Chapter 3: The Second Moment The Probabilistic Method Summer 2020 - - PowerPoint PPT Presentation

Chapter 3: The Second Moment

The Probabilistic Method Summer 2020 Freie Universität Berlin

Chapter Overview

• Introduce the second moment method
• Survey applications in graph theory and number theory

§1 Concentration Inequalities

Chapter 3: The Second Moment The Probabilistic Method

What Does the Expectation Mean?

Basic fact

• 𝑌 ≤ 𝔽 𝑌

and {𝑌 ≥ 𝔽 𝑌 } have positive probability

• Often want more quantitative information
• What are these positive probabilities?
• How much below/above the expectation can the random variable be?

Limit laws

• Law of large numbers
• Average of independent trials will tend to the expectation
• Central limit theorem
• Average will be normally distributed

Not always applicable

• We often only have a single instance, or lack independence
• Can still make use of more general bounds

Markov’s Inequality

Proof

• Let 𝑔 be the density function for the distribution of 𝑌
• 𝔽 𝑌 = ׬

∞ 𝑦𝑔 𝑦 ⅆ𝑦 = ׬ 𝑏 𝑦𝑔 𝑦 ⅆ𝑦 + ׬ 𝑏 ∞ 𝑦𝑔 𝑦 ⅆ𝑦

≥ ׬

𝑏 ∞ 𝑦𝑔 𝑦 ⅆ𝑦 ≥ ׬ 𝑏 ∞ 𝑏𝑔 𝑦 ⅆ𝑦 = 𝑏 ׬ 𝑏 ∞ 𝑔 𝑦 ⅆ𝑦 = 𝑏ℙ(𝑌 ≥ 𝑏)

∎

Moral: 𝔽 𝑌 small ⇒ 𝑌 typically small Theorem 3.1.1 (Markov’s Inequality) Let 𝑌 be a non-negative random variable, and let 𝑏 > 0. Then ℙ 𝑌 ≥ 𝑏 ≤ 𝔽 𝑌 𝑏 .

Theorem 3.1.2 (Chebyshev’s Inequality) Let 𝑌 be a random variable, and let 𝑏 > 0. Then ℙ 𝑌 − 𝔽 𝑌 ≥ 𝑏 ≤ Var 𝑌 𝑏2 .

Chebyshev’s Inequality

Converse? Does 𝔽 𝑌 large ⇒ 𝑌 typically large?

• Not necessarily; e.g. 𝑌 = 𝑜2 with probability 𝑜−1, 0 otherwise
• But such random variables have large variance…

Proof

• 𝑌 − 𝔽 𝑌

≥ 𝑏 = 𝑌 − 𝔽 𝑌

2 ≥ 𝑏2

• Let 𝑍 = 𝑌 − 𝔽 𝑌

2

• Then 𝔽 𝑍 = Var 𝑌
• Apply Markov’s Inequality

∎

Corollary 3.1.3 If Var 𝑌 = 𝑝 𝔽 𝑌 2 , then ℙ 𝑌 = 0 = 𝑝 1 .

Using Chebyshev

Moral

• 𝔽 𝑌 large and Var 𝑌 small ⇒ 𝑌 typically large
• Special case: showing 𝑌 nonzero

Proof

• 𝑌 = 0 ⊆

𝑌 − 𝔽 𝑌 ≥ 𝔽 𝑌

• Chebyshev ⇒ ℙ 𝑌 − 𝔽 𝑌

≥ 𝔽 𝑌 ≤ Var 𝑌

𝔽 𝑌 2 = 𝑝(1)

∎

• In fact, in this case 𝑌 = 1 + 𝑝 1

𝔽[𝑌] with high probability

Typical application

Set-up

• 𝐹𝑗 events, occurring with probability 𝑞𝑗
• 𝑌𝑗 = 1𝐹𝑗 their indicator random variables
• 𝑌 = σ𝑗 𝑌𝑗 their sum, the number of occurring events

Goal

• Show that with high probability, some event occurs

Applying Chebyshev

• Need to show Var 𝑌 = 𝑝 𝔽 𝑌 2

Expand the variance

• Var 𝑌 = Var σ𝑗 𝑌𝑗 = σ𝑗 Var(𝑌𝑗) + σ𝑗≠𝑘 Cov(𝑌𝑗, 𝑌

𝑘)

Corollary 3.1.4 Let 𝐹𝑗 be a sequence of events with probabilities 𝑞𝑗, and let 𝑌 count the number of events that occur. Write 𝑗 ∼ 𝑘 if the events 𝐹𝑗 and 𝐹

𝑘 are

not independent, and let Δ = σ𝑗∼𝑘 ℙ 𝐹𝑗 ∧ 𝐹

𝑘 . If 𝔽 𝑌 → ∞ and

Δ = 𝑝 𝔽 𝑌 2 , then 𝑄 𝑌 = 0 = 𝑝 1 .

