Chapter 5: Concentration The Probabilistic Method Summer 2020 Freie Universität Berlin
Chapter Overview • Prove some strong concentration inequalities • Improve bounds on Ramsey numbers • Study Hamiltonicity and chromatic number of 𝐻 𝑜, 𝑞
§1 Chernoff Bounds Chapter 5: Concentration The Probabilistic Method
Domination vs Minimum Degree Corollary 2.2.5 Let 𝐻 be an 𝑜 -vertex graph with 𝜀 𝐻 ≥ 𝜀 . Then 𝐻 has a dominating ln 𝜀+1 +1 set 𝑇 ⊆ 𝑊 𝐻 with 𝑇 ≤ 𝑜 . 𝜀+1 Homework exercise • Show bound is tight by consider 𝐻 𝑜, 𝑞 Degrees in 𝐻 𝑜, 𝑞 • Degree of a vertex ∼ Bin 𝑜 − 1, 𝑞 • ⇒ expected degree is 𝑜 − 1 𝑞 • Minimum degree: need to show not far from mean 1 • Suppose ℙ | deg 𝑤 − 𝑜 − 1 𝑞| ≥ 𝑏 < 2𝑜 1 • Union bound ⇒ ℙ 𝜀 𝐻 𝑜, 𝑞 ≥ 𝑜 − 1 𝑞 − 𝑒 > 2
Comparing Bounds Concentration inequalities 1 • Let 𝑌 ∼ Bin 𝑜 − 1, 𝑞 , 𝑞 ≤ 2 • 𝔽 𝑌 = 𝑜 − 1 𝑞 , Var 𝑌 = 𝑜 − 1 𝑞 1 − 𝑞 = Θ 𝑜𝑞 𝔽 𝑌 • Markov: ℙ 𝑌 ≥ 𝑏 ≤ 𝑏 1 • ⇒ error probability < 2𝑜 for 𝑏 = Ω 𝑜 Var 𝑌 • Chebyshev: ℙ 𝑌 − 𝔽 𝑌 ≥ 𝑏 ≤ 𝑏 2 1 • ⇒ error probability < 2𝑜 for 𝑏 = Ω 𝑜 𝑞 −𝑑𝑏 2 • Central Limit Theorem: ℙ 𝑌 − 𝔽 𝑌 ≥ 𝑏 ≤ exp Var 𝑌 1 • ⇒ error probability < 2𝑜 for 𝑏 = Ω 𝑜𝑞 log 𝑜
The Problem With CLT Asymptotics • Central Limit Theorem is asymptotic, valid as 𝑜 → ∞ • We would like a quantitative bound for some given 𝑜 Definition 5.1.1 𝑜 Let 𝑇 𝑜 = σ 𝑗=1 𝑌 𝑗 , where the 𝑌 𝑗 are independently and uniformly distributed on −1,1 . Binomial connection 1 𝑜 • We have 𝑇 𝑜 ∼ 2 Bin 𝑜, 2 − 2 • Convenient to translate so mean is zero Goal • Show 𝑇 𝑜 is exponentially unlikely to be far from zero
Chernoff Bounds Theorem 5.1.2 (Symmetric Chernoff Bound) 𝑏 2 For every 𝑏 > 0 , we have ℙ 𝑇 𝑜 ≥ 𝑏 ≤ exp − 2𝑜 . Remarks • Concrete bounds for all 𝑜, 𝑏 • Symmetry: same bound for ℙ(𝑇 𝑜 ≤ −𝑏) 1 1 • Bin 𝑜, 2 = 2 𝑇 𝑜 + 𝑜 • ⇒ concentration for binomial random variables Corollary 5.1.3 −2𝑏 2 1 𝑜 For every 𝑏 > 0 , we have ℙ Bin 𝑜, 2 − 2 ≥ 𝑏 ≤ 2 exp . 𝑜
Proving Chernoff Theorem 5.1.2 (Symmetric Chernoff Bound) 𝑏 2 For every 𝑏 > 0 , we have ℙ 𝑇 𝑜 ≥ 𝑏 ≤ exp − 2𝑜 . Proof • Exponential conversion 𝑇 𝑜 ≥ 𝑏 = 𝑓 𝑇 𝑜 ≥ 𝑓 𝑏 = 𝑓 𝜇𝑇 𝑜 ≥ 𝑓 𝜇𝑏 • • Concentration • 𝑓 𝜇𝑇 𝑜 a non-negative random variable • Markov: ℙ 𝑓 𝜇𝑇 𝑜 ≥ 𝑓 𝜇𝑏 ≤ 𝔽 𝑓 𝜇𝑇 𝑜 𝑓 −𝜇𝑏 • Expectation 𝑜 • Recall 𝑇 𝑜 = σ 𝑗=1 𝑌 𝑗 • ⇒ 𝑓 𝜇𝑇 𝑜 = ς 𝑗=1 𝑜 𝑓 𝜇𝑌 𝑗 𝑜 1 2 𝑓 𝜇 + 𝑓 −𝜇 = cosh 𝑜 𝜇 • Independence ⇒ 𝔽 𝑓 𝜇𝑇 𝑜 = ς 𝑗=1 𝑜 𝔽 𝑓 𝜇𝑌 𝑗 =
