Palm Measure Invariance and Exchangeability for Marked Point
Processes
Except where reference is made to the work of others, the work described in this
dissertation is my own or was done in collaboration with my advisory committee.
This dissertation does not include proprietary or classifled information.
Man Peng
Certiflcate of Approval:
Ming Liao
Professor
Mathematics and Statistics
Olav Kallenberg, Chair
Professor
Mathematics and Statistics
Jerzy Szulga
Professor
Mathematics and Statistics
Amnon J. Meir
Professor
Mathematics and Statistics
Thomas H. Pate
Professor
Mathematics and Statistics
George T. Flowers
Interim Dean
Graduate School
Palm Measure Invariance and Exchangeability for Marked Point
Processes
Man Peng
A Dissertation
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulflllment of the
Requirements for the
Degree of
Doctor of Philosophy
Auburn, Alabama
December 19, 2008
Palm Measure Invariance and Exchangeability for Marked Point
Processes
Man Peng
Permission is granted to Auburn University to make copies of this dissertation at its
discretion, upon the request of individuals or institutions and at
their expense. The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Vita
Man Peng, son of Pai Peng and Jurong Chen, was born on Nov 15, 1980 in Wuxue
City of Hubei Province, P.R. China. He graduated from Daqing Tieren High school in
Daqing City of Heilongjiang Province, in 1998. He then attended Tianjin University
in Tianjin City from 1998 to 2002, receiving his B.S. degree in Mathematics. In Fall
2002, he entered Ph.D. program at Auburn University.
iv
Dissertation Abstract
Palm Measure Invariance and Exchangeability for Marked Point
Processes
Man Peng
Doctor of Philosophy, December 19, 2008
(B.S., Tianjin University, 2002)
87 Typed Pages
Directed by Olav Kallenberg
A random measure ? on a real interval I is known to be exchangeable ifi suitably
reduced versions of the Palm distributions Qt are independent of t 2 I. In this
dissertation we prove a corresponding result where ? is a point process on I with
marks in some Borel space. For this case, the Palm distributions Qs;t depend on
parameters s 2 S and t 2 I, and we show that ? is exchangeable ifi the reduced
versions of Q0s;t are independent of t.
v
Acknowledgments
My greatest thanks extend to Dr. Olav Kallenberg for his patience and guidance
throughout this endeavor. Dr. Kallenberg took considerable time and energy to
further the progress in my studies of Probability Theory. I also wish to express my
appreciation for my parents and other family members Shuxin Yin, Pan Peng and
Zhen Peng who have given of their love and constant support throughout my life.
Dr. Ming Liao, Dr. Jerzy Szulga and Dr. Tin-Yau Tam are all deserving of many
thanks as well, as I have taken important classes and seminars from each of them.
I would also like to thank the professors on the committee for their contribution to
this dissertation and serving as the committee members.
vi
Style manual or journal used Transactions of the American Mathematical Society
(together with the style known as auphd).
Computer software used The document preparation package TEX(speciflcally
LATEX) together with the departmental style-flle auphd.sty.
vii
Table of Contents
List of Figures ix
1 Introduction 1
1.1 Motivation and History . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Organization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 Basic Notions 7
2.1 Random Measures and Point Processes . . . . . . . . . . . . . . . . . 7
2.2 Palm Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.3 Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3 Previous and New Results 23
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.2 Previously Known Results . . . . . . . . . . . . . . . . . . . . . . . . 23
3.3 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4 Proofs of Previously Known Results 29
4.1 Proof of Theorem 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.2 Proof of Theorem 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
5 Proof of Main Theorem 49
5.1 Some Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . 49
5.2 Exchangeability of MPP . . . . . . . . . . . . . . . . . . . . . . . . . 54
5.3 Proof of Theorem 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
Bibliography 76
viii
List of Figures
2.1 Point Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Exchangeable Point Process on S ?[0;1] . . . . . . . . . . . . . . . . 9
2.3 Transposition Ta;b operated on R+ . . . . . . . . . . . . . . . . . . . . 20
ix
Chapter 1
Introduction
1.1 Motivation and History
This dissertation deals with the relation between Palm measures and symmetry
properties of marked point processes. More precisely, we characterize the property
of exchangeability in terms of an invariance property of the Palm measures. In this
introduction, we will only discuss some of the basic notions and ideas in intuitive
terms. The precise deflnitions will be given in later sections.
The flrst studies of Palm measures date back to the work of Palm [25], Khinchin
([19], 1955), Kaplan (1955), Ryll-Nardzewski [28], Slivnyak [29], Matthes [22], and
Mecke [23]. Palm?s monograph ([25], 1943) deals with telephone tra?c intensity vari-
ations. This marks the beginning of queuing theory, which was later to be developed
more systematically by Khinchin [19] and others. Palm theory was originally used to
develop tools for formulating and studying the basic relationships between time and
event averages in queuing theory, but its applications have been explored in other
subjects in recent work. Most studies in the literature deal with Palm distributions
in the stationary case, but very few treat the exchangeable case.
The research on exchangeability started with de Finetti ([8], 1930). The char-
acterization of inflnite exchangeable sequences of random variables was established
1
by de Finetti ([9], 1937), and such sequences are also contractable, as noted by Ryll-
Nardzewski ([27], 1957). Hewitt and Savage ([12], 1955) extended the result to ran-
dom elements in a compact Hausdorfi space, and they also proved the celebrated
Hewitt-Savage zero-one law in that paper. B?uhlmann ([3], 1960) studied exchange-
able processes and showed that a process onR+ is exchangeable ifi it has conditional
i.i.d. increments. Exchangeable random measures on [0;1] were characterized by
Kallenberg ([13], 1975), and the corresponding results for random measures on prod-
uct spaces S?R+ and S?[0;1] were given by Kallenberg (1990). The fact that any
exchangeable, simple point process on [0, 1] is a mixed binomial process was noted
independently by Kallenberg (1973), Davidson ([7], 1974) and Matthes, Kerstan, and
Mecke (1974{82). The characterization of mixed Poisson processes by exchangeabil-
ity and stationarity with exchangeable spacing variables ([17], Proposition 1.28) was
proved by Nawrotzki (1962), Kallenberg (1975), Matthes (1978) and Freedman. For
more details on exchangeability, the reader may refer to the monographs of Aldous
([1], 1983) and Kallenberg ([17], 2005). The reader may also flnd more complete his-
torical and bibliographical remarks on exchangeability from Kallenberg ([17], 2005).
The adopted version of the celebrated result of Slivnyak ([29], 1962) states that
a point process is Poisson ifi the associated reduced Palm distributions agree with
the original distribution of the process. Slivnyak?s work linked the notions of ex-
changeability with Palm measures. Later, some people proved extensions in various
direction. Papangelou ([26], 1974) characterized the mixed Poisson processes through
2
the invariance of their reduced Palm distributions. The characterization of mixed bi-
nomial processes appeared in Kallenberg (1972), and the version for general random
measures was proved by Kallenberg ([13], 1975). The extension amounts to char-
acterizing exchangeability of a random measure by the invariance of the associated
Palm measures. More precisely, a random measure ? is exchangeable ifi the associ-
ated reduced Palm measures Q0s can be chosen to be independent of s. Inspired by
these previous characterizations, especially the result in Kallenberg ([13], 1975), we
characterize exchangeable marked point processes in terms of suitably deflned Palm
distributions. In the main result of this dissertation, we prove that a marked point
process ?, with  -flnite intensity measure E? admitting a factorization ? ??, is ex-
changeable ifi the associated reduced Palm measures Q0s;t can be chosen to depend
only on s 2 S. The factorization of the intensity measure E? is also necessary, due
to the translation invariance of E? when ? is exchangeable.
1.2 Organization
The basic notions used in this dissertation are deflned in Chapter 2. Here, we
begin in Section 2.1 with deflnitions of difierent types of random measures, such as
general random measures, point processes, marked point processes. The dissertation
relates the notions of Palm measures and exchangeability, which are the main top-
ics of Sections 2.2 and 2.3, respectively. In those sections we will also make some
bibliographical comments on Palm measures. This is followed by some other crucial
3
deflnitions concerning Palm measures in Section 2.2, such as Campbell measures, re-
duced Palm distributions. In Section 2.3, a brief summary of the classical theory
of exchangeable point processes and random measures is presented. In particular,
Section 2.3 discusses the relations between difierent symmetries for random measures
and gives a unqiue representation of an exchangeable random measure. This section
also collects some characterization results for exchangeable simple point processes
and marked point processes.
In Chapter 3, we begin with a discussion of previously known results relevant
to this dissertation. Former results characterize the exchangeability of a few special
types of random measure and general random measures in terms of Palm measures,
where Palm measures Qs for general random measures on S are assumed to depend
only on s 2 S. The discussion in Section 3.2 in the setting without marks provides a
motivation for the main result of this dissertation, Theorem 3.4, dealing with marked
point processes, where the Palm measures depend on two parameters s 2 S and t 2 I.
In Chapter 4 we prove some previously known results on this subject, in order
to provide some more details to the terse and technical proofs given in the references.
This makes this dissertation self-contained, and may help to explain some of the ideas
underlying the proof of our main result, Theorem 3.4, in Chapter 5.
Finally, the proof of our main result is given in Chapter 5. Since the proof is
quite complicated, we divide it into several steps, as follows: First we show that the
proof of Theorem 3.4 can be reduced to the case where the "time scale" I is bounded,
the total mass of the marked point process ? is flnite, and the projection of ? onto
4
the mark space is nonrandom. Then we provide an equivalence condition for the
exchangeability of ?. We complete the proof by using a connection between regular
condition probability and conditional reduced Palm distributions.
1.3 Notation
For convenience, we assume all random elements appearing in this dissertation to
be deflned on some abstract probability space (?;F;P) with associated expectation
operator E. For ease of reading, we list some symbols that will be used throughout
the dissertation:
5
Rd d-dimensional Euclidean space
R+ set of non-negative real numbers
Z;Z+ set of integers in R or R+
Q;Q+ set of rational numbers in R or R+
S
 -flnite state space of a point process or random measure ?,
often S =Rd
I interval [0;1] or R+
p permutation (p1;p2;:::) of integers (1;2;:::)
? ?p
sequence (?p1;?p2;:::), where ? is a sequence of random elements
and p is a permutation
A ? B A is deflned by B
B(R) Borel  -fleld on R
M(S) class of  -flnite measures on a Borel space S
N(S) class of locally flnite counting measures on a Borel space S
BM(S) Borel  -fleld on M(S)
????? ? and ? are conditionally independent given ?
Ta;b(?) transposition of a random measure ?
1B? restriction of random measure ? to a set B, i.e. 1B?(A) = ?(B \A)
? intensity of a random measure or Lebesgue measure on R
?t 1B(t) = ?t(B)
?B projection of a random measure ? on S ?I onto a set B
id id(x) = x
6
Chapter 2
Basic Notions
2.1 Random Measures and Point Processes
A random measure ? on a Borel state space S is deflned as a  -flnite kernel from
the basic probability space ? to Rd, i.e. ?(!;?) is a measure on Rd for any ! 2 ?.
Here the underlying probability space for the random measure is (?;F;P).
A point process is a random measure ? such that ?B is integer-valued for every
bounded Borel set B 2 S, in which case, we have a representation for the point
process ? = Pi ??i, where ??B = 1f? 2 Bg. In particular, if ?fsg ? 1 for every
s 2Rd outside a P-null kernel, then the point process ? is said to be simple.