Some Simplification

Estimating the summands

• Var 𝑌 = Var σ𝑗 𝑌𝑗 = σ𝑗 Var(𝑌𝑗) + σ𝑗≠𝑘 Cov(𝑌𝑗, 𝑌

𝑘)

• Var 𝑌𝑗 = 𝑞𝑗 1 − 𝑞𝑗 ≤ 𝑞𝑗
• ∴ σ𝑗 Var(𝑌𝑗) ≤ σ𝑗 𝑞𝑗 = σ𝑗 𝔽 𝑌𝑗 = 𝔽[𝑌]
• Cov 𝑌, 𝑍 = 𝔽 𝑌𝑍 − 𝔽 𝑌 𝔽 𝑍
• Cov 𝑌, 𝑍 = 0 if 𝑌 and 𝑍 are independent
• Otherwise Cov 𝑌𝑗, 𝑌

𝑘 ≤ 𝔽 𝑌𝑗𝑌 𝑘 = ℙ 𝐹𝑗 ∧ 𝐹 𝑘

Any questions?

§2 Thresholds

Chapter 3: The Second Moment The Probabilistic Method

Lemma 3.2.1 If 𝒬 is a monotone increasing graph property, then ℙ 𝐻 𝑜, 𝑞 ∈ 𝒬 is monotone increasing in 𝑞.

Monotone properties

Graph properties

• Say a graph 𝒬 is monotone (increasing) if adding edges preserves 𝒬
• e.g.: containing a subgraph 𝐼 ⊆ 𝐻, having 𝛽 𝐻 < 𝑙, connectivity, …

Proof (Coupling)

• Sampling 𝐻 𝑜, 𝑞
• Assign to each pair of vertices 𝑣, 𝑤 an independent uniform 𝑍

𝑣,𝑤~Unif 0,1

• Add edge 𝑣, 𝑤 to 𝐻 iff 𝑍

𝑣,𝑤 ≤ 𝑞

• Each edge appears independently with probability 𝑞
• If 𝑞 ≤ 𝑞′, then 𝐻 𝑜, 𝑞 ⊆ 𝐻 𝑜, 𝑞′ ⇒ if 𝐻 𝑜, 𝑞 ∈ 𝒬, then 𝐻 𝑜, 𝑞′ ∈ 𝒬

∎

Definition 3.2.2 (Thresholds) Given a nontrivial monotone graph property 𝒬, 𝑞0(𝑜) is a threshold for 𝒬 if ℙ 𝐻 𝑜, 𝑞 ∈ 𝒬 → ቊ0 if p ≪ 𝑞0 𝑜 , 1 if 𝑞 ≫ 𝑞0 𝑜 .

Thresholds

Transitions

• A monotone property 𝒬 is nontrivial if it is not satisfied by the edgeless graph,

and is satisfied by the complete graph

• ⇒ ℙ 𝐻 𝑜, 0 ∈ 𝒬 = 0 and ℙ 𝐻 𝑜, 1 ∈ 𝒬 = 1
• Lemma 3.2.1 ⇒ ℙ 𝐻 𝑜, 𝑞 ∈ 𝒬 increases from 0 to 1 as 𝑞 does
• How quickly does this increase happen?

Proposition 3.2.3 The threshold for 𝐻(𝑜, 𝑞) to contain a cycle is 𝑞0 𝑜 =

1 𝑜.

A Cyclic Example

Proof (lower bound)

• Let 𝑌 = # cycles in 𝐻(𝑜, 𝑞)
• For ℓ ≥ 3, let 𝑌ℓ = # 𝐷ℓ ⊆ 𝐻 𝑜, 𝑞
• ⇒ 𝑌 = σℓ=3

𝑜

𝑌ℓ

• Linearity of expectation: 𝔽 𝑌ℓ ≤ 𝑜ℓ𝑞ℓ
• ⇒ 𝔽 𝑌 ≤ σℓ=3

𝑜

𝑜𝑞 ℓ < 𝑜𝑞 3 σℓ=0

∞

𝑜𝑞 ℓ =

𝑜𝑞 3 1−𝑜𝑞

• ⇒ 𝔽 𝑌 = 𝑝(1) if 𝑞 ≪

1 𝑜

• Markov: ℙ 𝐻 𝑜, 𝑞 has a cycle = ℙ 𝑌 ≥ 1 ≤ 𝔽 𝑌 → 0

∎

Proposition 3.2.3 The threshold for 𝐻(𝑜, 𝑞) to contain a cycle is 𝑞0 𝑜 =

1 𝑜.

Cycles Continued

Proof (upper bound)

• Let 𝑞 =

4 𝑜−1 and set 𝑍 = 𝑓 𝐻 𝑜, 𝑞

• Then 𝑍 ∼ Bin

𝑜 2 , 𝑞

• ⇒ 𝔽 𝑍 =

𝑜 2 𝑞 = 2𝑜

• ⇒ Var 𝑍 =

𝑜 2 𝑞 1 − 𝑞 < 2𝑜

• ∴ Var 𝑍 = 𝑝 𝔽 𝑍 2
• Chebyshev: ℙ 𝑍 < 𝑜 → 0
• ℙ 𝐻 𝑜, 𝑞 has a cycle ≥ ℙ 𝑓 𝐻 𝑜, 𝑞

≥ 𝑜 → 1 ∎

Existence of Thresholds

Proof (upper bound)

