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Probability of any given neighbour- hood of the Probability of any given neighbourhood of root, conditioned on the root, conditioned on the tree being the tree being infinite infinite Moumanti Podder Moumanti Podder Courant Institute

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Probability of any given neighbourhood of the root, conditioned on the tree being infinite

Moumanti Podder

Courant Institute of Mathematical Sciences New York University

Logic and Random Graphs Lorentz Center, Leiden August 31, 2015

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Motivation

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Motivation

For every given finite subtree T0, a random infinite tree T will almost surely contain v such that T(v) ∼ = T0.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Motivation

For every given finite subtree T0, a random infinite tree T will almost surely contain v such that T(v) ∼ = T0. Almost surely there exists v ∈ T far from root R such that T(v) ∼ = UNIV(k).

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Motivation

For every given finite subtree T0, a random infinite tree T will almost surely contain v such that T(v) ∼ = T0. Almost surely there exists v ∈ T far from root R such that T(v) ∼ = UNIV(k). k-move Ehrenfeucht value of T thus determined by neighbourhood of R (of radius ≈ 4k). Need to compute probabilities of neighbourhoods conditioned on T infinite.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Motivation

For every given finite subtree T0, a random infinite tree T will almost surely contain v such that T(v) ∼ = T0. Almost surely there exists v ∈ T far from root R such that T(v) ∼ = UNIV(k). k-move Ehrenfeucht value of T thus determined by neighbourhood of R (of radius ≈ 4k). Need to compute probabilities of neighbourhoods conditioned on T infinite. Remark In fact, able to compute P[A|T is infinite] for any first order statement A.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First generation

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First generation

1

Ai = {R has i children}, i = 1, 2, . . . k − 1, many.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First generation

1

Ai = {R has i children}, i = 1, 2, . . . k − 1, many.

2

B = {T is finite}.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First generation

1

Ai = {R has i children}, i = 1, 2, . . . k − 1, many.

2

B = {T is finite}.

3

P[T is finite] = p ∈ (0, 1).

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SLIDE 11

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First generation

1

Ai = {R has i children}, i = 1, 2, . . . k − 1, many.

2

B = {T is finite}.

3

P[T is finite] = p ∈ (0, 1).

4

P[Ai ∩ B] = e−λ · λi

i! · pi,

i ∈ {1, . . . k − 1}.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First generation

1

Ai = {R has i children}, i = 1, 2, . . . k − 1, many.

2

B = {T is finite}.

3

P[T is finite] = p ∈ (0, 1).

4

P[Ai ∩ B] = e−λ · λi

i! · pi,

i ∈ {1, . . . k − 1}.

5

P[Ai ∩ Bc] =P[Ai] − P[Ai ∩ B] =e−λ · λi i! (1 − pi), i ∈ {1, . . . k − 1}.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First generation

1

Ai = {R has i children}, i = 1, 2, . . . k − 1, many.

2

B = {T is finite}.

3

P[T is finite] = p ∈ (0, 1).

4

P[Ai ∩ B] = e−λ · λi

i! · pi,

i ∈ {1, . . . k − 1}.

5

P[Ai ∩ Bc] =P[Ai] − P[Ai ∩ B] =e−λ · λi i! (1 − pi), i ∈ {1, . . . k − 1}.

6

P[Amany ∩ B] = ∞

j=k e−λ · λj j! · pj.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First generation

1

Ai = {R has i children}, i = 1, 2, . . . k − 1, many.

2

B = {T is finite}.

3

P[T is finite] = p ∈ (0, 1).

4

P[Ai ∩ B] = e−λ · λi

i! · pi,

i ∈ {1, . . . k − 1}.

5

P[Ai ∩ Bc] =P[Ai] − P[Ai ∩ B] =e−λ · λi i! (1 − pi), i ∈ {1, . . . k − 1}.

6

P[Amany ∩ B] = ∞

j=k e−λ · λj j! · pj.

