Brownian local minima, random dense countable sets and random equivalence classes

0.4 MB·English

by  Boris Tsirelson

Brownian local minima,
random dense countable sets
6 and random equivalence classes
0
0
2
Boris Tsirelson
n
a
J

7
2
Abstract
]
R
A random dense countable set is characterized (in distribution) by
P
. independenceandstationarity. TwoexamplesareBrownian local min-
h
ima and unordered infinite sample. They are identically distributed.
t
a
A framework for such concepts, proposed here, includes a wide class
m
of random equivalence classes.
[

1
v Introduction
3
7
6 Random countable sets arise naturally from the Brownian motion (local ex-
1
trema, see [5, 2.9.12]), percolation (double, or four-arm points, see [2]), ori-
0
6 ented percolation (points of type (2,1), see [3, Th. 5.15]) etc. They are
0
scarcely investigated, because they fail to fit into the usual framework. They
/
h
cannot be treated as random elements of ‘good’ (Polish, standard) spaces.
t
a The framework of adapted Poisson processes, used by Aldous and Barlow [1],
m
does not apply to the Brownian motion, since the latter cannot be correlated
:
v with a Poisson process adapted to the same filtration. The ‘hit-and-miss’
i
X framework used by Kingman [10] and Kendall [9] fails to discern the clear-
r cut distinction between Brownian local minima and, say, randomly shifted
a
set ofrationalnumbers. Anewapproachintroduced herecatches thisdistinc-
tion, does not use adapted processes, and shows that Brownian local minima
are distributed like an infinite sample in the following sense (see Theorem
6.11).
Theorem. There exists a probability measure P on the product space
C[0,1] (0,1)∞ such that
×
1Thisresearchwassupportedbytheisraelsciencefoundation(grantNo.683/05).
1
(a) the first marginal of P (that is, projection to the first factor) is the
Wiener measure on the space C[0,1] of continuous paths w : [0,1] R;
→
(b) the second marginal of P is the Lebesgue measure on the cube (0,1)∞
of infinite (countable) dimension;
(c) P-almost all pairs (w,u), w C[0,1], u = (u ,u ,...) (0,1)∞, are
1 2
∈ ∈
such that the numbers u ,u ,... are an enumeration of the set of all local
1 2
minimizers of the Brownian path w.
Thus, the conditional distribution of u ,u ,... given w provides a (ran-
1 2
domized) enumeration of Brownian minimizers by independent uniform ran-
dom variables.
The same result holds for every random dense countable set that satisfies
conditions of independence and stationarity, see Definitions 4.2, 6.8 and The-
orem 6.9. Two-dimensional generalizations, covering the percolation-related
models, are possible.
On a more abstract level the new approach is formalized in Sections 7, 8
in the form of ‘borelogy’ that combines some ideas of descriptive set theory
[6] and diffeology [4]. Random elements of various quotient spaces fit into
the new framework. Readers that like abstract concepts may start with these
sections.
1 Main lemma
Before the main lemma we consider an instructive special case.
1.1 Example. Let µ be the uniform distribution on the interval (0,1) and
ν the triangle distribution on the same interval, that is,
µ(B) = dx, ν(B) = 2xdx
ZB ZB
for all Borel sets B (0,1). On the space (0,1)∞ of sequences, the product
⊂
measure µ∞ is the joint distribution of uniform i.i.d. random variables, while
ν∞ is the joint distribution of triangular i.i.d. random variables. I claim
existence of a probability measure P on (0,1)∞ (0,1)∞ such that
×
(a)thefirstmarginalofP isequaltoµ∞ (thatis, P(B (0,1)∞) = µ∞(B)
×
for all Borel sets B (0,1)∞);
⊂
(b) the second marginal of P is equal to ν∞ (that is, P((0,1)∞ B) =
×
ν∞(B) for all Borel sets B (0,1)∞);
⊂
(c) P-almost all pairs (x,y), x = (x ,x ,...) (0,1)∞, y = (y ,y ,...)
