Probabilistic Methods for Complex Networks Lecture 5: Random - PowerPoint PPT Presentation

Probabilistic Methods for Complex Networks Lecture 5: Random Networks II - Linearity of Expectation Prof. Sotiris Nikoletseas University of Patras and CTI ΥΔΑ ΜΔΕ, Patras 2019 - 2020 Prof. Sotiris Nikoletseas Probabilistic Methods in Complex Networks ΥΔΑ ΜΔΕ, Patras 2019 - 2020 1 / 18

Summary of previous lecture Common, underlying concept of all techniques: “Non-constructive proof of existence of combinatorial structures that have certain desired properties. ” Method of “positive probability”: Construct (by using abstract random experiments) an appropriate probability sample Prove that the probability of the desired property in this space is positive (i.e. point) with the desired property. Prof. Sotiris Nikoletseas Probabilistic Methods in Complex Networks ΥΔΑ ΜΔΕ, Patras 2019 - 2020 2 / 18 space of combinatorial structures (points ↔ structures). non-zero). ⇒ There is at least one combinatorial structure (since there is at least one

Summary of this lecture i. Non-existence proofs using the Markov Inequality ii. Proofs of existence using the Linearity of Expectation method Prof. Sotiris Nikoletseas Probabilistic Methods in Complex Networks ΥΔΑ ΜΔΕ, Patras 2019 - 2020 3 / 18

(I) Markov Inequality Theorem 1 ΥΔΑ ΜΔΕ, Patras 2019 - 2020 Probabilistic Methods in Complex Networks Prof. Sotiris Nikoletseas 4 / 18 Proof : Let X be a non-negative random variable. Then: ∀ t > 0 : Pr { X ≥ t } ≤ E [ X ] t � � E [ X ] = x Pr { X = x } ≥ x Pr { X = x } x x ≥ t � � ≥ t Pr { X = x } = t Pr { X = x } = t · Pr { X ≥ t } x ≥ t x ≥ t ⇒ E [ X ] ≥ t · Pr { X ≥ t } ⇒ Pr { X ≥ t } ≤ E [ X ] t

(I) Markov Inequality Application It is actually a (weak) concentration inequality: ΥΔΑ ΜΔΕ, Patras 2019 - 2020 Probabilistic Methods in Complex Networks Prof. Sotiris Nikoletseas . . . 5 / 18 ≤ 1 � � Pr X ≥ 2 · E [ X ] 2 ≤ 1 � � Pr X ≥ 3 · E [ X ] 3 ≤ 1 � � Pr X ≥ k · E [ X ] k

A Basic Theorem Theorem 2 Proof : Using Markov’s inequality for t=1 we have that: Prof. Sotiris Nikoletseas Probabilistic Methods in Complex Networks ΥΔΑ ΜΔΕ, Patras 2019 - 2020 6 / 18 Let X be a non-negative integer random variable. Then if E [ X ] → 0 then Pr { X = 0 } → 1 Pr { X ≥ 1 } ≤ E [ X ] If E [ X ] → 0 then Pr { X = 0 } → 1 .

Non-Existence Proof Methodology ΥΔΑ ΜΔΕ, Patras 2019 - 2020 Probabilistic Methods in Complex Networks Prof. Sotiris Nikoletseas almost certainly there is no structure with the desired property. 4. c. b. otherwise the desired property holds a. 3. with the desired property. 2. sample space of combinatorial structures. Construct (by using abstract random experiments) an appropriate probability 1. 7 / 18 Defjne the random variable X that corresponds to the number of structures Express X as a sum of indicator variables X = X 1 + X 2 + · · · X n where � 1 X i = 0 Calculate E [ X ] using linearity of expectation. Prove that E [ X ] → 0 when X → ∞ Conclude by using theorem 2 that w.h.p. the r.v. X is limited to 0. Hence,

Random Graph Defjnition 3 (Random Graph) Prof. Sotiris Nikoletseas Probabilistic Methods in Complex Networks ΥΔΑ ΜΔΕ, Patras 2019 - 2020 8 / 18 A random graph is obtained by starting with a set of n isolated vertices and adding successive edges between them at random. In the G n,p model every possible edge occurs independently with probability 0 < p < 1 .

Example - Dominating Set Defjnition 4 (Dominating Set) Remark: The problem of fjnding a minimum dominating set is NP-hard. We will here address it by employing randomness. We will show that smaller than logarithmic-size dominating sets do not exist (w.h.p.) in dense random graphs. Theorem 5 Prof. Sotiris Nikoletseas Probabilistic Methods in Complex Networks ΥΔΑ ΜΔΕ, Patras 2019 - 2020 9 / 18 Given an undirected graph G = ( V, E ) , a dominating set is a subset S ⊆ V of its nodes such that for all nodes v ∈ V , either v ∈ S or a neighbor u of v is in S . For any k < ln n : Pr {∃ d.s. of size k in G n, 1 2 } → 0

Proof of theorem 5 (1/2) S is d.s. ΥΔΑ ΜΔΕ, Patras 2019 - 2020 Probabilistic Methods in Complex Networks Prof. Sotiris Nikoletseas 1. b. otherwise 10 / 18 3. a. 2. Let G be a graph generated using G n, 1 2 and S be any fjxed set of k vertices of G . Defjne r.v. X that corresponds to the number of dominating sets of size k . X = � S, | S | = k X S where X S are indicator variables � 1 X S = 0 Calculate E [ X ] : Using linearity of expectation:    � �  = E [ X ] = E X S E [ X S ] S, | S | = k S, | S | = k Calculate expectation of indicator variable X S : E [ X S ] = 1 · Pr { S is d.s. } + 0 · Pr { S is not d.s. } ⇒ E [ X S ] = Pr { S is d.s. }

