On the number of subtrees on the fringe of random trees

(1)

On the number of subtrees on the fringe of random trees

(partly joined with Huilan Chang)

Michael Fuchs

Institute of Applied Mathematics National Chiao Tung University

Hsinchu, Taiwan

April 16th, 2008

(2)

The number of subtrees

Xn,k=number of subtrees of sizek on the fringe of random binary search trees of size n.

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3 6 8

2 5

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(3)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3 6 8

2 5

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(4)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1

7

3 6 8

2 5

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(5)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1

7

3

6

8

2 5

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(6)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3

6

8

2 5

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(7)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3

6 8

2 5

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(8)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3

6 8

2

5

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(9)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3 6 8

2 1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(10)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3 6 8

2 5

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(11)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3 6 8

1

3 4

6 7

8

X8,1= 2

X_8,2= 2 X_8,3= 1 X8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(12)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3 6 8

2 5

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2

X_8,3= 1 X8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(13)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3 6 8

1

3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1

X8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(14)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3 6 8

2 5

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X_8,4= 0

X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

(15)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3 6 8

1

2 3

4

6 7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X_8,4= 0 X8,5= 1

X8,6= 0 X_8,7= 0 X_8,8= 1

(16)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3 6 8

2 5

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X_8,4= 0 X8,5= 1 X8,6= 0

X_8,7= 0 X_8,8= 1

(17)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3 6 8

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X_8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0

X_8,8= 1

(18)

The number of subtrees

Example: Input: 4,7,6,1,8,5,3,2

4

1 7

3 6 8

2 5

1

2 3

4

5 6

7

8

X8,1= 2 X_8,2= 2 X_8,3= 1 X_8,4= 0 X8,5= 1 X8,6= 0 X_8,7= 0 X_8,8= 1

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Mean value and variance

Xn,k satisfies

X_n,k =^d X_I_n_,k+X_n−1−I^∗ _n_,k, where X_k,k = 1,X_I_n_,k andX_n−1−I^∗

n,k are conditionally independent given In, andIn=Unif{0, . . . , n−1}.

This yields

µn,k :=E(Xn,k) = 2(n+ 1)

(k+ 1)(k+ 2), (n > k), and

σ²_n,k:= Var(Xn,k) = 2k(4k²+ 5k−3)(n+ 1) (k+ 1)(k+ 2)²(2k+ 1)(2k+ 3) for n >2k+ 1.

(20)

Mean value and variance

Xn,k satisfies

X_n,k =^d X_I_n_,k+X_n−1−I^∗ _n_,k, where X_k,k = 1,X_I_n_,k andX_n−1−I^∗

n,k are conditionally independent given In, andIn=Unif{0, . . . , n−1}.

This yields

µ_n,k:=E(X_n,k) = 2(n+ 1)

(k+ 1)(k+ 2), (n > k), and

σ_n,k² := Var(Xn,k) = 2k(4k²+ 5k−3)(n+ 1) (k+ 1)(k+ 2)²(2k+ 1)(2k+ 3) for n >2k+ 1.

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Some previous results

Aldous (1991): Weak law of large numbers X_n,k

µn,k

−→1 in probability.

Devroye (1991): Central limit theorem X_n,k−µ_n,k

σn,k

−→ Nd (0,1).

Flajolet, Gourdon, Martinez (1997):

Central limit theorem with optimal Berry-Esseen bound and LLT

−→ All the above results are for fixed k.

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Some previous results

µn,k

σn,k

−→ Nd (0,1).

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Some previous results

µn,k

σn,k

−→ Nd (0,1).

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Some previous results

µn,k

σn,k

−→ Nd (0,1).

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Results for k = k

_n

Theorem (Feng, Mahmoud, Panholzer (2008)) (i) (Normal range) Letk=o(√

n)and k→ ∞ as n→ ∞. Then,

Xn,k−µn,k

p2n/k²

−→ Nd (0,1).

(ii) (Poisson range) Let k∼c√

n asn→ ∞. Then,

X_n,k−→^d Poisson(2c⁻²).

(iii) (Degenerate range) Let k < nand√

n=o(k) as n→ ∞. Then, Xn,k

L1

−→0.

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Why are we interested in X

_n,k

?

Xn,k is a new kind of profile of a tree.

