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Chapter 12. Uniform measure versus product measure

Chapter 12. Uniform measure versus product measure

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70

12. Uniform measure versus product measure

There are two important measures in this setting. One is a uniform measure μk , where k is an integer with 0 ≤ k ≤ n. This is

deﬁned by a weight wk : 2[n] → [0, 1], where

n −1

k

wk (F ) =

0

if |F | = k,

if |F | = k,

for F ∈ 2[n] . Clearly we have μk (2[n] ) = μk (

[n]

k

) = 1.

The other is a product measure μp , where p is a ﬁxed real number

with 0 < p < 1. This is deﬁned by a weight wp : 2[n] → [0, 1], where

wp (F ) = p|F | (1 − p)n−|F | .

(12.1)

To understand the meaning of the weight, let us consider the following

random process. For each i = 1, 2, . . . , n we decide whether we pick i

with probability p or not pick i with probability 1 − p. We make each

decision independently. After n rounds we eventually obtain F with

probability precisely wp (F ). In view of this μp (2[n] ) = 1 is evident.

We can also show this from the deﬁnition. Indeed, we have

μp (2[n] ) =

p|F | (1 − p)n−|F |

wp (F ) =

F ∈2[n]

F ∈2[n]

n

n

p

=

|F |

(1 − p)

n−|F |

n k

p (1 − p)n−k

k

=

k=0 F ∈([n])

k=0

k

= (p + (1 − p))n = 1.

Next we compute the measures of the intersecting family

F = {F ∈ 2[n] : 1 ∈ F }.

For the uniform measure we note that |F ∩

n−1

n

/

k−1

k

μk (F) =

[n]

k

=

|=

n−1

k−1

. Then

k

.

n

For the product measure we note that F = {{1}∪G : G ∈ 2Y }, where

Y = [2, n]. By deﬁnition we have

p|{1}∪G| (1 − p)n−|{1}∪G|

μp (F) =

G∈2Y

p|G| (1 − p)|Y |−|G| = p.

=p

G∈2Y

12.3. Some extensions

71

This suggests a correspondence between μk and μp by relating k/n

with p.

12.2. Two versions of the EKR

With language from the previous section we can restate the Erd˝

os–

Ko–Rado Theorem (EKR) in the following form:

Theorem 12.1 (EKR reprise). If

k

n

≤ 12 , then Mn (μk ) =

k

n.

The above statement has a measure counterpart, which was ﬁrst

obtained by Ahlswede and Katona in [4]:

Theorem 12.2 (Product measure EKR). If p ≤ 12 , then Mn (μp ) = p.

We include a proof due to Filmus [38].

Proof. Let F be an intersecting family. Let C be a circle of circumference 1. By an arc we mean a directed arc on C, say in the clockwise

direction, of a ﬁxed length p. Choose an arc I at random, and pick

n independent random points x1 , . . . , xn uniformly distributed on C.

Let XI = {i ∈ [n] : xi ∈ I}. Then Pr(XI ∈ F) = μp (F).

Now ﬁx n points x1 , . . . , xn on C, and ﬁx an arc I0 such that

XI0 ∈ F. Then any arc I = I0 for which XI ∈ F intersects I0 . Since

p ≤ 12 , exactly one of the endpoints of I is in I0 . For a point x in

I0 \{x1 , . . . , xn } let Sx (resp. Ex ) be the arc starting (resp. ending) at

x. Then only at most one of XSx and XEx is in F. Thus there is an

injection from the set of arcs I with XI ∈ F to I0 . So, conditioned on

the positions of xi , Pr(XI ∈ F | x1 , . . . , xn ) ≤ p. This is true for every

conﬁguration of n points, and we conclude that Pr(XI ∈ F) ≤ p.

These two theorems seem to be “equivalent” in some sense.

Problem 12.3. Find a general theorem concerning Mn (μ) that includes both of the above two statements as special cases.

12.3. Some extensions

In the rest of this chapter we present some results and problems which

extend the above two theorems in various ways.

