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–
Khachatrian Theorem on k-uniform t-intersecting families. See also
[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.
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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.