Building Random Trees from Blocks Mohan Gopaladesikan Department of - - PowerPoint PPT Presentation

Building Random Trees from Blocks

Mohan Gopaladesikan

Department of Statistics, Purdue University

11th June 2013

Joint work with

• Dr. Hosam Mahmoud

Department of Statistics, The George Washington University

• Dr. Mark Daniel Ward

Department of Statistics, Purdue University 1

The Probability Model

We have a finite collection of unlabeled, rooted, nonplanar building blocks (trees) C = {T1, . . . , Tk}, that occur with respective probabilities p1, . . . , pk; k

j=1 pj = 1. 2

The Probability Model

We have a finite collection of unlabeled, rooted, nonplanar building blocks (trees) C = {T1, . . . , Tk}, that occur with respective probabilities p1, . . . , pk; k

j=1 pj = 1.

We use these as building blocks of an unlabeled, rooted, nonplanar tree called the “blocks tree”.

2

The Probability Model

We have a finite collection of unlabeled, rooted, nonplanar building blocks (trees) C = {T1, . . . , Tk}, that occur with respective probabilities p1, . . . , pk; k

j=1 pj = 1.

We use these as building blocks of an unlabeled, rooted, nonplanar tree called the “blocks tree”. Let Tn−1 be the blocks tree after inserting n − 1 blocks.

2

The Probability Model

We have a finite collection of unlabeled, rooted, nonplanar building blocks (trees) C = {T1, . . . , Tk}, that occur with respective probabilities p1, . . . , pk; k

j=1 pj = 1.

We use these as building blocks of an unlabeled, rooted, nonplanar tree called the “blocks tree”. Let Tn−1 be the blocks tree after inserting n − 1 blocks. At step n, with probability pi we choose the block Ti from C , and we adjoin it to the tree Tn−1 by choosing a “parent” node at random from Tn−1 (all nodes from Tn−1 being equally likely parents).

2

The Probability Model

We have a finite collection of unlabeled, rooted, nonplanar building blocks (trees) C = {T1, . . . , Tk}, that occur with respective probabilities p1, . . . , pk; k

j=1 pj = 1.

We use these as building blocks of an unlabeled, rooted, nonplanar tree called the “blocks tree”. Let Tn−1 be the blocks tree after inserting n − 1 blocks. At step n, with probability pi we choose the block Ti from C , and we adjoin it to the tree Tn−1 by choosing a “parent” node at random from Tn−1 (all nodes from Tn−1 being equally likely parents). We then attach the root of block Ti to the chosen parent.

2

The Probability Model

We have a finite collection of unlabeled, rooted, nonplanar building blocks (trees) C = {T1, . . . , Tk}, that occur with respective probabilities p1, . . . , pk; k

j=1 pj = 1.

We use these as building blocks of an unlabeled, rooted, nonplanar tree called the “blocks tree”. Let Tn−1 be the blocks tree after inserting n − 1 blocks. At step n, with probability pi we choose the block Ti from C , and we adjoin it to the tree Tn−1 by choosing a “parent” node at random from Tn−1 (all nodes from Tn−1 being equally likely parents). We then attach the root of block Ti to the chosen parent. For mathematical simplicity: We study the case where all members of C have the same size t.

2

The Probability Model

We have a finite collection of unlabeled, rooted, nonplanar building blocks (trees) C = {T1, . . . , Tk}, that occur with respective probabilities p1, . . . , pk; k

j=1 pj = 1.

We use these as building blocks of an unlabeled, rooted, nonplanar tree called the “blocks tree”. Let Tn−1 be the blocks tree after inserting n − 1 blocks. At step n, with probability pi we choose the block Ti from C , and we adjoin it to the tree Tn−1 by choosing a “parent” node at random from Tn−1 (all nodes from Tn−1 being equally likely parents). We then attach the root of block Ti to the chosen parent. For mathematical simplicity: We study the case where all members of C have the same size t. A special case: C consists of only one node; the block tree is isomorphic to the well-studied standard recursive tree.

2

Example

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

3

Example

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

Figure : A tree built from building blocks.

Step 1: The first block occurs with probability 1

3 ;

3

Example

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

Figure : A tree built from building blocks.

Step 1: The first block occurs with probability 1

3 ;

Step 2: The second block occurs with probability 2

3 , and the parent is chosen with probability 1 4 ;

3

Example

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

Figure : A tree built from building blocks.

Step 1: The first block occurs with probability 1

3 ;

Step 2: The second block occurs with probability 2

3 , and the parent is chosen with probability 1 4 ;

Step 3: The third block occurs with probability 2

3 , and the parent is chosen with probability 1 8 .

3

Motivation

The growth of large-scale software applications, which grow by the addition of smaller software modules.

4

Motivation

The growth of large-scale software applications, which grow by the addition of smaller software modules. The growth of wireless internet networks.

4

Motivation

The growth of large-scale software applications, which grow by the addition of smaller software modules. The growth of wireless internet networks. Hierarchical Bayesian models

4

Motivation

The growth of large-scale software applications, which grow by the addition of smaller software modules. The growth of wireless internet networks. Hierarchical Bayesian models Forensic science - some special kinds of probabilistic expert systems and Wigmorian evidence charts.