The phase change from normal to Poisson is a universal phenomenon expected to hold for many classes of random trees.

The methods for proving phase change results might be applicable to other parameters which are expected to exhibit the same phase change behavior as well.

X_n,k is related to parameters arising in genetics.

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Why are we interested in X

_n,k

?

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Why are we interested in X

_n,k

?

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Why are we interested in X

_n,k

?

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Yule generated random genealogical trees

Example:

5

4

2 3

1

1 2 3 4 5 6

5

4

2 3

1

1 2 3 4 5 6

genes

Random model: At every time point, two yellow nodes uniformly coalescent. Same model as random binary search tree model!

time

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Yule generated random genealogical trees

Example:

5

4

2 3

1

1 2 3 4 5 6

5

4

2 3

1

1 2 3 4 5 6 genes

Random model:

At every time point, two yellow nodes uniformly coalescent.

Same model as random binary search tree model!

time

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Yule generated random genealogical trees

Example:

5

4

2 3

1

1 2

3 4

5 6

5

4

2 3

1

1 2

3 4

5 6 genes

Random model:

time

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Yule generated random genealogical trees

Example:

5

4

2 3

1

1 2

3 4 5 6

5

4

2

3

1

1 2

3 4 5 6

genes

Random model:

time

(34)

Yule generated random genealogical trees

Example:

5

4

2 3

1

2 3 4 5 6

5

4

2 3

1

2 3 4 5 6

genes

Random model:

time

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Yule generated random genealogical trees

Example:

5

4

2 3

1

2 3 4 5 6

5

4

2 3

1

2 3 4 5 6

genes

Random model:

time

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Yule generated random genealogical trees

Example:

5

4

2 3

1

1 2 3 4 5 6

5

4

2 3

1

1 2 3 4 5 6

genes

Random model:

time

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Yule generated random genealogical trees

Example:

5

4

2 3

1

1 2 3 4 5 6

5

4

2 3

1

1 2 3 4 5 6 genes

Random model:

time

(38)

Yule generated random genealogical trees

Example:

5

4

2 3

1

1 2 3 4 5 6

5

4

2 3

1

1 2 3 4 5 6 genes

Random model:

time

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Shape parameters of genealogical trees

k-pronged nodes (Rosenberg 2006):

Nodes with an induced subtree with k−1internal nodes.

k-caterpillars (Rosenberg 2006):

Correspond to nodes whose induced subtree has k−1 internal nodes all of them with out-degree either0 or 1.

Nodes with minimal clade sizek (Blum and Fran¸cois (2005)): Ifk≥3, then they are internal nodes with induced subtree of size k−1 and either an empty right subtree or empty left subtree.

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Shape parameters of genealogical trees

Correspond to nodes whose induced subtree has k−1internal nodes all of them with out-degree either0 or 1.

Nodes with minimal clade sizek (Blum and Fran¸cois (2005)): Ifk≥3, then they are internal nodes with induced subtree of size k−1 and either an empty right subtree or empty left subtree.

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Shape parameters of genealogical trees

Correspond to nodes whose induced subtree has k−1internal nodes all of them with out-degree either0 or 1.

Nodes with minimal clade sizek (Blum and Fran¸cois (2005)):

Ifk≥3, then they are internal nodes with induced subtree of size k−1 and either an empty right subtree or empty left subtree.

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Counting pattern in random binary search trees

ConsiderX_n,k with

Xn,k

=d XIn,k+X_n−1−I^∗ _n_,k, where X_k,k =Bernoulli(p_k),X_I_n_,k andX_n−1−I^∗

n,k are conditionally independent givenIn, andIn=Unif{0, . . . , n−1}.

Then,

pk shape parameter

1 #ofk+ 1-pronged nodes

2/k #of nodes with minimal clade size k+ 1 2^k−1/k! #of k+ 1caterpillars

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Counting pattern in random binary search trees

ConsiderX_n,k with

Xn,k

=d XIn,k+X_n−1−I^∗ _n_,k, where X_k,k =Bernoulli(p_k),X_I_n_,k andX_n−1−I^∗

n,k are conditionally independent givenIn, andIn=Unif{0, . . . , n−1}.