• Let 𝑞 𝑜 = 𝑞0 be such that ℙ 𝐻 𝑜, 𝑞0 ∈ 𝒬 =

1 2

• Let 𝐻 ∼ 𝐻1 ∪ 𝐻2 ∪ ⋯ ∪ 𝐻𝑛, where each 𝐻𝑗 ∼ 𝐻 𝑜, 𝑞0 is independent
• ⇒ 𝐻 ∼ 𝐻 𝑜, 𝑞 for 𝑞 ≔ 1 − 1 − 𝑞0 𝑛 ≤ 𝑛𝑞0
• Property is monotone:
• ℙ 𝐻 ∈ 𝒬 ≥ ℙ ∪𝑗 𝐻𝑗 ∈ 𝒬

= 1 − ℙ ∩𝑗 𝐻𝑗 ∉ 𝒬

• Graphs are independent:
• ℙ ∩𝑗 𝐻𝑗 ∉ 𝒬

= ς𝑗 ℙ 𝐻𝑗 ∉ 𝒬

• Since 𝐻𝑗 ∼ 𝐻(𝑜, 𝑞0), ℙ 𝐻𝑗 ∉ 𝒬 =

1 2

• ∴ ℙ 𝐻 ∈ 𝒬 ≥ 1 − 2−𝑛 → 1 if 𝑛 → ∞ (or if 𝑞 ≫ 𝑞0)

∎

Theorem 3.2.4 (Bollobás-Thomason, 1987) Every nontrivial monotone graph property has a threshold.

Below the Threshold

Proof (lower bound)

• Let 𝐻 ∼ 𝐻1 ∪ 𝐻2 ∪ ⋯ ∪ 𝐻𝑛 as before, but with 𝐻𝑗 ∼ 𝐻 𝑜, 𝑞 for 𝑞 =

𝑞0 𝑛

• ⇒ 𝐻 ∼ 𝐻(𝑜, 𝑟) for 𝑟 = 1 − 1 − 𝑞 𝑛 ≤ 𝑛𝑞 = 𝑞0
• ⇒ ℙ 𝐻 ∉ 𝒬 ≥

1 2

• As before, ℙ 𝐻 ∉ 𝒬 ≤ ℙ 𝐻 𝑜, 𝑞 ∉ 𝒬 𝑛
• ⇒ ℙ 𝐻 𝑜, 𝑞 ∉ 𝒬 ≥

1 2 1/m

• ⇒ ℙ 𝐻 𝑜, 𝑞 ∈ 𝒬 ≤ 1 −

1 2 1/𝑛

→ 0 if 𝑛 → ∞ (or if 𝑞 ≪ 𝑞0) ∎

Theorem 3.2.4 (Bollobás-Thomason, 1987) Every nontrivial monotone graph property has a threshold.

Definition 3.2.5 (Sharp thresholds) We say 𝑞0(𝑜) is a sharp threshold for 𝒬 if there are positive constants 𝑑1, 𝑑2 such that ℙ 𝐻 𝑜, 𝑞 ∈ 𝒬 → ቊ0 if 𝑞 ≤ 𝑑1𝑞0 𝑜 , 1 if 𝑞 ≥ 𝑑2𝑞0 𝑜 .

Closing Remarks

Random graph theory

• Fundamental problem: given a graph property 𝒬, what is its threshold?

At the threshold

• We showed what happens for probabilities much smaller than the threshold,

and much larger than the threshold

• What if 𝑞 = Θ 𝑞0 𝑜

? Some properties have a much quicker transition

Any questions?

§3 Subgraphs of 𝐻 𝑜, 𝑞

Chapter 3: The Second Moment The Probabilistic Method

Returning to Ramsey

Choosing parameters

• Want to choose 𝑜 as large as possible
• Need to avoid large independent sets
• ⇒ would like to make edge probability 𝑞 large
• Limitation: need to avoid 𝐿ℓ

Question: What is the threshold for Kℓ ⊆ 𝐻(𝑜, 𝑞)? Theorem 1.5.7 Given ℓ, 𝑙, 𝑜 ∈ ℕ and 𝑞 ∈ 0,1 , if 𝑜 ℓ 𝑞

ℓ 2 + 𝑜

𝑙 1 − 𝑞

𝑙 2 < 1,

then 𝑆 ℓ, 𝑙 > 𝑜.

A Lower Bound

Goal

• Let 𝑌 count the number of 𝐿ℓ in 𝐻(𝑜, 𝑞)
• For which 𝑞 do we have ℙ 𝑌 ≥ 1 = 𝑝(1)?

First moment

• 𝔽 𝑌 =

𝑜 ℓ 𝑞

ℓ 2 = Θ 𝑜ℓ𝑞 ℓ 2

• Markov’s Inequality: ℙ 𝑌 ≥ 1 ≤ 𝔽 𝑌

Threshold bound

• 𝔽 𝑌 = Θ 𝑜ℓ𝑞

ℓ 2

≪ 1

• ⇔ 𝑞

ℓ 2 ≪ 𝑜−ℓ ⇔ 𝑞 ≪ 𝑜−2/(ℓ−1)

• ⇒ 𝑞0(𝑜) ≥ 𝑜−2/(ℓ−1)

Corollary 3.1.4 Let 𝐹𝑗 be a sequence of events with probabilities 𝑞𝑗, and let 𝑌 count the number of events that occur. Write 𝑗 ∼ 𝑘 if the events 𝐹𝑗 and 𝐹

𝑘 are

not independent, and let Δ = σ𝑗∼𝑘 ℙ 𝐹𝑗 ∧ 𝐹

𝑘 . If 𝔽 𝑌 → ∞ and

Δ = 𝑝 𝔽 𝑌 2 , then 𝑄 𝑌 = 0 = 𝑝 1 .