Figure 2.1: Point Process
A Poisson process on S with intensity measure ? 2M(R+) is a point process ?
onR+ with independent increments such that ?B is Poisson with mean ?B whenever
7
?B < 1. ACox process ? onspaceS isthepointprocesswhosedistributionisPoisson
with a random underlying intensity measure. A point process ? on S is a Cox process
directed by some random measure ? on S when, conditionally on ?, realizations of ?
are those of a Poisson process ?(?j?) on S with parameter measure ?. A Cox process,
also known as a doubly stochastic Poisson process, is a generalization of a Poisson
process. It was originally introduced by Cox ([5], 1955). For a Cox process on R+,
taking ? = ?? for some random variable ? ? 0 and  -flnite measure ? turns the Cox
process to a mixed Poisson process. We say that ? is a binomial process on [0;1] based
on k 2N if ? can be written as Pj?k ??j a.s. for some i.i.d. random variables ?1,:::,?k
with distribution U(0;1). A mixed binomial process on [0;1] is obtained by replacing
integer k with an integer-valued random variable ? independent of the ?j.
By a marked point process ? (or, MPP for short) on I with marks in S we mean
a simple point process ? such that ?fS ?ftgg? 1 for any t 2 I. If fl = Pj ?flj is an
arbitrary point process on S, then a uniform or ?-randomization of fl is deflned as a
point process on S?[0;1] of the form ? = Pj ?flj;?j, where the ?j are i.i.d. U(0;1) and
independent of fl. An instance of such an exchangeable marked point process is given
in Figure 2.2, where the projections of marks onto [0;1] are i.i.d. U(0;1). Since the
?j are also independent of fl, for any positive integer k ? ?(S ?[0;1]), the ?j sharing
the same flk are also i.i.d. U(0;1).
Here, we may present some results based on stronger conditions on the prob-
ability space. Readers may refer to David Vere-Jones [6] for more details. For a
probability space (?;F;P), let X be a locally compact, second countable Hausdorfi
8
Figure 2.2: Exchangeable Point Process on S ?[0;1]
space (abbreviated as lcscH), and let d be the metric such that the space (X;d) is
Polish. We denote by B the ring of relatively compact subsets of X. Let MX be the
space of Radon measures endowed with the vague topology, in other words MX is
the space of Borel nonnegative measures that are flnite on B. The vague topology
is the topology induced by functions ?f, where ?f(?) = ?f for any ? 2 MX and
f 2 CK(X), the space of real-valued functions on X with compact support. Denote
by M(X) the space of random measures on (?;F;P) taking values on MX.
The flnite-dimensional distributions of a random measure ? are the family of
proper distribution functions
P[?(Ai) 2 Bi;i = 1;:::;k] (2.1)
9
for all flnite families of bounded Borel sets A1;:::;Ak, and Borel sets Bi are chosen
from R+, and k = 1;2;:::
Proposition 2.1. The distribution of a random measure ? on an lcscH space X is
totally determined by its flnite-dimensional distributions.
10
2.2 Palm Measures
The deflnition of Palm measures requires a random measure ? on a measurable
space (S;S), along with a random element ? in a measurable space (T;T ), where ? is
also deflned as a  -flnite kernel from the basic probability space (?;F;P) to the space
S. Deflne the set of Palm distributions Qs of ? with respect to ? as Radon-Nikod?ym
densities, given by
Qs(A) = E[?(ds);? 2 A]E?(ds) ; s 2 S, A 2T , (2.2)
and regular version of Qs(A) is a measure in S for each s. In order to make sense of
this deflnition for every A, we need the intensity measure E? to be  -flnite. In order
to ensure the existence of Q as a probability kernel from S to T, we may also assume
the space T to be Borel. Rewriting the above equation of Palm distributions as a
disintegration, we get
Cf ? E
Z
f(s;?)?(ds) =
Z
E?(ds)
Z
f(s;t)Qs(dt); f ? 0 , (2.3)
where f is understood to be an arbitrary non-negative, measurable function on S?T.
Measure C is the corresponding Campbell measure of the pair (?;?) on S?T admitting
factorization C = E??Q, where this product measure is deflned in the sense of (2.3).
When ? is a simple point process, the Palm distribution Qs of ? with respect to
? is the conditional distribution of ?, given that ? has a unit mass (or point) at s. In
11
particular, if ? = ? for some random element  in S, then Q reduces to a regular
conditional distribution P[? 2 ?j ]. If E? is not  -flnite, then the denominator E?
in (2.2) needs to be replaced by a  -flnite supporting measure ? of ? on S such that
?B = 0 ifi ?B = 0 a.s. for every B 2S. The supporting measure ? is unique up to an
equivalence, in the sense of mutual absolute continuity. The Qs are ?-a.e. bounded
ifi the intensity measure E? is  -flnite, in which case we may choose ? = E? and
normalize the Qs to be probability measures on T if we compare (2.2) with (2.3). If
? itself is Borel, we may choose ? to be the identity mapping on ? (of course, ? = T
in this case), which makes Q a kernel from S to ?: Our main emphasis is on the case
when ? ? ?, S is Borel, and E? is  -flnite. In such a setting, Q becomes a kernel
from S to M(S), in which case the Qs are called the Palm measures of ?.
Similarly, we may also deflne the Palm measures Qs;t of the random measure ?
on a measurable space S?I by changing the S in (2.3) to S?I, where s 2 S, t 2 I.
When ? is a point process on S and ? = ?; the Palm measures Qs of ? are E?-a.e.
conflned to the set of measures ? 2N(S) with ?fsg? 1 for s 2 S a.e. ?: The reduced
Palm measures Q0s on N(S) are obtained by subtracting a trivial unit mass (or point)
at s from the point process ?, in which case, the formula of Q0s is given explicitly by
Q0s(B) =
Z
???s2B
Qs(d?) = Qsf?;???s 2 Bg; s 2 S. (2.4)
12
To justify this deflnition, we may introduce the reduced Campbell measure C0 = C0?
on S ?N(S); given by
C0f = E
Z
f(s;? ??s)?(ds); f ? 0,
where f is an arbitrary non-negative, measurable function on S ?N(S), and ?sA =
1A(s). A reduced Palm distribution Q0s of the simple point process ? with  -flnite
intensity measure E? is the conditional probability distribution of the simple point
process obtained by removing the point s from ? given that ? has a unit mass at
s 2 S. Again, if C0 is  -flnite, then C0 also admits a disintegration C0 = ? ? Q0,
where ? is the supporting measure of ?, and the product measure is also deflned in
the sense of (2.3). The measures Qs and Q0s are ?-a.e. related by (2.4).
A comprehensive introduction to Palm measures is given by Daley [6], and basic
facts and results on Palm measures associated with stationary point processes are
ofiered in Chapter 1 of Baccelli [2]. Palm probabilities give us a way of calculating
probabilities of events conditioning on sets of measure zero.
Next result (David Vere-Jones [6], section 12.1) states that the Campbell Measure
of random measure ? uniquely determines the distribution of ?, obviously, the converse
is also true.
Lemma 2.2. Let ? 2M(X) be a random measure on an lcscH space X , and deflne
? (A?B) = E[?A;? 2 B] (2.5)
13
for any measurable A ? X and B ? M(X). Then, ? is a  -flnite measure on
X ?BM(X) and uniquely determines L(?).
The following result shows how Palm measures can be described in terms of
ordinary conditional probabilities by introducing some auxiliary random element ?.
The result was flrst given by Kallenberg (2007) ([18], Proposition 4.1). Here we may
flll some details in the short proof given in this paper. For convenience, we write 1B?
for the restriction of ? to the set B, i.e. 1B?(A) = ?(B \A).
Proposition 2.3. Let ? be a random measure on a Borel state space S. For any set
B 2 S with ?B < 1 a.s. Consider a random element ? in T such that the Campbell
measure C of (?;?) is  -flnite. If a random element ? in S with ? 62 B whenever
?B = 0, and
P[? 2?j?;?] = 1B??B a.s. on f?B > 0g . (2.6)
Then, the Palm measures Qs of ? with respect to ? having supporting measure ? =
L(?) on B are given by
Q?(A) = E[?B;? 2 Aj?] a.s. on f? 2 Bg
for any A 2S
Proof: Taking expectation on both sides of (2.6), we have that on the set f?B >
0g
E[P[? 2?j?;?]] = P(? 2?) = E[?=?B].
14
Write ? = L(?) ? Pf? 2 ?g. Let us now check ? ? E?. Assume that E?B = 0 for
some B 2 S, then ?B = 0 a.s., so P(? 2 B) = 0 by the assumption of proposition.
Similarly, if ?B = 0, then
0 = P (? 2 B) = P(? 2 B;?B = 0)+P(? 2 B;?B > 0)
= P(? 2 Bj?B = 0)P(?B = 0)+P(? 2 Bj?B > 0)P(?B > 0)
= P(? 2 Bj?B = 0)P(?B = 0)+E[?B=?Bj?B > 0]P(?B > 0)
= P(? 2 Bj?B = 0)P(?B = 0)+P(?B > 0).
So, P(?B = 0) = 1?P(?B > 0) = 1, which gives E?B = 0. Hence ? ? E?.
From the introduction part of Palm measure, we know that E? is a supporting
measure on B since ?B < 1 a.s., therefore ? is also a supporting measure of ? on
B. To see C = ? ? Q on B with ?, Q as stated in the proposition, we shall apply
the disintegration theorem twice and use the deflnition of Campbell measure in the
calculations as follows. Let f ? 0 be an arbitrary S ?T -measurable function, then
15
on the set f?B > 0g
C(1Bf) = E
Z
B
f(s;?)?(ds) = E
?
?B ?
Z
B
f(s;?)?(ds)?B
?
= E
?
?B ?
Z
B
f(s;?)P[? 2 dsj?;?]
?
= E[?B ?E[f(?;?)j?;?]]
= E[E[?B ?f(?;?)j?;?]] = E[?B ?f(?;?)]
= E[E[?B ?f(?;?)j?];? 2 B] for ? 62 B when ?B = 0
=
Z
B
?(ds)
Z
f(s;t)Qs(dt),
which shows that the Qs are Palm measures (not necessary to be probability distri-
butions) of ? associated with the supporting measure ? of ?: The last step implies
Q?(dt) = E[?B;? 2 dtj?] a.s. on f? 2 Bg. ?
16
2.3 Symmetries
We turn to the brief summary of classical theory of exchangeable random mea-
sures, in particular, the exchangeable point processes. The detailed discussions of
difierent types of symmetries including exchangeability are ofiered in the book of
Kallenberg [17].
A flnite or inflnite sequence of random elements ? = (?1;?2;:::) in a measurable
space (S;S) is said to be exchangeable if
(?k1;:::;?km) d= (?1;:::;?m) (2.7)
for any sequence k1,:::;km of distinct elements in the index set of ?. We also say that
? is contractable if (2.7) holds whenever k1 < ??? < km. Note that any exchangeable
sequence is also contractable.
For a random measure ? on (0;1) or (0;1], it is exchangeable if the inflnite or
flnite sequence of random elements
(?(0;1=n];?(1=n;2=n];:::)
is exchangeable for any n 2N.
Assume that ?f0g = 0 a.s. or ?(S ?f0g) = 0 a.s. if ? is random measure on I or
S ?I, where I = [0;1) or [0;1]. For convenience, we write ?B (?) ? ?(??B) as the
projection of ? onto B 2 B(I) if the random measure ? is on S ?I. Let p denote a
17
permutation (p1;p2;:::) of the sequence (1;2;:::) such that only flnitely many integers
of sequence (1;2;:::) are rearranged, and L = (I1;I2;:::) be a sequence of disjoint
equal-length subintervals of I. Write L?p ? (Ip1;Ip2;:::) and ? ?L ? (?I1;?I2;:::). A
random measure ? on S ?I, where S is Borel and I = R+ or [0;1], is exchangeable
if ? ?L d= ? ?(L?p) for any such permutation p and sequence L.