72

12. Uniform measure versus product measure

Recall that a family F ⊂ 2[n] is t-intersecting if |F ∩ F | ≥ t for

all F, F ∈ F. For example,

F(n, t, i) := {F ⊂ [n] : |F ∩ [t + 2i]| ≥ t + i}

is a t-intersecting family. The Ahlswede–Khachatrian Theorem has

its measure version as well:

Theorem 12.4 (Product measure Ahlswede–Khachatrian Theorem).

Let n ≥ t ≥ 1 and p ∈ (0, 1/2]. Then

Mn (μp ) = max μp (F(n, t, i)).

i

The above result was ﬁrst obtained by Ahlswede and Khachatrian

[6] for rational p; see also Bey and Engel [11]. Then, Dinur and Safra

[26] and Tokushige [108] deduced Theorem 12.4 from the Ahlswede–

[39].

Exercise 12.5. Show that limn→∞ Mn (μp ) = 1 for p > 1/2. (Hint:

Consider the product measure of F = {F ⊂ [n] : |F | > n+t

2 }.)

Simple computation shows that if p ≤

1

t+1

then

max μp (F(n, t, i)) = μp (F(n, t, 0)) = pt .

i

So Theorem 12.4 in this case reads as follows.

Corollary 12.6. If n ≥ t ≥ 1 and p <

1

t+1 ,

then Mn (μp ) = pt .

1

Proof. Since p1 > t + 1 we can choose > 0 so that p+

> t + 1. Fix

1

this and let I = ((p − )n, (p + )n) ∩ N. Then n > p+ k > (t + 1)k

holds for all k ∈ I.

Let F ⊂ 2[n] be a t-intersecting family with the maximum measure. For a lower bound of the measure we clearly have

M (μp ) = μp (F) ≥ μp (F(n, t, 0)) = pt .

For an upper bound we have

(12.2) μp (F) ≤

F∩

k∈I

[n]

k

pk (1−p)n−k +

k∈I

n k

p (1−p)n−k .

k

12.3. Some extensions

73

n−t

Let k ∈ I. Then, by Theorem 10.2, we have |F ∩ [n]

k | ≤ k−t . We

n−t

n

t

t

also have that k−t / k ≤ (k/n) < (p + ) . So the ﬁrst term in the

RHS of (12.2) is at most

k∈I

n−t k

p (1 − p)n−k < (p + )t

k−t

k∈I

n k

p (1 − p)n−k ≤ (p + )t .

k

On the other hand, by the de Moivre–Laplace limit theorem, the

second term in the RHS of (12.2) goes to 0 as n → ∞. Thus we have

that limn→∞ Mn (μp ) < (p + )t . Since > 0 can be chosen arbitrarily

small and since Mn (μp ) ≥ pt we can conclude limn→∞ Mn (μp ) = pt .

Now we show that Mn+1 (μp ) ≥ Mn (μp ). To see this we just

notice that if F ⊂ 2[n] is t-intersecting, then

F := F ∪ {F

{n + 1} : F ∈ F} ⊂ 2[n+1]

is also t-intersecting, and these two families have the same measure.

In fact μp (F ) = μp (F)(1 − p) + μp (F)p = μp (F). Consequently we

have mn (μp ) = pt for all n ≥ t as needed.

Friedgut [68] found another proof of the above result, which is a

counterpart of Wilson’s spectral proof [115] of the Erd˝os–Ko–Rado

Theorem for the case n ≥ (t + 1)(k − t + 1). We will use one of his

ideas in Chapter 30.

Recall the deﬁnition of μp . One can rewrite the RHS of (12.1) as

i∈F p

j∈[n]\F (1 − p). Here we assigned the same probability p to

each vertex (an element in [n]). Fishburn et al. extended this deﬁnition in such a way that each vertex can have a diﬀerent probability as

follows. For a vector p = (p(1) , p(2) , . . . , p(n) ) ∈ (0, 1)n and a family

F ⊂ 2[n] , let us deﬁne

(1 − p(j) ).

p(i)

μp (F) :=

F ∈F i∈F

j∈[n]\F

If p(1) = · · · = p(n) = p, then μp coincides with the product measure

μp .