1 2 1 2
∈ ∈
(0,1)∞ are such that
x ,x ,... = y ,y ,... ;
1 2 1 2
{ } { }
2
b
10 10 10
b b b b b b b b
b b b
b b b b b b b
b b
b b b
b b b b b b
0 0 0
0 1 0 1 0 1
bb b bbb b bb b b bbb b bb bbb b b bbb b bb bb
x4 x2 x1 x3 y3 y1 y2 z2 z1 z3
y4 z4
(a) (b) (c)
Figure 1: different enumerations turn the same set into an infinite sample from:
(a) the uniform distribution, (b) the triangular distribution, (c) their mixture
(uniform z ,z ,... but triangular z ,z ,...).
1 3 2 4
inotherwords, thesequence y isapermutationofthesequence x. (Arandom
permutation, of course.)
A paradox: the numbers y are biased toward 1, the numbers x are not;
k k
nevertheless they are just the same numbers!
Anexplanation(andasketchyproof)isshownonFig.1(a,b). Acountable
subset of the strip (0,1) (0, ) is a realization of a Poisson point process.
× ∞
(The mean number of points in any domain is equal to its area.) The first 10
points of the same countable set are shown on both figures, but on Fig. 1(a)
the points areordered according to the vertical coordinate, while on Fig.1(b)
they are ordered according to the ratio of the two coordinates. We observe
that y ,...,y is indeed biased toward 1, while x ,...,x is not. On
1 10 1 10
{ } { }
the other hand, y is a permutation of x. (This time, y = x , y = x ,
1 1 2 3
y = x , ...)
3 2
A bit more complicated ordering, shown on Fig. 1(c), serves the measure
µ ν µ ν ...,thejointdistributionofindependent, differentlydistributed
× × × ×
random variables.
In every case we use an increasing sequence of (random) functions h :
n
(0,1) [0, ) such that for each n the graph of h contains a Poisson
n
→ ∞
point, while the region between the graphs of h and h does not. The
n−1 n
differences h h are constant functions on Fig. 1(a), triangular (that is,
n n−1
−
x const x) on Fig. 1(b), while on Fig. 1(c) they are constant for odd n
7→ ·
and triangular for even n.
Moreover, the same idea works for dependent random variables. In this
case h h is proportional to the conditional density, given the previous
n n−1
−
3
points. We only need existence of conditional densities and divergence of
their sum (in order to exhaust the strip).
Here is the main lemma.
1.2 Lemma. Let µ be a probability measure on (0,1)∞ such that
(a) for every n the marginal distribution of the first n coordinates is
absolutely continuous;
(b) for almost all x (0,1) and µ-almost all (x ,x ,...) (0,1)∞,
1 2
∈ ∈
∞
f (x ,...,x ,x)
n+1 1 n
= ;
f (x ,...,x ) ∞
n 1 n
n=1
X
here f is the density of the first n coordinates.
n
Letν beanotherprobabilitymeasureon(0,1)∞ satisfyingthesamecondi-
tions (a), (b). Then there exists a probability measure P on (0,1)∞ (0,1)∞
×
such that
(c) the first marginal of P is equal to µ;
(d) the second marginal of P is equal to ν;
(e) P-almost all pairs (x,y), x = (x ,x ,...) (0,1)∞, y = (y ,y ,...)
1 2 1 2
∈ ∈
(0,1)∞ are such that
x ,x ,... = y ,y ,... .
1 2 1 2
{ } { }
In other words, the sequence y is a permutation of the sequence x (since
x are pairwise different due to absolute continuity, as well as y ).
k k
The rest of the section is occupied by the proof of the main lemma.
Throughout the proof, Poisson point processes on the strip (0,1) [0, )
× ∞
(or its measurable part) are such that the mean number of points inany mea-
surable subset is equal to its two-dimensional Lebesgue measure. Random
variables (and random functions) are treated here as measurable functions
of the original Poisson point process on the strip.
We start with three rather general claims.
1.3 Claim. (a) A Poisson point process on the strip (0,1) [0, ) may be
× ∞
treatedasthesetΠofrandompoints(U ,T + +T )forn = 1,2,...,where
n 1 n
···
U ,T ,U ,T ,... are independent random variables, each U is distributed
1 1 2 2 k
uniformly on (0,1), and each T is distributed Exp(1) (that is, P T > t =
k k
e−t for t 0);
≥ (cid:0) (cid:1)
(b) conditionally on (U ,T ), the set Π = (U ,T + +T ) : n 2
1 1 1 n 1 n
{ ··· ≥ }
is (distributed as) a Poisson point process on (0,1) [T , ).
