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arXiv:math/0401389v2 [math.PR] 23 Jan 2006

Bivariate Uniqueness and Endogeny for

the Logistic Recursive Distributional

Equation

Antar Bandyopadhyay

Deaprtment of Mathematics

Chalmers University of Technology

SE - 412 96, G¨ oteborg

SWEDEN

E-Mail : antar@math.chalmers.se

February 1, 2008

Abstract

In this article we prove the bivariate uniqueness property for a particu-

lar “max-type” recursive distributional equation (RDE). Using the general

theory developed in [5] we then show that the corresponding recursive tree

process (RTP) has no external randomness, more preciously, the RTP is

endogenous. The RDE we consider is so called the Logistic RDE, which

appears in the proof of the ζ(2)-limit of the random assignment problem

[4] using the local weak convergence method. Thus this work provides a

non-trivial application of the general theory developed in [5].

AMS 2000 subject classification : 60E05, 60J80, 60K35, 62E10, 82B43.

Key words and phrases : Bivariate uniqueness, distributional identity, en-

dogeny, fixed point equations, Logistic distribution, random assignment prob-

lem, recursive distributional equations, recursive tree processes.

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1Introduction and the Main Result

Fixed-point equations or distributional identities have appeared in the probabil-

ity literature for quite a long time in a variety of settings. The recent survey of

Aldous and Bandyopadhyay [5] provides a general framework to study certain

type of distributional equations.

Given a space S write P (S) for the set of all probabilities on S. A recursive

distributional equation (RDE) [5] is a fixed-point equation on P (S) defined as

X

d= g(ξ;(Xj: 1 ≤ j ≤∗N)) on S,(1)

where it is assumed that (Xj)j≥1are i.i.d. S-valued random variables with

same distribution as X, and are independent of the pair (ξ,N). Here N is a

non-negative integer valued random variable, which may take the value ∞, and

g is a given S-valued function. (In the above equation by “≤∗N” we mean the

left hand side is “≤ N” if N < ∞, and “< N” otherwise). In (1) the distribution

of X is unknown, while the distribution of the pair (ξ,N) and the function g are

the known quantities. Perhaps a more conventional (analytic) way of writing

the equation (1) would be

µ = T (µ) ,(2)

where T : P → P (S) is a function defined on P ⊆ P (S) such that T (µ) is the

distribution of the right-hand side of the equation (1), when (Xj)j≥1are i.i.d.

µ ∈ P.

As outlined in [5] in many applications RDEs play a very crucial role. Ex-

amples include study of Galton-Watson branching processes and related ran-

dom trees, probabilistic analysis of algorithms with suitable recursive structure

[18, 10, 17], statistical physics models on trees [3, 2, 11, 6, 7, 8], and statistical

physics and algorithmic questions in the mean-field model of distance [1, 4, 2].

In many of these applications, particularly in the last two types mentioned

above, often one needs to construct a particular tree indexed stationary process

related to a given RDE, which is called a recursive tree process (RTP) [5]. More

precisely, suppose the RDE (1) has a solution, say µ. Then as shown in [5],

using the consistency theorem of Kolmogorov [9], one can construct a process,

say (Xi)i∈V, indexed by V :=

?

∪d≥1Nd?

∪ {∅}, such that

(i)

(ii)

(iii)

Xi∼ µ ∀ i ∈ V,

For each d ≥ 0,(Xi)|i|=d

Xi= g(ξi;(Xij: 1 ≤ j ≤∗Ni)) ∀ i ∈ V,

Xi is independent of

are independent,

(iv)

?

(ξi′,Ni′)

???|i′| < |i|

?

∀ i ∈ V,

(3)

where (ξi,Ni)i∈Vare taken to be i.i.d. copies of the pair (ξ,N), and by | · | we

mean the length of a finite word. The process (Xi)i∈Vis called an invariant

recursive tree process (RTP) with marginal µ.

(ξi,Ni)i∈Vare called the innovation process. In some sense an invariant RTP

with marginal µ, is an almost sure representation of a solution µ of the RDE

The i.i.d. random variables

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(1). Here we note that there is a natural tree structure on V. Taking V as the

vertex set, we join two words i,i′∈ V by an edge, if and only if, i′= ij or

i = i′j, for some j ∈ N. We will denote this tree by T∞. The empty-word ∅ will

be taken as the root of the tree T∞, and we will write ∅j = j for j ∈ N.

In the applications mentioned above the variables (Xi)i∈Vof a RTP are

often used as auxiliary variables to define or to construct some useful random

structures. In those cases typically the innovation process defines the “internal”

variables while the RTP is constructed “externally” using the consistency theo-

rem. It is then natural to ask whether the RTP is measurable only with respect

to the i.i.d. innovation process (ξi,Ni).