The celebrated de Finetti?s theorem ([9], 1937) states that the distribution of
an inflnite exchangeable sequence of random variables is a mixture of distributions
of i.i.d. sequences. A further extension ([17], Theorem 1.1) of his result shows that
the distribution of an inflnite sequence ? of random elements in a Borel space S is
exchangeable if ? is conditionally i.i.d. given some  -fleld F, i.e.
P[? 2?jF] = ?1 a.s. (2.8)
for some random probability measure ? on S ([17], section 1.1), which is a stronger
result than de Finetti?s. The counterpart of this extension result that a random
sequence ? = (?1;:::;?n) in a Borel space S is exchangeable ifi ? is conditionally
"urn", i.e.
P[? 2 BjF] = 1n!
X
p
1B(? ?p), B 2Sn, (2.9)
whereF can be taken to be the  -fleld generated by fl = Pnk=1 ??k, and the summation
extends over all permutations p = (p1;:::;pn) of f1;:::;ng, and the ??p ? (?p1;:::;?pn)
([17], section 1.2). Ryll-Nardzewski ([27]) proves that the exchangeability of an in-
flnite sequence ? of random elements in a Borel measurable space S is equivalent to
18
the contractibility of ?, and also equivalent to the condition that ? is conditionally
i.i.d in the sense of (2.8).
There are three common notions of symmetries considered in this section, and
we gave a short introduction to one type of symmetry for random sequence, the ex-
changeability. Formulas (2.8) and (2.9) characterize the exchangeabilities for inflnite
and flnite random sequences. We turn to the other two symmetries for random mea-
sures by introducing some simple transformations on I =R+ or [0;1], and give their
equivalence for random measures on I and product space S ?I.
For any a ? 0, the re ection Ra on I is deflned by
Ra(t) =
8
><
>:
a?t; t ? a
t; t > a
.
A random measure ? on I is said to be re ectable if ? ?R?1a d= ? for every a 2 I. In
the case when ? is a simple point process. For any flxed 0 ? a ? b, the contraction
Ca;b on I is deflned by
Ca;b(t) =
8
>>>
>><
>>>
>>:
t; t ? a
1; t 2 (a;b] .
t?b+a; t > b
Note that the efiect of Ca;b is to remove the interval (a;b] and join the remaining
two disconnected intervals together. Similarly, we say a random measure ? on R+ is
19
contractable if ? ?C?1a;b d= ? for any [a;b] ?R+, where ? ?C?1a;b is obtained by sticking
1[0;a]? and 1(b;1]? (or 1(b;1)?) together, but a random measure ? on [0;1] is said to be
contractable if ??C?1a;b d= ??C?1c;d whenever the lengths of intervals [a;b] and [c;d] are
equal. The transposition Ta;b on I is deflned for any 0 ? a ? b by
Ta;b(t) =
8
>>>>
><
>>>>
>:
t+b?a; t ? a
t?a; t 2 (a;b]
b; t > b
.
In other words, Ta;b switches the intervals [0;a] and (a;b], but the interval after b
remains the same. An example of such transposition is shown in the Figure 2.3.
Figure 2.3: Transposition Ta;b operated on R+
Informally, for any random measure ? on I, ? ?T?1a;b is a random measure on I
by switching two parts 1[0;a]? and 1(a;b]?. Intuitively, we may see from the deflnition
of Ta;b that a marked point process ??T?1a;b is obtained by switching two "slices" of ?
on S ?[0;a], S ?(a;b].
According to Theorem 1.15 in [17], the above three notions of symmetry deflned
by re ection, contraction and transposition are equivalent for a random measure ? on
20
R+. Indeed, this theorem still holds for random measures on S ?R+, where S is a
Borel space, but again, the notations ??R?1a , ??C?1a;b and ??T?1a;b are deflned under the
condition ?(S?f0g) = 0 a.s. We say f is a ?-preserving function on I if ? = ??f?1.
The next result summarizes the relations between the difierent symmetries quoted
from sections 1.4 and 1.5 in [17] for a random measure.
Lemma 2.4. (exchangeable random measures) Let ? be a random measure on I or
S ?I, where I =R+ and S is Borel. Then these conditions are equivalent:
(i) ? is contractable,
(ii) ? is exchangeable,
(iii) ? is re ectable,
(iv) ? ?f?1 d= ? for any ?-preserving function f on I,
If instead I = [0;1], then (i)((ii),(iii),(iv), and the exchangeability of ? is equiv-
alent to the condition
(v) ? has a representation
? = fi??+
X
j
flj ???j a.s.
for some i.i.d. U(0;1) random variables ?1,?2,... and an independent collection of
random measures fi and fl1,fl2,... on S.
Furthermore, if ? is a simple point process on R+ with inflnitely many points ?1 <
?2 < ???, then (i)?(iv) are also equivalent to
(vi) ? is stationary with exchangeable spacing variables ?k ??k?1, k 2N.
21
Of course, Lemma 2.4 also holds for marked point processes on S ?R+ or S ?
[0;1] and simple point processes on R+ or [0;1]. The following Lemma outlines the
exchangeability results in sections 1.4, 1.5 and 1.6 of [17] for simple point processes.
Lemma 2.5. (characterization of exchangeable simple and marked point processes)
(i) Let ? be a simple point process on R+ or [0;1]. Then ? is exchangeable ifi it
is a mixed Poisson or mixed binomial process.
(ii) Let ? be a marked point process on R+ with marks in Borel space S. Then ?
is exchangeable ifi it is a Cox process directed by ? ?? for some random measure ?
on S, where ? is Lebesgue measure.
(iii) Let ? be a marked point process on S ?[0;1], where mark space S is Borel.
Then ? is exchangeable ifi it is a ?-randomization of fl ? ?(??[0;1]).
22
Chapter 3
Previous and New Results
3.1 Introduction
Section 3.2 starts with Slivnyak?s theorem stating that the only point process
whose associated Palm distributions are the same as the distribution of this point
process is Poisson, which characterizes the Poisson process in terms of Palm measures.
Kallenberg and Papangelou ([17], section 2.7) provide the condition with explicit
formula to characterize the exchangeable point processes, in which case, the reduced
Palm distributions turn out to be invariant, so it naturally gives arise to conjecturing if
there is also an invariance property of reduced Palm distributions for the exchangeable
marked point processes. This is the origin of the ideas about the main results of this
dissertation presented in Section 3.3.
3.2 Previously Known Results
Pure Poisson processes were flrst characterized in terms of the Palm distribu-
tions by Slivnyak ([29], 1962). The following Theorem 3.2 ([17], section 2.7) extends
Slivnyak?s classical result and shows that the distribution of a Poisson process with
a flxed point added to s is the Palm distribution of this Poisson process at s. The
converse is also true.
23
Theorem 3.1. (Poisson criterion, Slivnyak 1962) Let ? be a point process on a Borel
space S with reduced Palm distributions Q0s; s 2 S. Then ? is a Poisson process ifi
Q0s = L(?) for E?-a.e. s.
Theorem 3.1 is a simple special case of the following Theorem 3.2 that completely
characterizes the exchangeable point processes through reduced Palm measures, and
also gives the explicit formula for the reduced Palm distributions of such exchange-
able simple point processes. There are several difierent approaches of the proof in
literature, but here we may take the approach in [17], Section 2.7. In Section 4.1 we
also add a few more details to that proof.
Recall from Lemma 2.5 that the only exchangeable simple point processes onR+
are mixed Poisson, whereas the only exchangeable simple point processes on [0;1] are
mixed binomial processes. However this is not true for point processes.
Consider a mixed Poisson process ? on S directed by ?? for some random variable
? ? 0 and  -flnite measure ? on a measurable space (S;S). Letting ?(t) = Ee?t?
denote the Laplace transform of ?, we have
P f?B = 0g = Ee???B = ?(?B); B 2S.
Next let the measure ? on S be such that 0 < ?S < 1: If ? is a mixed binomial
process based on the probability measure ?=?S and a Z+-valued random variable ?,
24
then the following formula holds.
P f?B = 0g = E
 
1? ?B?S
??
= ?(?B),
where ?(t) = E[1?(t=?S)]?. The following result was obtained by Papangelou
(1974), Kallenberg (1975) ([17], section 2.7).
Theorem 3.2. (reduced Palm measures) Let ? be a point process on a Borel space
S with reduced Palm measures Q0s, s 2 S. Then, ? is a mixed Poisson or binomial
process ifi Q0s are independent of s. Moreover, L(?) ? P ? ??1 = M(?;?) ifi Q0s =
M(?;??0) for ?-a.e. s.
A random measure ? on S may be decomposed into difiuse and atomic compo-
nents. Deflne ? 2 S1(?;fi;fl) if ? is symmetrically distributed associated with the
positive bounded measure ? on S. Then the difiuse component equals ?d = fi?=?S,
and the atomic component of ? is given byPj flj??j, where the flj are atom sizes at po-
sitions ?j, where the ?j are i.i.d. with the common distribution ?=?S and fl ?Pj ?flj.
Note that since ?d=fi = ?=?S, ? is difiuse when fi > 0.
Similarly, deflne ? 2 S1(?;fi;?) if ? is ?-symmetric and has conditionally inde-
pendent increments directed by the pair (fi;?), where the difiuse component of ? is
?d = fi? and atoms are given by a Cox process ????fi on S?(0;1) directed by ???.
The following Palm measure invariance result was extended to the general random
measures by Kallenberg ([13], 1975).
25
Theorem 3.3. (invariant Palm measures) Let ? be a random measure on a Borel
space S with supporting measure ? and associated reduced Palm measures Q0s. Then
Q0s = Q0 is a.e. independent of s ifi ? is ?-symmetric, in which case even the reduced
Palm measure Q0 is ?-symmetric, and ? is difiuse unless ? is a.s. degenerate.
A detailed proof of this result appears in [17], Section 2.7. See also Section 4.2
below.
3.3 Main Result
We are now ready to present the main result of this dissertation. Theorem 3.3
suggests the following result for characterizing the exchangeable marked point process
by using Palm measure invariance. Here ? denotes the Lebesgue measure, and the
Palm measures Qs;t for marked point processes are deflned on page 12.
Theorem 3.4. Let ? be a marked point process on Borel space S ?I, where I =R+
or [0;1]. Assume E? = ? ?? for some  -flnite measure ? on S. The following two
conditions are equivalent:
(i) ? is exchangeable.
(ii) Q0s;t has a version that is independent of t.
Note that the flrst statement of Theorem 3.2 is the special case where the mark
space S is replaced by a single point. The following flgure is displayed here for
elaborating the connection between Theorem 3.3 and our main result.
26
The atomic part of a random measure may be coded by a point process in a
product space S ? I, where the flrst component gives the location, and the second
component gives the size of an atom, which make it seemingly reasonable to give an
invariance result for Palm measures of an exchangeable marked point process. As the
above flgure shows, for a random measure without the difiuse component, we may
think of its atom sizes as the marks in the mark space if this random measure turns
out to be a marked point process. The atom positions of such a random measure
play a role similar to the times of marks on time scale I in the present case. The
flrst statement of Theorem 3.3 shows that the exchangeable random measure has
associated reduced Palm measures that are invariant only with respect to the atom
positions, which is the motivation behind Theorem 3.4. It naturally gives rise to
the extension of these results for random measures to marked point processes. The
intuition from Theorem 3.3 suggests that there is a strong connection between the
27
exchangeability of a marked point process and the invariance of its associated reduced
Palm measures with respect to the times of marks. However, the deflnitions of Palm
measures and reduced Palm measures for random measures are difierent from those
for marked point processes since, unlike random measures, the Palm measures depend
on both mark space S and time scale I in the case of marked point processes. This
major difierence explains why we need to impose some further conditions, such as
requiring E? to admit the stated factorization. Recall from Lemma 2.5 that a marked
point process on S ? [0;1] is exchangeable if and only if it is a ?-randomization of
point process fl on the mark space. Apparently this suggests the need of factoring
the intensity measure of an exchangeable marked point process on S ? [0;1] into a
product of measures on the mark space and the time scale.
