The Asymptotic Distribution of Symbols on Staircase Tableaux - - PowerPoint PPT Presentation

Definitions Connections/Motivation Previous Result Results

The Asymptotic Distribution of Symbols on Staircase Tableaux Diagonals

Amanda Lohss September 15, 2016

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Outline Definition of Staircase Tableaux Connections/Motivation Previous Results Results Open Problems

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Staircase Tableaux (Corteel-Williams (2010))

γ α δ β γ δ β γ β α

Figure: An example of a staircase tableau of size 7.

Definition A staircase tableau of size n is a Young diagram of shape (n, n-1, ..., 1) such that:

1 The boxes are empty or contain

an α, β, γ, or δ.

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Staircase Tableaux (Corteel-Williams (2010))

γ α δ β γ δ β γ β α

Figure: An example of a staircase tableau of size 7.

Definition A staircase tableau of size n is a Young diagram of shape (n, n-1, ..., 1) such that:

1 The boxes are empty or contain

an α, β, γ, or δ.

2 Every box on the diagonal

contains a symbol.

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Staircase Tableaux (Corteel-Williams (2010))

γ α δ β γ δ β γ β α

Figure: An example of a staircase tableau of size 7.

Definition A staircase tableau of size n is a Young diagram of shape (n, n-1, ..., 1) such that:

1 The boxes are empty or contain

an α, β, γ, or δ.

2 Every box on the diagonal

contains a symbol.

3 All boxes in the same column

and above an α or γ are empty.

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Staircase Tableaux (Corteel-Williams (2010))

γ α δ β γ δ β γ β α

Figure: An example of a staircase tableau of size 7.

Definition A staircase tableau of size n is a Young diagram of shape (n, n-1, ..., 1) such that:

1 The boxes are empty or contain

an α, β, γ, or δ.

2 Every box on the diagonal

contains a symbol.

3 All boxes in the same column

and above an α or γ are empty.

4 All boxes in the same row and to

the left of an β or δ are empty.

Amanda Lohss Purdue 2016

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Preliminaries

γ α δ β γ δ β γ β α

Figure: A staircase tableau with weight α2β3γ3δ2.

The rows and columns in a staircase tableau are numbered from 1 through n, beginning with the box in the NW-corner and continuing south and east respectively.

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Preliminaries

γ α δ β γ δ β γ β α

Figure: A staircase tableau with weight α2β3γ3δ2.

The rows and columns in a staircase tableau are numbered from 1 through n, beginning with the box in the NW-corner and continuing south and east respectively. Symmetric with respect to interchanging rows/columns, α/β, and γ/δ.

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Preliminaries

γ α δ β γ δ β γ β α

Figure: A staircase tableau with weight α2β3γ3δ2.

The rows and columns in a staircase tableau are numbered from 1 through n, beginning with the box in the NW-corner and continuing south and east respectively. Symmetric with respect to interchanging rows/columns, α/β, and γ/δ. The weight of a staircase tableau is the product of all its symbols.

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Preliminaries Cont’d

γ α δ β γ δ β γ β α

Figure: A staircase tableau with weight α2β3γ3δ2.

As proven by Corteel and Williams, summing over the weight of all staircase tableaux gives,

• S∈Sn

wt(S) =

n−1

• i=0

(α + β + δ + γ + i(α + γ)(β + δ)). and therefore the total number of staircase tableaux is 4n · n!.

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Connections

Introduced due to connections with the asymmetric simple exclusion process (ASEP), an important particle model with applications in physics, biology and biochemistry.

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Connections

Introduced due to connections with the asymmetric simple exclusion process (ASEP), an important particle model with applications in physics, biology and biochemistry. According to Yau, the ASEP is “the default stochastic model for transport phenomena.”

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Connections

Introduced due to connections with the asymmetric simple exclusion process (ASEP), an important particle model with applications in physics, biology and biochemistry. According to Yau, the ASEP is “the default stochastic model for transport phenomena.” Numerous other connections such as Askey-Wilson polynomials, tree–like tableaux, permutation tableaux, and permutations.

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

The ASEP

A Markov Chain with n sites.

• ◦ • • ◦ • ◦ ◦ •

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

The ASEP

A Markov Chain with n sites.