4

Motivation

The growth of large-scale software applications, which grow by the addition of smaller software modules. The growth of wireless internet networks. Hierarchical Bayesian models Forensic science - some special kinds of probabilistic expert systems and Wigmorian evidence charts. Biology/Chemistry - The growth of complex bacteria/organic molecules

4

Motivation

The growth of large-scale software applications, which grow by the addition of smaller software modules. The growth of wireless internet networks. Hierarchical Bayesian models Forensic science - some special kinds of probabilistic expert systems and Wigmorian evidence charts. Biology/Chemistry - The growth of complex bacteria/organic molecules Human Resources - the growth of large organizations

4

Scope of Work

We study the following asymptotic properties of the blocks tree:

5

Scope of Work

We study the following asymptotic properties of the blocks tree: Number of Leaves

5

Scope of Work

We study the following asymptotic properties of the blocks tree: Number of Leaves Three types of distances

5

Scope of Work

We study the following asymptotic properties of the blocks tree: Number of Leaves Three types of distances Depth of a node

5

Scope of Work

We study the following asymptotic properties of the blocks tree: Number of Leaves Three types of distances Depth of a node Total path length of the blocks tree

5

Scope of Work

We study the following asymptotic properties of the blocks tree: Number of Leaves Three types of distances Depth of a node Total path length of the blocks tree Height of the blocks tree

5

Leaves

Definition

A leaf is a node that has no children.

6

Leaves

Definition

A leaf is a node that has no children.

Theorem

Let Ln be the number of leaves in a random tree built from the building blocks T1, . . . , Tk, which are selected at each step with probabilities p1, . . . , pk. Let E[ΛC ] be the average number of leaves in the given collection, and Var[ΛC ] be the variance. Then, Ln − t E[ΛC ] t + 1 n √n

D

− → N

• 0, σ2

C

• ,

where σ2

C :=

Var[ΛC ] t + 2 + E[ΛC ](t + 1 − E[ΛC ]) (1 + t)2(2 + t)

• t.

6

Sketch of Proof.

We color the leaves and internal nodes with white(W) and blue(B) colors respectively.

7

Sketch of Proof.

We color the leaves and internal nodes with white(W) and blue(B) colors respectively. This will enable us to use the powerful theory of P´

• lya urns.

7

Sketch of Proof.

We color the leaves and internal nodes with white(W) and blue(B) colors respectively. This will enable us to use the powerful theory of P´

• lya urns.

Suppose Ti has ℓi leaves (and consequently it has t − ℓi internal nodes).

7

Sketch of Proof.

We color the leaves and internal nodes with white(W) and blue(B) colors respectively. This will enable us to use the powerful theory of P´

• lya urns.

Suppose Ti has ℓi leaves (and consequently it has t − ℓi internal nodes). Let ΛC be number of leaves in a randomly chosen block, i.e., ΛC has probability mass P(ΛC = ℓ) =

• j pj,

where the sum is taken over all j such that block Tj has ℓ leaves.

7

Sketch of Proof.

For example:

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

8

Sketch of Proof.

For example:

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

P(ΛC = 2) = 1 3, and P(ΛC = 3) = 2 3.

8

Sketch of Proof.

For example:

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

P(ΛC = 2) = 1 3, and P(ΛC = 3) = 2 3. The replacement matrix for this urn: A =    W B W ΛC − 1 t − ΛC + 1 B ΛC t − ΛC   

8

Sketch of Proof.

For example:

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

P(ΛC = 2) = 1 3, and P(ΛC = 3) = 2 3. The replacement matrix for this urn: A =    W B W ΛC − 1 t − ΛC + 1 B ΛC t − ΛC    Row sum for this matrix is a constant! Balanced Urns!

8

Sketch of Proof.

For balanced urns it is shown by [Athreya 1968] that Wn n

a.s.

− → λ1v1, where λ1 is the principal (largest real) eigenvalue of the average of the replacement matrix, and (v1, v2) is the corresponding left eigenvector of E[A].

9

Sketch of Proof.

For balanced urns it is shown by [Athreya 1968] that Wn n

a.s.

− → λ1v1, where λ1 is the principal (largest real) eigenvalue of the average of the replacement matrix, and (v1, v2) is the corresponding left eigenvector of E[A]. Calculating these for our blocks tree problem gives us: λ1 = t and λ2 = −1,

9

Sketch of Proof.

For balanced urns it is shown by [Athreya 1968] that Wn n

a.s.

− → λ1v1, where λ1 is the principal (largest real) eigenvalue of the average of the replacement matrix, and (v1, v2) is the corresponding left eigenvector of E[A]. Calculating these for our blocks tree problem gives us: λ1 = t and λ2 = −1, and hence, Wn n

a.s.

− → t t + 1 E[ΛC ].

9

Sketch of Proof.

Further we also notice that ℜλ2 < 1

2λ1.

[Smythe 1996] showed that Wn − λ1v1 √n

D

− → N(0, σ2), for some variance σ2 and states that σ2 is generally hard to compute, but we will obtain the exact variance of Wn.

10

Sketch of Proof.

Let I (W )

n

be the indicator of the event that a white ball is picked at the nth draw.

11

Sketch of Proof.

Let I (W )

n

be the indicator of the event that a white ball is picked at the nth draw. Wn = Wn−1 − I (W )

n

+ ΛC

11

Sketch of Proof.