Definition 1 An invariant RTP with marginal µ is called endogenous, if the

root variable X∅is almost surely measurable with respect to the σ-algebra

G := σ

??

(ξi,Ni)

???i ∈ V

??

.

This notion of endogeny has been the main topic of discussion in [5]. The

authors provide a necessary and sufficient condition for endogeny in the general

setup [5, Theorem 11]. Some other concepts similar to endogeny can be found

in [7].

In this article we provide a non-trivial application of the theory developed

in [5]. The example we consider here arise from the study of the asymptotic

limit of random assignment problem using local-weak convergence method [4].

A detailed background of this example is given in Section 2.

1.1Main Result

The following RDE plays the central role in deriving the asymptotic limit of the

random assignment problem [4],

X

d= min

j≥1(ξj− Xj) on R, (4)

where (Xj)j≥1are i.i.d with same law as X and are independent of (ξj)j≥1

which are points of a Poisson point process of rate 1 on (0,∞). It is known [4]

that the RDE (4) has a unique solution as the Logistic distribution, given by

P(X ≤ x) =

1

1 + e−x, x ∈ R. (5)

For this reason we will call RDE (4) the Logistic RDE. The following is our

main result.

Theorem 1 The invariant recursive tree process with Logistic marginals asso-

ciated with the RDE (4) is endogenous.

This result though looks technical but, provides a concrete example falling

under the general theory developed in [5]. The proof of Theorem 1 involves

analytic techniques, thus this work also demonstrate the need of developing

analytic tools for studying max-type RDEs in general.

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1.2Outline of Rest of the Paper

The next section provides the background and motivation for deriving our main

result. In Section 3 we review some of the concepts from [5] and state a version

of Theorem 11 of [5], which we will need to prove our main result. In Sections

4 and 5 we prove the main result. Finally Section 6 provides some further

discussion. Some known facts about Logistic distribution which are needed for

the proofs are given in the appendix.

2 Background and Motivation for Logistic RDE

For a given n × n matrix of costs (Cij), consider the problem of assigning n

jobs to n machines in the most “cost effective” way. Thus the task is to find a

permutation π of {1,2,...,n}, which solves the following minimization problem

An:= min

π

n

?

i=1

Ci,π(i). (6)

This problem has been extensively studied in literature for a fixed cost matrix,

and there are various algorithms to find the optimal permutation π. A prob-

abilistic model for the assignment problem can be obtained by assuming that

the costs are independent random variables each with Uniform[0,1] distribution.

Although this model appears to be quite simple, careful investigations of it in

the last few decades have shown that it has enormous richness in its structure.

See [20, 2] for survey and other related works.

Our interest in this problem is from another perspective. In 2001 Aldous [4]

showed

n→∞E[An] = ζ(2) =π2

lim

6, (7)

confirming the earlier work of M´ ezard and Parisi [13], where they computed

the same limit using some non-rigorous arguments based on the replica method

[14]. In an earlier work Aldous [1] showed that the limit of E[An] as n → ∞

exists for any i.i.d. cost distribution. He also proved that the final limit does

not depend on the specifics of the cost distribution, except only on the value of

the density at 0, provided it exists and is strictly positive. So for calculation

of the limiting constant one can assume that Cij’s are i.i.d. with Exponential

distribution with mean n. Then we can redefine the objective function An in

the normalized form,

1

n

An:= min

π

n

?

i=1

Ci,π(i). (8)

From historical perspective it is worth mentioning that in 1998 Parisi [16] con-

jectured that in this case the following exact formula holds

E[An] = 1 +1

4+ ··· +

1

n2, ∀ n ≥ 1.

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Recently two separate groups Linusson and W¨ astlund [12] and C. Nair, B. Prab-

hakar and M. Sharma [15] have independently proved this conjecture using

combinatorial techniques. Thus also proving the limit. However Aldous [4] used

local-weak convergence techniques to identify the limit constant ζ(2) in terms

of an optimal matching problem on an infinite tree with random edge weights,

described as follows

Let T∞ := (V,E) be the canonical infinite rooted labeled tree, as

before, where ∅ is the root. For every vertex i ∈ V, let (ξij)j≥1be

points of a Poisson point process of rate 1 on (0,∞), and they are

independent as i varies. Define the weight of the edge e = (i,ij) ∈ E

as ξij.

This structure is called Poisson weighted infinite tree and henceforth abbrevi-

ated as PWIT.