7

P[Amany ∩ Bc] =P[Amany] − P[Amany ∩ B] =

∞

  • j=k

e−λ · λj j! (1 − pj).

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Second generation

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Second generation

We will do this in two parts:

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Second generation

We will do this in two parts:

1

First find P[A ∩ B] where

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Second generation

We will do this in two parts:

1

First find P[A ∩ B] where

A is the event that the root R has a given 2-generation neighbourhood;

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Second generation

We will do this in two parts:

1

First find P[A ∩ B] where

A is the event that the root R has a given 2-generation neighbourhood; B is the event that the tree is finite.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Second generation

We will do this in two parts:

1

First find P[A ∩ B] where

A is the event that the root R has a given 2-generation neighbourhood; B is the event that the tree is finite.

2

Then find P[A]. Finally find P[A ∩ Bc] = P[A] − P[A ∩ B].

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First step

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First step

Definition A child v of the root R is of type:

1

αi, i ∈ {0, 1, . . . k − 1, many}, if v has i children and T(v) finite;

2

O otherwise.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First step

Definition A child v of the root R is of type:

1

αi, i ∈ {0, 1, . . . k − 1, many}, if v has i children and T(v) finite;

2

O otherwise. Xi, i ∈ {0, 1, . . . k − 1, many}, number of children of R

  • f type αi;
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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First step

Definition A child v of the root R is of type:

1

αi, i ∈ {0, 1, . . . k − 1, many}, if v has i children and T(v) finite;

2

O otherwise. Xi, i ∈ {0, 1, . . . k − 1, many}, number of children of R

  • f type αi;

Y number of children of R of type O.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

First step

Definition A child v of the root R is of type:

1

αi, i ∈ {0, 1, . . . k − 1, many}, if v has i children and T(v) finite;

2

O otherwise. Xi, i ∈ {0, 1, . . . k − 1, many}, number of children of R

  • f type αi;

Y number of children of R of type O. Lemma Xi ∼ Poi(λpi), i ∈ {0, . . . many}, Y ∼ Poi(λ(1−p0−. . . pmany)) and all mutually independent.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

What are these pi’s?

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

What are these pi’s?

Given R has n children, (X0, . . . Xmany, Y) ∼ Mult(n, p0, . . . pmany, 1−p0−. . . pmany).

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

What are these pi’s?

Given R has n children, (X0, . . . Xmany, Y) ∼ Mult(n, p0, . . . pmany, 1−p0−. . . pmany). For i ∈ {0, 1, . . . k − 1},

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

What are these pi’s?

Given R has n children, (X0, . . . Xmany, Y) ∼ Mult(n, p0, . . . pmany, 1−p0−. . . pmany). For i ∈ {0, 1, . . . k − 1}, pi = e−λ · λi i! · pi, and

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

What are these pi’s?

Given R has n children, (X0, . . . Xmany, Y) ∼ Mult(n, p0, . . . pmany, 1−p0−. . . pmany). For i ∈ {0, 1, . . . k − 1}, pi = e−λ · λi i! · pi, and pmany =

∞

  • j=k

e−λ · λj j! · pj =e−λ(1−p) − e−λ

k−1

  • j=0

(λp)j j! .

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Some notations for convenience

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Some notations for convenience

Definition Functions fi : [0, ∞) → [0, ∞), for 0 ≤ i ≤ k − 1, defined as: fi(λ) = λi i! .

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Some notations for convenience

Definition Functions fi : [0, ∞) → [0, ∞), for 0 ≤ i ≤ k − 1, defined as: fi(λ) = λi i! . Thus P[Poi(λ) = i] = e−λfi(λ).

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Some notations for convenience

Definition Functions fi : [0, ∞) → [0, ∞), for 0 ≤ i ≤ k − 1, defined as: fi(λ) = λi i! . Thus P[Poi(λ) = i] = e−λfi(λ). Definition Function fmany : [0, ∞) → [0, ∞), defined as: fmany(λ) =

∞

  • j=k

λj j! .