Then,

pk shape parameter

1 # ofk+ 1-pronged nodes

2/k #of nodes with minimal clade size k+ 1 2^k−1/k! #of k+ 1caterpillars

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Underlying recurrence and solution

All (centered or non-centered) moments satisfy an,k= 2

n

n−1

X

j=0

aj,k+bn,k,

where ak,k is given and an,k = 0 forn < k.

We have

a_n,k = 2(n+ 1)

(k+ 1)(k+ 2)a_k,k+ 2(n+ 1) X

k<j<n

b_j,k

(j+ 1)(j+ 2)+b_n,k, where n > k.

(45)

Underlying recurrence and solution

n

n−1

X

j=0

aj,k+bn,k,

We have

a_n,k= 2(n+ 1)

(k+ 1)(k+ 2)a_k,k+ 2(n+ 1) X

k<j<n

b_j,k

(j+ 1)(j+ 2)+b_n,k, where n > k.

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Mean value and variance

We have

E(X_n,k) = 2(n+ 1)

(k+ 1)(k+ 2)p_k, (n > k), and

Var(X_n,k) = 2p_k(4k³+ 16k²+ 19k+ 6−(11k²+ 22k+ 6)p_k)(n+ 1) (k+ 1)(k+ 2)²(2k+ 1)(2k+ 3)

for n >2k+ 1.

Note that

E(X_n,k)∼Var(X_n,k)∼ 2p_k k² n for n >2k+ 1andk→ ∞ as n→ ∞.

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Mean value and variance

We have

E(X_n,k) = 2(n+ 1)

(k+ 1)(k+ 2)p_k, (n > k), and

Var(X_n,k) = 2p_k(4k³+ 16k²+ 19k+ 6−(11k²+ 22k+ 6)p_k)(n+ 1) (k+ 1)(k+ 2)²(2k+ 1)(2k+ 3)

for n >2k+ 1.

Note that

E(X_n,k)∼Var(X_n,k)∼ 2p_k k² n for n >2k+ 1andk→ ∞ as n→ ∞.

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Higher moments

Denote by

A^(m)_n,k :=E(X_n,k−E(X_n,k))^m.

Then,

A^(m)_n,k = 2 n

n−1

X

j=0

A^(m)_j,k +B_n,k^(m),

where

B_n,k^(m):= X

i1+i2+i3=m 0≤i₁,i2<m

m i₁, i₂, i₃

1 n

n−1

X

j=0

A⁽ⁱ_j,k¹⁾A⁽ⁱ_n−1−j,k²⁾ ∆ⁱ_n,j,k³

and

∆n,j,k =E(Xj,k) +E(Xn−1−j,k)−E(Xn,k).

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Higher moments

Denote by

A^(m)_n,k :=E(X_n,k−E(X_n,k))^m.

Then,

A^(m)_n,k = 2 n

n−1

X

j=0

A^(m)_j,k +B_n,k^(m),

where

B_n,k^(m):= X

i1+i2+i3=m 0≤i₁,i2<m

m i₁, i₂, i₃

1 n

n−1

X

j=0

A⁽ⁱ_j,k¹⁾A⁽ⁱ_n−1−j,k²⁾ ∆ⁱ_n,j,k³

and

∆n,j,k =E(Xj,k) +E(Xn−1−j,k)−E(Xn,k).

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Methods of moments

Theorem

Let Zn, Z be random variables. Assume that, asn→ ∞, E(Z_n^m)−→E(Z^m)

for allm≥1andZ is uniquely determined by its moment sequence. Then, Zn

−→d Z.

Inductive approach based on asymptotic transfers was extensively used in the AofA to prove numerous result (“moment pumping”).

Fuchs, Hwang, Neininger (2007): variation of the above scheme to study the profile of random binary search trees and random recursive trees.

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Methods of moments

Theorem

−→d Z.

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Methods of moments

Theorem

−→d Z.

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Normal range

Proposition

Uniformly for n, k, m≥1andn > k

A^(m)_n,k = max

(2pkn k² ,

2pkn k²

m/2) .

Proposition

For E(X_n,k)→ ∞ asn→ ∞,

A^(2m−1)_n,k =o

2p_kn k²

m−1/2!

, A^(2m)_n,k ∼g_m

2p_kn k²

m

,

where

g_m = (2m)!/(2^mm!).

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Normal range

Proposition

Uniformly for n, k, m≥1andn > k

A^(m)_n,k = max

(2pkn k² ,

2pkn k²

m/2) .