Let Kr

n,nbe the complete graph on n vertices with a root selected uniformly

at random. Suppose we also equip it with i.i.d. Exponential edge weights with

mean n. Then one can show [4, 2] that in the sense of Aldous-Steel local weak

convergence Kr

n,nconverges to the PWIT. Moreover heuristically the random

assignment problem on Kr

n,nhas a “natural” analog to the limit structure, which

is to consider the “optimal” (in sense of minimizing the “total cost”) matching

problem on PWIT. Naturally PWIT being an infinite graph with edge weights

each having mean at least 1, the “total cost” of any matching is infinite a.s.,

and hence minimizing “total cost” is not quite meaningful. However Aldous [4]

showed that it is possible to make a sensible definition of “optimal matching”

on PWIT which is invariant with respect to the automorphism of the tree T∞,

and minimizes the “average edge weight”. This construction is quite hard, and

we refer the readers to [4, 2] for the technical details. Here we only provide the

basic essentials to understand the motivation for our work.

Consider the heuristic description of the “optimal” matching problem on

PWIT and suppose we define variables Xifor each vertex i as follows

Xi

=Total cost of a maximal matching on the subtree Ti

−Total cost of a maximal matching on the forest Ti

∞

∞\ {i}, (9)

where Ti

the sum total of all the edge weights in the matching. As noted above, both

the “total costs” appearing in (9) are infinity almost surely. Thus rigorously

speaking Xi is not well defined. But at the heuristic level if we forget this

important issue, and work with these Xi-variables as if they are well defined,

then simple manipulation yields that they must satisfy the following recurrence

relation (see Section 4.2 of [4])

∞is the subtree rooted at the vertex i. Here by “total cost” we mean

Xi= min

j≥1(ξij− Xij). (10)

This is of course the recurrence relation for a RTP associated with the Logistic

RDE (4). Having observe that one can now construct the Xi-variables externally

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as the RTP associated with the Logistic RDE, and use them to redefine the

optimal matching on PWIT. This is preciously what Aldous did in [4], and later

on referred as 540-degree argument by Aldous and Bandyopadhyay in [5]. This

construction also provides a characterization of the optimal matching on the

PWIT. Finally one can then derive the ζ(2)-limit for the random assignment

problem.

Once again a natural question would be to figure out whether the random

variables Xi’s are truly external or not, in other words to see whether the RTP

is endogenous or not (see remarks (4.2.d) and (4.2.e) in [4]). This is our main

motivation for this work. Theorem 1 proves that the Xi-variables can be defined

using only the edge-weights and hence they have no external randomness in

them.

Other significance of this result has been pointed out in Section 7.5 of [5].

We would like to note that the endogeny of the Logistic RTP helps to define

approximately feasible solution for the finite n-matching problem by using the

optimal solution of the matching problem on PWIT. Thus with the help of

endogeny one can write a possibly simpler proof of Aldous’ original argument

for the ζ(2)-limit of the random assignment problem. But such derivation for

this particular problem is not quite illuminating, and hence we do not pursue

in that direction. As indicated in Section 7.5 of [5] in general endogeny is an

essential ingredient to make rigorous argument for the cavity method, and this

work is only to illustrate one such non-trivial proof of endogeny.

3 Review of Bivariate Uniqueness and Endogeny

In this section we review some of the concepts from [5] which will be needed to

prove our main result, Theorem 1.

In the general setting of equation (1) the question of endogeny is quite ab-

stract. Aldous and Bandyopadhyay in [5] introduces a concept called bivariate

uniqueness for an invariant RTP, and showed under certain conditions that is

equivalent to endogeny. In the general setting bivariate uniqueness is defined as

follows.

Consider a general RDE given by (1) and let T : P → P (S) be the induced

operator. We will consider a bivariate version of it. Write P(2)for the space of

probability measures on S2= S × S, with marginals in P. We can now define

a map T(2): P(2)→ P?S2?as follows

Definition 2 For a probability µ(2)∈ P(2), T(2)?µ(2)?is the joint distribution

g

?

where we assume

of

g

?

ξ,X(1)

j,1 ≤ j ≤∗N

ξ,X(2)

?

?

j,1 ≤ j ≤∗N

1.

?

X(1)

j,X(2)

j

?

j≥1are independent with joint distribution µ(2)on S2;

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2. the family of random variables

?

X(1)

j,X(2)

j

?

j≥1are independent of the

innovation pair (ξ,N).

We note that we use the same realization of the pair (ξ,N) in both components.

Immediately from the definition we have

(a) If µ is a solution of the RDE then the associated diagonal measure µրis

a fixed-point for the operator T(2), where

µր:= dist(X,X), (11)

where X ∼ µ.

(b) If µ(2)is a fixed-point of the operator T(2)then each marginal is a solution

of the original RDE.

