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

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Introduction Random Labels Random Trees Key Lemma

Inversions in randomly labelled trees

Xing Shi Cai, Cecilia Holmgren, Svante Janson, Tony Johansson and Fiona Skerman

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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

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Introduction Random Labels Random Trees Key Lemma

Inversions in a Permutation/Labelled Tree

inv(352146)

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Introduction Random Labels Random Trees Key Lemma

Inversions in a Permutation/Labelled Tree

inv(352146)

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Introduction Random Labels Random Trees Key Lemma

Inversions in a Permutation/Labelled Tree

π ∈ Sn inv(π) = #inversions in π inv(352146) = 6

ρ 3 2 4 6 u2 u3 u4 u5 u6

a a

5 1

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Introduction Random Labels Random Trees Key Lemma

Inversions in a Permutation/Labelled Tree

π ∈ Sn inv(π) = #inversions in π inv(352146) = 6

ρ 3 2 4 6 u2 u3 u4 u5 u6 5 1

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Introduction Random Labels Random Trees Key Lemma

Inversions in a Permutation/Labelled Tree

inv(352146) = 6 5 > 1 u2 an ancestor of u4

ρ 3 2 4 6 u2 u3 u4 u5 u6 5 1

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Introduction Random Labels Random Trees Key Lemma

Inversions in a Permutation/Labelled Tree

inv(352146) = 6 5 > 1 u2 an ancestor of u4 T rooted tree with n nodes, write u < v if u is an ancestor of v π : V (T) → [n]

ρ 3 2 4 6 u2 u3 u4 u5 u6 5 1

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Introduction Random Labels Random Trees Key Lemma

Inversions in a Permutation/Labelled Tree

inv(352146) = 6 5 > 1 u2 an ancestor of u4 T rooted tree with n nodes, write u < v if u is an ancestor of v π : V (T) → [n] inv(T, π) =

u<v

1[π(u) > π(v)]

ρ 3 2 4 6 u2 u3 u4 u5 u6 5 1

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Introduction Random Labels Random Trees Key Lemma

Inversions in a Permutation/Labelled Tree

inv(352146) = 6 T rooted tree with n nodes, write u < v if u is an ancestor of v π : V (T) → [n] inv(T, π) =

u<v

1[π(u) > π(v)] Pn path with n vetices, inv(Pn, π) = inv(π)

ρ 3 2 4 6 u2 u3 u4 u5 u6

a a

5 1

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Introduction Random Labels Random Trees Key Lemma

Inversions in a Permutation/Labelled Tree

inv(352146) = 6 T rooted tree with n nodes, write u < v if u is an ancestor of v π : V (T) → [n] inv(T, π) =

u<v

1[π(u) > π(v)] Pn path with n vetices, inv(Pn, π) = inv(π)

ρ

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Introduction Random Labels Random Trees Key Lemma

Inversions in a Permutation/Labelled Tree

inv(352146) = 6 T rooted tree with n nodes, write u < v if u is an ancestor of v π : V (T) → [n] inv(T, π) =

u<v

1[π(u) > π(v)] Pn path with n vetices, inv(Pn, π) = inv(π)

ρ 2 9 11 8 12 3 10 4 6 5 1 7

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Introduction Random Labels Random Trees Key Lemma

Inversions in a Permutation/Labelled Tree

inv(352146) = 6 T rooted tree with n nodes, write u < v if u is an ancestor of v π : V (T) → [n] inv(T, π) =

u<v

1[π(u) > π(v)] Pn path with n vetices, inv(Pn, π) = inv(π) # inversions = 8

ρ 2 9 11 8 12 3 10 4 6 5 1 7

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Introduction Random Labels Random Trees Key Lemma

Random Node Labels: Expectation

Start with fixed tree T, |T| = n Choose π : V (T) → [n] uniformly

ρ

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Introduction Random Labels Random Trees Key Lemma

Random Node Labels: Expectation

Start with fixed tree T, |T| = n Choose π : V (T) → [n] uniformly

ρ 2 9 11 8 12 3 10 4 6 5 1 7

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Introduction Random Labels Random Trees Key Lemma

Random Node Labels: Expectation

Start with fixed tree T, |T| = n Choose π : V (T) → [n] uniformly Inv(T, π) =

u<v

1[π(u) > π(v)]

ρ u v

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Introduction Random Labels Random Trees Key Lemma

Random Node Labels: Expectation

Start with fixed tree T, |T| = n Choose π : V (T) → [n] uniformly Inv(T, π) =

u<v

1[π(u) > π(v)] E[Inv(T)] =

u<v

1 2 = 1 2Υ(T) Υ(T) is the total path length Υ(T) =

v h(v), height of v is distance from ρ

ρ

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Introduction Random Labels Random Trees Key Lemma

