The Probabilistic Method - Probabilistic Techniques Lecture 7: The Janson Inequality Sotiris Nikoletseas Associate Professor Computer Engineering and Informatics Department 2014-2015 Sotiris Nikoletseas, Associate Professor The Probabilistic Method 1 / 17
Άδειες Χρη σης Το παρο ν εκπαιδευτικο υλικο υπο κειται σε α δειες χρη σης Creative Commons. Για εκπαιδευτικο υλικο, ο πως εικο νες, που υπο κειται σε α λλου τυ που α δεια χρη σης, η α δεια χρη σης αναφε ρεται ρητω ς. Sotiris Nikoletseas, Associate Professor The Probabilistic Method 2 / 17
Χρηματοδο τηση Το παρο ν εκπαιδευτικο υλικο ε χει αναπτυχθει στα πλαι σια του εκπαιδευτικου ε ργου του διδα σκοντα. Το ε ργο Ανοικτα Ακαδημαι κα Μαθη ματα για το Πανεπιστη μιο Πατρω ν ε χει χρηματοδοτη σει μο νο την αναδιαμο ρφωση του εκπαιδευτικου υλικου. Το ε ργο υλοποιει ται στα πλαι σια του επιχειρισιακου προγρα μματος Εκπαι δευση και Δια Βι ου Μα θηση και συγχρηματοδοτει ται απο την Ευρωπαι κη Ένωση (Ευρωπαι κο Κοινοτικο Ταμειο) και απο εθνικου ς πο ρους. Sotiris Nikoletseas, Associate Professor The Probabilistic Method 3 / 17
Summary of previous lecture 1. The Lova sz Local Lemma 2. Example - Diagonal Ramsey Numbers Sotiris Nikoletseas, Associate Professor The Probabilistic Method 4 / 17
Summary of this lecture 1) On the importance of stochastic independence 2) The Janson Inequality 3) Example - Triangle-free sparse Random Graphs 4) Example - Paths of length 3 in G n,p Sotiris Nikoletseas, Associate Professor The Probabilistic Method 5 / 17
1) On the importance of stochastic independence The local Lemma demonstrates that rare dependencies yield results similar to the case of stochastic independence. The Janson inequality actually does the same, but for the case when the total amount of dependencies is rather small. Sotiris Nikoletseas, Associate Professor The Probabilistic Method 6 / 17
Intuition 1 Let B i be the undesired events. 2 non-trivial dependence : i j i j and B i, B j dependent 3 = i j Pr{B i B j }: measure of dependencies. 4 If the events were independent then the probability of the desired property is de ined as follows: Pr{ B i } = i Pr{B i } = M 5 The Janson inequality shows that Pr{ B i } remains very close to M if the dependencies are small. Sotiris Nikoletseas, Associate Professor The Probabilistic Method 7 / 17
2) The Janson Inequality Theorem 1 Let B i be undesired events. De ine 1) = i j Pr{B i B j } and 2) M = i Pr{B i} If Pr{B i } ϵ then M Pr { i } ( 1 B i M exp 1 ϵ ) 2 Remark: If αʹ. ϵ is small (e.g. ϵ is constant or smaller i.e. undesired events are not very probable) and βʹ. is small e.g. o(1) (i.e. there are small dependencies) Then, { } e ( 1 ϵ 1 2 ) 1 Pr B i M = Pr{B i } i i Sotiris Nikoletseas, Associate Professor The Probabilistic Method 8 / 17