So if µ is a solution of the RDE (1) then µրis a fixed point of T(2)and there

may or may not be other fixed points of T(2)with marginals µ.

Definition 3 An invariant RTP with marginal µ has the bivariate uniqueness

property if µրis the unique fixed point of the operator T(2)with marginals µ.

Sometimes with slight abuse of terminology we will say that a solution of the

RDE (1) has bivariate uniqueness property, or even the RDE has bivariate

uniqueness property, if it has unique solution, meaning that the invariant RTP

associated with the solution has the bivariate uniqueness property.

abuse will be done for the term endogeny also.

Theorem 11 of [5] shows that under appropriate assumptions the two con-

cepts, namely bivariate uniqueness and endogeny are equivalent. Rather than

stating this general equivalence theorem, we here only state the part we will

need to prove endogeny for the Logistic RDE.

Similar

Theorem 2 (Theorem 11(b) of [5]) Let S be a Polish space. Consider an

invariant RTP with marginal distribution µ. Suppose the bivariate uniqueness

property holds. If also T(2)is continuous with respect to the weak convergence on

the set of bivariate distributions with marginals µ, then the endogenous property

holds.

Thus to prove endogenous property for the Logistic RDE (4) we will show

that the bivariate uniqueness property holds and also establish the technical

condition of Theorem 2, these are done in the following two sections.

4 Bivariate Uniqueness for the Logistic RDE

In this section we prove the bivariate uniqueness property for the Logistic RDE

(4).

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Theorem 3 Consider the following bivariate RDE

?

X

Y

?

d=

min

j≥1(ξj− Xj)

min

j≥1(ξj− Yj)

,(12)

where (Xj,Yj)j≥1are i.i.d. pairs with same joint distribution as (X,Y ) and

are independent of (ξj)j≥1which are points of a Poisson process of rate 1 on

(0,∞). Then the unique solution of this RDE is given by the diagonal measure

µրwhere µ is the Logistic distribution.

4.1Proof of Theorem 3

First observe that if the equation (12) has a solution then, the marginal dis-

tributions of X and Y solve the Logistic RDE (4), and hence they are both

Logistic. Further by inspection µրis a solution of (12). So it is enough to

prove that µրis the only solution of (12).

Let µ(2)be a solution of (12). Notice that the points {(ξj;(Xj,Yj))| j ≥ 1}

form a Poisson point process, say P, on (0,∞) × R2, with mean intensity

ρ(t;(x,y))dtd(x,y) := dtµ(2)(d(x,y)). Thus if G(x,y) := P(X > x,Y > y),

for x,y ∈ R, then

G(x,y)=P

?

?

min

j≥1(ξj− Xj) > x, and, min

j≥1(ξj− Yj) > y

???t − u ≤ x, or, t − v ≤ y

ρ(t;(u,v))dtd(u,v)

?

=P No points of P are in

?

(t;(u,v))

??

= exp

?

−

−

? ? ?

t−u≤x, or, t−v≤y

?∞

H(x)H(y) exp

= exp

0

?H(t − x) + H(t − y) − G(t − x,t − y)?dt

??∞

?

=

0

G(t − x,t − y)dt

?

, (13)

where H is the right tail of Logistic distribution, defined as H(x) = e−x/(1 + e−x)

for x ∈ R. The last equality follows from properties of the Logistic distribution

(see Fact 3 of appendix). For notational convenience in this paper we will write

F (·) := 1 − F (·), for any distribution function F.

The following simple lemma reduces the bivariate problem to a univariate

problem.

Lemma 4 For any two random variables U and V , U = V a.s. if and only if

U

d= V

d= U ∧ V .

Proof : First of all if U = V a.s. then U ∧ V = U a.s.

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Conversely suppose that U

our assumption,

d= V

d= U ∧ V . Fix a rational q, then under

P(U ≤ q < V )=P(V > q) − P(U > q, V > q)

P(V > q) − P(U ∧ V > q)

0

=

=

A similar calculation will show that P(V ≤ q < U) = 0. These are true for any

rational q, thus P(U ?= V ) = 0.

Thus if we can show that X ∧ Y also has Logistic distribution, then from

the lemma above we will be able to conclude that X = Y a.s., and hence the

proof will be complete. Put g(·) := P(X ∧ Y > ·), we will show g = H. Now,

for every fixed x ∈ R, by definition g(x) = G(x,x). So using (13) we get

g(x) = H

2(x) exp

??∞

−x

g(s)ds

?