Let I (W )

n

be the indicator of the event that a white ball is picked at the nth draw. Wn = Wn−1 − I (W )

n

+ ΛC W 2

n = W 2 n−1 + I (W ) n

+ Λ2

C + 2ΛC Wn−1 − 2ΛC I (W ) n

− 2Wn−1I (W )

n

11

Sketch of Proof.

Let I (W )

n

be the indicator of the event that a white ball is picked at the nth draw. Wn = Wn−1 − I (W )

n

+ ΛC W 2

n = W 2 n−1 + I (W ) n

+ Λ2

C + 2ΛC Wn−1 − 2ΛC I (W ) n

− 2Wn−1I (W )

n

Solving these recurrences, we get

11

Sketch of Proof.

Let I (W )

n

be the indicator of the event that a white ball is picked at the nth draw. Wn = Wn−1 − I (W )

n

+ ΛC W 2

n = W 2 n−1 + I (W ) n

+ Λ2

C + 2ΛC Wn−1 − 2ΛC I (W ) n

− 2Wn−1I (W )

n

Solving these recurrences, we get E

• Wn
• ∼ t E[ΛC ]

(t + 1) n + O

• n−1/t

,

11

Sketch of Proof.

Let I (W )

n

be the indicator of the event that a white ball is picked at the nth draw. Wn = Wn−1 − I (W )

n

+ ΛC W 2

n = W 2 n−1 + I (W ) n

+ Λ2

C + 2ΛC Wn−1 − 2ΛC I (W ) n

− 2Wn−1I (W )

n

Solving these recurrences, we get E

• Wn
• ∼ t E[ΛC ]

(t + 1) n + O

• n−1/t

, Var[Wn] ∼ Var[ΛC ] t + 2 + E[ΛC ](t + 1 − E[ΛC ]) (1 + t)2(2 + t)

• tn

+ O(n1−ǫ) := σ2

C n.

11

Sketch of Proof.

Let I (W )

n

be the indicator of the event that a white ball is picked at the nth draw. Wn = Wn−1 − I (W )

n

+ ΛC W 2

n = W 2 n−1 + I (W ) n

+ Λ2

C + 2ΛC Wn−1 − 2ΛC I (W ) n

− 2Wn−1I (W )

n

Solving these recurrences, we get E

• Wn
• ∼ t E[ΛC ]

(t + 1) n + O

• n−1/t

, Var[Wn] ∼ Var[ΛC ] t + 2 + E[ΛC ](t + 1 − E[ΛC ]) (1 + t)2(2 + t)

• tn

+ O(n1−ǫ) := σ2

C n.

Hence the theorem follows!

11

Depth

Definition

The depth of a node is the length of the shortest path from the node to the root.

12

Depth

Definition

The depth of a node is the length of the shortest path from the node to the root.

Theorem

Let Dn be the depth of the root of the nth inserted block in a random tree built from the building blocks T1, . . . , Tk, which are selected at each step with probabilities p1, . . . , pk. Let E[∆C ] be the average depth of a node in the given collection, and Var[∆C ] be the variance of that depth. Then, Dn − (E[∆C ] + 1) ln n √ ln n

D

− → N

• 0, Var[∆C ] + (E[∆C ] + 1)2

.

12

Sketch of Proof.

At step n, the newcomer can join any of the n − 1 existing blocks.

13

Sketch of Proof.

At step n, the newcomer can join any of the n − 1 existing blocks. The root of the nth block inherits the depth of any of the existing blocks, adjusted by the depth of the node it is choosing as parent within its block plus an extra 1.

13

Sketch of Proof.

At step n, the newcomer can join any of the n − 1 existing blocks. The root of the nth block inherits the depth of any of the existing blocks, adjusted by the depth of the node it is choosing as parent within its block plus an extra 1. We call the block to which the parent belongs the “parent block”, which can be any member i of the collection, with probability pi.

13

Sketch of Proof.

At step n, the newcomer can join any of the n − 1 existing blocks. The root of the nth block inherits the depth of any of the existing blocks, adjusted by the depth of the node it is choosing as parent within its block plus an extra 1. We call the block to which the parent belongs the “parent block”, which can be any member i of the collection, with probability pi. δn: be the random depth at which the nth parent node appears in its own block.

13

Sketch of Proof.

At step n, the newcomer can join any of the n − 1 existing blocks. The root of the nth block inherits the depth of any of the existing blocks, adjusted by the depth of the node it is choosing as parent within its block plus an extra 1. We call the block to which the parent belongs the “parent block”, which can be any member i of the collection, with probability pi. δn: be the random depth at which the nth parent node appears in its own block. δ1, δ2, . . . are equidistributed.

13

Sketch of Proof.

At step n, the newcomer can join any of the n − 1 existing blocks. The root of the nth block inherits the depth of any of the existing blocks, adjusted by the depth of the node it is choosing as parent within its block plus an extra 1. We call the block to which the parent belongs the “parent block”, which can be any member i of the collection, with probability pi. δn: be the random depth at which the nth parent node appears in its own block. δ1, δ2, . . . are equidistributed. ∆C (representing the depth of a parent node), which is completely determined by the structure of the trees in the collection;

13

Sketch of Proof.