Over the Cosh Recall • ℙ 𝑇 𝑜 ≥ 𝑏 ≤ 𝔽 𝑓 𝜇𝑇 𝑜 𝑓 −𝜇𝑏 • 𝔽 𝑓 𝜇𝑇 𝑜 = cosh 𝑜 𝜇 A little calculus 1 2 𝑓 𝑦 + 𝑓 −𝑦 • cosh 𝑦 = 𝑦 2 𝑦 3 𝑦 4 • Taylor series: 𝑓 𝑦 = 1 + 𝑦 + 2 + 6 + 24 + ⋯ 𝑦2 𝑦 2 𝑦 4 𝑦 6 𝑦 2 𝑦 4 𝑦 6 • ⇒ cosh 𝑦 = 1 + 2 + 24 + 720 + ⋯ ≤ 1 + 2 + 8 + 48 + ⋯ = 𝑓 2 Finishing the proof 1 2 𝑜𝜇 2 − 𝜇𝑏 • ∴ ℙ 𝑇 𝑜 ≥ 𝑏 ≤ exp 𝑏 2 𝑏 • Minimise: 𝜇 = 𝑜 ⇒ ℙ 𝑇 𝑜 ≥ 𝑏 ≤ exp − ∎ 2𝑜
The General Setting Shortcomings • Required each 𝑌 𝑗 to be uniform on −1,1 Wider Framework • 𝑞 1 , 𝑞 2 , … , 𝑞 𝑜 ∈ 0,1 , and 𝑞 = 𝑜 −1 σ 𝑗=1 𝑜 𝑞 𝑗 • 𝑌 𝑗 independent with ℙ 𝑌 𝑗 = 1 − 𝑞 𝑗 = 𝑞 𝑗 and ℙ 𝑌 = −𝑞 𝑗 = 1 − 𝑞 𝑗 𝑜 • 𝑌 = σ 𝑗=1 𝑌 𝑗 Theorem 5.1.4 (Asymmetric Chernoff Bound) Let 𝑏 > 0 and let 𝑌 and 𝑞 be as above. Then 𝑏 2 −𝑏 2 𝑏 3 ℙ 𝑌 ≤ −𝑏 ≤ exp − and ℙ 𝑌 ≥ 𝑏 ≤ exp 2𝑞𝑜 + 2 𝑞𝑜 2 . 2𝑞𝑜
An Asymmetric Chernoff Bound Theorem 5.1.4 (Asymmetric Chernoff Bound) Let 𝑏 > 0 and let 𝑌 and 𝑞 be as above. Then −𝑏 2 −𝑏 2 𝑏 3 ℙ 𝑌 ≤ −𝑏 ≤ exp and ℙ 𝑌 ≥ 𝑏 ≤ exp 2𝑞𝑜 + 2 𝑞𝑜 2 . 2𝑞𝑜 Special case • 𝑞 𝑗 = 𝑞 for all 𝑗 ⇒ 𝑌 + 𝑜𝑞 ~ Bin 𝑜, 𝑞 −𝑏 2 𝑏 3 • ⇒ ℙ Bin 𝑜, 𝑞 − 𝑜𝑞 ≥ 𝑏 ≤ 2 exp 2𝑞𝑜 + 2 𝑞𝑜 2 Corollary 5.1.5 For every 𝜁 > 0 there is some 𝑑 𝜁 > 0 such that, if 𝑍 is the sum of mutually independent indicator random variables and 𝜈 = 𝔽 𝑍 , then ℙ 𝑍 − 𝜈 ≥ 𝜁𝜈 ≤ 2 exp −𝑑 𝜁 𝜈 .
Any questions?
§2 Returning to Ramsey Chapter 5: Concentration The Probabilistic Method
The Story So Far Goal • Determine the order of magnitude of 𝑆(3, 𝑙) Upper bound 𝑙+1 = O 𝑙 2 • Erd ő s-Szekeres (1935): 𝑆 3, 𝑙 ≤ 2 Lower bounds • First moment, Mantel: 𝑆 3, 𝑙 = Ω(𝑙) 3/2 𝑙 • Alterations: 𝑆 3, 𝑙 = Ω log 𝑙 2 𝑙 • Lovász Local Lemma: 𝑆 3, 𝑙 = Ω log 𝑙
Alterations Revisited Theorem 2.1.2 ( ℓ = 3 ) For every 𝑜, 𝑙 ∈ ℕ and 𝑞 ∈ [0,1] , we have 𝑆 3, 𝑙 > 𝑜 − 𝑜 3 𝑞 3 − 𝑜 𝑙 2 . 1 − 𝑞 𝑙 Proof sketch • Take 𝐻 ∼ 𝐻 𝑜, 𝑞 • Remove one vertex from each triangle, independent set of size 𝑙 • Resulting graph 𝐻′ is Ramsey • First moment ⇒ with positive probability 𝐻 ′ has many vertices Optimisation 3/2 𝑙 • Largest right-hand side can be is O log 𝑙
Alternative Alterations Vertex removal • Wasteful operation • To fix a single, small triangle, we make Ω(𝑜) changes to the graph • Shrinks our resulting Ramsey graph too much Edge removal • More efficient fix • To fix a triangle, need only remove a single edge • Problematic • Being triangle-free and having small independence numbers are in conflict • Need to ensure we can destroy all triangles without creating large independent sets • A new hope • Can our more advanced probabilistic tools help?