• ◦ • • ◦ • ◦ ◦ •

Transition Probabilities:

• A to • A :

α n + 1

• A to ◦ A :

γ n + 1

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The ASEP

A Markov Chain with n sites.

• ◦ • • ◦ • ◦ ◦ •

Transition Probabilities:

• A to • A :

α n + 1

• A to ◦ A :

γ n + 1 A ◦ to A• : δ n + 1 A • to A◦ : β n + 1

Amanda Lohss Purdue 2016

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The ASEP

A Markov Chain with n sites.

• ◦ • • ◦ • ◦ ◦ •

Transition Probabilities:

• A to • A :

α n + 1

• A to ◦ A :

γ n + 1 A ◦ to A• : δ n + 1 A • to A◦ : β n + 1 A • ◦B to A ◦ •B : u n + 1 A ◦ •B to A • ◦B : q n + 1

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Connection with the ASEP

Type of a staircase tableaux:

• for each α or δ on diagonal.
• for each β or γ on diagonal.

γ α δ β γ δ β γ β α

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Connection with the ASEP

Type of a staircase tableaux:

• for each α or δ on diagonal.
• for each β or γ on diagonal.

γ α δ β γ δ β γ β α

• Amanda Lohss

Purdue 2016

Definitions Connections/Motivation Previous Result Results

Connection with the ASEP

Type of a staircase tableaux:

• for each α or δ on diagonal.
• for each β or γ on diagonal.

Filling rules for u’s and q’s:

1 u’s in all boxes east of a β and

q’s in all boxes east of a δ.

2

γ α δ β γ δ β γ β α

• Amanda Lohss

Purdue 2016

Definitions Connections/Motivation Previous Result Results

Connection with the ASEP

Type of a staircase tableaux:

• for each α or δ on diagonal.
• for each β or γ on diagonal.

Filling rules for u’s and q’s:

1 u’s in all boxes east of a β and

q’s in all boxes east of a δ.

2

γ α δ β γ δ β γ β α

• u u

u u

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Connection with the ASEP

Type of a staircase tableaux:

• for each α or δ on diagonal.
• for each β or γ on diagonal.

Filling rules for u’s and q’s:

1 u’s in all boxes east of a β and

q’s in all boxes east of a δ.

2

γ α δ β γ δ β γ β α

• u u

u u q q q q q q q q

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Connection with the ASEP

Type of a staircase tableaux:

• for each α or δ on diagonal.
• for each β or γ on diagonal.

Filling rules for u’s and q’s:

1 u’s in all boxes east of a β and

q’s in all boxes east of a δ.

2 u’s in all boxes north of a α or δ

and q’s in all boxes north of a β

• r γ.

γ α δ β γ δ β γ β α

• u u

u u q q q q q q q q

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Connection with the ASEP

Type of a staircase tableaux:

• for each α or δ on diagonal.
• for each β or γ on diagonal.

Filling rules for u’s and q’s:

1 u’s in all boxes east of a β and

q’s in all boxes east of a δ.

2 u’s in all boxes north of a α or δ

and q’s in all boxes north of a β

• r γ.

γ α δ β γ δ β γ β α

• u u

u u q q q q q q q q u u

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Connection with the ASEP

Type of a staircase tableaux:

• for each α or δ on diagonal.
• for each β or γ on diagonal.

Filling rules for u’s and q’s:

1 u’s in all boxes east of a β and

q’s in all boxes east of a δ.

2 u’s in all boxes north of a α or δ

and q’s in all boxes north of a β

• r γ.

γ α δ β γ δ β γ β α

• u u

u u q q q q q q q q u u q q q q

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Steady State Probability of the ASEP

Theorem (Corteel and Williams) The steady state probability that the ASEP is in state η is:

• T∈T wt(T)
• S∈Sn wt(S).

γ α δ β γ δ β γ β α

• u u

u u q q q q q q q q u u q q q q

Figure: A staircase tableau and its type

• • ◦ • ◦ ◦ ◦

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Random α/β-Staircase Tableaux

For combinatorical considerations, let u = q = 1. Then W.L.O.G. we can study α/β-staircase tableaux. It was shown that

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Random α/β-Staircase Tableaux

For combinatorical considerations, let u = q = 1. Then W.L.O.G. we can study α/β-staircase tableaux. It was shown that

• S∈Sn

wt(S) = αnβn(a + b)n where a := α−1 and b := β−1.