, x ∈ R.(14)

Notice that from (A1) (see Fact 3 of appendix) g = H is a solution of this non-

linear integral equation (14), which corresponds to the solution µ(2)= µրof

the original equation (12). To complete the proof of Theorem 3 we need to show

that this is the only solution. For that we will prove that the operator associated

with (14) (defined on an appropriate space) is monotone and has unique fixed-

point as H. The techniques we will use here are similar to Eulerian recursion

[19], and are heavily based on analytic arguments.

Let F be the set of all functions f : R → [0,1] such that

• H

2(x) ≤ f(x) ≤ H(x), ∀ x ∈ R,

• f is continuous and non-increasing.

Observe that by definition H ∈ F. Further from (14) it follows that g(x) ≥

H

also that g being the tail of the random variable X ∧Y , is continuous (because

both X and Y are continuous random variables) and non-increasing. So it is

appropriate to search for solutions of (14) in F.

Let T : F → F be defined as

2(x), as well as, g(x) = P(X ∧ Y > x) ≤ P(X > x) = H(x), ∀ x ∈ R. Note

T(f)(x) := H

2(x) exp

??∞

−x

f(s)ds

?

, x ∈ R. (15)

Note that this operator T is not same as the general operator defined in Section

1, henceforth by T we will mean the specific operator defined above. Proposition

9 of Section 4.2 shows that T does indeed map F into itself. Observe that

the equation (14) is nothing but the fixed-point equation associated with the

operator T, that is,

g = T(g) on F. (16)

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We here note that using (A1) (see Fact 3 of appendix) T can also be written as

T(f)(x) := H(x) exp

?

−

?∞

−x

?H(s) − f(s)?ds

?

, x ∈ R, (17)

which will be used in the subsequent discussion.

Define a partial order ? on F as, f1? f2 in F if f1(x) ≤ f2(x), ∀ x ∈ R,

then the following result holds.

Lemma 5 T is a monotone operator on the partially ordered set (F,?).

Proof : Let f1? f2be two elements of F, so from definition f1(x) ≤ f2(x), ∀ x ∈

R. Hence

∞ ?

⇒

T(f1)(x)

≤

⇒

T(f1)

?

−x

f1(s)ds

≤

∞ ?

T(f2)(x),

T(f2).

−x

f2(s)ds,

∀ x ∈ R

∀ x ∈ R

Put f0= H

Now from Lemma 5 we get that if g is a fixed-point of T in F then,

2, and for n ∈ N, define fn∈ F recursively as, fn= T(fn−1).

fn? g, ∀ n ≥ 0. (18)

If we can show fn→ H pointwise, then using (18) we will get H ? g, so from

definition of F it will follow that g = H, and our proof will be complete. For

that, the following lemma gives an explicit recursion for the functions {fn}n≥0.

Lemma 6 Let β0(s) = 1 − s, 0 ≤ s ≤ 1. Define recursively

βn(s) :=

?1

s

1

w

?

1 − e−βn−1(1−w)?

dw, 0 < s ≤ 1.(19)

Then for n ≥ 1,

fn(x) = H(x) exp?−βn−1(H(x))?, x ∈ R.

Proof : We will prove this by induction on n. Fix x ∈ R, for n = 1 we get

(20)

f1(x)=T(f0)(x)

=

H(x) exp

?

?

?

?

−

?∞

?∞

?∞

?∞

−x

?

H(s)?1 − H(s)?ds

H(s)H(s)ds

H(s) − H

2(s)

?

ds

?

?

[using (17)]

=

H(x) exp

−

−x

=H(x) exp

−

−x

?

=

H(x) exp

−

−x

H′(s)ds

?

[using Fact 1 of appendix]

=

=

H(x) exp(−H(x))

H(x) exp?−β0(H(x))?

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Now, assume that the assertion of the Lemma is true for n ∈ {1,2,...,k},

for some k ≥ 1, then from definition we have

fk+1(x)=T(fk)(x)

=

H(x) exp

?

?

?

−

?∞

?∞

?1

−x

?H(s) − fk(s)?ds

H(s)

?

1

w

?

[using (17)]

=

H(x) exp

−

−x

1 − e−βk−1(H(s))?

1 − e−βk−1(1−w)?

ds

?

=

H(x) exp

−

H(x)

?

dw

?

(21)

The last equality follows by substituting w = H(s) and thus from Fact 1 and

Fact 2 of the appendix we get thatdw

w= H(s)ds and H(−x) = H(x). Finally

by definition of βn’s and using (21) we get fk+1= T(fk).

To complete the proof it is now enough to show that βn→ 0 pointwise, which

will imply by Lemma 6 that fn→ H pointwise, as n → ∞. Using Proposition

10 (see Section 4.2) we get the following characterization of the pointwise limit

of these βn’s.