Plan of Attack Detriangulation • Need to remove at least one edge from each triangle • Let 𝒰 be a maximal set of edge-disjoint triangles in 𝐻 • If 𝑈 is a triangle in 𝐻 , maximality ⇒ must share an edge with some 𝑈 ′ ∈ 𝒰 • Remove all edges of all triangles in 𝒰 • Removes 3 𝒰 edges • Need to remove at least 𝒰 edges Independent sets • Cannot let a set 𝑇 of 𝑙 vertices become independent • Would help if 𝐻 𝑇 had many edges to begin with • Expect to see 𝑙 2 𝑞 edges • Chernoff ⇒ very unlikely to see many fewer • Can afford a union bound over all such sets 𝑇
Local Edge Distribution Local edge counts • Fix a set 𝑇 of 𝑙 vertices 𝑙 • 𝑓 𝐻 𝑇 ∼ Bin 2 , 𝑞 • Expect 𝑙 2 𝑞 edges, how likely are we to see at least half of that? Theorem 5.1.4 (Asymmetric Chernoff Bound) −𝑏 2 Let 𝑏 > 0 and let 𝑌 and 𝑞 be as before. Then ℙ 𝑌 ≤ −𝑏 ≤ exp 2𝑞𝑜 . Applying Chernoff 1 𝑙 𝑙 𝑙 • Set 𝑌 = Bin 2 , 𝑞 − 2 𝑞 , and let 𝑏 = 2 𝑞 2 1 1 𝑙 𝑙 • ⇒ ℙ 𝑓 𝐻 𝑇 ≤ 2 𝑞 ≤ exp − 2 𝑞 2 8
Local Properties Globally Recall 1 1 𝑙 𝑙 • ℙ 𝑓 𝐻 𝑇 ≤ 2 𝑞 ≤ exp − 2 𝑞 2 8 Union bound • We need every 𝑙 -set to have many edges • Apply a union bound over choice of 𝑇 1 1 1 𝑙 𝑜 𝑙 𝑙 • ℙ ∃𝑇: 𝑓 𝐻 𝑇 ≤ 2 𝑞 ≤ 𝑙 exp − 2 𝑞 ≤ exp 𝑙 ln 𝑜 − 2 𝑞 2 8 8 Setting parameters 1 𝑙 • Small if 𝑙 ln 𝑜 ≤ 2 𝑞 , say 10 20 ln 𝑜 • ⇔ 𝑞 ≥ 𝑙−1 • To avoid too many triangles, take equality above 1 𝑙 • Then with high probability each 𝑙 -set spans at least 2 𝑞 = 5𝑙 ln 𝑜 edges 2
A Tangle of Triangles Recall 20 ln 𝑜 • Setting 𝑞 = 𝑙−1 ⇒ almost surely, every 𝑙 -set has at least 5𝑙 ln 𝑜 edges New independent sets • Remove all edges from a maximal set 𝒰 of edge-disjoint triangles • Need to avoid creating an independent set of 𝑙 vertices • Fix a set 𝑇 of 𝑙 vertices How many edges do we lose? • Only triangles with an edge in 𝑇 are relevant • Number of potential such triangles: 𝑙 𝑙 • 3 + 𝑜 − 𝑙 2 • Expected number of relevant triangles 4000 𝑜−𝑙 𝑜 𝑞 3 ≈ ln 3 𝑜 + 4000 𝑙 ln 3 𝑜 ≈ 4000 𝑙 ln 3 𝑜 𝑙 𝑙 • 3 + 𝑜 − 𝑙 2 3
Accounting for Triangles Recall • With high probability, each 𝑙 -set 𝑇 spans at least 5𝑙 ln 𝑜 edges 𝑜 𝑙 ln 3 𝑜 triangles with an edge in 𝑇 • Expect there to be at most 4000 Setting more parameters 𝑜 𝑙 ln 3 𝑜 ≤ 𝑑𝑙 ln 𝑜 • In order to ensure 𝑇 does not become independent, need • 𝑑 > 0 some small constant 2 2 𝑙 𝑙 = 𝑑 ′ • Solving: 𝑜 ≤ 𝑑 ln 𝑜 log 𝑙 Large deviations • Need to ensure that no set 𝑇 sees too many triangles • Union bound over 𝑜 𝑙 many sets • ⇒ need the probability that we get more triangles than expected to be small
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