3) Example - Triangle-free sparse Random Graphs Theorem 2 Consider the G n,p graph space. For every constant c, If p = c n Pr{ K 3} e c3 /6 (G n,p with p = c n Proof: is sparse because the connectivity threshold is p = clogn n.) Let S be any ixed set of 3 vertices ( S = 3). We de ine the event B S = {S is K 3 (triangle)}. Pr{B S } = p 3 = c3 n 3 We want to prove that { Pr } B S e c3 /6 S, S =3 Sotiris Nikoletseas, Associate Professor The Probabilistic Method 9 / 17
Proof of theorem 2 Intuition: If events B S were independent, then: { } M = Pr B S = Pr { } B S S S, S =3 = (1 p 3 ) (n 3) e c3 n 3 n3 3! = e c3 /6 Since events are dependent, we will show that dependencies in this sparse graph are small, to get a very similar result, via applying the Janson inequality: M = S, S =3 ϵ = Pr{B S } = c3 n 3 0 Pr{B S } = (1 p 3 ) (n 3) e c 3 /6 = S T Pr{B S B T } Sotiris Nikoletseas, Associate Professor The Probabilistic Method 10 / 17
Proof of theorem 2 Non-trivial dependence: { } S T S T S T 2 S T = 2 Pr{B S B T } = p 5 ( )( )( ) n 3 n 3 = p 5 3 2 1 ) ( ) = O(n 4 p 5 ) = O (n 4 c5 1 n 5 = Θ = o(1) n Since ϵ = o(1) and = o(1) we apply the basic case of Janson inequality that gives us: { } lim Pr B s = lim M = /6 n n e c3 S Sotiris Nikoletseas, Associate Professor The Probabilistic Method 11 / 17
4) Example - Paths of length 3 in G n,p Theorem 3 De ine the event B={ a path of length 3 between any pair of vertices in G n,p }. For every constant c 2: ( ) 1/3 c ln n If p = Pr{B} 1 n 2 Proof: We de ine the event: B u,v = { a path of length 3 between vertices u and v}. { } Pr{B} = Pr B u,v u,v u,v Pr{B u,v } = O(n 2 ) Pr{B u,v } Sotiris Nikoletseas, Associate Professor The Probabilistic Method 12 / 17
Proof of theorem 3 In order to prove that Pr{B} 0 or, equivalently Pr{B} 1, we must prove that Pr{B u,v } = o(n 2 ) We apply Janson Inequality to prove that Pr{B u,v } = o(n 2 ). We de ine the undesired event: A w1,w 2 = {the edges (u, w 1 ), (w 1, w 2 ), (w 2, v) exist} and express the event B u,v as follows: { } B u,v = w 1,w 2 A w1,w 2 Sotiris Nikoletseas, Associate Professor The Probabilistic Method 13 / 17
Proof of theorem 3 } ϵ = Pr {A w1,w2 = p 3 = ( (c ) ) ln n 1/3 3 0 M = } Pr {A w1,w2 = ( 1 p 3) (n 2)(n 3) w 1,w 2 n 2 e p3 n 2 = e c ln n n 2 n2 = n c = o ( n 2) So, the result would held if there were no dependencies. { } = Pr Aw1,w 2 A w 1,w 2 (w 1,w 2 ) (w 1,w 2 ) Sotiris Nikoletseas, Associate Professor The Probabilistic Method 14 / 17
Proof of theorem 3 case 1: u w 1 w 2 = w 2 w 1 v Contribution to : case 1: ( ) n 2 p 5 3 case 2: u w 1 = w 1 w 2 v case 2: ( ) n 2 p 5 3 case 3: u w 2 w 1 = w 2 w 2 = w 1 v case 3: ( ) n 2 p 5 2 Sotiris Nikoletseas, Associate Professor The Probabilistic Method 15 / 17
Proof of theorem 3 = O(n 3 )p 5 + O(n 3 )p 5 + O(n 2 )p 5 ( c ln n = O(n 3 )p 5 = O(n 3 ) n ( ) 2 (c ln n) 5/3 = O = o(1) n 1/3 ) 5/3 Since ϵ = o(1) and = o(1) we apply the basic case of Janson inequality that gives us: { } lim Pr = lim M = o(n 2 ) n n w 1,w 2 A w1,w 2 Sotiris Nikoletseas, Associate Professor The Probabilistic Method 16 / 17
Sotiris Nikoletseas, Associate Professor The Probabilistic Method 17 / 17