An Upper Bound

Goal

• For which 𝑞 do we have ℙ 𝑌 = 0 = 𝑝 1 ?

Our parameters

• Let 𝐻 ∼ 𝐻(𝑜, 𝑞) and, for 𝑇 ∈

𝑜 ℓ , let 𝐹𝑇 = 𝐻 𝑇 ≅ 𝐿ℓ

• 𝔽 𝑌 =

𝑜 ℓ 𝑞

ℓ 2 → ∞ for 𝑞 ≫ 𝑜−2/(ℓ−1)

• Suffices to show Δ = 𝑝 𝔽 𝑌 2

Clique Dependencies

Independent events

• 𝐹𝑗 occurs ⇔ all edges in 𝑗th clique present
• Edges appear independently
• ∴ 𝑇 ∩ 𝑈 ≤ 1 ⇒ 𝐹𝑇, 𝐹𝑈 independent

Dependent events

• Suppose 𝑇 ∩ 𝑈 = 𝑡 ≥ 2
• ⇒ 𝑇 ∼ 𝑈
• 𝐹𝑇 ∧ 𝐹𝑈: 𝐻 𝑇 , 𝐻[𝑈] both ℓ-cliques, sharing 𝑡 vertices
• Number of prescribed edges: 2 ℓ

2 − 𝑡 2

• ⇒ ℙ 𝐹𝑇 ∧ 𝐹𝑈 = 𝑞2 ℓ

2 − 𝑡 2

Computing Δ

Recall

• 𝑇 ∼ 𝑈 ⇔ 𝑡 ≔ 𝑇 ∩ 𝑈 ≥ 2
• ℙ 𝐹𝑇 ∧ 𝐹𝑈 = 𝑞2 ℓ

2 − s 2

Substituting terms

Δ = σ𝑇∼𝑈 ℙ 𝐹𝑇 ∧ 𝐹𝑈 = σ 𝑇∩𝑈 ≥2 ℙ 𝐹𝑇 ∧ 𝐹𝑈 = σ𝑇 σ𝑈: 𝑇∩𝑈 ≥2 ℙ 𝐹𝑇 ∧ 𝐹𝑈 = σ𝑇 σ𝑡=2

ℓ−1 σ𝑈: 𝑇∩𝑈 =𝑡 ℙ 𝐹𝑇 ∧ 𝐹𝑈

= σ𝑇 σ𝑡=2

ℓ−1 σ𝑈: 𝑇∩𝑈 =𝑡 𝑞2 ℓ

2 − 𝑡 2

⇒ Δ =

𝑜 ℓ σ𝑡=2 ℓ−1 ℓ 𝑡 𝑜−ℓ ℓ−𝑡 𝑞2 ℓ

2 − 𝑡 2

Bounding Δ

Recall

• Δ =

𝑜 ℓ σ𝑡=2 ℓ−1 ℓ 𝑡 𝑜−ℓ ℓ−𝑡 𝑞2 ℓ

2 − 𝑡 2

Goal

• Show Δ = 𝑝 𝔽 𝑌 2

Estimates

• ℓ

𝑡 ≤ 2ℓ

• 𝑜−ℓ

ℓ−𝑡

≤ 𝑜ℓ−𝑡 = Θ

𝑜 ℓ 𝑜−𝑡

Bound

Δ ≤

𝑜 ℓ σ𝑡=2 ℓ−1 2ℓΘ 𝑜 ℓ 𝑜−𝑡 𝑞2 ℓ

2 − 𝑡 2

=

𝑜 ℓ 2𝑞2 ℓ

2 σ𝑡=2

ℓ−1 Θ 𝑜−𝑡𝑞− 𝑡

2

= 𝔽 𝑌 2 σ𝑡=2

ℓ−1 Θ 𝑜−𝑡𝑞− 𝑡

2

Theorem 3.3.1 For ℓ ≥ 2, the threshold for Kℓ ⊆ 𝐻(𝑜, 𝑞) is 𝑞0 𝑜 = 𝑜−2/(ℓ−1).

Completing the Calculation

Recall

• Δ = 𝔽 𝑌 2 σ𝑡=2

ℓ−1 Θ 𝑜−𝑡𝑞− 𝑡

2

Substituting 𝑞

• 𝑜−𝑡𝑞− 𝑡

2 = 𝑜𝑞 𝑡−1 /2 −𝑡

• We took 𝑞 ≫ 𝑜−2/(ℓ−1)
• ⇒ 𝑜−𝑡𝑞− 𝑡

2 ≪ n1−(s−1)/(ℓ−1) −s

• For 2 ≤ 𝑡 ≤ ℓ − 1, this is 𝑝(1)
• ⇒ Δ = 𝑝 1

An Incomplete Result

Lower bound

• Let 𝑌 be the number of copies of 𝐼 in 𝐻 𝑜, 𝑞
• Markov: 𝔽 𝑌 = 𝑝 1 ⇒ 𝑞0 𝑜 ≫ 𝑞

Expectation

• Number of possible copies
• Specify vertices of 𝐼 – at most 𝑜𝑤 𝐼 possibilities
• Probability of appearance
• Each edge of 𝐼 must be present – probability is 𝑞𝑓 𝐼
• ⇒ 𝔽 𝑌 ≤ 𝑜𝑤 𝐼 𝑞𝑓 𝐼

Conclusion: 𝑞0 𝑜 ≥ 𝑜−𝑤 𝐼 /𝑓(𝐼) Problem Given a graph 𝐼, what is the threshold 𝑞0

𝐼(𝑜) for 𝐼 ⊆ 𝐻 𝑜, 𝑞 ?

