u ) + 1, So Υ 1 ( T ) = Υ( T ) + n . Υ( T ) is the total path length Υ( T ) = � v h ( v ), height of v is distance from ρ

Introduction Random Labels Random Trees Key Lemma Results: Combinatorial Interpretation Theorem CHJJS ’17+ Let T be a fixed tree with n vertices. Write X = Inv ( T , π ). Let κ k ( X ) be the k -th cumulant of X . Then E [ X ] = 1 ( z v − 1) = 1 � 2Υ( T ) 2 v ∈ V v − 1) = 1 � ( z 2 V [ X ] = 12(Υ 2 ( T ) − n ) v ∈ V and more generally, κ k ( X ) = B k k ( − 1) k (Υ k ( T ) − n ) .

Introduction Random Labels Random Trees Key Lemma Galton Watson Trees Begin with one node. Recursively, each node has a random number of children. Number of children drawn independently from offspring distribution ξ .

Introduction Random Labels Random Trees Key Lemma Galton Watson Trees � For a unit Brownian excursion e ( u ), η = [0 , 1] 2 min s ≤ u ≤ t e ( u ) . Theorem CHJJS ’17+ Suppose T n is a conditional Galton-Watson tree with offspring distribution ξ such that E [ ξ ] = 1 and V [ ξ ] = σ 2 ∈ (0 , ∞ ), and define X n = Inv ( T n , π ) − Υ( T n ) / 2 . n 5 / 4 Then we have X n → d X ∼ 12 σ 1 / 2 √ η N , where N is a standard normal random variable, independent from the random variable η . This strengthen results of Panholtzer and Seitz 2012.

Introduction Random Labels Random Trees Key Lemma Key Lemma Let Z u = � v > u 1 [ π ( u ) > π ( v )]. ρ Inv ( T , π ) = � u Z u . u

Introduction Random Labels Random Trees Key Lemma Key Lemma Let Z u = � v > u 1 [ π ( u ) > π ( v )]. ρ Inv ( T , π ) = � u Z u . u Lemma For each node u , Z u ∼ Unif { 0 , 1 , . . . , z u − 1 } and furthermore the Z u ’s are independent.