At step n, the newcomer can join any of the n − 1 existing blocks. The root of the nth block inherits the depth of any of the existing blocks, adjusted by the depth of the node it is choosing as parent within its block plus an extra 1. We call the block to which the parent belongs the “parent block”, which can be any member i of the collection, with probability pi. δn: be the random depth at which the nth parent node appears in its own block. δ1, δ2, . . . are equidistributed. ∆C (representing the depth of a parent node), which is completely determined by the structure of the trees in the collection; Each δn has the same random distribution as ∆C.

13

For example:

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

14

For example:

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

14

For example:

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

∆C =        0, with probability 3/12; 1, with probability 7/12; 2, with probability 2/12.

14

For example:

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

∆C =        0, with probability 3/12; 1, with probability 7/12; 2, with probability 2/12. Dn = Di + δn + 1 with probability 1/(n − 1)

14

For example:

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

∆C =        0, with probability 3/12; 1, with probability 7/12; 2, with probability 2/12. Dn = Di + δn + 1 with probability 1/(n − 1) Hence the moment generating function of Dn conditioned on Fn−1, the sigma field generated by the first n − 1 insertions,would be,

14

For example:

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

∆C =        0, with probability 3/12; 1, with probability 7/12; 2, with probability 2/12. Dn = Di + δn + 1 with probability 1/(n − 1) Hence the moment generating function of Dn conditioned on Fn−1, the sigma field generated by the first n − 1 insertions,would be, E

• eDnu | Fn−1
• = E

n−1

• i=1

e(Di +δn+1)u 1 n − 1

• Fn−1
• =

1 n − 1E

• e(δn+1)u n−1
• i=1

eDi u.

14

Sketch of Proof.

Taking expectation again, φDn(u) = euψC(u) n − 1

n−1

• i=1

φDi (u),

15

Sketch of Proof.

Taking expectation again, φDn(u) = euψC(u) n − 1

n−1

• i=1

φDi (u), valid for n ≥ 2, with the initial condition φD1(u) = 1.

15

Sketch of Proof.

Taking expectation again, φDn(u) = euψC(u) n − 1

n−1

• i=1

φDi (u), valid for n ≥ 2, with the initial condition φD1(u) = 1. Solving this full history recurrence and after appropriate centering and scaling we get E

• exp

Dn − (E[∆C] + 1) ln n √ ln n

• u
• → e

1 2

• Var[∆C ]+(E[∆C ]+1)2

u2. 15

Sketch of Proof.

Taking expectation again, φDn(u) = euψC(u) n − 1

n−1

• i=1

φDi (u), valid for n ≥ 2, with the initial condition φD1(u) = 1. Solving this full history recurrence and after appropriate centering and scaling we get E

• exp

Dn − (E[∆C] + 1) ln n √ ln n

• u
• → e

1 2

• Var[∆C ]+(E[∆C ]+1)2

u2.

Therefore we have the theorem, Dn − (E[∆C] + 1) ln n √ ln n

D

− → N

• 0, Var[∆C] + (E[∆C] + 1)2

.

15

Sketch of Proof.

Taking expectation again, φDn(u) = euψC(u) n − 1

n−1

• i=1

φDi (u), valid for n ≥ 2, with the initial condition φD1(u) = 1. Solving this full history recurrence and after appropriate centering and scaling we get E

• exp

Dn − (E[∆C] + 1) ln n √ ln n

• u
• → e

1 2

• Var[∆C ]+(E[∆C ]+1)2

u2.

Therefore we have the theorem, Dn − (E[∆C] + 1) ln n √ ln n

D

− → N

• 0, Var[∆C] + (E[∆C] + 1)2

.

• Remark. The expressions for mean and variance are valid, even if t = t(n)

grows with n. However, in the asymptotic derivations of the central limit theorem we have to keep t relatively small, compared to n.

15

Total path length

Definition

The total path length of a tree is the sum of the depths of all the nodes.

16

Total path length

Definition

The total path length of a tree is the sum of the depths of all the nodes.

Theorem

Let Xn be the total path length of a tree built from the blocks of a collection C . Then, there is an absolutely integrable random variable X, such that Xn/n −

• t + E[χC ]
• Hn + t converges to X,

both in L2 and almost surely.

16

Sketch of Proof.

Let xi be the total path length of a block Ti.

17

Sketch of Proof.

Let xi be the total path length of a block Ti. Let χC be a discrete random variable that assumes the value xi, with probability

j pj,

where the sum is taken over every block Tj with total path length xi.

17

Sketch of Proof.

Let xi be the total path length of a block Ti. Let χC be a discrete random variable that assumes the value xi, with probability

j pj,

where the sum is taken over every block Tj with total path length xi. For example:

Figure : A collection of building blocks of size 4, with probabilities 1

3 , and 2 3 , respectively.

P(χC = 5) = 1 3, and P(χC = 3) = 2 3.

17

Sketch of Proof.

Let Xn be the total path length of the entire tree Tn built from the first n block insertions.

18

Sketch of Proof.

Let Xn be the total path length of the entire tree Tn built from the first n block insertions. If the nth block is adjoined to a node v ∈ Tn−1, at depth ˜ D(v) in the tree Tn−1 then each node in the last inserted block appears at distance equal to ˜ D(v) + 1, plus its own depth in the last block.

18

Sketch of Proof.