Theorem 12.7 (Fishburn et al. [40]). Let μp be the extended product

measure deﬁned above, and assume that p(1) = max{p(l) : l ∈ [n]}

and that p(l) ≤ 1/2 for l ≥ 2. If F ⊂ 2[n] is intersecting, then

74

12. Uniform measure versus product measure

μp (F) ≤ p(1) . Moreover, if p(1) > p(l) for l ≥ 2, then equality holds

if and only if F = {F ⊂ [n] : 1 ∈ F }.

We will deduce Theorem 12.7 from a stronger result concerning

cross intersecting families in Chapter 30.

Let [m]n denote the set of integer sequences (chosen from [m])

of length n, that is, [m]n = {(a1 , . . . , an ) : ai ∈ [m]}. We say that a

family of integer sequences A ⊂ [m]n is t-intersecting if

#{i : ai = ai } ≥ t

uredi used a down

for all (a1 , . . . , an ), (a1 , . . . , an ) ∈ A. Frankl and Fă

shift introduced by Kleitman [83] to show the following.

Theorem 12.8 ([55]). The maximum size of t-intersecting families

(of integer sequences) in [m]n is given by the maximum of mn μp (F),

where p = 1/m and F runs over all t-intersecting families (of subsets)

in 2[n] .

Combining the above result with Theorem 12.4, one can obtain the maximum size of t-intersecting families in [m]n . In part−1

, then the maximum size is

ticular, if n ≥ t + 2i, where i = m−2

n

m μ1/m (F(n, t, i)); see [6, 11, 64].

Let r ≥ 2 and t ≥ 1 be integers. We say that r families of subsets

F1 , . . . , Fr ⊂ 2[n] are r-cross t-intersecting if

|F1 ∩ F2 ∩ · · · ∩ Fr | ≥ t

for all Fi ∈ Fi , 1 ≤ i ≤ r. If, moreover, F1 = F2 = · · · = Fr =: F,

then we say that F is r-wise t-intersecting. For simplicity we omit

r and/or t in the case of r = 2 or t = 1. So “cross t-intersecting”

means “2-cross t-intersecting,” and “r-wise intersecting” means “rwise 1-intersecting.” We introduce the family F(n, r, t, i) deﬁned by

F(n, r, t, i) := {F ⊂ 2[n] : |F ∩ [t + ri]| ≥ t + (r − 1)i}

and its k-uniform subfamily F (k) (n, r, t, i) deﬁned by

F (k) (n, r, t, i) := F(n, r, t, i) ∩

[n]

k

.

Exercise 12.9. Show that F(n, r, t, i) is r-wise t-intersecting.

12.3. Some extensions

75

These families F(n, r, t, i) and F (k) (n, r, t, i) play an important role

when we construct a large intersecting structure in general.

Here is one of the most fundamental problems on intersecting

families:

Problem 12.10. Let k1 , . . . , kr be positive integers. Determine

r

|Fi |,

max

i=1

where the maximum is taken over all r-cross t-intersecting families

Fi ⊂ [n]

ki , 1 ≤ i ≤ r.

The corresponding measure counterpart is as follows.

Problem 12.11. Let p1 , . . . , pr ∈ (0, 1) be given. Determine

r

μpi (Fi ),

max

i=1

where the maximum is taken over all r-cross t-intersecting families

F1 , . . . , Fr ⊂ 2[n] .

For the case t = 1 we conjecture as follows.

Conjecture 12.12. Let (r − 1)/r ≥ max{k1 /n, . . . , kr /n}, and let

Fi ⊂ [n]

ki for 1 ≤ i ≤ r. If F1 , . . . , Fr are r-cross intersecting, then

r

r

μk (Fi ) ≤

i=1

(ki /n).

i=1

Conjecture 12.13. Let (r − 1)/r ≥ max{p1 , . . . , pr }, and let Fi ⊂

2[n] for 1 ≤ i ≤ r. If F1 , . . . , Fr are r-cross intersecting, then

r

r

μpi (Fi ) ≤

i=1

pi .

i=1

We have some partial answers to these conjectures and will discuss

some of them in later chapters.