1
× ∞
The proof is left to the reader.
4
1.4 Claim. Let f : (0,1) [0, ) be a measurable function satisfying
→ ∞
1
f(x)dx = 1, and Π be a Poisson point process on (0,1) [0, ). Then
0 × ∞
(a) the minimum
R y
t = min
1
(x,y)∈Π f(x)
(where y/0 = ) is reached at a single point (x ,y ) Π;
1 1
∞ ∈
(b) x and t = y /f(x ) are independent, t is distributed Exp(1), and
1 1 1 1 1
the distribution of x has the density f;
1
(c) conditionally on (x ,y ), the set Π = Π (x ,y ) is (distributed
1 1 1 1 1
\ { }
as) a Poisson point process on (x,y) : 0 < x < 1, t f(x) < y < .
1
{ ∞}
Proof. A special case, f(x) = 1 for all x, follows from 1.3.
A more general case, f(x) > 0 for all x, results from the special case by
the transformation (x,y) (F(x),y/f(x)) where F(x) = xf(x′)dx′. The
7→ 0
transformation preserves Lebesgue measure on (0,1) [0, ), therefore it
× R∞
preserves also the Poisson point process.
In the general case the same transformation sends A [0, ) to (0,1)
× ∞ ×
[0, ), where A = x : f(x) > 0 . Conditionally on (x ,t ) we get a Poisson
1 1
∞ { }
point process on (x,y) : x A, t f(x) < y < independent of the
1
{ ∈ ∞}
Poisson point process on (x,y) : x / A, 0 < y < .
{ ∈ ∞}
1.5 Claim. Let f : (0,1) [0, ) and g : (0,1) (0,1) [0, ) be
→ ∞ × → ∞
1 1
measurable functions satisfying f(x)dx = 1 and g(x ,x )dx = 1 for
0 0 1 2 2
almost all x . Let Π be a Poisson point process on (0,1) [0, ), while
1
R R × ∞
(x ,y ), t and Π be as in 1.4. Then
1 1 1 1
(a) the minimum
y t f(x)
1
t = min −
2
(x,y)∈Π1 g(x1,x)
is reached at a single point (x ,y ) Π ;
2 2 1
∈
(b) conditionally on (x ,y ) we have: x and t = (y t f(x ))/g(x ,x )
1 1 2 2 2 1 2 1 2
−
are independent, t is distributed Exp(1), and the distribution of x has the
2 2
density g(x , ) (conditional distributions are meant);
1
·
(c) conditionally on (x ,y ),(x ,y ), the set Π = Π (x ,t ),(x ,y ) is
1 1 2 2 2 1 1 2 2
\{ }
a Poisson point process on (x,y) : 0 < x < 1,t f(x)+t g(x ,x) < y < .
1 2 1
{ ∞}
Proof. After conditioning on (x ,y ) we apply 1.4 to the Poisson point pro-
1 1
cess (x,y t f(x)) : (x,y) Π on (0,1) [0, ) and the function
1 1
{ − ∈ } × ∞
g(x , ) : (0,1) [0, ).
1
· → ∞
Equipped with these claims we prove the main lemma as follows. Intro-
ducing conditional densities
f (x ,...,x ,x)
n+1 1 n
g (x x ,...,x ) = , g (x) = f (x)
n+1 1 n 1 1
| f (x ,...,x )
n 1 n
5
and a Poisson point process Π on the strip (0,1) [0, ), we construct a
× ∞
sequence of points (X ,Y ) of Π, random variables T and random functions
n n n
H as follows:
n
y Y
1
T = min = ;
1
(x,y)∈Π g1(x) g1(X1)
H (x) = T g (x);
1 1 1
y H (x) Y H (X )
1 2 1 2
T = min − = − ;
2
(x,y)∈Π1 g2(x X1) g2(X2 X1)
| |
H (x) = H (x)+T g (x X );
2 1 2 2 1
|
...