Proposition

For E(Xn,k)→ ∞ asn→ ∞,

A^(2m−1)_n,k =o

2p_kn k²

m−1/2!

, A^(2m)_n,k ∼g_m

2p_kn k²

m

,

where

g_m= (2m)!/(2^mm!).

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Poisson range

Consider

A¯^(m)_n,k =E(Xn,k(Xn,k−1)· · ·(Xn,k−m+ 1)).

Then, similarly as before: Proposition

(i) Uniformly for n, k, m≥1 andn > k

A¯^(m)_n,k = max 2p_kn

k² , 2p_kn

k² m

. (ii) For E(X_n,k)→cand k < nasn→ ∞,

A¯^(m)_n,k −→c^m.

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Poisson range

Consider

A¯^(m)_n,k =E(Xn,k(Xn,k−1)· · ·(Xn,k−m+ 1)).

Then, similarly as before:

Proposition

(i) Uniformly for n, k, m≥1 andn > k

A¯^(m)_n,k = max 2p_kn

k² , 2p_kn

k² m

. (ii) For E(X_n,k)→cand k < nasn→ ∞,

A¯^(m)_n,k −→c^m.

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The phase change

Theorem

(i) (Normal range) LetE(X_n,k)→ ∞ andk→ ∞ asn→ ∞. Then, X_n,k−E(X_n,k)

p2p_kn/k²

−→ Nd (0,1).

(ii) (Poisson range) Let E(X_n,k)→c >0and k < nasn→ ∞. Then,

X_n,k −→^d Poisson(c).

(iii) (Degenerate range) Let E(X_n,k)→0as n→ ∞. Then,

X_n,k−→^L¹ 0.

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A comparison of the phase change

For k-caterpillars, we have

E(X_n,k) = 2^k−1n (k+ 2)!. Note that either

E(X_n,k)→ ∞ or E(X_n,k)→0.

So, there is no Poisson range.

shape parameter location phase change

k-pronged nodes √

n normal - poisson - degenerate minimal clade sizek √³

n normal - poisson - degenerate k-caterpillars lnn/(ln lnn) normal - degenerate

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A comparison of the phase change

For k-caterpillars, we have

E(X_n,k) = 2^k−1n (k+ 2)!. Note that either

E(X_n,k)→ ∞ or E(X_n,k)→0.

So, there is no Poisson range.

shape parameter location phase change

k-pronged nodes √

n normal - poisson - degenerate minimal clade size k √³

n normal - poisson - degenerate k-caterpillars lnn/(ln lnn) normal - degenerate

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Refined results (for # of subtrees)

Define

φ_n,k(y) =e^−σ²^n,k^y²^/2E

e^(X^n,k^−µ^n,k^)y

. and

φ^(m)_n,k = d^mφ_n,k(y) dy^m

_y=0

.

Proposition

Uniformly for n, k≥1 andm≥0

|φ^(m)_n,k| ≤m!A^mmax n

k²,n k²

m/3

for a suitable constant A.

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Refined results (for # of subtrees)

Define

φ_n,k(y) =e^−σ²^n,k^y²^/2E

e^(X^n,k^−µ^n,k^)y

. and

φ^(m)_n,k = d^mφ_n,k(y) dy^m

_y=0

.

Proposition

Uniformly for n, k≥1 andm≥0

|φ^(m)_n,k| ≤m!A^mmax n

k²,n k²

m/3

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Characteristic function

Let ϕn,k(y) =E(exp{(X_n,k−µn,k)iy/σn,k}).

Proposition Let 1≤k=o(√

n).

(i) For nlarge

ϕn,k(y) =e^−y²^/2

1 +O

|y|³ k

√n

,

uniformly for y with |y| ≤n^1/6/k^1/3. (ii) For nlarge

|ϕ_n,k(y)| ≤e^−y²^/2, where|y| ≤πσ_n,k.

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Berry-Esseen bound and LLT for the normal range

Theorem (Rate of convergency) For 1≤k=o(√

n)as n→ ∞,

sup

x∈R

P Xn,k−µn,k

σ_n,k < x

!

−Φ(x)

=O k

√n

.