Let Xn be the total path length of the entire tree Tn built from the first n block insertions. If the nth block is adjoined to a node v ∈ Tn−1, at depth ˜ D(v) in the tree Tn−1 then each node in the last inserted block appears at distance equal to ˜ D(v) + 1, plus its own depth in the last block. So we have the following stochastic recurrence for n ≥ 2: E[Xn | Fn−1] = Xn−1 + t

• 1

t(n − 1)

• v∈Tn−1

˜ D(v) + 1

• + E
• χC | Fn−1
• = Xn−1 + Xn−1

n − 1 + t + E[χC].

18

Sketch of Proof.

Let Xn be the total path length of the entire tree Tn built from the first n block insertions. If the nth block is adjoined to a node v ∈ Tn−1, at depth ˜ D(v) in the tree Tn−1 then each node in the last inserted block appears at distance equal to ˜ D(v) + 1, plus its own depth in the last block. So we have the following stochastic recurrence for n ≥ 2: E[Xn | Fn−1] = Xn−1 + t

• 1

t(n − 1)

• v∈Tn−1

˜ D(v) + 1

• + E
• χC | Fn−1
• = Xn−1 + Xn−1

n − 1 + t + E[χC]. Taking expectations again, E[Xn] = n n − 1E[Xn−1] + t + E[χC].

18

Sketch of Proof.

Let Xn be the total path length of the entire tree Tn built from the first n block insertions. If the nth block is adjoined to a node v ∈ Tn−1, at depth ˜ D(v) in the tree Tn−1 then each node in the last inserted block appears at distance equal to ˜ D(v) + 1, plus its own depth in the last block. So we have the following stochastic recurrence for n ≥ 2: E[Xn | Fn−1] = Xn−1 + t

• 1

t(n − 1)

• v∈Tn−1

˜ D(v) + 1

• + E
• χC | Fn−1
• = Xn−1 + Xn−1

n − 1 + t + E[χC]. Taking expectations again, E[Xn] = n n − 1E[Xn−1] + t + E[χC]. Solving we get E[Xn] =

• t + E[χC]
• n Hn − nt ∼
• t + E[χC]
• n ln n,

18

Sketch of Proof.

If ˇ Dn be the depth of the node chosen as parent for the root of the nth block.

19

Sketch of Proof.

If ˇ Dn be the depth of the node chosen as parent for the root of the nth block. Then, we have the stochastic recurrence Xn = Xn−1 + t( ˇ Dn + 1) + χC.

19

Sketch of Proof.

If ˇ Dn be the depth of the node chosen as parent for the root of the nth block. Then, we have the stochastic recurrence Xn = Xn−1 + t( ˇ Dn + 1) + χC. Squaring and taking double expectation we get, E[X 2

n ] = n + 1

n − 1 E[X 2

n−1] +

2n n − 1

• E[χC] + t
• E[Xn−1]

+ t2E[ ˇ D2

n] + t2 + 2tE[χC] + E[χ2 C]. 19

Sketch of Proof.

If ˇ Dn be the depth of the node chosen as parent for the root of the nth block. Then, we have the stochastic recurrence Xn = Xn−1 + t( ˇ Dn + 1) + χC. Squaring and taking double expectation we get, E[X 2

n ] = n + 1

n − 1 E[X 2

n−1] +

2n n − 1

• E[χC] + t
• E[Xn−1]

+ t2E[ ˇ D2

n] + t2 + 2tE[χC] + E[χ2 C].

We develop a separate recurrence for E[ ˇ D2

n] and solving for E[X 2 n ] we get,

Var[Xn] ∼

• t2

E[∆2

C] + 2(E[∆C])2 + 4E[∆C] + 4

• + E[χ2

C] + 4tE[χC]

−

• E[χC] + t

2 2 + π2 6

• n2.

19

Sketch of Proof.

We know the conditional relation: E[Xn | Fn−1] = Xn−1 + Xn−1 n − 1 + t + E[χC],

20

Sketch of Proof.

We know the conditional relation: E[Xn | Fn−1] = Xn−1 + Xn−1 n − 1 + t + E[χC], Hence we see that X ∗

n = Xn n −

• t + E[χC]
• Hn + t is a martingale.

20

Sketch of Proof.

We know the conditional relation: E[Xn | Fn−1] = Xn−1 + Xn−1 n − 1 + t + E[χC], Hence we see that X ∗

n = Xn n −

• t + E[χC]
• Hn + t is a martingale.

From the variance expression we see that E

• (X ∗

n )2

n2 = c + o(1).

20

Sketch of Proof.

We know the conditional relation: E[Xn | Fn−1] = Xn−1 + Xn−1 n − 1 + t + E[χC], Hence we see that X ∗

n = Xn n −

• t + E[χC]
• Hn + t is a martingale.

From the variance expression we see that E

• (X ∗

n )2

n2 = c + o(1). Hence, we have supn≥1 E

• (X ∗

n )2

< ∞, and the theorem follows from Doob’s martingale convergence theorem.

20

Sketch of Proof.

We know the conditional relation: E[Xn | Fn−1] = Xn−1 + Xn−1 n − 1 + t + E[χC], Hence we see that X ∗

n = Xn n −

• t + E[χC]
• Hn + t is a martingale.