28
Chapter 4
Proofs of Previously Known Results
In this chapter, we show some technical proofs of the classical results shown in
section 3.3.2 basically following the ideas of proofs given by Kallenberg ([17], section
2.7). The main reasons for including the following technical proofs are to make this
dissertation self-contained and to explain some basic ideas that will be used again in
the proofs of the main results for this dissertation. Meanwhile, the proofs presented in
this chapter provide more details that will be helpful in the next chapter for proving
the main result, and also unify notations used later on.
4.1 Proof of Theorem 3.2
Here we may state Theorem 3.2 again to save time of turning pages back and
forth for readers.
Theorem3.2 Let ? be a point process on a Borel space S with reduced Palm mea-
sures Q0s, s 2 S. Then, ? is a mixed Poisson or binomial process ifi Q0s are inde-
pendent of s. Moreover, L(?) ? P ???1 = M(?;?) ifi Q0s = M(?;??0) for ?-a.e.
s.
29
First we need the following two auxiliary lemmas to prove Proposition 4.3 that
is important for proving equivalence between two equations in this Theorem. For ele-
mentary proofs of those lemmas and references, see Lemmas 12.2 and 12.4 in Kallen-
berg ([16], 2002). We start with the Lemma that characterizes Poisson processes by
their unique Laplace functionals.
Lemma 4.1. ? is Poisson with intensity measure E? = ? ifi it has Laplace functional
Ee??f = exp'??(1?e?f)?; f ? 0 measurable.
Next Lemma provides the Laplace functional for a mixed binomial process.
Lemma 4.2. If ? is a mixed binomial process based on the probability measure ?=?S
with 0 < ?S < 1 and an integer-valued random variable ?, then ? has Laplace
functional
Ee??f = E??e?f=?S?? ; f ? 0 measurable.
The following proposition gives the Laplace functional for an exchangeable simple
point process using the function ? deflned in Section 3.2. (refer to [17], section 2.7)
Proposition 4.3. A point process ? is either mixed Poisson or mixed binomial ifi it
has Laplace functional
Ee??f = ????1?e?f??; f ? 0 measurable, (4.1)
30
where ?, ? in both mixed Poisson and mixed binomial cases are deflned as in Section
3.2. Hence, by the uniqueness of Laplace functional, we may write L(?) ? P ???1 =
M(?;?), a function of ? and ?.
Proof of Proposition 4.3: When ? is a mixed Poisson process directed by ?? for
some random variable ? ? 0 and  -flnite measure ? on the measurable space (S;S),
then by Lemma 4.1, we have
Ee??f = E?E?e??fj??? = E?E?exp'???(1?e?f)?j??? = ???(1?e?f)? .
If ? is a mixed binomial process based on the probability measure ?=?S with
0 < ?S < 1 and a random variable ?; then by Lemma 4.2, we obtain
Ee??f = E??e?f=?S?? = E?1??(1?e?f)=?S?? = ???(1?e?f)? .
The distribution of the point process ? is uniquely determined by the Laplace
functional as in (4.1), so we may write the distribution of ? as a function of ?;?, in
other words, L(?) = M(?;?) for some function M. ?
Now we are ready to prove Theorem 3.2. This is a more detailed version of the
proof in Kallenberg (2005) (refer to [17], section 2.7).
Proof of Theorem 3.2: Assume that ? is a mixed Poisson or binomial process.
Proposition 4.3 shows that L(?) = M(?;?) for some function M, in which case
Ee??f = ????1?e?f?? for any measurable f ? 0. If ? is a mixed Poisson process
31
directed by directed ?? for some random variable ? ? 0 and  -flnite measure ? on
the measurable space (S;S), then ? is a supporting measure of ? for ? ? E? as an
observation from E? = E???.
Fix a measurable function f ? 0 on S with ?f > 0 and a set B 2 S with
?B < 1. Then
Ee??f?t?B = ????1?e?f?t1B??; t ? 0.
Taking derivatives with respect to t on both sides, and by dominated convergence
theorem together with the equation
d
dt
Z
B
g(s;t)?(ds) =
Z
B
@g(s;t)
@t ?(ds) if jgj? 2;?B < 1, (4.2)
we have
E???B ?e??f?t?B? = ?0???1?e?f?t1B?????1Be?f?t1B?.
Let t = 0, then
E??B ?e??f? = ??0???1?e?f?????1Be?f?:
Let Qs denote the Palm measures of ? corresponding to the supporting measure
?. Note that
E??B ?e??f? = E
Z
B
e??f?(ds) = C(1Be??f) =
Z
B
?(ds)
Z
e??fQs(d?).
32
Therefore, we get
Z
e??fQs(d?)
Z
B
?(ds) = ??0???1?e?f???
Z
B
e?f(s)?(ds).
For any ?-a.e. s 2 S, we have
Z
e??fQs(d?) = ??0???1?e?f???e?f(s),
because B is arbitrary, which implies
Z
e??fef(s)Qs(d?) =
Z
e?(???s)fQs(d?)
=
Z
e??fQ0s(d?)
= ??0???1?e?f??,
where R e??fQ0s(d?) is the Laplace functional of ? corresponding to the reduced Palm
measures. To extend the result to an arbitrary measurable function f ? 0, we simply
take non-negative measurable functions 0 ? fn " f. Note that ?e??fn "?e??f, hence
?0???1?e?fn??" ?0???1?e?f??
by monotone convergence theorem.
33
Comparing the following equation
Z
e??fQ0s(d?) = ??0???1?e?f??
with (4.1), we conclude that Q0s = M(?;??0) for ?-a.e. s. In particular, Q0s = Q0 is
independent of s 2 S a.e. ?.
Let us now prove the theorem in the converse direction. Choose the support-
ing measure ? and the associated reduced Palm measures Q0s such that Q0s = Q0 is
independent of s 2 S.
First, assume that ?S is bounded a.s. and Pf? 6= 0g > 0. Introduce a random
element ? in S satisfying
P[? 2?j?] = ?=?S a.s. on f?S > 0g. (4.3)
By the deflnition of Palm measures, for an arbitrary set B 2S, we have
P [? 2 Bj?S = n] = E
??B
?S
flfl
fl?S = n
?
(4.4)
= E[?B;?S = n]nPf?S = ng =
R
B Qsf?S = ng?(ds)
nPf?S = ng (4.5)
=
R
B Qsf(???s)S = n?1g?(ds)
nPf?S = ng
=
R
B Q
0
sf?S = n?1g?(ds)
nPf?S = ng
= Q
0f?S = n?1g?B
nPf?S = ng .
34
In particular, taking B = S implies 0 < ?S < 1 since ?S is a.s. bounded by
assumption.
If Pf?S = ng > 0 for some n 2Z+, then by Proposition 2.3
P [? ??? 2 Bj?S = n;? 2 ds]
= E[1f?S = ng?1B(? ???)j? 2 ds]=Pf?S = ng
=
Z
1f?S = ng?1B(???s)Qs(d?)=Pf?S = ng
= Qs[???s 2 Bj?S = n],
which implies ????S(? ???) on the set f?S > 0g by the invariance of reduced Palm
measures. Since (4.4) shows that P [? 2 Bj?S = n] is proportional to ?B for a flxed
n 2N, we get that
P[? 2 Bj?S;? ???] = P[? 2 Bj?S] = ?B=?S a.s. on f?S > 0g. (4.6)
Since ? is a point process and S is Borel, we may write ? = Pj?? ? j, where
 1;:::; ? are listed by a suitable ordering and ? ? ?S: In order to generate an ex-
changeable sequence ?1;:::;?? by permuting the sequence  1;:::; ?, we may introduce
a sequence of independent integer-valued random variables ?1;:::;?? with (?i)?? (?;?)
and 1 ? ?i ? i. The distributions of ?i are given by
Pf?i = jg = j?1; 1 ? j ? i, j 2Z+, 1 ? i ? ?. (4.7)
35
Deflne ?1 ?  ??, and rearrange the remains f 1;:::; ?gnf ??g by the same or-
dering as we used for ordering the set f 1;:::; ?g. Pick the ?k?1-th element as ?2.
Continue to pick the ?i recursively until the whole sequence ?1;:::;?? is constructed.
It is easy to see that the sequence ?1;:::;?? is exchangeable as ? = ?S is given (a short
proof of this conclusion is given on page 49), which shows that (4.3) follows with ?
replaced by ?1.
Note that ? = Pj?? ? j = Pj?? ??j: Then by (4.6) and independence of (?i), we
get
P[?1 2 Bj?;?2;:::;??] = ?B=?S a.s. on f? > 0g, (4.8)
which extends by the exchangeability of the sequence ?1;:::;?? for a given ? to
P[?i 2 Bj?;f?1;:::;??gnf?ig] = ?B=?S a.s. on f? > 0g;1 ? i ? ?. (4.9)
Equation (4.9) shows that the random elements (?i) are i.i.d. with distribution ?=?S
a.s. onf?S > 0g, which means that ? = Pj?? ??j is a mixed binomial process based on
the probability measure ?=?S: It follows by Proposition 4.3 that L(?) = M(?;?) for
some completely monotone function ?(t) = E[1?(t=?S)]? on the bounded interval
[0;?S].
Next, we consider the case when ?S is unbounded by borrowing the result ob-
tained from the previous argument for the bounded case. Choose some sequence of
36
sets Bn " S with ?Bn < 1 a.s. for every n. By the previous argument, we see that
L(1Bn?) = M(1Bn?;?n); n 2N
for some completely monotone functions ?n(t) = E[1?(t=?S)]?Bn on bounded inter-
vals [0;?Bn]. Fix an arbitrary measurable function f ? 0 on S and put fn ? 1Bnf.
Then fn " f. Note that
Ee??fn = ?(?(1?e?fn)), n 2N,
where ? on [0;?S) is obtained by the uniqueness of ?n on the bounded intervals
[0;?Bn]. Therefore, ? is unique since the extension of ?n on the increasing intervals
[0;?Bn] to [0;?S) is unique.
Note that e??fn and e??f are bounded by 1, and ? is decreasing and continuous,
we get
Ee??f = limn!1Ee??fn = limn!1?(?(1?e?fn)) = ?(?(1?e?f)),
where the flrst convergence is by dominated convergence and the second convergence
is due to monotone convergence and the continuity of ?:
Summarizing the previous arguments together with Theorem 12.4 in [16] (2002),
we conclude that ? is a mixed binomial process when ?S is a.s. bounded or a mixed
Poisson process when ?S is a.s. unbounded. ?
37
4.2 Proof of Theorem 3.3
For convenience, we may again state Theorem 3.3 in the previous chapter.
Theorem3.3 Let ? be a random measure on a Borel space S with supporting mea-
sure ? and associated reduced Palm measures Q0s. Then Q0s = Q0 is a.e. independent
of s ifi ? is ?-symmetric, in which case even the reduced Palm measure Q0 is ?-
symmetric, and ? is difiuse unless ? is a.s. degenerate.