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Some notations for convenience

Definition Functions fi : [0, ∞) → [0, ∞), for 0 ≤ i ≤ k − 1, defined as: fi(λ) = λi i! . Thus P[Poi(λ) = i] = e−λfi(λ). Definition Function fmany : [0, ∞) → [0, ∞), defined as: fmany(λ) =

∞

  • j=k

λj j! . Thus P[Poi(λ) = many] = e−λfmany(λ).

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A ∩ B]

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A ∩ B]

Consider the root having xi ∈ {0, 1 . . . k − 1, many} number of children of type αi, and the tree being finite.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A ∩ B]

Consider the root having xi ∈ {0, 1 . . . k − 1, many} number of children of type αi, and the tree being finite. Equivalent to event A ∩ B =

k−1

  • i=0

{Xi = xi}

  • {Xmany = xmany}
  • {Y = 0}.
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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A ∩ B]

Consider the root having xi ∈ {0, 1 . . . k − 1, many} number of children of type αi, and the tree being finite. Equivalent to event A ∩ B =

k−1

  • i=0

{Xi = xi}

  • {Xmany = xmany}
  • {Y = 0}.

Probability of this event P[A ∩ B] = e−λ ·

k−1

  • i=0

fxi(λpi) · fxmany(λpmany).

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Example 1 Illustrated

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Example 1 Illustrated

R α3 α3 α2

Figure: The neighbourhood of the root up to 2 generations, with the tree finite. This means X3 = 2, X2 = 1 and all else 0.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Example 1

Suppose k = 5.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Example 1

Suppose k = 5. A = {R has 2 children each having 3 children each, and 1 child having 2 children}.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Example 1

Suppose k = 5. A = {R has 2 children each having 3 children each, and 1 child having 2 children}. A ∩ B = {X3 = 2, X2 = 1, everything else 0}.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Example 1

Suppose k = 5. A = {R has 2 children each having 3 children each, and 1 child having 2 children}. A ∩ B = {X3 = 2, X2 = 1, everything else 0}. Then P[A ∩ B] =e−λp3 (λp3)2 2! · e−λp2(λp2) · e−λ(1−p2−p3) =e−λ · f2(λp3) · f1(λp2).

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Example 2

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Example 2

A = {R has many children with many children, everything else 0}.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Example 2

A = {R has many children with many children, everything else 0}. A ∩ B = {Xmany ≥ k, everything else 0}.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Example 2

A = {R has many children with many children, everything else 0}. A ∩ B = {Xmany ≥ k, everything else 0}. Then P[A ∩ B] =   

∞

  • j=k

e−λpmany · (λpmany)j j!    · e−λ(1−pmany) =e−λfmany(λpmany).

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Second step: calculating probabilities without restriction of finiteness

Definition A child v of the root R is of type:

1

βi, i ∈ {0, 1, . . . k − 1, many}, if v has i children.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Second step: calculating probabilities without restriction of finiteness

Definition A child v of the root R is of type:

1

βi, i ∈ {0, 1, . . . k − 1, many}, if v has i children. Yi, i ∈ {0, 1, . . . k − 1, many}, number of children of R

  • f type βi.
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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Second step: calculating probabilities without restriction of finiteness

Definition A child v of the root R is of type:

1

βi, i ∈ {0, 1, . . . k − 1, many}, if v has i children. Yi, i ∈ {0, 1, . . . k − 1, many}, number of children of R

  • f type βi.

Lemma Then (Y0, . . . Yk−1, Ymany) ∼ (Poi(λq0), . . . Poi(λqk−1), Poi(λqmany) and mutually independent.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

What are the qi’s?

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

What are the qi’s?

Given R has n children, (Y0, . . . Ymany) ∼ Mult(n, q0, . . . qmany).

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SLIDE 55

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

What are the qi’s?

Given R has n children, (Y0, . . . Ymany) ∼ Mult(n, q0, . . . qmany). For i ∈ {0, 1, . . . k − 1},

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SLIDE 56

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

What are the qi’s?