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Random α/β-Staircase Tableaux

For combinatorical considerations, let u = q = 1. Then W.L.O.G. we can study α/β-staircase tableaux. It was shown that

• S∈Sn

wt(S) = αnβn(a + b)n where a := α−1 and b := β−1. The total number of α/β-staircase tableaux is |Sn| = 2n = (n + 1)!

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

Random α/β-Staircase Tableaux

For combinatorical considerations, let u = q = 1. Then W.L.O.G. we can study α/β-staircase tableaux. It was shown that

• S∈Sn

wt(S) = αnβn(a + b)n where a := α−1 and b := β−1. The total number of α/β-staircase tableaux is |Sn| = 2n = (n + 1)! Random α/β-staircase tableaux: P(Sn,α,β = S) = wt(S) αnβn(a + b)n .

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Distribution of Symbols

Because of the ASEP, the following random variables are interesting:

1 Ak

n, the number of α’s along the kth

diagonal.

2 Bk

n , the number of β’s along the kth

diagonal.

3 X k

n , the number of non-empty boxes

along the kth diagonal.

• Amanda Lohss

Purdue 2016

Definitions Connections/Motivation Previous Result Results

Some Previous Results (Hitczenko-Janson)

Previous Result: A1

n − n/2

√n

d

→ N(0, 1/12)

Amanda Lohss Purdue 2016

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Some Previous Results (Hitczenko-Janson)

Previous Result: A1

n − n/2

√n

d

→ N(0, 1/12) Conjecture: Ak

n and Bk n are asymptotically Poisson (k ≥ 2).

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Results

Theorem Let Pois(λ) be a Poisson random variable with parameter λ. Then as n → ∞: Ak

n d

→ Pois 1 2

• Bk

n d

→ Pois 1 2

• X k

n d

→ Pois (1)

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Outline of Proof

Theorem (Method of Factorial Moments) If a sequence of random variables {Xk}n

k=1 is such that

lim

n→∞ E(Xk)r → λr,

r = 0, 1, . . . then, Xn

d

→ Pois(λ) For this calculation, one needs to calculate P(αj1 ∩ αj2 · · · ∩ αjr ).

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2nd and 3rd Diagonals (with Pawe l Hitczenko)

Method: conditional probability and induction P(αj1 ∩ αj2 · · · ∩ αjr ) = P(αj2 · · · ∩ αjr |αj1) · P(αj1). For the second and third diagonal, P(αj2 · · · ∩ αjr |αj1) can be computed by considering all cases. α α β

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2nd and 3rd Diagonals (with Pawe l Hitczenko)

Method: conditional probability and induction P(αj1 ∩ αj2 · · · ∩ αjr ) = P(αj2 · · · ∩ αjr |αj1) · P(αj1). For the second and third diagonal, P(αj2 · · · ∩ αjr |αj1) can be computed by considering all cases. α α β

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2nd and 3rd Diagonals (with Pawe l Hitczenko)

Method: conditional probability and induction P(αj1 ∩ αj2 · · · ∩ αjr ) = P(αj2 · · · ∩ αjr |αj1) · P(αj1). For the second and third diagonal, P(αj2 · · · ∩ αjr |αj1) can be computed by considering all cases. α α β α β

Amanda Lohss Purdue 2016

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2nd and 3rd Diagonals (with Pawe l Hitczenko)

Method: conditional probability and induction P(αj1 ∩ αj2 · · · ∩ αjr ) = P(αj2 · · · ∩ αjr |αj1) · P(αj1). For the second and third diagonal, P(αj2 · · · ∩ αjr |αj1) can be computed by considering all cases. α α β α β

Amanda Lohss Purdue 2016

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2nd and 3rd Diagonals (with Pawe l Hitczenko)

Method: conditional probability and induction P(αj1 ∩ αj2 · · · ∩ αjr ) = P(αj2 · · · ∩ αjr |αj1) · P(αj1). For the second and third diagonal, P(αj2 · · · ∩ αjr |αj1) can be computed by considering all cases. α α β β α

