CS 591-06, Lecture 10

Jared Saia University of New Mexico

Class Outline

A reasonable probability is the only certainty. - E.W. Howe

• Majority-Preserving Property

• Chernoff Bounds

• In-Class Exercise

• Proof for Spam Resistant P2P network

Majority-Preserving Property

A bipartite graph G = (L, R) where |L| = |R| = n is said to have the majority preserving property iff for some ϵ > 0,

• For all sets L′ ⊂ L such that |L′| ≥ 2n/3, the number of nodes in R which have a majority of their neighbors in the set L′ is at least n − ϵn/ log n.

Claim:

• A bipartite graph where every node in L has edges to C log n random nodes in R has the majority preserving property. Further the constant C depends only on ϵ.

Chernoff Bounds

• Let X = ∑_{i=1}^{m} X_{i} where the X_{i} are random variables which

are: independent and all take on values between 0 and 1. then for any 0 < δ ≤ 1,

• P (|X − E(X)| ≥ δE(X)) ≤ 2e^{\frac{−E(X)δ2}{4}}

Review: Linearity of Expectation

Linearity of Expecation Fact: If there are random variable X, X_{1}, …, X_{n}, such that X = ∑_{i=1}^{n} X_{i}, then

• E(X) = ∑_{i=1}^{n} E(X_{i})

Example:

Let the variable X_{i} be 1 if the i-th flip of a coin is heads and 0 otherwise and say that E(X_{i}) = 1/2. Let X = ∑_{i=1}^{n} X_{i}. Then:

E(X) = n/2

Chernoff Example

• Let X = X, where each X_{i} is 1 if the i-th flip of a coin flip is heads tails otherwise. Let the coin turn up heads with probability 1/2.

• Q: What is a bound on the probability that either the number of heads exceeds (3/4)m or the number of heads is less than (1/4)m?

Majority Preserving Property

• Assume we’ve got a biparitite graph G = (L, R), where each vertex in L has C log n edges to random nodes in R. (The edges are chosen uniformly and independently at random),

• Consider some fixed sets L′ ⊂ L, R′ ⊂ R such that |L′| = 2n/3 and |R′| = ϵn/ log n, for some fixed ϵ > 0

• Question: What is the probability that no node in R′ has a majority of its edges to nodes in L′?

Expander Picture

Probability of Failure

7 • Question: What is the probability that no node in R′ has a

majority of its edges to nodes in L′?

• This bad event occurs only if the number of edges between L′ and R′ is no more than the number of edges between L−L′ and R′

• Q1: What is the expected number of edges between L′ and R′?

• Q2: What is the expected number of edges between L − L′ and R′?

• Question: What is the probability that no node in R′ has a majority of its edges to nodes in L′?

• This bad event occurs only if the number of edges between L′ and R′ is no more than the number of edges between L−L′ and R′

• Q1: What is the expected number of edges between L′ and R′? A: (2/3)ϵnC

• Q2: What is the expected number of edges between L − L′ and R′? A: (1/3)ϵnC • Bad Event 1: The number of edges between L′ and R′ is less

than 9/10 ∗ ((2/3)ϵnC)

• Bad Event 2: The number of edges between L − L′ and R′ is greater than 10/9 ∗ ((1/3)ϵnC)

• We can use a union bound to bound the probability that either bad event occurs.

Bad Event 1

• Let X be the number of edges between L′ and R′

• Note that X = sum_{i=1}^{m} where X_{i} is 1 if the i-th edge from L′ lands in R′ and is 0 otherwise and m = (2/3)nC log n

• From before, E(X) = ((2/3)ϵnC)

• By Chernoff bounds:

P (|X − E(X)| > (1/10)E(X))  ≤ 2e^{\frac{−E(X)(1/10)2}{4}}  (1)
≤ 2e^{\frac{−(2/3)ϵnC}{400}}  (2)
= 2e^{\frac{−ϵnC}{600}}  (3)

(4)

Bad Event 2

• Let Y be the number of edges between L − L′ and R′

• Note that Y = ∑_{i=1}^{l} Y_{i} where Y_{i} is 1 if the i-th edge from L − L′ lands in R′ and is 0 otherwise, and l = (1/3)nC log n

• From before, E(Y) = (1/3)ϵnC

• By Chernoff bounds:

P (|Y − E(Y)| > (1/10)E(Y))  ≤ 2e^{\frac{−E(Y)(1/10)2}{4}}  (5)
≤ 2e^{\frac{−(1/3)ϵnC}{400}}  (6)
= 2e^{\frac{−ϵnC}{1200}}  (7)

• Let γ_{1} be bad event 1 and γ_{2} be bad event 2

• Then:

P (γ_{1} ⋃ γ_{2}) ≤ P (γ_{1}) + P (γ_{2})  (8)
≤ 2e^{\frac{−ϵnC}{600}} + 2e^{\frac{−ϵnC}{1200}}  (9)
≤ 4e^{\frac{−ϵnC}{1200}}  (10)

• This is a bound on the probability that the majority preserving property fails for a fixed set L′ and a fixed set R′.

• It’s very small, but is it small enough???

Prob of Any Failure

• We know that the prob. of failure for a fixed set L′ and fixed set R′ is no more than 4e^{\frac{−ϵnC}{1200}}

• There are \binom{n}{n/2} ways to choose a set L′

• There are \binom{n}{ϵn/ log n} ways to choose a set R′

• \binom{n}{n/2} and \binom{n}{ϵn/ log n} are both no more than e^{n}

Prob of Any Failure

• For any set L′ and R′ of appropriate size, let γ_{L′,R′} be the probability that the bad event occurs for the sets L′ and R′

• We want P (⋃_{L′⊂L,R′⊂R s.t. |L′|=n/2,|R′|=ϵn/ log n} , γ_{L′,R′})

• Use the union bound:

P (_{L′,R′}^{⋃} γ_{L′,R′}) ≤ _{L′,R′}^{∑} P (γ_{L′,R′})  (11)
= \binom{n}{n/2} ∗ \binom{n}{ϵn/ log n} ∗ (4e^{\frac{−ϵnC}{1200}}) (12)
≤ e^{n} ∗ e^{n} ∗ 4e^{\frac{−ϵnC}{1200}}  (13)
= e^{\frac{−ϵnC+2400n+2400}{1200}}  (14)

Analysis

• For C chosen dependent only on ϵ, this probability can be to go to 0 exponentially quickly as n grows large.

• This means that with high probability, for appropriate constant C, a random bipartite graph will have the majority preserving property.

• Note: Our proof made use of the union bound and Chernoff bounds.

Some Useful Bounds

• Stirlings Formula: n! = √2πn (\frac{n}{e})^{n} (1 + \frac{1}{12n} + O(\frac{1}{n^{2}}))

• n! ≥ (\frac{n}{e})^{n}

• \binom{n}{k} ≤ \frac{n^{k}}{k!}

• \binom{n}{k} ≤ (\frac{ne}{k})^{k}

In Class Exercise

• Consider a directed random graph with possible loops on n vertices where each of the n^{2} possible edges is included with probability p.

• Q1: What is the probability that some fixed set of n/2 vertices share more than n^{2}p/4 edges?

• Q2: What is an upperbound on the probability that the graph contains some set of n/2 vertices which share n^{2}p/4 edges?