The following lemmas are required to reduce the proof of Theorem 3.3.
Lemma 4.4. Let ? be a random measure on a Borel space S with 0 < ?S < 1 a.s.
If the associated reduced Palm measures Q0s are a.e. independent of s, then for any
flxed x > 0, the measures
Qs[(?fsg;???fsg) 2?j?S 2 dx] (4.10)
are also a.e. independent of s.
Proof: Fix any measurable set B ? S ?M(S) and C 2B(R+), then
Qs[(?fsg;???fsg) 2 B;?S 2 C] = Q0s[(y;?) 2 B;?S +y 2 C]. (4.11)
We note that the right-hand side of (4.11) are a.e. independent of s since reduced
Palm measures Q0s are a.e. independent of s. Write (4.10) as the Radon-Nikod?ym
38
derivatives, i.e.
Qs [(?fsg;???fsg) 2?j?S 2 dx] = Qs[(?fsg;???fsg) 2?;?S 2 dx]Q
sf?S 2 dxg
. (4.12)
Then from (4.11) we see that (4.10) are a.e. independent of s if we take C = fxg and
B = S ?M(S). ?
Next proposition shows that some properties of a random measure ? such as
exchangeability hold locally so as to reduce the proof of Theorem 3.3 by assuming
the total mass of the random measure ? to be flnite.
Proposition 4.5. Let ? be a random measure on a Borel space S with  -flnite in-
tensity measure E? and Palm distributions Qs. If Bn " S, then
(i) ? is exchangeable ifi 1Bn? is exchangeable for every n 2N.
(ii) limn!1Q(n)s = Qs for a.e. s, where Q(n)s are the Palm distribution associated with
the random measure 1Bn?.
Proof: The proof is very similar to that of Lemma 5.6. Readers may refer to
page 64 of this dissertation for details. ?
Let us now prove Theorem 3.3 by some reduction.
Proof of Theorem 3.3: To prove the flrst assertion, it su?ces to just consider the
restrictions of ? to the sets B 2S with ?B < 1 a.s. by Proposition 4.5. So, we may
henceforth assume ?S < 1 a.s.
39
Assume that Q0s = Q0 are a.e. independent of s. Let ? be the random variable
with conditional distribution
P[? 2?j?] = ?=?S on f?S > 0g, (4.13)
and set (?;?) ? (?f?g;? ??f?g??). By Proposition 2.3, we get that any x > 0 and
s 2 S
P [(?;?) 2?j?S 2 dx;? 2 ds] (4.14)
= P[(?f?g;? ??f?g??) 2?j?S 2 dx;? 2 ds] (4.15)
= Qs[(?fsg;???fsg?s) 2?j?S 2 dx].
Since Q0s = Q0 are a.e. independent of s, the right-hand side of (4.14) is independent
of s by Lemma 4.4, which gives the conditional independence
????S(?;?) on f?S > 0g. (4.16)
Therefore, we have
P [? 2?j?S;?;?] = P[? 2?j?S] = E[?=?Sj?S] (4.17)
= E[?j?S]=?S a.s. on f?S > 0g.
40
Recall that the atom sizes flj of random measure ? can be given by the point process
fl = Pj ?flj. Also note that fl is a measurable function of (?;?). So, combining the
following equation
P[? 2?j?;fl] = P[? 2?j?]
with
P[? 2?j?S;?;?;fl] = P[? 2?j?S],
we may henceforth assume fl to be non-random. A detailed proof of the su?ciency
of similar reduction is given on page 51.
Let ? be the supporting measure of ?. To see that ? is ?-symmetric, we may
note that for any B 2S and A 2B(R+),
E[?B;?S 2 A] =
Z
B
Qsf?S 2 Ag?(ds)
=
Z
B
Q0sf(x;?);x+?S 2 Ag?(ds)
= Q0f(x;?);x+?S 2 Ag?B
by the assumption that Q0s = Q0 are a.e. independent of s 2 S. In particular, taking
B = S. gives
E[?B;?S 2 A]=E[?S;?S 2 A] = ?B=?S.
So,
E[E[?Bj?S];?S 2 A] = E[?B;?S 2 A] = E[?S;?S 2 A]?B=?S (4.18)
41
shows that E[?Bj?S] = ?S ? ?B=?S a.s. since A is arbitrary. Comparing it with
(4.17), we get
P[? 2?j?S;?f?g;? ??f?g??] = P[? 2?j?S] = ?=?S a.s. on f?S > 0g. (4.19)
By the same means of generating a random permutation as the previous proof on page
35 shows, positions of the atoms are enumerated in exchangeable order when atom
sizes at those positions are the same, therefore, we obtain a sequence of positions
?1,?2,... of the atoms of sizes fl1,fl2,...
Let n be the number of distinct atom sizes and write fli as the i-th smallest atom
size. Construct a random variable $ independent of ?, ? such that
P ($ = i) = fl
i ?fl(fli)
?S .
Write j$ = minfj : ?f?jg = fl$g. Deflne ? ? ?j$. Let us now show that such ? sat-
isfles the condition P[? 2 ?j?] = ?=?S a.s. on f?S > 0g by the following calculations.
By the exchangeability of atom positions corresponding to the same atom size, we
42
have a.s.
P [? 2 A;$ = ij?] = 1fl(fli)
X
j2fk:?f?kg=flig
P [?j 2 A;$ = ij?]
= 1fl(fli)
X
j2fk:?f?kg=flig
E[1f?j 2 Ag?1f$ = igj?]
= 1fl(fli)
X
j2fk:?f?kg=flig
1f?j 2 Ag?P[$ = ij?]
= 1fl(fli)
X
j2fk:?f?kg=flig
1f?j 2 Ag?Pf$ = ig
=
X
j2fk:?f?kg=flig
fli ?1f?j 2 Ag=?S,
hence,
P [? 2 Aj?] =
nX
i=1
P [? 2 A;$ = ij?]
=
nX
i=1
X
j2fk:?f?kg=flig
fli ?1f?j 2 Ag=?S
= ?A=?S,
which shows that the constructed ? satisfles the condition P[? 2 ?j?] = ?=?S a.s. on
f?S > 0g.
Onthesetf? = flkg, (4.19)togetherwiththeexchangeabilityoff?i : ?f?ig = flk;i 2Ng
implies
P [?k 2?j?f?kg;? ??f?kg??k;?j;j 6= k] = ?=?S a.s.
43
by a similar argument as in the previous proof on page 35, which shows that ?1,?2,...
are i.i.d. with distribution ?=?S. It proves that ? is ?-symmetric. Here, ? is difiuse
unless ? is a.s. degenerate, since otherwise, say ?ftg > 0 for some t, then we have
P (?1 = ?2) > P(?1 = t;?2 = t) = (?ftg)2 > 0,
a contradiction to the fact that the ?i are distinct.
Conversely, suppose that ? is ?-symmetric, and ? is difiuse unless ? is a.s. de-
generate. Again by Proposition 4.5, it is enough to consider the restrictions of ? on
the sets B 2 S with 0 < ?B < 1, so we may assume that ? 2 S1(?;fi;fl) with
0 < ?S < 1. As before, we may also assume that fl = Pi ?fli and fi are non-random.
Note that the ?-symmetry implies E[?j?S] = (?S)??=?S a.s. Fix an arbitrary mea-
surable set B 2S, by calculations similar to (4.18) we obtain that
Z
B
Qsf?S 2?g?(ds)
= E[?B;?S 2?] = E[E[?Bj?S];?S 2?]
= E[?S;?S 2?]?B=?S,
which shows that for a.e. s 2 S,
Qsf?S 2?g = E[?S;?S 2?]=?S. (4.20)
Hence, Qsf?S 2?g are a.e. independent of s.
44
Next, we may construct a random variable ? satisfying P[? 2?j?] = ?=?S a.s. and
????S(?;?) so that (4.14) holds and gives the invariance property for Palm measures,
here ? ? ?f?g and ? ? ? ??f?g?? as before.
By the assumption that ? is ?-symmetric, ? has an a.s. representation
? = fi ??S +
X
j
flj??j, (4.21)
for some i.i.d. random elements ?1,?2,... with distribution ?=?S and independent of
fi,fl1,fl2,....
Choose ? to be independent of ?1,?2,... with distribution
P f? = 0g = fi?S,
and
P f? = flkg = fl(flk)?flk?S , k ? 1.
Let jk = minfj : ?f?jg = flkg for k ? 1. Choose ? = ?jk if ? = flk as k ? 1, and
set ? to be an independent random variable with distribution ?=?S if ? = 0. Let
?jk;1,?jk;2,... denote the positions of atoms with jk ? jk;1 ? jk;2 ???? such that
?f?jk;1g = ?f?jk;2g = ??? = flk:
45
For any measurable A 2S and k ? 1, letting k0 = jk for convenience, we have
P [? 2 A;? = flkj?]
= P [? 2 A;? = flk0j?] = P
h
?jk0;1 2 A;? = flk0j?
i
= P
h
?jk0;2 2 A;? = flk0j?
i
= ??????
= 1fl(fl
k)
X
n=1
P
h
?jk0;n 2 A;? = flk0j?
i
= 1fl(fl
k)
X
n=1
1f?jk0;n 2 AgP [? = flk0j?]
= flk?S
X
n=1
1f?jk0;n 2 Ag
= flk?S
X
i=1
1ffli = flkg1f?i 2 Ag
by ?-symmetry of ?.
The calculations for verifying P[? 2 ?j?] = ?=?S a.s. are as follows: For any
measurable A 2S, by ?-symmetry of ? we have
P [? 2 Aj?] = P [? 2 A;? = 0j?]+
X
j=1
P ?? 2 A;? = fljj??
= fi?S ?A?S +
X
j=1
flj
?S
X
i=1
1f?i 2 Ag1ffli = fljg
= 1?S
"
fi?A?S +
X
k=1
flk ???k(A)
#
= ?A=?S.
Note that ? = [k=1(? = flk)[(? = 0) and recall that ?S is flxed at the beginning
of the proof. Then, on the set (? = flk) for some k ? 1, by the independence of ? and
46
?1,?2,... we have
P [? 2?j?;?] = P [? 2?j?;? = flk]
= P [?jk 2?j?i;i 6= jk;? = flk]
= P f?jk 2?g = ?=?S a.s.
Similarly, since ? is deflned to be independent of ? with distribution ?=?S and ? = ?
on the set (? = 0), we have
P [? 2?j?;?] = P [? 2?j?;? = 0] = P [? 2?j? = 0] = ?=?S a.s.
on the set (? = 0). Therefore, ????S(?;?) with ? satisfying (4.13), so (4.14) shows
that the conditional distributions
Qs[(?fsg;???fsg?s)j?S], s 2 S (4.22)
are a.e. independent of s. Combining it with (4.20), we have Q0s = Q0 are a.e.
independent of s.
The ? constructed above is assumed to be independent of ?, so on the set (? = 0)
we have
P [? 2?j?] = P[? 2?j? = 0] = P f? 2?g. (4.23)
47
Similarly, on the set (? = flk) for some k ? 1, we get
P [? 2?j?] = P[? ?flk?jk 2?j? = flk] = P f? ?flk?jk 2?g. (4.24)
Note that the ?-symmetry of ? also implies that ? ? flk?jk is ?-symmetric because
of (4.21), the representation of ?-symmetric random measures. This together with
(4.23) and (4.24) implies ? is conditionally ?-symmetric given ?, hence (?;?) is also
?-symmetric. Then we see from (4.14) that the conditional distributions in (4.22) are
also ?-symmetric. This together with (4.20) implies that Q0 is ?-symmetric. ?