y H (x) Y H (X )
n−1 n n−1 n
T = min − = − ;
n
(x,y)∈Πn−1 gn(x X1,...,Xn−1) gn(Xn X1,...,Xn−1)
| |
H (x) = H (x)+T g (x X ,...,X );
n n−1 n n 1 n−1
|
...
here Π stands for Π (X ,Y ),...,(X ,Y ) . By 1.4(b), X and T are
n 1 1 n n 1 1
\ { }
independent, T Exp(1) and X g . By 1.4(c) and 1.5(b), conditionally
1 1 1
∼ ∼
on X and T , Π is a Poisson point process on (x,y) : y > H (x) , while
1 1 1 1
{ }
X and T are independent, T Exp(1) and X g ( X ). It follows that
2 2 2 2 2 1
∼ ∼ ·|
T , T and (X ,X ) are independent, and the joint distribution of X ,X
1 2 1 2 1 2
has the density g (x )g (x x ) = f (x ,x ). By 1.5(c), Π is a Poisson point
1 1 2 2 1 2 1 2 2
|
process on (x,y) : y > H (x) conditionally, given (X ,Y ) and (X ,Y ).
2 1 1 2 2
{ }
The same arguments apply for any n. We get two independent sequences,
(T ,T ,...) and (X ,X ,...). Random variables T are independent, dis-
1 2 1 2 n
tributed Exp(1) each. The joint distribution of X ,X ,... is equal to µ,
1 2
since for every n the joint distribution of X ,...,X has the density f .
1 n n
Also, Π is a Poisson point process on (x,y) : y > H (x) conditionally,
n n
{ }
given (X ,Y ),...,(X ,Y ).
1 1 n n
1.6 Claim. H (x) for almost all pairs (Π,x).
n
↑ ∞
Proof. By 1.2(b), g (x x ,...,x ) = for almost all x (0,1) and
n n+1 | 1 n ∞ ∈
µ-almost all (x ,x ,...). Therefore g (x X ,...,X ) = for almost
1 2P n n+1 | 1 n ∞
allx (0,1)andalmostallΠ. Itiseasytoseethat c T = a.s.foreach
∈ P n n n ∞
sequence (c ) such that c = . Taking intoaccount that(T ,T ,...)is
n n n n ∞ P 1 2
independent of(X ,X ,...)weconcludethat T g (x X ,...,X ) =
1 2 P n n n | 1 n−1 ∞
for almost all x and Π. The partial sums of this series are nothing but
P
H (x).
n
6
Still, we have to prove that the points (X ,Y ) exhaust the set Π. Of
n n
course, a non-random negligible set does not intersect Π a.s.; however, the
negligible set x : lim H (x) < is random.
n n
{ ∞}
1.7 Claim. The set Π is empty a.s.
n n
∩
Proof. It is sufficient to prove that Π = a.s. for every M (0, ),
n n,M
∩ ∅ ∈ ∞
where Π = (x,y) Π : y < M . Conditionally, given (X ,Y ),...,
n,M n 1 1
{ ∈ }
(X ,Y ), the set Π is a Poisson point process on (x,y) : H (x) < y <
n n n,M n
{
M ; the number Π of points in Π satisfies
n,M n,M
} | |
1
E Π (X ,Y ),...,(X ,Y ) = (M H (x))+dx.
n,M 1 1 n n n
| | −
Z0
(cid:0) (cid:12) (cid:1)
Therefore (cid:12)
1
E Π = E (M H (x))+dx.
n,M n
| | −
Z0
By 1.6 and the monotone convergence theorem, E 1(M H (x))+dx 0
0 − n →
as n . Thus, lim H = 0 a.s.
n n,M
→ ∞ | | R
Now we are in position to finish the proof of the main lemma. Applying
our construction twice (for µ and for ν) we get two enumerations of a single
Poisson point process on the strip,
(X ,Y ) : n = 1,2,... = Π = (X′,Y′) : n = 1,2,... ,
{ n n } { n n }
such that the sequence (X ,X ,...) is distributed µ and the sequence
1 2
(X′,X′,...) isdistributed ν. The joint distribution P of these two sequences
1 2
satisfies the conditions 1.2(c,d,e).
2 Random countable sets
Following the ‘constructive countability’ approach of Kendall [9, Def. 3.3] we
treat a random countable subset of (0,1) as
(2.1) ω X (ω),X (ω),...