An Illustrated Example

Graph statistics

• Let 𝐼 be 𝐿4 with a pendant edge
• Statistics:
• 𝑤 𝐼 = 5
• 𝑓 𝐼 = 7
• ⇒ 𝑞0

𝐼 𝑜 ≥ 𝑜−5/7

An issue

• 𝐿4 ⊆ 𝐼
• ⇒ if 𝐼 ⊆ 𝐻(𝑜, 𝑞), then 𝐿4 ⊆ 𝐻 𝑜, 𝑞
• ⇒ 𝑞0

𝐿4 𝑜 ≤ 𝑞0 𝐼 𝑜

• But we showed 𝑞0

𝐿4 𝑜 = 𝑜−2/3 ≫ 𝑜−5/7

Definition 3.3.2 (Maximum density) Given a graph 𝐼, define d 𝐼 =

𝑓 𝐼 𝑤 𝐼 , and let

𝑛 𝐼 = max {ⅆ 𝐺 : 𝐺 ⊆ 𝐼}.

Monotonicity and Density

General lower bound

• 𝑞0

𝐼 𝑜 ≥ max 𝑞0 𝐺 𝑜 : 𝐺 ⊆ 𝐼

• Can substitute first moment bound
• ⇒ 𝑞0

𝐼 𝑜 ≥ max 𝑜−𝑤 𝐺 /𝑓(𝐺): 𝐺 ⊆ 𝐼, 𝑓 𝐺 ≥ 1

Remarks

• We have 𝑞0

𝐼 𝑜 ≥ 𝑜−1/𝑛(𝐼)

• Say 𝐼 is balanced if d 𝐼 = 𝑛 𝐼
• 𝐼 is strictly balanced if ⅆ 𝐺 < 𝑛(𝐼) for all 𝐺 ⊂ 𝐼

Expected Subgraph Counts

Boundless expectations

• Let 𝑌𝐼 be the number of copies of 𝐼 in 𝐻(𝑜, 𝑞)
• Total # possible copies = Θ 𝑜𝑤 𝐼
• Probability of each copy: 𝑞𝑓 𝐼
• ⇒ 𝔽 𝑌𝐼 = Θ 𝑜𝑤 𝐼 𝑞𝑓 𝐼
• ∴ 𝔽 𝑌𝐼 → ∞ when 𝑞 ≫ 𝑜−𝑤 𝐼 /𝑓 𝐼

Guaranteeing subgraph existence

• Goal: to show ℙ 𝑌𝐼 = 0 = 𝑝(1) for 𝑞 ≫ 𝑞0

𝐼(𝑜)

• Apply second moment: need to show Δ = o 𝔽 𝑌𝐼 2
• Edge-disjoint copies are independent

Dependent Subgraphs

Common subgraphs

• Let 𝐼1, 𝐼2 be two copies of 𝐼 sharing an edge
• 𝐹𝐼1 ∧ 𝐹𝐼2 = 𝐼1 ∪ 𝐼2 ⊆ 𝐻 𝑜, 𝑞
• Let 𝐺 ≔ 𝐼1 ∩ 𝐼2 be the common subgraph
• 𝑤 𝐼1 ∪ 𝐼2 = 2𝑤 𝐼 − 𝑤 𝐺
• 𝑓 𝐼1 ∪ 𝐼2 = 2𝑓 𝐼 − 𝑓 𝐺

Counting pairs

• Group dependent pairs 𝐼1, 𝐼2 by common subgraphs 𝐺 = 𝐼1 ∩ 𝐼2
• At most 2𝑓 𝐼 possible subgraphs 𝐺
• For each 𝐾, O 𝑜2𝑤 𝐼 −𝑤 𝐺

pairs (𝐼1, H2)

• For each such pair, ℙ 𝐹𝐼1 ∧ 𝐹𝐼2 = 𝑞2𝑓 𝐼 −𝑓 𝐺

Bounding Δ

Recall

Δ = σ𝑗∼𝑘 ℙ 𝐹𝐼𝑗 ∧ 𝐹𝐼𝑘

Group by common subgraph

Δ = σ𝑗∼𝑘 ℙ 𝐹𝐼𝑗 ∧ 𝐹𝐼𝑘 = σ𝐺⊂𝐼 σ 𝑗,𝑘 :𝐼𝑗∩𝐼𝑘=𝐺 ℙ 𝐹𝐼𝑗 ∧ 𝐹𝐼𝑘

Substitute estimates

Δ = σ𝐺⊂𝐼 𝑃 𝑜2𝑤 𝐼 −𝑤 𝐺 𝑞2𝑓 𝐼 −𝑓 𝐺 ⇒ Δ = 𝑜𝑤 𝐼 𝑞𝑓 𝐼

2 σ𝐺⊂𝐼 𝑃 𝑜−𝑤 𝐺 𝑞−𝑓 𝐺

⇒ Δ = 𝔽 𝑌𝐼 2 σ𝐺⊂𝐼 𝑃 𝑜−𝑤 𝐺 𝑞−𝑓 𝐺

Theorem 3.3.3 Given a graph 𝐼, the threshold for 𝐼 ⊆ 𝐻(𝑜, 𝑞) is 𝑞0

𝐼 𝑜 = 𝑜−1/𝑛(𝐼),

where 𝑛 𝐼 = max 𝑓 𝐺 𝑤 𝐺 : 𝐺 ⊆ 𝐼 .