48
Chapter 5
Proof of Main Theorem
5.1 Some Auxiliary Results
The most important results for proving Theorem 3.4 are Propositions 5.4 and
5.5. In Proposition 5.4 we will establish an equivalence result for connecting the
exchangeability of a marked point process on S?[0;1] with some independence result
by introducing a so-called average random variable ? in [0;1]. One assumption of
Proposition 5.5 is the average property of this ? introduced in Proposition 5.4. This
equivalence result together with Proposition 5.5 are very handy when dealing with the
Palm distributions through a simple calculation of regular conditional probabilities.
The following Lemma provides a construction of an exchangeable sequence of
random variables by picking integers 1;:::;n one by one at random. This exchangeable
sequence is used in Proposition 5.4 to generate an exchangeable sequence of times with
the same mark for a marked point process.
Lemma 5.1. Let ?1;:::;?n be independent random elements such that P(?i = j) =
i?1, where 1 ? j ? i ? n. Deflne ?1 ? ?n,, and let ?2 be the ?n?1-st smallest integer
in the remaining set f1;...,ngnf?1g. Continue recursively to construct the sequence
(?1;:::;?n). Then (?1;:::;?n) is an exchangeable permutation of (1;:::;n).
Proof: We use induction. Fix a positive integer k ? n ? 1 and assume that
the sequence (?1;:::;?k) is exchangeable. For any distinct integers a1;:::;ak 2 [1;n],
49
2 ? k ? n ? 1, we deflne A ? f! : (?1(!) = a1;:::;?k(!) = ak)g ? ?. Note that
?1(A),...,?k(A) are flxed integers due to the construction of the sequence (?1;:::;?k),
i.e. (?1 = a1;:::;?k = ak) is equivalent to B ? (?1 = n1;:::;?k = nk) for some integers
n1,...,nk.
Deflne b1,...,bn?k to be the integers in the set f1;:::;ngnfa1;:::;akg listed in in-
creasing order. Since ?k+1 is independent of random elements ?1;...,?k, for any positive
integer i ? n?k, we see from the exchangeability of the sequence (?1;:::;?n) that
P [?k+1 = bi;?1 = a1;:::;?k = ak] = P [?k+1 = bijA]P(A)
= P [?k+1 = ijB] (n?k)!n! = P(?k+1 = i)(n?k)!n! = [n?(k +1)]!n! ,
It follows by induction that (?1;:::;?n) is an exchangeable sequence. ?
We continue with an elementary result about the existence of probability kernels.
Lemma 5.2. Let ? be a random element in a Borel space (S;S), and fl be a random
element in a Borel space T. Then there exists a regular conditional distribution of ?
with respect to some  -fleld F generated by fl. Furthermore, if ? is such a probability
kernel from T to S satisfying ?(fl;?) = P[? 2 ?jfl] a.s., then ??f?1 is also a regular
conditional distribution of f(?) with respect to F for any measurable function f : S !
S.
50
Proof: The existence of ? is clear from Theorem 6.3 in [16]. For any measurable
set A 2S, we have
P[f(?) 2 Ajfl] = P[? 2 f?1(A)jfl] = ?fl(f?1A) a.s.
This completes the proof. ?
If two random elements ?, fl are related by fl = h(?) for some measurable function
h, then by saying fl is invariant under f we mean h = h?f for some transformation
function f on ?. The following Lemma 5.3 helps reduce the proof of Proposition 5.4.
Lemma 5.3. Fix two measurable Borel spaces (S;S) and (T;T ). Let ? be a random
measure on S, and ? be random elements in T. Let fl be a ?-measurable random
element in a Borel space (U;U). Let ? be a probability kernel satisfying ?fl [(?;?) 2?] =
P[(?;?) 2 ?jfl] a.s. Let f be a measurable transformation on S such that fl remains
invariant under f. Then,
(i) Fix two measurable sets A 2S, B 2T . For an arbitrary L(fl)-a.e. b, we have
?b [? 2 B;? 2 A] = E
h~
?B;? 2 Ajfl = b
i
a.s.,
where the random variable ~?B = P [? 2 Bj?] a.s.
(ii) ? and ? are conditionally independent given fl ifi
?b ?(?;?)?1 = ??b ???1??(?b ???1)
51
for an arbitraryL(fl)-a.e. b, where ?b?(?;?)?1 denotes the joint probability distribution
of (?;?) under probability measure ?b.
(iii) f is L(?)-preserving ifi f is ?b ???1-preserving for L(fl)-a.e. b.
Proof: (i) Since fl is ?-measurable, by the assumption P [? 2 Bj?] = ~?B a.s., we
have, for any measurable C 2U,
Z
b2C
?b [? 2 B;? 2 A]P ffl 2 dbg
E[?fl [? 2 B;? 2 A];fl 2 C]
= P [? 2 B;? 2 A;fl 2 C]
= E
h~
?B;? 2 A;fl 2 C
i
=
Z
b2C
E
h~
?B;? 2 Ajfl = b
i
P ffl 2 dbg
=
Z
b2C
E?b
h~
?B;? 2 A
i
P ffl 2 dbg,
which shows that, for an L(fl)-a.e. b,
?b [? 2 B;? 2 A] = E
h~
?B;? 2 Ajfl = b
i
.
(ii) Assume that ? and ? are conditionally independent given fl, then we have
?fl f(?;?) 2 A?Bg = P[(?;?) 2 A?Bjfl] = ?fl f? 2 Ag??fl f? 2 Bg a.s.
52
for any measurable sets A 2 S, B 2 T . Since S and T are Borel, there exits a
measure-determing class fAi ?Bjg1i=1 of (S ?T;S ?T ) satisfying
P
1\
i;j=1
[?fl f(?;?) 2 Ai ?Bjg = ?fl f? 2 Aig??fl f? 2 Bjg] = 1. (5.1)
Write C = fb 2 U;?b?(?;?)?1 = (?b ???1)?(?b???1)g, and note that P(fl 2 C) = 1
from (5.1). Then, for L(fl)-a.e. b, we have
?b ?(?;?)?1 = ??b ???1??(?b ???1): (5.2)
Conversely, if (5.2) holds for L(fl)-a.e. b, then
?b ?(?;?)?1 = ??b ???1??(?b ???1) a.s.
which shows that ? and ? are conditionally independent given fl.
(iii) Assume f isL(?)-preserving. Write fl = h(?) since fl is a measurable function
of ?. Since fl is invariant under f, we get
? d= f(?) ) (?;h(?)) d= (f(?);h(f(?)) = (f(?);h(?)),
and it implies
P[? 2 Ajfl] = P[f(?) 2 Aj(h?f)(?)] = P ?? 2 f?1Ajfl? a.s., (5.3)
53
Therefore, f is ?b???1-preserving for L(fl)-a.e. b by a measure-determing class argu-
ment as in the proof of (ii).
Conversely, if f is ?b???1-preserving for L(fl)-a.e. b, then (5.3) holds by reversing
the previous argument. Hence, f is L(?)-preserving by taking expectation on both
sides of (5.3). ?
5.2 Exchangeability of MPP
Let us flrst state the following Proposition that plays a crucial role in the proof
of our main result. In particular, it provides a condition equivalent to exchangeability
for marked point processes on S ?[0;1] in terms of a pair of random elements. This
pair connects regular conditional probabilities with conditional Palm distributions in
the sense of Proposition 5.5.
Proposition 5.4. Let ? be a marked point process on a Borel space S ? I with
?(S ?I) < 1 a.s., where I = [0;1]. Let ( ;?) be a random element in S ?I such
that
P[( ;?) 2?j?] = ?=?(S ?I) a.s. on f? 6= 0g. (5.4)
Then, ? is ?-symmetric ifi ? ???6=0 ( ;? ?? ;?) with distribution ?.
To make this long proof easier to read, we prove the proposition in three steps.
The proof is organized as follows:
Proof: Step 1: Here we consider the special case when fl is non-random and
prove that ?-symmetry of ? implies ? ???6=0 ( ;? ?? ;?).
54
We may assume that fl = Pj?N ?flj. Here N = fl(S) is a constant since fl is
non-random. Since f? 6= 0g = fN > 0g, and ?(S ?I) < 1 a.s. by assumption, we
henceforth assume that N > 0 and ? is bounded. Recall that a ?-symmetric ? has
an a.s. representation ? = Pj?N ?flj;?j for some i.i.d. random variables ?1;?2;::: with
distribution ?.
To construct a pair ( 0;?0) satisfying the condition
P[( 0;?0) 2?j?] = ?=?(S ?I) a.s.,
we choose a random variable  0 independent of ?1,?2,... with distribution fl=N. Note
that ? = Pi ?fli;?i is a function of (?1;?2;:::) since the fli are non-random. Then  0???.
Deflne i 0;j to be the j-th smallest number in the index set fi ? 1;fli =  0g. Simi-
larly, let ik;j be the j-th smallest number in the set fi ? 1;fli = flkg. For convenience,
write ?ik;1 ? ?k;1 and mk ? fl(flk). Deflne ?0 ? ? 0;1.
The exchangeability of ?1;?2;::: implies
(?p1;:::;?pN) d= (?1;:::;?N), (5.5)
for any permutation (p1;:::;pN) of 1;:::;N. For any sets A 2 [0;1]N ; B ? N(S) and
C ?N(S ?I), since the fli are non-random, (5.5) implies
P
n
(?p1;:::;?pN) 2 A;
X
1?i?N??i 2 B
o
= P
n
(?1;:::;?N) 2 A;
X
1?i?N??i 2 B
o
.
55
Therefore,
P[(?p1;:::;?pN) 2 A;? 2 C]
= P[(?p1;:::;?pN) 2 A;
X
1?i?N?flpi;?pi 2 C]
= P[(?1;:::;?n) 2 A;
X
1?i?N?fli;?i 2 C]
= P[(?1;:::;?n) 2 A;? 2 C],
which gives
P[(?p1;:::;?pN) 2?j?] = P[(?1;:::;?N) 2?j?] a.s.
Thus, for any positive integer k ? N and an arbitrary interval A ? [0;1], we have
P[?k;1 2 Aj?] = P[?k;2 2 Aj?].
Note that  0??? implies  0?? (?k;1;?k;2). Therefore, for any C ?N(S ?I), we get
P[?k;1 2 A; 0 = flkj?] = P[?k;2 2 A; 0 = flkj?] a.s.
56
Hence, we get
P[?0 2 A; 0 = flkj?]
= P[?k;1 2 A; 0 = flkj?]
= P[?k;2 2 A; 0 = flkj?] = ???
= P[?k;mk 2 A; 0 = flkj?]
= m?1k
X
i?mkP[?k;i 2 A; 
0 = flkj?]
= m?1k E
hX
i?mk1A(?k;i)?1f 
0 = flkg
flfl
fl?
i
= m?1k ?
X
i?mk1A(?k;i)?P [ 
0 = flkj?]
= m?1k ?
X
i?mk1A(?k;i)?Pf 
0 = flkg
= m?1k ??(fflkg?A)?mk=N = ?(fflkg?A)N .
Then for any measurable B ?C ? S ?I, we have
P[( 0;?0) 2 B ?Cj?] = ?(B ?C)=?(S ?I) a.s.
By a monotone-class argument, we get
P[( 0;?0) 2?j?] = ?=?(S ?I) a.s., (5.6)
which proves that the constructed pair ( 0;?0) satisfles condition (5.4).
57
By the exchangeability of the i.i.d. sequence (?k;1;?k;2;:::;?k;mk) and the condition
 0???k;1, for any measurable M ?M(S ?I), A ? I, we have
P f?0 2 Ag =
X
kP f 
0 = flk;?k;1 2 Ag = ?A,
and so
Pf? ?? 0;?0 2 M;?0 2 Ag
=
X
kP
'? ??
flk;?k;1 2 M;?k;1 2 A;
?