Amanda Lohss Purdue 2016

Definitions Connections/Motivation Previous Result Results

2nd and 3rd Diagonals (with Pawe l Hitczenko)

Method: conditional probability and induction P(αj1 ∩ αj2 · · · ∩ αjr ) = P(αj2 · · · ∩ αjr |αj1) · P(αj1). For the second and third diagonal, P(αj2 · · · ∩ αjr |αj1) can be computed by considering all cases. α α β β α

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Kth Diagonal

α α β

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Kth Diagonal

α α β α β

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Kth Diagonal

α α β α β β α

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D-connected symbols

Definition Define a symbol to be D-connected if it is not on the first diagonal and one of the following two conditions hold:

1 The symbol lies on D but

is not on the kth diagonal.

2 There exists a symbol

above or to the left that is D-connected or lies on D. α α β β α β α β α β β β α β α β α α β α β

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D-connected symbols

Lemma Properties of D-connected symbols:

1 Any symbol in the same column

as a D-connected α or the same row as a D-connected β is also D-connected.

2 There are at most k − 2

D-connected symbols.

3 Each D-connected symbol can

be paired uniquely with an

• pposite symbol on the first

diagonal. α α β β α β α β α β β β α β α β α α β α β

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Sub-tableaux Decomposition

β α β α β β β α α α β α β β β α α β β α α β α β β β α α − → α β

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Sub-tableaux Decomposition

β α β α β β β α α α β α β β β α α β β α α β α β β β α α − → α β

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Sub-tableaux Decomposition

β α β α β β β α α α β α β β β α α β β α α β α β β β α α − → α β

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Sub-tableaux Decomposition

β α β α β β β α α α β α β β β α α β β α α β α β β β α α − → α β

Notice that the weight of the larger tableau is α2βα2β2 times the weight

• f the smaller tableau.

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Arrangements of D-connected symbols

Define Ak,h to be the the set of all possible arrangements of h D-connected symbols in a tableau of size k. Example: A4,2 = {β2

1 ∩ β3 1, β2 1 ∩ α2 3, β3 1 ∩ α3 2, β2 2 ∩ α3 2, β3 1 ∩ α2 2, α3 2 ∩ α2 3}

Note that Ak,0 = {∅} for all k ≥ 1.

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Lemma There exists a bijection between tableaux of size n with α1 and triples (h, a, T) where 0 ≤ h ≤ k − 2, a ∈ Ak,h and T is a tableaux

• f size n − h − 2

β α β α β β β α α α β α β β β α α β β α α β α β β β α α − → α β

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An example of the bijection

− →

• 0, ∅, β
• β

β α α − →

• 0, ∅, α
• β

α α α − →

• 1, α2

2, ∅

• β

β α α − →

• 1, β2

1, ∅

• β

α α α β α

Figure: The bijection when n = 3 and k = 3.

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Proof: Surjectivity

β β β β α α α α ↓ ↓ ↓ ↓ ↓ β β α α α * * * * * * * * * * * * * * * β β β α α α β β β α α α α α − → β α β α

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Key Result

If Ck,h :=|Ak,h|,

• S∈Tn,α

wt(S) =

k−2

• h=0

Ck,hαh+2βh+1

• T∈Tn−h−2,α,h

wt(T). Therefore, Pn,α,β(αk

1, αk j2, . . . , αk jr )

=

k−2

• h=0

Ck,h b (n + a + b − 1)h+2 Pn−h−2,α,β(αj2−h−2, . . . , αjr−h−2).

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The Distribution of Alpha Boxes

Theorem Let 1 ≤ j1 < ... < jr ≤ n − k + 1. If jl ≤ jl+1 − k, ∀l = 1, 2, ..., r − 1 then Pn,α,β(αj1, ..., αjr ) =

r

• l=1

jr−l+1 n2 + O

• 1

nr+1

• .

Otherwise, Pn,α,β(αj1, ..., αjr ) = O 1 nr

• .

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Open Problems

What about if k is not fixed? What about other regions? Asymptotic joint distribution of symbols on different diagonals?

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Thank you!

Amanda Lohss Purdue 2016