1 2
7→ { }
where X ,X , : Ω (0,1) are random variables. (To be exact, it would
1 2
··· →
be called a random finite or countable set, since X (ω) need not be pairwise
n
distinct.) It may happen that X (ω),X (ω),... = Y (ω),Y (ω),... for
1 2 1 2
{ } { }
almost all ω (the sets are equal, multiplicity does not matter); then we say
that the two sequences (X ) , (Y ) of random variables represent the same
k k k k
7
random countable set, and write X ,X ,... = Y ,Y ,... . On the other
1 2 1 2
{ } { }
hand it may happen that the joint distribution of X ,X ,... is equal to the
1 2
joint distribution of some random variables X′,X′, : Ω′ (0,1) on some
1 2 ··· →
probability space Ω′; then we may say that the two random countable sets
X ,X ,... , X′,X′,... are identically distributed. We combine these
{ 1 2 } { 1 2 }
two ideas as follows.
2.2 Definition. Two random countable sets X ,X ,... , Y ,Y ,...
1 2 1 2
{ } { }
are identically distributed (in other words, Y ,Y ,... is distributed like
1 2
{ }
X ,X ,... ), if there exists a probability measure P on the space (0,1)∞
1 2
{ } ×
(0,1)∞ such that
(a) the first marginal of P is equal to the joint distribution of X ,X ,...;
1 2
(b)thesecondmarginalofP isequaltothejointdistributionofY ,Y ,...;
1 2
(c) P-almost all pairs (x,y), x = (x ,x ,...) (0,1)∞, y = (y ,y ,...)
1 2 1 2
∈ ∈
(0,1)∞ are such that
x ,x ,... = y ,y ,... .
1 2 1 2
{ } { }
A sufficient condition is given by Main lemma 1.2: if Conditions (a), (b)
of Main lemma are satisfied both by the joint distribution of X ,X ,... and
1 2
by the joint distribution of Y ,Y ,... then X ,X ,... and Y ,Y ,... are
1 2 1 2 1 2
{ } { }
identically distributed.
2.3 Remark. The relation defined by 2.2 is transitive. Having a joint dis-
tribution of two sequences (X ) and (Y ) and a joint distribution of (Y )
k k k k k k
and (Z ) we may construct an appropriate joint distribution of three se-
k k
quences (X ) , (Y ) and (Z ) ; for example, (X ) and (Z ) may be made
k k k k k k k k k k
conditionally independent given (Y ) .
k k
2.4 Definition. A random countable set X ,X ,... has the uniform dis-
1 2
{ }
tribution (in other words, is uniform), if X ,X ,... and Y ,Y ,... are
1 2 1 2
{ } { }
identically distributed for some (therefore, all) Y ,Y ,... whose joint distri-
1 2
bution satisfies Conditions (a), (b) of Main lemma 1.2.
The joint distribution of X ,X ,... may violate Condition 1.2(b), see
1 2
9.8.
If X ,X ,... are i.i.d. random variables then the random countable set
1 2
X ,X ,... may be called an unordered infinite sample from the corre-
1 2
{ }
sponding distribution. If the latter has a non-vanishing density on (0,1) then
X ,X ,... has the uniform distribution. A paradox: the distribution of
1 2
{ }
the sample does not depend on the underlying one-dimensional distribution!
(See also 9.9.)
8
2.5 Remark. If the joint distribution of X ,X ,... satisfies Condition (a) of
1 2
Mainlemma(butmayviolate(b))thentherandomcountableset X ,X ,...
1 2
{ }
may be treated as a part (subset) of a uniform random countable set, in the
following sense.
We say that X ,X ,... is distributed as a part of Y ,Y ,... if there
1 2 1 2
{ } { }
exists P satisfying (a), (b) of 2.2 and (c) modified by replacing the equality
x ,x ,... = y ,y ,... with the inclusion x ,x ,... y ,y ,... .
1 2 1 2 1 2 1 2
{ } { } { } ⊂ { }
Indeed, the proof of Main lemma uses Condition (b) only for exhausting
all elements of the Poisson random set.
3 Selectors
A single-element part of a random countable set is of special interest.