= Pf?1;1 2 Ag
X
kP
'? ??
flk;?k;1 2 M
?
= Pf? ?? 0;?0 2 MgPf?0 2 Ag,
where the flrst step is due to the fact that ???flk;?k;1 and ?k;1 are independent. Since
 0 is a measurable function of fl and ? ?? 0;?0, where fl is non-random, we have
P [?0 2?j 0;? ?? 0;?0] = P[?0 2?jfl;? ?? 0;?0]
= P[?0 2?j? ?? 0;?0]
= ?.
Therefore ?0??( 0;? ?? 0;?0) with distribution ?. Note that the joint distribution of
(?; 0;?0) is determined by (5.6). Likewise, (5.4) determines the joint distribution of
58
(?; ;?), and so (?; 0;?0) d= (?; ;?). Hence, the independence ???( ;? ? ? ;?) also
holds since the independence only depends on the joint distribution of (?; ;?).
Step 2: Here we prove the converse assertion. Assume that ???( ;??? ;?) with
distribution ? conditionally on f? 6= 0g. We may again assume that ? is bounded and
non-zero and fl is non-random as before.
Fix a positive integer k ? N. Let ?k1 < ??? < ?kmk be the times with a common
mark flk. Consider a sequence of independent random elements ?k1,..., ?kmk that are
independent of ? with distributions
Pf?ki = jg = i?1, 1 ? j ? i ? mk ; i;j 2N. (5.7)
Choose ?k;1 to be the ?kmk-th smallest time among ?k1;?k2;:::;?kmk. Then let ?k;2 be the
?kmk?1-st smallest time in the remaining set f?ki ;i 6= ?kmkg. Continue to construct the
sequence (?k;i)mki=1 recursively. By Lemma 5.1 the constructed sequence (?k;i)i?mk is
exchangeable. Construct sequences (?k;i) for all other k in the same way. Since the
space S ?I is Borel, we may write ? = Pi?N ? i;?i by choosing
 i = flk if
X
n?k?1mn +1 ? i ?
X
n?kmn,
and
?i = ?k;j if  i = flk and i = j +
X
n?k?1mn.
59
Deflne  0 to be a random variable independent of ? and all the random elements
(?kj) with distribution fl=N as before. Let ?0 ? ? 0;1. By computations similar to
those proving (5.6), we see that the pair ( 0;?0) satisfles the same relation. Write
?k ? ~? ? ?k;1 for convenience. Then ?k ? ~? and ?kmk are uniquely determined by each
other.
Fix an arbitrary measurable set A ? I and an integer k ? N. By Lemma 6.2
in Kallenberg [16], we see that the assumption ?0??( 0;? ?? 0;?0) with distribution ?
implies
?A = P[?0 2 Aj 0;? ?? 0;?0]
= P[?k ? ~? 2 Aj 0 = flk;? ??flk;?k?~?]
= P[?k ? ~? 2 Aj? ??flk;?k?~?] a.s. on ( 0 = flk),
where the last step is due to the fact that  0 is a measurable function of ? ? ? 0;?0
and the non-random fl. Note that (?i)i?Nn(?k ? ~?) and ??ij?n?kmk are enough to
determine ? ??flk;?k?~?, i.e., they contain complete information about ? ??flk;?k?~?, by
the construction of the sequences (?i) and (?ij). Therefore, by the independence of
the random elements ?ij, we have
? = P[?k ? ~? 2?j? ??flk;?k?~?]
= P[?k ? ~? 2?j(?i)i?Nn(?k ? ~?);??ij?n?kmk]
= P[?k;1 2?j(?i)n?k;1] a.s. on ( 0 = flk).
60
This, together with the exchangeability of the sequence (?k;i) with a common mark
position flk, implies that the times (?k;i)k;i are i.i.d. with the common distribution
? a.s. Hence, if any pair ( ;?) satisfles both (5.4) and ???( ;? ? ? ;?), then the
?-symmetry of ? follows. This completes the proof for the case of non-random fl.
Step 3: We now consider the case for a general fl. By Lemma 5.3 (i), we get
E?b (?=?S) = ?b [( ;?) 2?j?] a.s.
for L(fl)-a.e. b, where ? in Lemma 5.3 is replaced by the pair ( ;?). Combining
Lemma 5.2 with Lemma 5.3 (i)-(ii), we get ???( ;??? ;?) with distribution ? if and
only if
?b ?(?; ;? ?? ;?)?1 = ??b ???1????b ?( ;? ?? ;?)?1?.
for L(fl)-a.e. b. By Proposition 5.3 (iii), we see that ? is ?-symmetric ifi ?b ???1 is
invariant under an arbitrary transposition Tc;d for L(fl)-a.e. b, where c;d 2 Q+ \ I.
Therefore, for L(fl)-a.e. b, the assumption ???( ;? ? ? ;?) with distribution ?, the
?-symmetry of ?, and equation (5.4) all remain true simultaneously under probability
measures P and ?b. This shows that the proof for non-random fl, using ?b, is enough
to prove the general case by a change of probability measures. ?
In many cases, the direct calculations of Palm distributions can be di?cult. The
following result gives a way of using regular conditional probabilities to do calculations
involving Palm distributions.
61
Proposition 5.5. Let ? be a marked point process on a Borel space S?I with Palm
distribution Qs;t, where (s;t) 2 S ? I and ? = ?(S ? I) < 1 a.s. Let the random
element ( ;?) in S ?I have the conditional distribution
P[( ;?) 2?j?] = ?=?(S ?I) a.s. on f? > 0g.
Then for any measurable function f ? 0 , n 2N, we have a.s. on f? > 0g
E[f(?; ;?)j?(S ?I) = n] =
Z
f(?; ;?)Q ;?[d?j?(S ?I) = n] .
Proof: We may assume ?(S ? I) > 0 without loss of generality. Let ? ? E?
be the supporting measure associated with the probability kernel Q from S ? I to
N(S ?I). By Proposition 7.26 in FMP[16], the following formula
Qs;t [d?;?(S ?I) = n]
Qs;tf?(S ?I) = ng
62
has a version of probability kernel, say q, from (S ?I)?N to N(S ?I). Then, for
any non-negative product measurable function f, we have
E[f(?; ;?)] = EE[f(?; ;?)j?]
= E
Z
f(?;s;t)P[( ;?) 2 dsdtj?]
= E
Z
f(?;s;t)??(dsdt)=?(S ?I)
=
Z
?(dsdt)
Z
N(S?I)
f(?;s;t)?Qs;t(d?)=?(S ?I)
=
Z
?(dsdt)
X
k2N
k?1Qs;t [?(S ?I) = k]
Z
f(?;s;t)Qs;t [d?j?(S ?I) = k].
Writing rs;t(k) ? k?1Qs;t [?(S ?I) = k], qs;t;k(d?) ? Qs;t[d?j?(S ?I) = k], we get
Lf( ;?);?(S ?I);?g = ? ?rs;t ?qs;t;k,
and so,
Lf( ;?);?(S ?I)g = ? ?rs;t
since q is a probability kernel. Thus, it follows that
E[f(?; ;?);?(S ?I) = n;( ;?) 2 dsdt]
= ?(dsdt)?rs;t(n)
Z
f(?;s;t)qs;t;n(d?)
=
Z
f(?;s;t)Qs;t[d?j?(S ?I) = n]?P [?(S ?I) = n;( ;?) 2 dsdt],
63
which shows
E[f(?; ;?)j?(S ?I) = n] =
Z
f(?; ;?)Q ;?[d?j?(S ?I) = n] a.s.
?
5.3 Proof of Theorem 3.4
Let us recall our main result of this dissertation, Theorem 3.4 on page 26. Here,
? denotes Lebesgue measure as before.
Theorem3.4 Let ? be a marked point process on Borel space S?I, where I =R+
or [0;1]. Assume E? = ? ?? for some  -flnite measure ? on S. The following two
conditions are equivalent:
(i) ? is exchangeable.
(ii) Q0s;t has a version that is independent of t.
Our plan is to prove the statement in the special case when I = [0;1], and ? is
a.s. bounded and non-zero. We may need the following auxiliary results to show that
it is enough to reduce the proof of Theorem 3.4 in this special case.
Lemma 5.6. It is enough to prove Theorem 3.4 in the case when I = [0;1].
Proof: Write In = [0;n]. Assume that Theorem 3.4 holds when I = I1. Then,
for every n 2 N, Theorem 3.4 also holds for In by changing every 1 in [0;1] to n in
the proof for the case when I = I1.
64
Let I =R+. Write ?n for the restriction of ? on S ?In, i.e. ?n = 1S?In?.
Assume E? = ? ??. Then, for any n 2 N, E?n also admits such factorization
since ? = ?n on S ?In. Likewise, if E?n = ? ?? for every n, then E? = ? ?? since,
for any A ? B ? S ?R+, we get E?(A?B) = E?n (A?B) = ?A ? ?B for some
su?ciently large n.
Assume that ? is exchangeable (or ?-symmetric). Deflne ?n by
?n? = 1S?In?; ? 2N (S ?I).
Fix an arbitrary n 2N. For any a;b 2 In, we have Ta;b?n = ?nTa;b. Thus
Ta;b?n = Ta;b (?n?) = ?n (Ta;b?) d= ?n? = ?n,
which shows that ?n is also exchangeable.
Assume that ?n is exchangeable for each n. Let D be the class of measurable
sets A ? N(S ? I) satisfying P f? 2 Ag = P fTa;b? 2 Ag for any [a;b] ? Q+. Let
B1;B2,... be a semi-ring generating  -algebra BN(S?I). Also let C be the class of
consisting of sets
Cni = Bi \N(S ?In), i;n 2N.
65
Clearly ? 2D. For any [a;b] ?Q+, if A;B 2D with A ? B, then
P f? 2 BnAg = P f? 2 Bg?P f? 2 Ag
= P fTa;b? 2 Bg?P fTa;b? 2 Ag
= P fTa;b? 2 BnAg.
If A1;A2;::: 2D with An " A, then A 2D by the continuity of probability measures.
This shows that D is a ?-system. If Cn1i , Cn2j 2C, then
Cn1i \Cn2j = (Bi \Bj)\N(S ?In1^n2) 2C,
which shows that C is a ?-system. For any Cni 2C, the assumption Ta;b?n d= ? implies
P f? 2 Cni g = P f?n 2 Cni g = P fTa;b?n 2 Cni g = P fTa;b? 2 Cni g,
which shows that C ?D. Therefore, we see that BN(S?I) =  (C) = D by a standard
monotone-class theorem. Together with the previous result, this shows that the ?-
symmetry of ?n for any n implies the ?-symmetry of ?. This proves that ? is ?-
symmetric ifi ?n is ?-symmetric for each n.
Write Qs;t and Q(n)s;t for the Palm distributions of the marked point processes ?
and ?n, respectively. Let C0, C0n be the reduced Campbell measures associated with ?
and ?n, respectively. Assume that the reduced Palm distribution Q(n)0s;t has a version
independent of t for any n 2N. Note that Qs;t = Q(n)s;t onN(S?In). For any bounded
66
measurable function f ? 0 on S ?R+ ?N(S ?R+), we have
C0?1S?In?N(S?In)f? =
Z
S?In
E?(dsdt)
Z
N(S?In)
f (s;t;???s;t)Qs;t (d?)
=
Z
S?In
E?n (dsdt)
Z
N(S?In)
f (s;t;???s;t)Q(n)s;t (?nd?)