A Complete Solution

Recall

• Δ = 𝔽 𝑌𝐼 2 σ𝐺⊂𝐼 𝑃 𝑜−𝑤 𝐺 𝑞−𝑓 𝐺

Choice of 𝑞

• We have 𝑞 ≫ 𝑜−1/𝑛(𝐼)
• ⇒ 𝑞 ≫ 𝑜−𝑤 𝐺 /𝑓(𝐺) for all nonempty 𝐺 ⊂ 𝐼
• ⇒ 𝑜−𝑤 𝐺 𝑞−𝑓 𝐺 = 𝑝(1)
• ⇒ Δ = 𝑝(1)

Any questions?

§4 Prime Factors

Chapter 3: The Second Moment The Probabilistic Method

Theorem 3.4.1 (Hadamard, De la Vallée Poussin, 1896) The number 𝜌(𝑜) of prime numbers in [𝑜] satisfies 𝜌 𝑜 = 1 + 𝑝 1 𝑜 ln 𝑜 .

Time For Primes

Fun facts

• There are infinitely many primes (Euclid, -300)
• The primes contain arbitrarily long arithmetic progressions (Green–Tao, 2004)
• Infinitely many pairs of primes are at most 70000000 apart (Zhang, 2014)

Central problem

• How are the primes distributed in ℕ?

Definition 3.4.2 Given 𝑦 ∈ ℕ, let 𝜉 𝑦 denote the number of distinct prime factors of 𝑦.

Prime Factorisation

The funnest of facts

• Every natural number is the product of primes

Our goal

• To understand what these factorisations look like

Examples

• 𝜉 19 = ?
• 𝜉 210 = ?
• 𝜉 256 = ?
• 𝜉 2020 = ?

The Average Case

Proof

• Express 𝜉(𝑦) in terms of indicator random variables:
• 𝜉 𝑦 = σ𝑞≤𝑜 1{𝑞|𝑦}
• Exchange order of summation
• 1

𝑜 σ𝑦∈ 𝑜 𝜉 𝑦 = 1 𝑜 σ𝑞≤𝑜 σ𝑦∈ 𝑜 1{𝑞|𝑦}

• Count multiples
• σ𝑦∈ 𝑜 1{𝑞|𝑦} =

𝑜 𝑞 = n p + O 1

• ⇒

1 𝑜 σ𝑦∈ 𝑜 𝜉 𝑦 = σ𝑞≤𝑜 1 𝑞 + 𝑃 1 = ln ln 𝑜 + 𝑃(1)

∎

Proposition 3.4.3 The average number of distinct prime factors of a number 𝑦 ∈ [𝑜] is ln ln 𝑜 + 𝑃(1).

A Harmonic Digression

“Proof”

• Let 𝑛 = 𝜌 𝑜 ∼

𝑜 ln 𝑜

• σ𝑞≤𝑜

1 𝑞 = σ𝑙=1 𝑛 1 𝑞𝑙

• Prime Number Theorem ⇒ 𝑞𝑙 ∼ 𝑙 ln 𝑙
• ⇒ σ𝑞≤𝑜

1 𝑞 ∼ σ𝑙=2 𝑛 1 𝑙 ln 𝑙

• Approximate by an integral:
• σ𝑙=2

𝑛 1 𝑙 ln 𝑙 ∼ ׬ 𝑦=2 𝑛 1 𝑦 ln 𝑦 ⅆ𝑦 ∼ ln ln 𝑛 ∼ ln ln 𝑜

∎

Theorem 3.4.4 (Mertens, 1874) As 𝑜 → ∞, we have σ𝑞≤𝑜

1 𝑞 = ln ln 𝑜 + 𝑃 1 .

Theorem 3.4.5 (Hardy-Ramanujan, 1920) As 𝑜 → ∞, we have 𝜉 𝑦 = 1 + 𝑝 1 ln ln 𝑜 for all but 𝑝 𝑜 integers 𝑦 ∈ 𝑜 .

The Typical Case

Variation in 𝜉(𝑦), 𝑦 ∈ 𝑜

• Minimum:

1

• Average:

ln ln 𝑜 + 𝑃 1

• Maximum:

1 + 𝑝 1

ln 𝑜 ln ln 𝑜

• Product of first 𝑛 primes ∼ ς𝑙=1

𝑛

𝑙 ln 𝑙 ∼ 𝑛! ln 𝑛 𝑛 ≤ 𝑜 for 𝑛 ∼

ln 𝑜 ln ln 𝑜

What can we say about the distribution of 𝜉 𝑦 ?

Theorem 3.4.5 (Hardy-Ramanujan, 1920) As 𝑜 → ∞, we have 𝜉 𝑦 = 1 + 𝑝 1 ln ln 𝑜 for all but 𝑝 𝑜 integers 𝑦 ∈ 𝑜 .