Lemma 7 There exists a function L : [0,1] → [0,1] with L(1) = 0, such that

L(s) =

?1

s

1

w

?

1 − e−L(1−w)?

dw, ∀s ∈ [0,1), (22)

and L(s) = lim

n→∞βn(s), ∀ 0 ≤ s ≤ 1.

Proof : From the Proposition 10 we know that for any s ∈ [0,1] the sequence

{βn(s)} is decreasing, and hence ∃ a function L : [0,1] → [0,1] such that L(s) =

lim

n→∞βn(s). Now observe that βn(1 − w) ≤ β0(1 − w) = w, ∀ 0 ≤ w ≤ 1, and

hence

0 ≤1

w

Thus by taking limit as n → ∞ in (19) and using the dominated convergence

theorem along with part (a) of Proposition 10 we get that

?

1 − e−βn(1−w)?

≤βn(1 − w)

w

≤ 1, ∀ 0 ≤ w ≤ 1.

L(s) =

?1

s

1

w

?

1 − e−L(1−w)?

dw, ∀ 0 ≤ s < 1.

The above lemma basically translates the non-linear integral equation (14)

to the non-linear integral equation (22), where the solution g = H of (14) is

given by the solution L ≡ 0 of (22). So at first sight this may not lead us to the

conclusion. But fortunately, something nice happens for equation (22), and we

have the following result which is enough to complete the proof of Theorem 3.

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Lemma 8 If L : [0,1] → [0,1] is a function which satisfies the non-linear

integral equation (22), namely,

L(s) =

?1

s

1

w

?

1 − e−L(1−w)?

dw, ∀ 0 ≤ s < 1,

and if L(1) = 0, then L ≡ 0.

Proof : First note that L ≡ 0 is a solution. Now let L be any solution of

(22), then L is infinitely differentiable on the open interval (0,1), by repetitive

application of Fundamental Theorem of Calculus.

Consider,

η(w) := (1 − w)eL(1−w)+ we−L(w)− 1, w ∈ [0,1]. (23)

Observe that η(0) = η(1) = 0 as L(1) = 0. Now, from (22) we get that

L′(w) = −1

w

?

1 − e−L(1−w)?

, w ∈ (0,1).(24)

Thus differentiating the function η we get

η′(w) = e−L(w)?

2 −

?

eL(1−w)+ e−L(1−w)??

≤ 0, ∀ w ∈ (0,1). (25)

So the function η is decreasing in (0,1) and is continuous in [0,1] with boundary

values as 0, hence η ≡ 0 Thus we must have η′≡ 0, so from equation (25) we

get that

eL(s)+ e−L(s)= 2 for all s ∈ (0,1).

This implies L ≡ 0 on [0,1].

4.2Some Technical Details

This section provides some of the technical results which were needed in the

previous section.

Proposition 9 The operator T maps F into F.

Proof : First note that if f ∈ F, then by definition T(f)(x) ≥ H

Next by definition of F we get that f ∈ F ⇒ f ? H, thus

2(x), ∀ x ∈ R.

?∞

T(f)(x) ≤ H

−x

f(s)ds ≤

?∞

2(x) exp

−x

H(s)ds, ∀ x ∈ R

⇒

??∞

−x

H(s)ds

?

= H(x), ∀ x ∈ R

The last equality follows from (A1) (see Fact 3 of appendix). So,

H

2(x) ≤ T(f)(x) ≤ H(x), ∀ x ∈ R. (26)

12

Page 13

Now we need to show that for any f ∈ F we must have T(f) continuous

and non-increasing. From the definition T(f) is continuous (in fact, infinitely

differentiable). Moreover if x ≤ y be two real numbers, then

?∞

−x

?H(s) − f(s)?ds ≤

?∞

−y

?H(s) − f(s)?ds,

because f ? H. Also H(x) ≥ H(y), thus using (17) we get

T(f)(x) ≥ T(f)(y) (27)

So using (26) and (27) we conclude that T(f) ∈ F if f ∈ F.

Proposition 10 The following are true for the sequence of functions {βn}n≥0

defined in (19).

(a) For every fixed s ∈ (0,1], the sequence {βn(s)} is decreasing.

(b) For every n ≥ 1, lim

s→0+βn(s) exists, and is given by

?1

0

1

w

?

1 − e−βn−1(1−w)?

dw,

we will write this as βn(0).

(c) The sequence of numbers {βn(0)} is also decreasing.

Proof : (a) Notice that β0(s) = 1 − s for s ∈ [0,1], thus

β1(s) =

?1

s

1 − e−w

w

dw < 1 − s = β0(s), ∀ s ∈ (0,1].