=
Z
S?R+
E?n (dsdt)
Z
N(S?In)
f (s;t;???s;0)Qs;0 (d?).
By dominated convergence, we get
Z
S?R+
E?(dsdt)
Z
N(S?R+)
f (s;t;???s;t)Qs;t (d?)
= C0f ? C0?1S?In?N(S?In)f?
=
Z
S?R+
E?n (dsdt)
Z
N(S?In)
f (s;t;???s;0)Qs;0 (d?)
!
Z
S?R+
E?(dsdt)
Z
N(S?R+)
f (s;t;???s;0)Qs;0 (d?).
We conclude that Q0s;t has a version independent of t since f is arbitrary.
Conversely, assume that Q0s;t has a version independent of t. For any measurable
f ? 0 on S ?In ?N(S ?In), we obtain
Z
S?In
E?n (dsdt)
Z
N(S?In)
f (s;t;???s;t)Q(n)s;t (d?)
=
Z
S?In
E?(dsdt)
Z
N(S?In)
f (s;t;?n???s;t)Qs;t (d?)
=
Z
S?In
E?(dsdt)
Z
N(S?In)
f (s;t;?n???s;0)Qs;0 (d?)
=
Z
S?In
E?n (dsdt)
Z
N(S?In)
f (s;t;???s;0)Q(n)s;0 (d?),
67
which shows that Q(n)0s;t has a version independent of t for any n 2 N. This proves
that the Palm-measure invariance property holds for ? and all ?n simultaneously.
Write (F) for the statement E? = ???. Statements (i) and (ii) refer to those in
Theorem 3.4. For an arbitrary n 2N, also write (F)n, (i)n and (ii)n for the versions
of statements (F), (i) and (ii) with ? replaced by ?n. By previous discussions, if
statements (F) and (i) hold for any n 2 N, then (F)n and (i)n also hold, which
implies that (ii)n holds. So does (ii). Likewise, statements (F) and (ii) imply (i).
This shows that it is enough to prove Theorem 3.4 in the case when I = [0;1]. ?
The next lemma shows another reduction of the proof of our main theorem.
Lemma 5.7. Let I = [0;1]. Assume that Theorem 3.4 holds in the special case
when ?(S?I) < 1 a.s. Then it is enough to prove Theorem 3.4 in this special case.
Proof: Since S is Borel, we can choose Sn " S with ?(Sn ?I) < 1 a.s. For
convenience, let 1Sn?I? denote the restriction of ? to the set Sn ?I.
Since ? = 1Sn?I? on Sn ?I, the intensity measure E(1Sn?I?) admits the same
factorization ? ? ? on Sn ? I. If E(1Sn?I?) = ? ? ? on any Sn ? I, then for any
measurable A ? B ? S ? I, there exists an n 2 N such that A ? Sn, and so
E?(A?B) = E(1Sn?I?)(A?B) = ?A??B. Thus, E? = ? ?? ifi E(1Sn?I?) also
admits such factorization on Sn ?I for any n 2N.
Next, we show that ? is ?-symmetric if and only if 1Sn?I? is ?-symmetric for any
n 2N.
68
Assume that ? is ?-symmetric. Then Ta;b? d= ? for any [a;b] ? I. For any n 2N,
deflne a surjective mapping pn by
pn? = 1Sn?I? , ? 2N(S ?I).
Note that pnTa;b = Ta;bpn. Hence,
pn? d= pnTa;b? = Ta;bpn?,
which gives the ?-symmetry of 1Sn?I? since [a;b] is an arbitrary interval.
Assume that pn? is ?-symmetric for any n 2N. By a similar argument as in the
proof of Lemma 5.6, we conclude that the ?-symmetry of pn? for any n implies the
?-symmetry of ?. This proves that ? is ?-symmetric ifi 1Sn?I? is ?-symmetric for any
n 2N.
Write (]) for the following equivalence statement
Q0s;t has a version independent of t
, Q(n)0s;t has a version independent of t.
By changing ?n and S?In in the proof of the statement (]) in Lemma 5.6 to pn and
Sn ?I, respectively, we see that the statement (]) in the current lemma is also true.
Therefore, it is enough to prove Theorem 3.4 in the case when ?(S ?I) < 1 a.s.by
the same argument as in the last paragraph of the proof for Lemma 5.6. ?
69
We also need the following result to reduce the proof of Theorem 3.4 to the case
when ? is non-zero.
Lemma 5.8. Let I = [0;1]. Assume that Theorem 3.4 holds in the special case when
?(S ?I) 6= 0 a.s. Then it is enough to prove Theorem 3.4 in this special case.
Proof: If P(? = 0) = p 2 [0;1), then by transfer theorem we may flnd a
marked point process ? on S ? I satisfying P(? 2 ?) = P[? 2 ?j? 6= 0]. Since
E? = (1?p)?1 E? = (1?p)?1 ? ??, even E? admits such factorization.
Assume that ? is ?-symmetric. Then for any transposition Ta;b and A ?N(S?I),
P '? 2 T?1a;b A? = P[? 2 T?1a;b Aj? 6= 0] = P[Ta;b? 2 AjTa;b? 6= 0]
= P[? 2 Aj? 6= 0] = P f? 2 Ag,
where the second equality holds since fTa;b? 6= 0g = f? 6= 0g. So, the ?-symmetry of
? implies the ?-symmetry of ?. Furthermore, if ? is ?-symmetric, then
P fTa;b? 2?g = (1?p)P fTa;b? 2?g+p1f0 2?g
= (1?p)P f? 2?g+p1f0 2?g
= P f? 2?g.
Therefore, ? is ?-symmetric ifi ? is ?-symmetric.
70
Write (s;t) = u. By simple calculations, we get
Qs;t(A) = E[?(ds?dt);? 2 A]E?(ds?dt) = E[?(du);? 2 A]E?(du)
= E[?(du);? 2 A;? 6= 0]E[?(du);? 6= 0]
= E[?(du);? 2 Aj? 6= 0]E[?(du)j? 6= 0]
= E[?(du);? 2 A]E?(du) .
This shows that the Palm distributions are the same for ? and ?. Thus, it is enough
to further assume ? is non-zero a.s. ?
Let us now prove Theorem 3.4 in some special case.
Proof of Theorem 3.4: By Lemmas 5.6, 5.7 and 5.8, we may henceforth assume
that I = [0;1] and 0 < ?(S ?I) < 1 a.s.
Let ? be a canonical space, i.e. ? = S ? I ?N(S ? I). Assume that ? is ?-
symmetric. Deflne ( ;?) to be the pair of random variables satisfying P[( ;?) 2?j?] =
~? a.s., where ~? = ?=?(S ?I). By Proposition 5.4 we have ???( ;? ?? ;?) due to the
?-symmetry of ?. Hence, ???( ;??? ;?;?(S?I)) with distribution ? since ?(S?I) is
a measurable function of ??? ;?. Fix an arbitrary measurable set M ?N(S?I) and
let f be a measurable function on ? satisfying f(x;y;z) ? 1z??x;y(M). By Proposition
5.5 we get, for any (s;t) 2 S ?I
P[? ?? ;? 2 Mj?(S ?I) = n;( ;?) 2 dsdt] (5.8)
= Qs;t[???s;t 2 Mj?(S ?I) = n] a.s. (5.9)
71
Combining (5.8) with the condition ???( ;? ?? ;?;?(S ?I)), we have
P[? ?? ;? 2 Mj?(S ?I) = n;( ;?) 2 dsdt]
= P[? ?? ;? 2 Mj?(S ?I) = n; 2 ds],
which is a.e. independent of t. Then the right-hand side of (5.8) is also a.e. indepen-
dentoftforanymeasurablesetM.Therefore, theexpressionsQs;t[???s;t 2?j?(S ?I) = n]
are a.e. independent of t.
Since ? is a ?-symmetric marked point process on the Borel space S?I, we may
write ? = P?i=1 ?fli;?i and fl = ?(??I) = P?i=1 ?fli, where the ?i are i.i.d. U(0;1) and
independent of marks fl1,fl2,... For any measurable set A 2 S and B ? [0;1], by the
independence between ?i and fli, we have
E?(A?B) = E
X
i??1ffli 2 A;?i 2 Bg =
X
i??P(fli 2 A)P(?i 2 B)
= ?B ?
X
i??P(fli 2 A) = ?B ?Efl(A) = ?A??B,
which shows that the intensity measure of the ?-symmetric marked point process ?
on S ?I has a version of factorization ? ??.
72
Fix an arbitrary n 2 N. For any measurable sets A 2 S; B ? [0;1], by the
deflnition of Palm distributions and the ?-symmetry of ?, we have
Z
A?B
Qs;tf?(A?I) = ng? ??(dsdt)
= E[?(A?B);?(A?I) = n]
= E[E[?(A?B)j?(A?I)];?(A?I) = n]
= E[?B ??(A?I);?(A?I) = n]
= ?B ?nP[?(A?I) = n].
Since B is arbitrary, for a ?-a.e. t, we have
Z
A
Qs;tf?(A?I) = ng?(ds) = nPf?(A?I) = ng. (5.10)
Hence, Qs;tf?(A?I) = ngisa.e. independentoftforanyn:SinceQs;t[???s;t 2?j?(S ?I) = n]
is a.e. independent of t, we conclude that Q0s;t ? Qs;tf???s;t 2?g is a.e. independent
of t by Lemma 4.4, i.e., Q0s;t is independent of t a.e. ?.
Conversely, suppose that Q0s;t is independent of t a.e. ?, and E? = ??? for some
measure ?. Deflne ( ;?) as in Proposition 5.4 (see page 59) satisfying
P [( ;?) 2?j?] = ?=?S a.s.
73
Then, Proposition 5.5 shows that, for any (s;t) 2 S ? I, (5.8) also holds for an
arbitrary measurable set M ?N(S ?I), and so,
P [? ?? ;? 2?j?(S ?I) = n;( ;?) 2 dsdt] (5.11)
= Q0s;t[? 2?j?(S ?I) = n?1] a.s. (5.12)
Clearly the right-hand side of (5.11) is independent of t by assumption. So is the
left-hand side, which implies
????(S?I); ? ?? ;?.
To show that ? is ?-symmetric, we just need to establish the independence
???( ;? ? ? ;?) with distribution ?, in order to apply Proposition 5.4. Note that
?(S ?I) is a measurable function of ? ?? ;?, it is enough to prove ??? ( ;?(S ?I))
with distribution ?, and so, the independence ???( ;? ? ? ;?) follows by chain rule
(Proposition 6.6, [16]). By the deflnition of Palm distributions and the assumption
74
E? = ? ??, for any measurable sets A 2S and B ? [0;1], and any n 2N, we have
P [? 2 Bj 2 A;?(S ?I) = n]
= P[( ;?) 2 A?Bj?(S ?I) = n]P[ 2 Aj?(S ?I) = n]
= E[P[( ;?) 2 A?Bj?]j?(S ?I) = n]P[ 2 Aj?(S ?I) = n]
= E[?(A?B);?(S ?I) = n]E[?(A?I);?(S ?I) = n]
=
R
A?B Qs;t[?(S ?I) = n]?E?(ds?dt)R
A?I Qs;t[?(S ?I) = n]?E?(ds?dt)
=
R
B ?(dt)
R
A Q
0
s;0[?(S ?I) = n?1]?(ds)R
I ?(dt)
R
A Q
0
s;0[?(S ?I) = n?1]?(ds)
= ?B.
Therefore, ???( ;?(S ?I)). This completes the proof of theorem for the case when
I = [0;1]. ?
75
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