Chapter 13

Kleitman’s correlation

inequality

Let p be a ﬁxed real number with 0 < p < 1. For a family F ⊂ 2X

let μX

p (F) denote the product measure of F, that is,

p|F | (1 − p)|X|−|F | .

μX

p (F) =

F ∈F

We often write μp instead of μX

p when X is ﬁxed.

n

If p = 1/2 and X = [n], then μX

p (F) is just |F|/2 . This can

be viewed as the probability of F in the following sense. Consider

the uniform distribution on 2X where each subset of X has the same

probability 1/2n . In this probability space the probability that a

randomly chosen set is in F is exactly |F|/2n .

In a probability space two events with probabilities p1 and p2 are

positively (resp. negatively) correlated if the probability that they

occur simultaneously is at least (resp. at most) p1 p2 .

Recall that we say that F ⊂ 2X is an upset (resp. a downset) if

F ∈ F and F ⊂ F (resp. F ⊂ F ) imply F ∈ F. The following

lemma is an extension of a result from [84].

Lemma 13.1 (Kleitman’s correlation inequality). Let F, G ⊂ 2[n] .

If F is an upset and G is a downset, then

μp (F ∩ G) ≤ μp (F)μp (G).

77

78

13. Kleitman’s correlation inequality

To prove the lemma we list some easy facts which immediately

follow from the deﬁnition of the product measure. We note that

if F ⊂ 2X is an upset (resp. downset) then F(x) ⊃ F(x) (resp.

F(x) ⊂ F(x)) for x ∈ X, where F(x) and F(x) are deﬁned by (0.1)

and (0.2).

[n−1]

Claim 13.2. Let F ⊂ 2[n] , fn = μp

[n−1]

(F(n)), and fn¯ = μp

(F(¯

n)).

[n]

(i) μp (F) = pfn + (1 − p)fn¯ .

(ii) If F is an upset then fn ≥ fn¯ .

(iii) If F is a downset then fn ≤ fn¯ .

Proof of Lemma 13.1. We prove the statement by induction on n.

It is easy to check for the case n = 1. Let n ≥ 2. Since F is an upset,

F(n) and F(¯

n) are both upsets. Similarly, G(n) and G(¯

n) are both

downsets. We deﬁne fn and fn¯ as in the previous claim, and deﬁne

gn and gn¯ for G similarly. Let q = 1 − p.

It follows from (i) of the claim that

(13.1)

[n−1]

μ[n]

((F ∩ G)(n)) + qμ[n−1]

((F ∩ G)(¯

n)).

p (F ∩ G) = pμp

p

Since (F ∩ G)(n) = F(n) ∩ G(n) and (F ∩ G)(¯

n) = F(¯

n) ∩ G(¯

n),

it follows from the induction hypothesis that the RHS of (13.1) is

≤ pfn gn + qfn¯ gn¯ .

On the other hand, we have (fn − fn¯ )(gn − gn¯ ) ≤ 0 by (ii) and

(iii) of the claim. Thus it follows that

[n]

μ[n]

¯ )(pgn + qgn

¯)

p (F)μp (G) = (pfn + qfn

≥ (pfn + qfn¯ )(pgn + qgn¯ ) + pq(fn − fn¯ )(gn − gn¯ )

= pfn gn + qfn¯ gn¯ ,

[n−1]

and, by the induction hypothesis, the RHS is ≥ pμp

[n−1]

[n]

qμp

(F(n) ∩ G(n)) = μp (F ∩ G).

(F(n)∩G(n))+

Exercise 13.3. Show that if F, G ⊂ 2[n] are both upsets, then

μp (F ∩ G) ≥ μp (F)μp (G).

(Hint: Note that G = 2[n] \ G is a downset, and apply Lemma 13.1

to F and G .)

13. Kleitman’s correlation inequality

79

Kleitman invented the correlation inequality in order to prove the

following result, conjectured by Erd˝os.

Theorem 13.4. Let 0 < p ≤ 1/2 be a real number and q = 1 − p,

and let n ≥ r be positive integers. Suppose that F1 , . . . , Fr ⊂ 2[n] are

intersecting families. Then

μp (F1 ∪ · · · ∪ Fr ) ≤ 1 − q r .