Given R has n children, (Y0, . . . Ymany) ∼ Mult(n, q0, . . . qmany). For i ∈ {0, 1, . . . k − 1}, qi = e−λ · λi i! , and

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

What are the qi’s?

Given R has n children, (Y0, . . . Ymany) ∼ Mult(n, q0, . . . qmany). For i ∈ {0, 1, . . . k − 1}, qi = e−λ · λi i! , and qmany =

∞

  • j=k

e−λ · λj j! =e−λ  1 −

k−1

  • j=0

(λp)j j!   .

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A] and hence P[A ∩ Bc]

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A] and hence P[A ∩ Bc]

Consider the root having xi ∈ {0, 1 . . . k − 1, many} number of children of type βi.

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SLIDE 60

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A] and hence P[A ∩ Bc]

Consider the root having xi ∈ {0, 1 . . . k − 1, many} number of children of type βi. Equivalent to event A =

k−1

  • i=0

{Yi = xi}

  • {Ymany = xmany}.
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SLIDE 61

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A] and hence P[A ∩ Bc]

Consider the root having xi ∈ {0, 1 . . . k − 1, many} number of children of type βi. Equivalent to event A =

k−1

  • i=0

{Yi = xi}

  • {Ymany = xmany}.

Probability of this event P[A] = e−λ ·

k−1

  • i=0

fxi(λqi) · fxmany(λqmany).

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SLIDE 62

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A] and hence P[A ∩ Bc]

Consider the root having xi ∈ {0, 1 . . . k − 1, many} number of children of type βi. Equivalent to event A =

k−1

  • i=0

{Yi = xi}

  • {Ymany = xmany}.

Probability of this event P[A] = e−λ ·

k−1

  • i=0

fxi(λqi) · fxmany(λqmany). Finally P[A ∩ Bc] = P[A] − P[A ∩ B].

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SLIDE 63

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Interesting observations so far

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SLIDE 64

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Interesting observations so far

Probabilities of atomic events with restrictions of finiteness

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SLIDE 65

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Interesting observations so far

Probabilities of atomic events with restrictions of finiteness

will contain polynomials in λ · p, e−λ and e−λ(1−p) when X0, X1, . . . Xk−1, Xmany all take values in {0, 1, . . . k − 1};

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SLIDE 66

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Interesting observations so far

Probabilities of atomic events with restrictions of finiteness

will contain polynomials in λ · p, e−λ and e−λ(1−p) when X0, X1, . . . Xk−1, Xmany all take values in {0, 1, . . . k − 1}; will additionally contain double exponential terms when at least one of X0, . . . Xk−1, Xmany equals many.

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SLIDE 67

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Interesting observations so far

Probabilities of atomic events with restrictions of finiteness

will contain polynomials in λ · p, e−λ and e−λ(1−p) when X0, X1, . . . Xk−1, Xmany all take values in {0, 1, . . . k − 1}; will additionally contain double exponential terms when at least one of X0, . . . Xk−1, Xmany equals many.

Probabilities without restrictions on finiteness will have similar expressions, but without p.

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SLIDE 68

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Interesting observations so far

Probabilities of atomic events with restrictions of finiteness

will contain polynomials in λ · p, e−λ and e−λ(1−p) when X0, X1, . . . Xk−1, Xmany all take values in {0, 1, . . . k − 1}; will additionally contain double exponential terms when at least one of X0, . . . Xk−1, Xmany equals many.

Probabilities without restrictions on finiteness will have similar expressions, but without p. Example: Consider event A that root has many children each with 1 child, when k = 2. P[A] =e−λq0e−λqmany[1 − e−λq1 − λq1e−λq1] =e−λ(1−q1) − e−λ(1 + λq1).

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SLIDE 69

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Generalizing to n-generation-neighbourhoods

  • f the root
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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Generalizing to n-generation-neighbourhoods

  • f the root

Again denote any number ≥ k as many.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Generalizing to n-generation-neighbourhoods

  • f the root

Again denote any number ≥ k as many. Σi = finite set of i-generation-neighbourhoods of R, for all i ∈ N.