The Probabilistic Approach

Probabilistic proof (Turán, 1934)

• Choose 𝑦 ∈ 𝑜 uniformly at random
• Interested in the random variable 𝑌 = 𝜉 𝑦
• Proposition 3.4.3 ⇒ 𝔽 𝑌 = ln ln 𝑜 + 𝑃 1

Corollary 3.1.3’ If Var 𝑌 = 𝑝 𝔽 𝑌 2 , then 𝑌 = 1 + 𝑝 1 𝔽 𝑌 with high probability.

Expressing the Variance

Recall

• 𝑦 ∈ 𝑜 uniformly random
• 𝑌 = 𝜉(𝑦) number of distinct prime factors
• Goal: show Var 𝑌 = 𝑝 𝔽 𝑌 2

Indicator random variables

• For a prime 𝑞, let 𝑌𝑞 = 1{𝑞|𝑦}, Bernoulli random variable
• ℙ 𝑌𝑞 = 1 =

𝑜/𝑞 n

∈

1 𝑞 − 1 𝑜 , 1 𝑞

• 𝑌 = σ𝑞≤𝑜 𝑌𝑞

Our friend the variance

• Var 𝑌 = σ𝑞 Var 𝑌𝑞 + σ 𝑞,𝑟 :𝑞≠𝑟 Cov 𝑌𝑞, 𝑌𝑟
• σ𝑞 Var 𝑌𝑞 ≤ σ𝑞 𝔽 𝑌𝑞 = 𝔽 𝑌

Computing Covariances

Pairs 𝑞 ≠ 𝑟

• Cov 𝑌𝑞, 𝑌𝑟 = 𝔽 𝑌𝑞𝑌𝑟 − 𝔽 𝑌𝑞 𝔽 𝑌𝑟
• 𝔽 𝑌𝑞 ≥

1 𝑞 − 1 𝑜, 𝔽 𝑌𝑟 ≥ 1 𝑟 − 1 n

• 𝔽 𝑌𝑞𝑌𝑟 = ℙ 𝑞𝑟 𝑦 ≤

1 𝑞𝑟

• ⇒ Cov 𝑌𝑞, 𝑌𝑟 ≤

1 𝑞𝑟 − 1 𝑞 − 1 𝑜 1 𝑟 − 1 𝑜 ≤ 1 𝑜 1 𝑞 + 1 𝑟

Bounding the sum

• ⇒ σ 𝑞,𝑟 :𝑞≠𝑟 Cov 𝑌𝑞, 𝑌𝑟 ≤

1 𝑜 σ 𝑞,𝑟 :𝑞≠𝑟 1 𝑞 + 1 𝑟 ≤ 2𝜌 𝑜 𝑜

σ𝑞≤𝑜

1 𝑞

• 𝜌 𝑜 = 1 + 𝑝 1

𝑜 ln 𝑜 and σ𝑞≤𝑜 1 𝑞 = ln ln 𝑜 + 𝑃 1 = 𝔽 𝑌

• ⇒ σ 𝑞,𝑟 :𝑞≠𝑟 Cov 𝑌𝑞, 𝑌𝑟 = 𝑝 𝔽 𝑌

A Final Flourish

The variance

• Var 𝑌 = σ𝑞 Var 𝑌𝑞 + σ𝑞≠𝑟 Cov 𝑌𝑞, 𝑌𝑟
• σ𝑞 Var 𝑌𝑞 ≤ 𝔽 𝑌

and σ𝑞≠𝑟 Cov 𝑌𝑞, 𝑌𝑟 = 𝑝 𝔽 𝑌

• ⇒ Var 𝑌 = 1 + 𝑝 1

𝔽 𝑌 = 1 + 𝑝 1 ln ln 𝑜

Applying Chebyshev

• ℙ 𝜉 𝑦 − ln ln 𝑜 > 𝜇 ln ln 𝑜 ≤ Var 𝑌

𝜇2 ln ln 𝑜 = 1 𝜇2 + 𝑝(1)

• ⇒ ℙ 𝜉 𝑦 ≠ 1 + 𝑝 1

ln ln 𝑜 = 𝑝 1

• 𝑦 uniform in [𝑜] ⇒ 𝑝(𝑜) such integers

∎

Remark

• Most 𝑦 ∈ 𝑜 satisfy 𝜉 𝑦 = ln ln 𝑜 + 𝑃

ln ln 𝑜

Any questions?

§5 Distinct Sums

Chapter 3: The Second Moment The Probabilistic Method

Mathemagic

An illusion

• You have a deck of cards, with each card bearing a number
• You invite your friend to select as many cards from the deck as they like
• They add the numbers and only tell you the sum
• The chosen cards are then shuffled back into the deck
• You then go through the deck, and magically pick out your friend’s cards

The secret

• Cards labelled with powers of two: 1,2,4,8,16, …
• Each number 𝑦 ∈ ℕ has a unique binary expansion, 𝑦 = σ𝑘 2𝑗𝑘
• ⇒ given the sum 𝑦, can recover the labels 2𝑗𝑘 of the chosen cards

A Little Showmanship

Obstacles

• Mathematician friends will see through the illusion
• Non-mathematician friends may not be able to add well
• Card labels shouldn’t be larger than 𝑜
• Binary labels ⇒ log 𝑜 cards
• Small deck is not so impressive

Better decks

• Can we replace the binary labels?
• Suppose we have labels 𝑇 = 𝑡1, 𝑡2, … , 𝑡𝑙
• Key property:
• distinct sums – no two subsets should have the same total
• Extremal problem
• How large can a subset 𝑇 ⊆ [𝑜] with distinct sums be?