The above inequality is sharp. To see this let

Fi = {F ⊂ [n] : i ∈ F }.

Then this is an intersecting family as well as an upset. Since

2[n] \ (F1 ∪ · · · ∪ Fr ) = 2[r+1,n] ,

it follows that μp (F1 ∪ · · · ∪ Fr ) = 1 − q r .

Proof of Theorem 13.4. If we replace F with the upset generated

by F, that is, {G ⊂ [n] : F ⊂ G for some F ∈ F}, then the product

measure only increases while the intersection property is preserved.

So we may assume that all Fi are upsets.

We prove the statement by induction on r. The case r = 1 follows

from Theorem 12.2. Let r ≥ 2. Suppose that we know the statement

is true for r − 1 and let us prove it for r. Let G = F1 ∪ · · · ∪ Fr−1 .

Applying Lemma 13.1 to G and Fr , we obtain

μp (G ∩ Fr ) ≥ μp (G)μp (Fr ).

Thus

1 − μp (G ∪ Fr ) = 1 − (μp (G) + μp (Fr ) − μp (G ∩ Fr ))

≥ 1 − μp (G) − μp (Fr ) + μp (G)μp (Fr )

= (1 − μp (G))(1 − μp (Fr )).

Then, using the induction hypothesis μp (G) ≤ 1 − q r−1 and μp (Fr ) ≤

1 − q, we ﬁnally get (1 − μp (G))(1 − μp (Fr )) ≥ q r−1 q = q r .

A family F ⊂ 2X is called intersecting-union if F is intersecting

and also union, that is, F ∩ F = ∅ and F ∪ F = X for all F, F ∈ F.

Exercise 13.5. Let F ⊂ 2[n] be an intersecting-union family.

80

13. Kleitman’s correlation inequality

(1) Let p = 1/2. Show that μp (F) ≤ 1/4. (Hint: Deﬁne an

intersecting upset F ∗ and a union downset F∗ by

F ∗ = {G ⊂ [n] : F ⊂ G for some F ∈ F},

F∗ = {G ⊂ [n] : G ⊂ F for some F ∈ F},

and apply Lemma 13.1 to these families.)

(2) Let p = 1/2. Show that

lim max μp (F) = min{p, 1 − p},

n→∞

where the maximum is taken over all intersecting-union

families F ⊂ 2[n] . (Hint: Let p < 1/2. For the lower

bound consider the family {F ⊂ [n] : 1 ∈ F, |F | < n/2}.

For the upper bound let Fk be the k-uniform subfamily

of F, where k ∈ (p − , p + ) and p + < 1/2. Then

|Fk | ≤ n−1

k−1 by the EKR. Note that the diﬀerence between μp (F) and k∈(p− ,p+ ) μp (Fk ) can be made arbitrarily small by choosing n large enough.)

We say that a family F ⊂ 2X is r-wise intersecting-union if F1 ∩

· · ·∩Fr = ∅ and F1 ∪· · ·∪Fr = X for all F1 , . . . , Fr ∈ F. In this case it

is known that μp (F) ≤ p(1−p) provided r ≥ 2 and 1/r ≤ p ≤ 1−1/r;

see [101].

Let us restate Theorem 4.5 in its original, intersecting form. Deﬁne G(n, t) by

G(n, t) =

{G ⊂ [n] : |G| ≥

n+t

2 }

{G ⊂ [n] : |G ∩ [n − 1]| ≥

if n + t is even,

(n−1)+t

}

2

if n + t is odd.

Exercise 13.6. Show that the family G(n, t) is isomorphic to the

complement family of K(n, n − t) deﬁned in (4.2).

Exercise 13.7. Verify that G(n, t) is a t-intersecting family.

Theorem 13.8. If F ⊂ 2[n] is t-intersecting, then |F| ≤ |G(n, t)|.

Exercise 13.9. Deduce the above theorem from Theorem 4.5.

Considering G(n, t) as a family in 2[n+1] , it is both t-intersecting

and union. Katona conjectured that it is maximal for this property.

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