Proof of theorem 5 (2/2) So, ΥΔΑ ΜΔΕ, Patras 2019 - 2020 Probabilistic Methods in Complex Networks Prof. Sotiris Nikoletseas Using theorem 2 we prove that almost certainly there are no dominating sets 4. It holds that c. 11 / 18 assume a vertex v �∈ d.s.: Pr { ∄ ( v, u ) : u ∈ S } = ( 1 2 ) k ⇒ Pr {∃ ( v, u ) : u ∈ S } = 1 − ( 1 2 ) k 2 k ) n − k 1 ⇒ Pr {∀ v out of S, ∃ ( v, u ) : u ∈ S } = (1 − 2 k ) n − k 1 ⇒ E [ X S ] = Pr { S is d.s. } = (1 − � � � � n − k n 1 − 1 � E [ X ] = E [ X S ] = k 2 k S, | S | = k − n − k 2 k ) n − k < e � n � ≤ n k and (1 − 1 2 k k ⇒ E [ X ] ≤ n k e − n − k k k ln n − n � � ≤ e 2 k e 2 k 2 k n If k ln n − 2 k → −∞ then E [ X ] → 0 , So, if k < ln n ⇒ E [ X ] → 0 of size k < ln n

(II) Linearity of Expectation Method where ΥΔΑ ΜΔΕ, Patras 2019 - 2020 Probabilistic Methods in Complex Networks Prof. Sotiris Nikoletseas 4. otherwise the desired property holds Basic methodology(1/2) 12 / 18 3. characteristics (e.g. the number or the size of the structures). 2. sample space of combinatorial structures. Construct (by using abstract random experiments) an appropriate probability 1. Defjne a random variable X that corresponds to the desired quantitative Express X as a sum of random indicator variables: X = X 1 + X 2 + · · · X n � 1 X i = 0 Calculate E [ X i ] = Pr { X i = 1 } .

(II) Linearity of Expectation Method 6. ΥΔΑ ΜΔΕ, Patras 2019 - 2020 Probabilistic Methods in Complex Networks Prof. Sotiris Nikoletseas Proof by contradiction: Basic methodology(2/2) Obvious observation: 13 / 18 5. Linearity of Expectation: �� � � E [ X ] = E = E [ X i ] X i i i even when X i are dependent. a random variable gets at least one value ≤ E [ X ] and at least one value ≥ E [ X ] . � � � µ = E [ X ] = x · f ( x ) > µ · f ( x ) = µ f ( x ) = µ x x x ⇒ ∃ at least one point (a structure) in the sample space for which X ≥ E [ X ] and at least one point (a structure) for which X ≤ E [ X ]

(II) Linearity of Expectation Method method’s capabilities and limitations ΥΔΑ ΜΔΕ, Patras 2019 - 2020 Probabilistic Methods in Complex Networks Prof. Sotiris Nikoletseas independence (less generic methods) technical diffjculties: linearity of variance generally requires stochastic inequalities with higher moments e.g. Chebyshev’s inequality : for more powerful results: is associated with fjrst moment Markov inequality: linearity does not require stochastic independence (generic method) estimation of expectation suffjces. 14 / 18 technically easy (indicator random variable ⇔ probability of property) Pr { X ≥ t } ≤ E [ X ] ⇔ Pr { X ≥ t · E [ X ] } ≤ 1 t t Pr {| X − µ | ≥ λσ } ≤ 1 λ 2

Example - Tournament with many Hamiltonian Cycles equiprobably for the two directions and independently for every edge. ΥΔΑ ΜΔΕ, Patras 2019 - 2020 Probabilistic Methods in Complex Networks Prof. Sotiris Nikoletseas otherwise (1/2) 2. 3. random tournaments by choosing the direction of each edge at random, We construct a probability sample space with points corresponding to 1. Proof : For every positive integer n, there exists a tournament on n vertices with at least Theorem 6 (Szele, 1943) 15 / 18 ( n − 1)!2 − ( n − 1) Hamiltonian cycles. We defjne the r.v. X that corresponds to the number of Hamiltonian Cycles. Let σ be a permutation of the vertices of the tournament. We have that X = � σ X σ where: � 1 σ leads to a Hamiltonian Cycle X σ = 0

Example - Tournament with many Hamiltonian Cycles (2/2) ΥΔΑ ΜΔΕ, Patras 2019 - 2020 Probabilistic Methods in Complex Networks Prof. Sotiris Nikoletseas Thus, there must exist at least one tournament on n vertices which has at 6. cycles. So, we have that: 5. 16 / 18 4. direction. A permutation σ leads to a Hamiltonian Cycle only if all edges have the same � n � n � n � 1 � 1 � 1 = 2 − ( n − 1) E [ X σ ] = Pr { X σ = 1 } = + = 2 2 2 2 By linearity of Expectation E [ X ] = E ( � σ X σ ) = � σ E [ X σ ] There are n ! n = ( n − 1)! permutations of n vertices that create difgerent E [ X ] = ( n − 1)!2 − ( n − 1) least ( n − 1)! · 2 − ( n − 1) Hamiltonian cycles.

Example - Bipartite Subgraphs Defjnition 7 (Bipartite Graph) Theorem 8 edges. Prof. Sotiris Nikoletseas Probabilistic Methods in Complex Networks ΥΔΑ ΜΔΕ, Patras 2019 - 2020 17 / 18 A bipartite graph is a graph whose vertices can be divided into two disjoint sets V 1 and V 2 such that every edge connects a vertex in V 1 to one in V 2 . Every graph G=(V,E) has a bipartite subgraph with at least | E | 2