Theorem (LLT) For 1≤k=o(√

n)as n→ ∞,

P(X_n,k=bµ_n,k+xσ_n,kc) = e^−x²^/2

√2πσ_n,k

1 +O

(1 +|x|³) k

√n

,

uniformly in x=o(n^1/6/k^1/3).

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Berry-Esseen bound and LLT for the normal range

Theorem (Rate of convergency) For 1≤k=o(√

n)as n→ ∞,

sup

x∈R

P Xn,k−µn,k

σ_n,k < x

!

−Φ(x)

=O k

√n

.

Theorem (LLT) For 1≤k=o(√

n)as n→ ∞,

P(X_n,k=bµ_n,k+xσ_n,kc) = e^−x²^/2

√ 2πσn,k

1 +O

(1 +|x|³) k

√n

,

uniformly in x=o(n^1/6/k^1/3).

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LLT for the Poisson range

Define

φ¯_n,k(y) =e^−µ^n,k^(y−1)E y^X^n,k . and

φ^(m)_n,k = d^mφ¯_n,k(y) dy^m

y=1

.

Proposition

Uniformly for n > k andm≥0

|φ¯^(m)_n,k| ≤m!A^mn k³

m/2

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LLT for the Poisson range

Define

φ¯_n,k(y) =e^−µ^n,k^(y−1)E y^X^n,k . and

φ^(m)_n,k = d^mφ¯_n,k(y) dy^m

y=1

.

Proposition

Uniformly for n > k andm≥0

|φ¯^(m)_n,k| ≤m!A^m n

k³ m/2

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Poisson approximation

Theorem (LLT)

For k < nandn→ ∞,

P(X_n,k =l) =e^−µ^n,k(µ_n,k)^l

l! +On k³

uniformly in l.

Theorem (Poisson approximation)

Let k < nandk→ ∞ asn→ ∞. Then,

dT V(Xn,k,Poisson(µn,k))−→0. Remark: A rate can be given as well.

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Poisson approximation

Theorem (LLT)

For k < nandn→ ∞,

P(X_n,k =l) =e^−µ^n,k(µ_n,k)^l

l! +On k³

uniformly in l.

Theorem (Poisson approximation)

Let k < nandk→ ∞ asn→ ∞. Then,

dT V(Xn,k,Poisson(µn,k))−→0.

Remark: A rate can be given as well.

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Other types of random trees

Random recursive trees

Non-plane, labelled trees with every label sequence from the root to a leave increasing; random model is the uniform model.

Methods works as well (with minor modifications) and similar results can be proved.

Plane-oriented recursive trees (PORTs)

Plane, labelled trees with every label sequence from the root to a leave increasing; random model is the uniform model.

Method works as well, but details more involved.

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Other types of random trees

Random recursive trees

Non-plane, labelled trees with every label sequence from the root to a leave increasing; random model is the uniform model.

Methods works as well (with minor modifications) and similar results can be proved.

Plane-oriented recursive trees (PORTs)

Plane, labelled trees with every label sequence from the root to a leave increasing; random model is the uniform model.

Method works as well, but details more involved.

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# of subtrees for PORTs

X_n,k satisfies

X_n,k=^d

N

X

i=1

X_I⁽ⁱ⁾

i,k, where X_k,k = 1 andX_I⁽ⁱ⁾

i,k are conditionally independent given (N, I₁, I₂, . . .).

This can be simplified to

X_n,k =^d X_I_n_,k+X_n−I^∗ _n_,k−1_{n−I_n_=k}, whereX_k,k = 1,X_I_n_,k andX_n−I^∗

n,k are conditionally independent givenIn

and

P(I_n=j) = 2(n−j)C_jCn−j

nC_n .

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# of subtrees for PORTs

X_n,k satisfies

X_n,k=^d

N

X

i=1

X_I⁽ⁱ⁾

i,k, where X_k,k = 1 andX_I⁽ⁱ⁾

i,k are conditionally independent given (N, I₁, I₂, . . .).

This can be simplified to

X_n,k =^d X_I_n_,k+X_n−I^∗ _n_,k−1_{n−I_n_=k}, whereX_k,k = 1,X_I_n_,k andX_n−I^∗

n,k are conditionally independent givenIn

and

P(I_n=j) = 2(n−j)C_jCn−j

nC_n .