From the variance expression we see that E

• (X ∗

n )2

n2 = c + o(1). Hence, we have supn≥1 E

• (X ∗

n )2

< ∞, and the theorem follows from Doob’s martingale convergence theorem.

20

Sketch of Proof.

We know the conditional relation: E[Xn | Fn−1] = Xn−1 + Xn−1 n − 1 + t + E[χC], Hence we see that X ∗

n = Xn n −

• t + E[χC]
• Hn + t is a martingale.

From the variance expression we see that E

• (X ∗

n )2

n2 = c + o(1). Hence, we have supn≥1 E

• (X ∗

n )2

< ∞, and the theorem follows from Doob’s martingale convergence theorem.

• Remark. The expressions for mean and variance are valid, even when t = t(n)

is no longer fixed but grows with n.

20

Height

Definition

The height of the tree is the maximum depth among all the existing nodes. Hn = max

v∈Tn D(v).

21

Height

Definition

The height of the tree is the maximum depth among all the existing nodes. Hn = max

v∈Tn D(v).

Theorem

Let Hn be the height of a random tree built from the building blocks T1, . . . , Tk, which are selected at each step with probabilities p1, . . . , pk. We then have Hn ln n

a.s.

− → e

• E[∆C ] + 1
• .

21

Sketch of Proof.

Recall a standard recursive trees:

22

Sketch of Proof.

Recall a standard recursive trees: grows out of a root node in steps.

22

Sketch of Proof.

Recall a standard recursive trees: grows out of a root node in steps. At each step, a new node is added by choosing a parent node from the existing tree at random (all nodes are equally likely).

22

Sketch of Proof.

Recall a standard recursive trees: grows out of a root node in steps. At each step, a new node is added by choosing a parent node from the existing tree at random (all nodes are equally likely). If ˆ Hn be the height of the recursive tree, [Pittel 1994] showed that

ˆ Hn ln n a.s.

− → e.

22

Sketch of Proof.

Recall a standard recursive trees: grows out of a root node in steps. At each step, a new node is added by choosing a parent node from the existing tree at random (all nodes are equally likely). If ˆ Hn be the height of the recursive tree, [Pittel 1994] showed that

ˆ Hn ln n a.s.

− → e. Bursting argument:

22

Sketch of Proof.

Recall a standard recursive trees: grows out of a root node in steps. At each step, a new node is added by choosing a parent node from the existing tree at random (all nodes are equally likely). If ˆ Hn be the height of the recursive tree, [Pittel 1994] showed that

ˆ Hn ln n a.s.

− → e. Bursting argument: The blocks tree can be obtained from a recursive tree by “bursting” its nodes.

22

Sketch of Proof.

Recall a standard recursive trees: grows out of a root node in steps. At each step, a new node is added by choosing a parent node from the existing tree at random (all nodes are equally likely). If ˆ Hn be the height of the recursive tree, [Pittel 1994] showed that

ˆ Hn ln n a.s.

− → e. Bursting argument: The blocks tree can be obtained from a recursive tree by “bursting” its nodes. Each node in the recursive tree is replaced with (bursts into) a building block, with block Ti being chosen with probability pi

22

Sketch of Proof.

Recall a standard recursive trees: grows out of a root node in steps. At each step, a new node is added by choosing a parent node from the existing tree at random (all nodes are equally likely). If ˆ Hn be the height of the recursive tree, [Pittel 1994] showed that

ˆ Hn ln n a.s.

− → e. Bursting argument: The blocks tree can be obtained from a recursive tree by “bursting” its nodes. Each node in the recursive tree is replaced with (bursts into) a building block, with block Ti being chosen with probability pi Then each child of that node in the recursive tree independently chooses a parent in the parent block at random, with all nodes of that parent being equally likely (each may be taken as parent with probability 1/t).

22

Sketch of Proof.

Recall a standard recursive trees: grows out of a root node in steps. At each step, a new node is added by choosing a parent node from the existing tree at random (all nodes are equally likely). If ˆ Hn be the height of the recursive tree, [Pittel 1994] showed that

ˆ Hn ln n a.s.

− → e. Bursting argument: The blocks tree can be obtained from a recursive tree by “bursting” its nodes. Each node in the recursive tree is replaced with (bursts into) a building block, with block Ti being chosen with probability pi Then each child of that node in the recursive tree independently chooses a parent in the parent block at random, with all nodes of that parent being equally likely (each may be taken as parent with probability 1/t). When a child of a node bursts into a block, it is its root that gets joined to a randomly chosen parent in the parent block.

22

Sketch of Proof.

We now connect Hn and ˆ Hn.

23

Sketch of Proof.

We now connect Hn and ˆ Hn. Suppose v1, . . . , v ˆ

Hn is a path in the recursive tree leading from the root

to a node at the highest level (of depth ˆ Hn).

23

Sketch of Proof.

We now connect Hn and ˆ Hn. Suppose v1, . . . , v ˆ

Hn is a path in the recursive tree leading from the root

to a node at the highest level (of depth ˆ Hn). v1 is necessarily the root and when v2 bursts into a block, the root of that block appears in the blocks tree at distance 1 + ˆ δ1 from the root of that tree.

23

Sketch of Proof.