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Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Generalizing to n-generation-neighbourhoods

  • f the root

Again denote any number ≥ k as many. Σi = finite set of i-generation-neighbourhoods of R, for all i ∈ N. Definition Call a child v of R of type

1

ασ, σ ∈ Σn−1, if v has its (n − 1)-neighbourhood ∼ = σ and T(v) finite;

2

O otherwise.

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SLIDE 73

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Generalizing to n-generation-neighbourhoods

  • f the root

Again denote any number ≥ k as many. Σi = finite set of i-generation-neighbourhoods of R, for all i ∈ N. Definition Call a child v of R of type

1

ασ, σ ∈ Σn−1, if v has its (n − 1)-neighbourhood ∼ = σ and T(v) finite;

2

O otherwise. Lemma Xσ number of children of R of type ασ, σ ∈ Σ, then (Xσ : σ ∈ Σ) ∼ (Poi(λpσ) : σ ∈ Σ) and mutually independent.

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SLIDE 74

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A ∩ B], with B indicating finiteness

Consider the root having xσ ∈ {0, 1 . . . k − 1, many} number of children of type ασ, σ ∈ Σn−1, and the tree being finite.

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SLIDE 75

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A ∩ B], with B indicating finiteness

Consider the root having xσ ∈ {0, 1 . . . k − 1, many} number of children of type ασ, σ ∈ Σn−1, and the tree being finite. Equivalent to event A =

  • σ∈Σn−1

{Xσ = xσ}.

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SLIDE 76

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A ∩ B], with B indicating finiteness

Consider the root having xσ ∈ {0, 1 . . . k − 1, many} number of children of type ασ, σ ∈ Σn−1, and the tree being finite. Equivalent to event A =

  • σ∈Σn−1

{Xσ = xσ}. Probability of this event P[A ∩ B] = e−λ ·

  • σ∈Σ

fxσ(λpσ).

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SLIDE 77

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A], without restriction of finiteness

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SLIDE 78

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A], without restriction of finiteness

Yσ = number of children of R having (n − 1)-generation-neighbourhood ∼ = σ ∈ Σn−1.

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SLIDE 79

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A], without restriction of finiteness

Yσ = number of children of R having (n − 1)-generation-neighbourhood ∼ = σ ∈ Σn−1. Yσ ∼ Poi(λqσ), and mutually independent over σ ∈ Σn−1.

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SLIDE 80

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A], without restriction of finiteness

Yσ = number of children of R having (n − 1)-generation-neighbourhood ∼ = σ ∈ Σn−1. Yσ ∼ Poi(λqσ), and mutually independent over σ ∈ Σn−1. Then P[A] = P[

  • {Yσ = xσ}] = e−λ ·
  • σ∈Σ

fxσ(λqσ).

slide-81
SLIDE 81

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A], without restriction of finiteness

Yσ = number of children of R having (n − 1)-generation-neighbourhood ∼ = σ ∈ Σn−1. Yσ ∼ Poi(λqσ), and mutually independent over σ ∈ Σn−1. Then P[A] = P[

  • {Yσ = xσ}] = e−λ ·
  • σ∈Σ

fxσ(λqσ). Get P[A ∩ Bc] = P[A] − P[A ∩ B].

slide-82
SLIDE 82

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Calculating P[A], without restriction of finiteness

Yσ = number of children of R having (n − 1)-generation-neighbourhood ∼ = σ ∈ Σn−1. Yσ ∼ Poi(λqσ), and mutually independent over σ ∈ Σn−1. Then P[A] = P[

  • {Yσ = xσ}] = e−λ ·
  • σ∈Σ

fxσ(λqσ). Get P[A ∩ Bc] = P[A] − P[A ∩ B]. Finally, P[A|Bc] = P[A ∩ Bc]/P[Bc].

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SLIDE 83

Probability of any given neighbour- hood of the root, conditioned on the tree being infinite Moumanti Podder

Thank you.