3.1 Definition. A selector of a random countable set X ,X ,... is a
1 2
{ }
probability measure P on the space (0,1)∞ (0,1) such that
×
(a) the first marginal of P is equal to the joint distribution of X ,X ,...;
1 2
(b) P-almost all pairs (x,z), x = (x ,x ,...) (0,1)∞, z (0,1) satisfy
1 2
∈ ∈
(3.2) z x ,x ,... .
1 2
∈ { }
The second marginal of P is called the distribution of the selector.
Less formally, a selector is a randomized choice of a single element. The
conditional distribution P of z given x is a probability measure concentrated
x
on x ,x ,... . This measure may happen to be a single atom, which leads
1 2
{ }
to a non-randomized selector
(3.3) z = x ,
N(x1,x2,...)
where N is a Borel map (0,1)∞ 1,2,... . (See also 3.9.)
→ { }
In order to prove existence of selectors with prescribed distributions we
use a deep duality theory for measures with given marginals, due to Kellerer.
It holds for a wide class of spaces , , but we need only two. Below, in
1 2
X X
3.4, 3.5 and 3.6 we assume that
is either (0,1) or (0,1)∞,
1
X
is either (0,1) or (0,1)∞.
2
X
Here is the result used here and once again in Sect. 8.
9
3.4 Theorem. (Kellerer) Let µ ,µ be probability measures on , re-
1 2 1 2
X X
spectively, and B a Borel set. Then
1 2
⊂ X ×X
S (B) = I (B),
µ1,µ2 µ1,µ2
where S (B) is the supremum of µ(B) over all probability measures µ on
µ1,µ2
withmarginalsµ ,µ , andI (B)istheinfimumofµ (B )+µ (B )
X1×X2 1 2 µ1,µ2 1 1 2 2
over all Borel sets B , B such that B (B ) ( B ).
1 1 2 2 1 2 1 2
⊂ X ⊂ X ⊂ ×X ∪ X ×
See [8, Corollary 2.18 and Proposition 3.3].
Note that B need not be closed.
In fact, the infimum I (B) is always reached [8, Prop. 3.5], but the
µ1,µ2
supremum S (B) is not always reached [8, Example 2.20].
µ1,µ2
3.5 Remark. (a) If µ ,µ are positive (not just probability) measures such
1 2
that µ ( ) = µ ( ) then still S (B) = I (B).
1 X1 2 X2 µ1,µ2 µ1,µ2
(b) S (B) is equal to the supremum of ν(B) over all positive (not just
µ1,µ2
probability) measures ν on B such that ν µ and ν µ , where ν ,ν
1 1 2 2 1 2
≤ ≤
are the marginals of ν. This new supremum in ν is reached if and only if the
original supremum in µ is reached.
3.6 Lemma. Let µ ,µ be probability measures on , respectively and
1 2 1 2
X X
B a Borel set such that
1 2
⊂ X ×X
S (B) = S (B) ν(B)
µ1−ν1,µ2−ν2 µ1,µ2 −
for every positive measure ν on B such that ν µ and ν µ , where
1 1 2 2
≤ ≤
ν ,ν are the marginals of ν. Then the supremum S (B) is reached.
1 2 µ1,µ2
(See also 8.12.)
Proof. First, taking a positive measure ν on B such that ν µ , ν µ
1 1 2 2
≤ ≤
and ν(B) 1S (B) we get
≥ 2 µ1,µ2
1
S (B) = S (B) ν(B) S (B).
µ1−ν1,µ2−ν2 µ1,µ2 − ≤ 2 µ1,µ2
Second, taking apositive measure ν′ onB such that ν′ µ ν , ν′ µ ν
1 ≤ 1− 1 2 ≤ 2− 2
and ν′(B) 1S (B) we get (ν+ν′) µ and (ν+ν′) µ , thus,
≥ 2 µ1−ν1,µ2−ν2 1 ≤ 1 2 ≤ 2
1
Sµ1−ν1−ν1′,µ2−ν2−ν2′(B) = Sµ1,µ2(B)−ν(B)−ν′(B) ≤ 4Sµ1,µ2(B).
Continuing this way we get a convergent series of positive measures, ν +
ν′ +ν′′ +...; its sum is a measure that reaches the supremum indicated in
3.5(b).
10