We now connect Hn and ˆ Hn. Suppose v1, . . . , v ˆ

Hn is a path in the recursive tree leading from the root

to a node at the highest level (of depth ˆ Hn). v1 is necessarily the root and when v2 bursts into a block, the root of that block appears in the blocks tree at distance 1 + ˆ δ1 from the root of that tree. ˆ δ1 is distributed like ∆C.

23

Sketch of Proof.

We now connect Hn and ˆ Hn. Suppose v1, . . . , v ˆ

Hn is a path in the recursive tree leading from the root

to a node at the highest level (of depth ˆ Hn). v1 is necessarily the root and when v2 bursts into a block, the root of that block appears in the blocks tree at distance 1 + ˆ δ1 from the root of that tree. ˆ δ1 is distributed like ∆C. Similarly when v2 bursts into a block, the root of that block appears in the blocks tree at distance 1 + ˆ δ2 from the root of that tree.

23

Sketch of Proof.

We now connect Hn and ˆ Hn. Suppose v1, . . . , v ˆ

Hn is a path in the recursive tree leading from the root

to a node at the highest level (of depth ˆ Hn). v1 is necessarily the root and when v2 bursts into a block, the root of that block appears in the blocks tree at distance 1 + ˆ δ1 from the root of that tree. ˆ δ1 is distributed like ∆C. Similarly when v2 bursts into a block, the root of that block appears in the blocks tree at distance 1 + ˆ δ2 from the root of that tree. Again ˆ δ2 is distributed like ∆C, and so forth along that path.

23

Sketch of Proof.

We now connect Hn and ˆ Hn. Suppose v1, . . . , v ˆ

Hn is a path in the recursive tree leading from the root

to a node at the highest level (of depth ˆ Hn). v1 is necessarily the root and when v2 bursts into a block, the root of that block appears in the blocks tree at distance 1 + ˆ δ1 from the root of that tree. ˆ δ1 is distributed like ∆C. Similarly when v2 bursts into a block, the root of that block appears in the blocks tree at distance 1 + ˆ δ2 from the root of that tree. Again ˆ δ2 is distributed like ∆C, and so forth along that path. Hence

23

Sketch of Proof.

We now connect Hn and ˆ Hn. Suppose v1, . . . , v ˆ

Hn is a path in the recursive tree leading from the root

to a node at the highest level (of depth ˆ Hn). v1 is necessarily the root and when v2 bursts into a block, the root of that block appears in the blocks tree at distance 1 + ˆ δ1 from the root of that tree. ˆ δ1 is distributed like ∆C. Similarly when v2 bursts into a block, the root of that block appears in the blocks tree at distance 1 + ˆ δ2 from the root of that tree. Again ˆ δ2 is distributed like ∆C, and so forth along that path. Hence Hn ≥ (1 + ˆ δ1) + (1 + ˆ δ2) + · · · + (1 + ˆ δ ˆ

Hn) = ˆ

Hn + ˆ

Hn i=1 ˆ

δi, where ˆ δi, for i = 1, . . . , ˆ Hn are all independent.

23

Sketch of Proof.

By the strong law of large numbers

1 ˆ Hn

ˆ

Hn i=1 ˆ

δi

a.s.

− → E[∆C].

24

Sketch of Proof.

By the strong law of large numbers

1 ˆ Hn

ˆ

Hn i=1 ˆ

δi

a.s.

− → E[∆C].

Hn ln n ≥ ˆ Hn ln n +

• 1

ˆ Hn

ˆ

Hn i=1 ˆ

δi

• ×

ˆ Hn ln n a.s.

− → e + E[∆C] e.

24

Sketch of Proof.

By the strong law of large numbers

1 ˆ Hn

ˆ

Hn i=1 ˆ

δi

a.s.

− → E[∆C].

Hn ln n ≥ ˆ Hn ln n +

• 1

ˆ Hn

ˆ

Hn i=1 ˆ

δi

• ×

ˆ Hn ln n a.s.

− → e + E[∆C] e. This gives us an a.s. lower bound.

24

Sketch of Proof.

By the strong law of large numbers

1 ˆ Hn

ˆ

Hn i=1 ˆ

δi

a.s.

− → E[∆C].

Hn ln n ≥ ˆ Hn ln n +

• 1

ˆ Hn

ˆ

Hn i=1 ˆ

δi

• ×

ˆ Hn ln n a.s.

− → e + E[∆C] e. This gives us an a.s. lower bound. We label the nodes of the bursting recursive tree according to their time

• rder of appearance. For example, the root is labeled 1, the second node

is labeled 2 and so on...

24

Sketch of Proof.

By the strong law of large numbers

1 ˆ Hn

ˆ

Hn i=1 ˆ

δi

a.s.

− → E[∆C].

Hn ln n ≥ ˆ Hn ln n +

• 1

ˆ Hn

ˆ

Hn i=1 ˆ

δi

• ×

ˆ Hn ln n a.s.

− → e + E[∆C] e. This gives us an a.s. lower bound. We label the nodes of the bursting recursive tree according to their time

• rder of appearance. For example, the root is labeled 1, the second node

is labeled 2 and so on... Suppose node i in the recursive tree is at depth ˆ Di, the jth node in the path from the root to node i in the recursive trees bursts into a tree in which the next node down the same path is adjoined to a node at depth δ(i)

j . 24

Sketch of Proof.

