Brun’s Sieve

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Brun’s Sieve Let B 1 , …, B m be events, X i the indicator random variable for Bi and X = X 1 + … + X m the number of Bi that hold. Let there be a hidden parameter n (so that actually m = m ( n ), B i = B i ( n ), X = X( n )) which will define the following o , O notation. - PowerPoint PPT Presentation

Transcript of Brun’s Sieve

Page 1: Brun’s Sieve
Page 2: Brun’s Sieve

Brun’s Sieve Let B1, …, Bm be events, Xi the indicator random varia

ble for Bi and X = X1 + … + Xm the number of Bi that hold.

Let there be a hidden parameter n (so that actually m = m(n), Bi = Bi(n), X = X(n)) which will define the following o, O notation.

Define S(r) = ∑ Pr[Bi1 Λ…Λ Bir

], the sum over all sets {i

1,…,ir} {1,…,m}.

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Theorem 8.3.1 Suppose there is a constant μso that

E[X] = S(1) → μ and such that for every fixed r, E[X(r) / r!] = S(r) → μr / r!.

Then Pr[X = 0] → and indeed for every t Pr[X = t] →

e

e

!t

t e

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Pr[X = r] ≤ S(r) = ∑ Pr[ ], where {i1,…,ir} {1,…,m}.

The Inclusion-Exclusion Principle gives that Pr[X = 0] = Pr[ ] = 1 – S(1) + S(2) -…+(-1)rS(r)…

Bonferroni’s inequality: Let P(Ei) be the probability that Ei is true, and be the probability that at least one of Ei,…, En is true. Then

mBB ...1

rii BB ...1

]P[1 Ei

n

i

n

i

n

iEiEi

1 1 ]P[]P[

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Proof. We do only the case t = 0. Fix є> 0. Choose s so that

The Bonferroni Inequalities states that, in general, the inclusion-exclusion formula alternatively over and underestimates Pr[X = 0]. In particular,

Select no(the hidden variable) so that for n

no,

for 0 ≤ r ≤ 2s.

2

2

0! |)1(|

es

rrur r

s

r

rr S2

0

)()1(]0XPr[

)12(2!)( || sr

ur r

S

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Proof(cont.) For such n

Pr[X = 0] ≤ + є Similarly, taking the sum to 2s+1 we find no s

o that for n no

Pr[X = 0] ≤ - є

As є was arbitrary Pr[X = 0] →

e

e

e

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Let G ~ G(n,p), the random graph and let EPIT represent the statement that every vertex lies in a triangle.

Theorem 8.3.2 Let c > 0 be fixed and let p = p(n),μ=μ(n) satisfy

p3 = μ, =

Then Pr[G(n,p) |= EPIT] =

2

1n e nc

limn

ce

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Proof. First fix xV(G). For each unordered y, z V(G) – {x} let Bxyz

be the event that {x,y,z} is a triangle of G. Let Cx be the event and Xx be the cor

responding indicator random variable. We use Janson’s Inequality to bound E[Xx]

= Pr[Cx]. Here p = o(1) so є = o(1).

as defined above.

xyzB

xyzB

]Pr[ xyzB

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Proof(cont.) Dependency xyz ~ xuv occurs if and only if th

e sets overlap (other than x). Hence

Since . Thus Now define

the number of vertices x not lying in a triangle. Then from Linearity of Expectation,

ncex ~]X[E

)1()(]Pr[ 53 opnOBB zxyxyz

)1(3/2 onp ncex ~]X[E

)(

XXGVx

x

cGVx

x

]E[XE[X])(

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Proof(cont.) We need to show that the Poisson Paradigm

applies to X. Fix r. Then

the sum over all sets of vertices {x1,…,xr}. All r-sets look alike so

where x1,…,xr are some particular vertices. But

the conjunction over 1 ≤ i ≤ r and all y,z.

]...Pr[]!/X[E 1)((r)

rxxr CCSr

!~

!)(

~!/X[E (r)

rc

rne

rrr

yzxxx ir BCC ...1

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Proof(cont.) We apply Janson’s Inequality to this conjunction.

Again є = p3 = o(1). The number of {xi,y,z} is , the overc

ount coming from those triangles containing two(or three of the xi). (Here it is crucial that r is fixed.) Thus

As before Δ is p5 times the number of pairs xiyz~ xj

y’z’. There are O(rn3) = O(n3) terms with i = j and O(r2n2) = O(n2) terms with i j so again Δ = o(1). Therefore

and

)(2

1nO

nr

)()(2

1]Pr[ )1(13 o

yzx nOrnOn

rpB i

eCC rxx ~]...Pr[ 1

!~]...Pr[]!/X[E 1

(r)

rn

CCr

nr

r

xx r

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Large Deviations Given a point in the probability space(i.e., a

selection of R) we call an index set J I a disjoint family (abbreviated disfam) if Bj for every j J. For no j, j’ J is j ~ j’.

If, in addition, If j’ J and Bj’ then j ~ j’ for some j J.

Then we call J a maximal disjoint family

(abbreviated maxdisfam).

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Lemma 8.4.1 With the above notation and for any integer s,

Pr[there exists a disfam J, |J| = s] ≤

Proof. Let denote the sum over all s-sets J I with

no j ~ j’. Let denote the sum over ordered s-tuples

(j1 ,…, js) with {j1 ,…, js} forming such a J. Let denote the sum over all ordered s-tuples

(j1,…, js).

!s

s

o

a

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Proof(cont.) Pr[there exists a disfam J, |J| = s]

!!1

!1

!1

]]Pr[[

]Pr[]...Pr[

]Pr[]...Pr[

]Pr[

]Pr[

1

1

ss

iIis

jja

s

jjo

s

jJj

jJj

s

s

s

B

BB

BB

B

B

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For smaller s we look at the further condition of J being a maxidisfam. To that end we let μs denote the minimum, over all j1, … , js of

,the sum taken over all i I except those i with

i ~ jl for some 1≤ l ≤ s. In application s will be small (otherwise we use

Lemma 8.4.1) and μs will be close to μ. For some applications it is convenient to set

and note that μs >= μ – sv.

]Pr[ iB

ji

iIj

Bv~

]Pr[max

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Lemma 8.4.2 With the above notation and for any integer s, Pr[there exists a maxdisfam J,

|J| = s] ≤

Proof. As in Lemma 8.4.1 we bound this probability

by of J = {j1, … , js} being a maxdisfam. For this to occur J must first be a disfam and then , where is the conjunction over all i I except those with i ~ jl for some 1 ≤ l ≤ s.

svss

ees

ees

s

!!

2

iB

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Proof(cont.) We apply Janson’s Inequality to give an

upper bound to .The associated values satisfy

the latter since has simply fewer addends. Thus

and

]Pr[ iB

!/

]Pr[]disfammaxPr[

2

2

see

BeeJ

s

jJj

s

s

,

,

,

s

2]Pr[ eeB s

i

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When Δ = o(1) and vμ = o(1) or, more generally, μ3μ = μ + o(1), then Lemma 8.4.2 gives a close approximation to the Poisson Distribution since

Pr[there exists a maxdisfam J, |J| = s]

For s ≤ 3μ and the probability is quite small for

larger s by Lemma 8.4.1

eo s

s

!))1(1(

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