Now assume that for some n ≥ 1 we have βn(s) ≤ βn−1(s) ≤ ··· ≤ β0(s), ∀ s ∈

(0,1], if we show that βn+1(s) ≤ βn(s), ∀ s ∈ (0,1] then by induction the proof

will be complete. For that, fix s ∈ (0,1] then

βn+1(s)=

?1

?1

βn(s).

s

1

w

?

?

1 − e−βn(1−w)?

1 − e−βn−1(1−w)?

dw

≤

s

1

w

dw

=

This proves the part (a).

(b, c) First note that by trivial induction βn(s) ≥ 0 for every s ∈ (0,1], n ≥ 0.

Thus from definition for every n ≥ 0, the limit lim

s→0+βn(s) exists in [0,∞] and

is given by

?1

0

1

w

?

1 − e−βn−1(1−w)?

dw.

13

Page 14

Now using (a) above we conclude

βn+1(0) = lim

s→0+βn+1(s) ≤ lim

s→0+βn(s) = βn(0),(28)

for every n ≥ 0. Since β0(0) = 1, so we get βn(0) < ∞ for all n ≥ 0, and the

sequence is decreasing. Proving parts (b) and (c).

5 Proof of Theorem 1

Once again we will use the general Theorem 11(b) of [5], stated here as Theorem

2. We note that by Theorem 3 the Logistic RDE (4) has bivariate uniqueness

property and hence all remains is to check the technical continuity condition.

Proposition 11 Let S be the set of all probabilities on R2and let Γ : S → S

be the operator associated with the RDE (12), that is,

Γ

?

µ(2)?

d=

min

j≥1(ξj− Xj)

min

j≥1(ξj− Yj)

,(29)

where (Xj,Yj)j≥1are i.i.d with joint law µ(2)∈ S and are independent of

(ξj)j≥1which are points of a Poisson point process of rate 1 on (0,∞). Then

Γ is continuous with respect to the weak convergence topology when restricted to

the subspace S∗defined as

S∗:=

?

µ(2)???both the marginals of µ(2)are Logistic distribution

is not continuous with respect to the weak convergence topology on the whole

space S. In fact, as it turns out it is every where discontinuous on S (see

Section 6). But fortunately for applying Theorem 2 we only need the continuity

of Γ when restricted to the subspace S∗.

Proof of Proposition 11 : Let

?

µ(2)∈ S⋆. We will show that Γ(µ(2)

n )

−→ Γ(µ(2)).

Let (Ω,F,P) be a probability space such that, ∃ {(Xn,Yn)}∞

random vectors taking values in R2, with (Xn,Yn) ∼ µ(2)

µ(2). Notice that by definition Xn

distribution.

Fix x,y ∈ R, then using similar calculations as in (13) we get

?

. (30)

Before we prove this proposition, it is worth mentioning that the operator Γ

µ(2)

n

?∞

n=1⊆ S⋆and suppose that µ(2)

d

n

d

−→

n=1and (X,Y )

n , n ≥ 1, and (X,Y ) ∼

d= Y , and each has Logistic

d= Yn

d= X

Gn(x,y) := Γ(µ(2)

n)((x,∞) × (y,∞))

=

H(x)H(y) exp

?

?

−

?∞

?∞

0

P(Xn> t − x,Yn> t − y) dt

?

=

H(x)H(y) exp

−

0

P((Xn+ x) ∧ (Yn+ y) > t) dt

?

=

H(x)H(y) exp?−E?(Xn+ x)+∧ (Yn+ y)+??,

14

(31)

Page 15

and a similar calculation will also give that

G(x,y):=

=

Γ(µ(2))((x,∞) × (y,∞))

H(x)H(y) exp?−E?(X + x)+∧ (Y + y)+??. (32)

Now to complete the proof all we need is to show

E?(Xn+ x)+∧ (Yn+ y)+?−→ E?(X + x)+∧ (Y + y)+?.

Since we assumed that (Xn,Yn)

−→ (X,Y ) thus

d

(Xn+ x)+∧ (Yn+ y)+

d

−→ (X + x)+∧ (Y + y)+, ∀ x,y ∈ R. (33)

Fix x,y ∈ R, define Zx,y

(Y + y)+. Observe that

n

:= (Xn+ x)+∧(Yn+ y)+, and Zx,y:= (X + x)+∧

0 ≤ Zx,y

n

≤ (Xn+ x)+≤ |Xn+ x|, ∀ n ≥ 1. (34)

But, |Xn+ x|

grable. Hence we conclude (using Theorem 25.12 of Billingsley [9]) that

d= |X + x|, ∀ n ≥ 1. So clearly {Zx,y

n}∞

n=1is uniformly inte-

E[Zx,y

n] −→ E[Zx,y].