By the strong law of large numbers

1 ˆ Hn

ˆ

Hn i=1 ˆ

δi

a.s.

− → E[∆C].

Hn ln n ≥ ˆ Hn ln n +

• 1

ˆ Hn

ˆ

Hn i=1 ˆ

δi

• ×

ˆ Hn ln n a.s.

− → e + E[∆C] e. This gives us an a.s. lower bound. We label the nodes of the bursting recursive tree according to their time

• rder of appearance. For example, the root is labeled 1, the second node

is labeled 2 and so on... Suppose node i in the recursive tree is at depth ˆ Di, the jth node in the path from the root to node i in the recursive trees bursts into a tree in which the next node down the same path is adjoined to a node at depth δ(i)

j .

Then, the height of the blocks tree is bounded above: Hn ≤ max

1≤i≤n

• (1 + ˆ

δ(i)

1 ) + (1 + ˆ

δ(i)

2 ) + · · · + (1 + ˆ

δ(i)

ˆ Di )

• .

24

Sketch of Proof.

Hn ≤ max

1≤i≤n

ˆ Di + ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• 25

Sketch of Proof.

Hn ≤ max

1≤i≤n

ˆ Di + ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ max

1≤i≤n

ˆ Di + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• 25

Sketch of Proof.

Hn ≤ max

1≤i≤n

ˆ Di + ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ max

1≤i≤n

ˆ Di + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ ˆ

Hn + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Hn

• ,

25

Sketch of Proof.

Hn ≤ max

1≤i≤n

ˆ Di + ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ max

1≤i≤n

ˆ Di + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ ˆ

Hn + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Hn

• ,

where ˆ δ(i)

ˆ Di +1, . . . , δ ˆ H(i)

n

are additional independent random variable padded at the end to make all the expressions of the same length.

25

Sketch of Proof.

Hn ≤ max

1≤i≤n

ˆ Di + ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ max

1≤i≤n

ˆ Di + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ ˆ

Hn + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Hn

• ,

where ˆ δ(i)

ˆ Di +1, . . . , δ ˆ H(i)

n

are additional independent random variable padded at the end to make all the expressions of the same length. By the strong law of large numbers, we have

25

Sketch of Proof.

Hn ≤ max

1≤i≤n

ˆ Di + ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ max

1≤i≤n

ˆ Di + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ ˆ

Hn + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Hn

• ,

where ˆ δ(i)

ˆ Di +1, . . . , δ ˆ H(i)

n

are additional independent random variable padded at the end to make all the expressions of the same length. By the strong law of large numbers, we have Hn ln n ≤ ˆ Hn ln n + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Hn

ln n

• 25

Sketch of Proof.

Hn ≤ max

1≤i≤n

ˆ Di + ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ max

1≤i≤n

ˆ Di + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ ˆ

Hn + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Hn

• ,

where ˆ δ(i)

ˆ Di +1, . . . , δ ˆ H(i)

n

are additional independent random variable padded at the end to make all the expressions of the same length. By the strong law of large numbers, we have Hn ln n ≤ ˆ Hn ln n + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Hn

ln n

• ≤ e + eE[∆C] + o(1),

a.s.

25

Sketch of Proof.

Hn ≤ max

1≤i≤n

ˆ Di + ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ max

1≤i≤n

ˆ Di + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ ˆ

Hn + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Hn

• ,

where ˆ δ(i)

ˆ Di +1, . . . , δ ˆ H(i)

n

are additional independent random variable padded at the end to make all the expressions of the same length. By the strong law of large numbers, we have Hn ln n ≤ ˆ Hn ln n + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Hn

ln n

• ≤ e + eE[∆C] + o(1),

a.s.

a.s.

− → e

• 1 + E[∆C]),

25

Sketch of Proof.

Hn ≤ max

1≤i≤n

ˆ Di + ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ max

1≤i≤n

ˆ Di + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Di

• ≤ ˆ

Hn + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Hn

• ,

where ˆ δ(i)

ˆ Di +1, . . . , δ ˆ H(i)

n

are additional independent random variable padded at the end to make all the expressions of the same length. By the strong law of large numbers, we have Hn ln n ≤ ˆ Hn ln n + max

1≤i≤n

ˆ δ(i)

1 + ˆ

δ(i)

2 + · · · + ˆ

δ(i)

ˆ Hn

ln n

• ≤ e + eE[∆C] + o(1),

a.s.

a.s.

− → e

• 1 + E[∆C]),

Combining the bounds the result follows!

25

Future Work and Challenges

The distribution of the number of nodes of outdegree j > 0.

26

Future Work and Challenges

The distribution of the number of nodes of outdegree j > 0. The joint distribution or atleast covariances of number nodes

• f outdegree i and j.

26

Future Work and Challenges

The distribution of the number of nodes of outdegree j > 0. The joint distribution or atleast covariances of number nodes

• f outdegree i and j.

We can not handle a collection of blocks of different sizes, which would be more general.

26

Future Work and Challenges

The distribution of the number of nodes of outdegree j > 0. The joint distribution or atleast covariances of number nodes

• f outdegree i and j.

We can not handle a collection of blocks of different sizes, which would be more general. The underlying structure here is a recursive tree. We could try to build other types of tree from blocks.

26

Thank You! Questions are welcome!

27