Claim 3.5.1 The greedy algorithm returns the set of powers of two.

The Greedy Magician

Greedy algorithm

• Start with 𝑇 = ∅
• Go through elements in 𝑜 one at a time
• Add to 𝑇 if they preserve distinct sums property

Proof

• After the first step, we have 𝑇 = 1
• Suppose we have 𝑇 = 1,2, … , 2𝑠 at some stage in the algorithm
• We can write every number up to 2𝑠+1 − 1 as a sum of these elements
• None of these added to 𝑇
• Next available number to be added: 2𝑠+1

∎

The Extremal Function

Notation

• Let 𝑔 𝑜 = max

𝑇 : 𝑇 ⊆ 𝑜 has distinct sums

Lower bound

• Binary set ⇒ 𝑔 𝑜 ≥ log 𝑜 + 1
• Is this best possible?

Counterexamples

• 𝑇 = {11,17,20,22,23,24} has distinct sums
• ⇒ 𝑔 𝑜 ≥ log 𝑜 + 2 for 24 ≤ 𝑜 ≤ 31
• If a set 𝑇 has distinct sums, so does 𝑇′ = 2𝑇 ∪ {1}
• Iterating → infinite sequence of counterexamples

An Upper Bound

Proof

• Let 𝑙 = 𝑔(𝑜) and let 𝑇 ⊆ [𝑜] be a largest set with distinct sums
• For each 𝑈 ⊆ 𝑇, we have 0 ≤ σ𝑡∈𝑈 𝑡 < 𝑙𝑜
• Distinct sums ⇒ each of these 2𝑙 sums is distinct
• ⇒ 2k ≤ 𝑙𝑜

⇒ 𝑙 ≤ log 𝑜 + log 𝑙 ⇒ 𝑙 ≤ log 𝑜 + log log 𝑜 + log 𝑙 ≤ log 𝑜 + log 2 log 𝑜 = log 𝑜 + log log 𝑜 + 1 ∎

Proposition 3.5.2 As 𝑜 → ∞, we have 𝑔 𝑜 ≤ log 𝑜 + log log 𝑜 + 1.

Theorem 3.5.3 As 𝑜 → ∞, 𝑔 𝑜 ≤ log 𝑜 +

1 2 log log 𝑜 + 𝑃(1).

An Improved Upper Bound

Flawed argument

• Wasteful in estimating range of sums
• Max sum ∼ 𝑙𝑜 ⇒ all members of 𝑇 ∼ 𝑜
• In that case, few small numbers will be sums

Fix

• Try to find a smaller interval still containing many sums
• Chebyshev ⇒ sums may concentrate around the average

Probabilistic Framework

Random variables

• Let 𝑔 𝑜 = 𝑙, let 𝑇 = 𝑡1, 𝑡2, … , 𝑡𝑙 ⊆ [𝑜] be a largest set with distinct sums
• Let 𝑌 be a uniformly random sum from 𝑇
• ⇒ 𝑌 = σ𝑗=1

𝑙

𝜁𝑗𝑡𝑗, where each 𝜁𝑗 is independent, uniform on 0,1

Expectation

• Let 𝜈 ≔ 𝔽 𝑌 = σ𝑗=1

𝑙

𝔽 𝜁𝑗𝑡𝑗 =

1 2 σ𝑗=1 𝑙

𝑡𝑗

• Actual value is unimportant

Variance

• Variables 𝜁𝑗 are independent
• ⇒ Var 𝑌 = Var σ𝑗=1

𝑙

𝜁𝑗𝑡𝑗 = σ𝑗=1

𝑙

Var 𝜁𝑗 𝑡𝑗

2 = 1 4 σ𝑗=1 𝑙

𝑡𝑗

2 ≤ 1 4 𝑜2𝑙

Concentrated Sums

Recall

• Var 𝑌 ≤

1 4 𝑜2𝑙

Applying Chebyshev

• ℙ 𝑌 − 𝜈 ≥ 𝑜 𝑙 ≤ Var 𝑌

𝑜2𝑙

≤

1 4

• ⇒ ℙ 𝑌 − 𝜈 < 𝑜 𝑙 ≥

3 4

Distinct sums

• Each value comes from at most one sum ⇒ ℙ 𝑌 = 𝑦 ∈ 0, 2−𝑙
• ∴ ℙ 𝑌 − 𝜈 < 𝑜 𝑙 = ℙ 𝜈 − 𝑜 𝑙 < 𝑌 < 𝜈 + 𝑜 𝑙 ≤ 2𝑜 𝑙 ⋅ 2−𝑙

Bounding 𝑙

• 2𝑙 ≤

8 3 𝑜 𝑙 ⇒ 𝑙 ≤ log 𝑜 + 1 2 log 𝑙 + log 8 3 ≤ log 𝑜 + 1 2 log log 𝑜 + 𝑃(1)

∎

Any questions?