This completes the proof.

6 Final Remarks

(a) Intuitively, a natural approach to show that the fixed-point equation Γ(µ(2)) =

µ(2)on S has unique solution, would be to specify a metric ρ on S such that

the operator Γ becomes a contraction with respect to it. Unfortunately, this

approach seems rather hard or may even be impossible. Perhaps the reason

being the Logistic RDE (4) itself does not have a contractive property, in fact,

it does not have a full domain of attraction (see [5]). However its exact domain

of attraction is not yet known (see open problem 62 of [5]). On the other hand

from the proof of Theorem 3 it is clear that equation (14) has the whole of F

within its domain of attraction. So it is possible to have a suitable metric of

contraction for T but, we have been unable to find it.

(b) Although at first glance it seems that the operator T as defined in (15) is

just an analytic tool to solve the equation (14) but, it has a nice interpretation

through Logistic RDE (4). Suppose A is the operator associated with Logistic

RDE, that is,

A(µ)

j≥1(ξj− Xj),

d= min (35)

15

Page 16

where (ξj)j≥1are points of a Poisson point process of mean intensity 1 on (0,∞),

and are independent of (Xj)j≥1, which are i.i.d with distribution µ on R. It is

easy to check that the domain of definition of A is the space

A :=

?

F

???F is a distribution function on R and

Note that the condition?∞

extended on whole of A. In that case the following identity holds

?∞

0

F(s)ds < ∞

?

. (36)

0F(s)ds < ∞ means EF[X+] < ∞. Now it is easy

to see that F can be embedded into A and definition of T can be naturally

T(µ)(·)

H(·)

×A(µ)(·)

H(·)

= 1, ∀ µ ∈ A. (37)

This at least explains the monotonicity of T through anti-monotonicity property

of the Logistic operator A (easy to check).

(c) It is interesting to note that the operator A is every where discontinuous

with respect to the weak convergence topology on A. This is because, given any

distribution F0∈ A, we can construct a sequence of distributions {Fn}n≥1⊆ A

converging in distribution to F0, such that

EFn

?

(x + Xn)+?

(x + X0)+?

→ ∞, for all x ∈ R.

Note F0∈ A ⇒ EF0

know that for any distribution function F,

?

< ∞, for all x ∈ R. On the other hand we

A(F)(x) = 1 − exp

?

−EF

?

(x + X)+??

, for all x ∈ R

(see the proof of Fact 3 in the appendix). Thus for every x ∈ R,

A(Fn)(x) ? A(F0)(x).

So A is discontinuous at F0for every F0∈ A with respect to the weak conver-

gence topology. This also indicates that the same phenomenon is true for the

bivariate operator Γ.

Appendix

Here we provide some known facts about the Logistic distribution which are

used in the Sections 4 and 5. First recall that we say a real valued random

variable X has Logistic distribution if its distribution function is given by (5),

namely,

H(x) = P(X ≤ x) =

1

1 + e−x, x ∈ R.

The following facts hold for the function H.

16

Page 17

Fact 1 H is infinitely differentiable, and H′(·) = H(·)H(·), where H(·) = 1 −

H(·).

Proof : From the definition it follows that H is infinitely differentiable on R.

Further,

H′(x)=

1

1 + e−x×

e−x

1 + e−x

=H(x)H(x) ∀ x ∈ R

Fact 2 H is symmetric around 0, that is, H(−x) = H(x) ∀ x ∈ R.

Proof : From the definition we get that for any x ∈ R,

H(−x) =

1

1 + ex=

e−x

1 + e−x= H(x).

Fact 3 H is the unique solution of the non-linear integral equation

H(x) = exp

?

−

?∞

−x

H(s)ds

?

, ∀ x ∈ R. (A1)

Proof : Notice that the equation (A1) is nothing but Logistic RDE, this is

because

P

?

min

j≥1(ξj− Xj) > x

?

= exp

?

−

?∞

−x

H(s)ds

?

, ∀ x ∈ R

where (Xj)j≥1are i.i.d. with distribution function H and are independent of

(ξj)j≥1, which are points of a Poisson point process of rate 1 on (0,∞). Thus

from the fact that H is the unique solution of Logistic RDE (Lemma 5 of [4])

we conclude that H is unique solution of equation (A1).

Acknowledgments

This work was done in University of California, Berkeley, as a part of the au-

thor’s doctoral dissertation, written under guidance of Professor David J. Al-

dous, whom the author would like to thank for suggesting the problem and for

many illuminating discussion. The author would also like to thank an anony-

mous referee for some useful comments on an earlier version of the paper.

17

Page 18

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