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Underlying recurrence and solution

n−1

X

j=1

CjCn−j

Cn

aj,k+bn,k,

We have

a_n,k= C_k(n+ 1−k)Cn+1−k

C_n a_k,k+ X

k<j≤n

C_j(n+ 1−j)Cn+1−j

C_n b_n,k, where n > k.

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Underlying recurrence and solution

n−1

X

j=1

CjCn−j

Cn

aj,k+bn,k,

We have

a_n,k= C_k(n+ 1−k)Cn+1−k

C_n a_k,k+ X

k<j≤n

C_j(n+ 1−j)Cn+1−j

C_n b_n,k, where n > k.

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Mean value and variance of PORTs

We have,

µ_n,k:=E(X_n,k) = 2n−1

4k²−1, (n > k).

Moreover, for fixed k asn→ ∞,

Var(X_n,k)∼c_kn, where

c_k = 8k²−4k−8

(4k²−1)² − ((2k−3)!!)²

((k−1)!)²4^k−1k(2k+ 1), and, for k < nandk→ ∞ asn→ ∞,

E(X_n,k)∼Var(X_n,k)∼ n 2k².

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Mean value and variance of PORTs

We have,

µ_n,k:=E(X_n,k) = 2n−1

4k²−1, (n > k).

Moreover, for fixed k asn→ ∞,

Var(X_n,k)∼c_kn, where

c_k = 8k²−4k−8

(4k²−1)² − ((2k−3)!!)²

((k−1)!)²4^k−1k(2k+ 1), and, for k < nandk→ ∞ asn→ ∞,

E(X_n,k)∼Var(X_n,k)∼ n 2k².

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The phase change

Theorem

(i) (Normal range) Letk=o(√

X_n,k−µ_n,k pn/(2k²)

−→ Nd (0,1).

n asn→ ∞. Then, Xn,k d

−→Poisson((2c²)⁻¹).

n=o(k) as n→ ∞. Then, X_n,k−→^L¹ 0.

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More results and future research

Parameters of genealogical trees under different random models

Universality of the phase change for the number of subtrees Very simple classes of increasing trees and more general classes of increasing trees (polynomial varieties, mobile trees, etc.)

Phase change results for the number of nodes with out-degree k Important in computer science.

A phase change from normal to degenerate is expected (no Poisson range).

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More results and future research

Parameters of genealogical trees under different random models Universality of the phase change for the number of subtrees Very simple classes of increasing trees and more general classes of increasing trees (polynomial varieties, mobile trees, etc.)

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Polynomial varieties

Bergeron, Flajolet, Salvy (1992): classes of increasing trees with degree functionφ(ω) under the uniform model.

Polynomial varieties: class of increasing tree with φ(ω) =φdω^d+· · ·+φ0,

where d≥2 andφ_d, φ₀ 6= 0.

For mean value and variance of the number of subtrees, E(Xn,k)∼Var(Xn,k)∼ d

d−1 · n k², where k→ ∞ as n→ ∞.

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Polynomial varieties

Polynomial varieties: class of increasing tree with φ(ω) =φdω^d+· · ·+φ0, where d≥2 andφ_d, φ₀6= 0.

d−1 · n k², where k→ ∞ as n→ ∞.

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Polynomial varieties

Polynomial varieties: class of increasing tree with φ(ω) =φdω^d+· · ·+φ0,

where d≥2 andφ_d, φ₀6= 0.

d−1 · n k², where k→ ∞ asn→ ∞.

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Polynomial varieties - phase change

Theorem

(i) (Normal range) Letk=o(√

X_n,k−µ_n,k pnd/((d−1)k²)

−→ Nd (0,1).

n asn→ ∞. Then, Xn,k d

−→Poisson(d/((d−1)c²)).

n=o(k) as n→ ∞. Then, X_n,k−→^L¹ 0.

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Mobile trees - phase change

Theorem

(i) (Normal range) Letk=o p

n/lnn

andk→ ∞ asn→ ∞. Then,

Xn,k−µn,k

pn/(k²lnk)

−→ Nd (0,1).

(ii) (Poisson range) Let k∼cp

n/lnn asn→ ∞. Then,

X_n,k−→^d Poisson(2c⁻²).

(iii) (Degenerate range) Let k < nandp

n/lnn=o(k) as n→ ∞.

Then,

X_n,k−→^L¹ 0.

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More results and future research

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