Random Node Labels: Expectation

Start with fixed tree T, |T| = n Choose π : V (T) → [n] uniformly Inv(T, π) =

u<v

1[π(u) > π(v)] E[Inv(T)] =

u<v

1 2 = 1 2Υ(T) Υ(T) is the total path length Υ(T) =

v h(v), height of v is distance from ρ

Υ = 24

ρ

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Introduction Random Labels Random Trees Key Lemma

Results on Fixed Trees

For a path Pn, asymptotic normality. Theorem Feller ’68, Let π be uniformly random permutation. Moment generating function E[etInv(π)] =

n

j=1

ejt − 1 j(et − 1), and Inv(π) − E(Inv(π)) V(Inv(π)) → N(0, 1). For node u, let zu denote the number of nodes in the tree rooted at u. Cumulant moments of r.v. X, κk(X): ln E[etX] = ∞

k=0 κk tk k!

Bk denotes the k-th Bernoulli number.

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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 κk(X) be the k-th cumulant of X. Then E[X] = 1 2

v∈V

(zv − 1) V[X] =

v∈V

(z2

v − 1)

and more generally, κk(X) = Bk k (−1)k

v∈V

(zk

v − 1)

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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 κk(X) be the k-th cumulant of X. Then E[X] = 1 2

v∈V

(zv − 1) V[X] =

v∈V

(z2

v − 1)

and more generally, κk(X) = Bk k (−1)k

v∈V

(zk

v − 1)

E[etX] =

v∈V

ezvt − 1 zv(et − 1).

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Introduction Random Labels Random Trees Key Lemma

Results: Combinatorial Interpretation

To express the sum

v zk v .

Denote number of common ancestors c(v1, . . . , vk) = |{u : u ≤ vi, ∀i}|

ρ

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Introduction Random Labels Random Trees Key Lemma

Results: Combinatorial Interpretation

To express the sum

v zk v .

Denote number of common ancestors c(v1, . . . , vk) = |{u : u ≤ vi, ∀i}|

ρ

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Introduction Random Labels Random Trees Key Lemma

Results: Combinatorial Interpretation

To express the sum

v zk v .

Denote number of common ancestors c(v1, . . . , vk) = |{u : u ≤ vi, ∀i}| Υk(T) :=

v1,...,vk

c(v1, . . . , vk)

ρ

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Introduction Random Labels Random Trees Key Lemma

Results: Combinatorial Interpretation

To express the sum

v zk v .

Denote number of common ancestors c(v1, . . . , vk) = |{u : u ≤ vi, ∀i}| Υk(T) :=

v1,...,vk

c(v1, . . . , vk) =

v

zk

v

For one vertex c(u) = h(u) + 1, So Υ1(T) = Υ(T) + n. Υ(T) is the total path length Υ(T) =

v h(v), height of v is distance from ρ

ρ

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Introduction Random Labels Random Trees Key Lemma

Results: Combinatorial Interpretation

To express the sum

v zk v .

Denote number of common ancestors c(v1, . . . , vk) = |{u : u ≤ vi, ∀i}| Υk(T) :=

v1,...,vk

c(v1, . . . , vk) =

v

zk

v

For one vertex c(u) = h(u) + 1, So Υ1(T) = Υ(T) + n. Υ(T) is the total path length Υ(T) =

v h(v), height of v is distance from ρ

ρ

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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 κk(X) be the k-th cumulant of X. Then E[X] = 1 2

v∈V

(zv − 1) = 1 2Υ(T) V[X] =

v∈V

(z2

v − 1) = 1

12(Υ2(T) − n) and more generally, κk(X) = Bk k (−1)k(Υk(T) − n).

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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.

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Introduction Random Labels Random Trees Key Lemma

Galton Watson Trees

This strengthen results of Panholtzer and Seitz 2012. Theorem CHJJS ’17+ Suppose Tn is a conditional Galton-Watson tree with offspring distribution ξ such that E[ξ] = 1 and V[ξ] = σ2 ∈ (0, ∞), and define Xn = Inv(Tn, π) − Υ(Tn)/2 n5/4 . Then we have Xn →d X ∼ 12σ1/2√η N, where N is a standard normal random variable, independent from the random variable η. For a unit Brownian excursion e(u), η =

[0,1]2 mins≤u≤t e(u).

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Introduction Random Labels Random Trees Key Lemma

Key Lemma

Let Zu =

v>u 1[π(u) > π(v)].

Inv(T, π) =

u Zu.

ρ

u

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Introduction Random Labels Random Trees Key Lemma