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"WI/II III/III

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This is to certify that the

dissertation entitled

Markov Properties of Measure-indexed
Gaussian Random Fields

presented by
Sixiang Zhang

has been accepted towards fulfillment
of the requirements for

 

Ph.D. degreein Statistics

 

Major professor

Date July 30, 1990

 

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MSU Is An Affirmative Action/Equal Opportunity Institution
cAeIIchnG-pd

MARKOV PROPERTIES OF MEASURE-INDEXED

GAUSSIAN RANDOM FIELDS

Sixiang Zhang

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
Department of Statistics and Probability

1990

Abstract

Markov Properties of Measure - indexed Gaussian Random Fields
By

Sixiang Zhang

We consider the Gaussian random field {X n, p E M (E )},where M (E) is
a vector space of signed Radon measures with compact support on a separable
locally compact Hausdorff space E. We assume that the covariance C (p, V) =
E(X,,X,,)(p, V E M(E)) is bilinear. The Markov properties of {Xm p E
M (E)} are defined. The necessary and sufficient conditions for {XM 11 E
M (E )} to have the Markov property in terms of the geometric and analytic
structure of the reproducing kernel Hilbert space of C(p, u) are given under
some assumptions on the index set M (E) We also define the concept of dual
process and in the case that a Gaussian random field {X p, p E M (E)} has a
dual, we can simplify the necessary and sufficient conditions. Applications to
generalized Gaussian random fields, to the Gaussian fields related to Dirichlet

forms and to the ordinary Gaussian processes are derived.

To my parents

iii

Acknowledgements

I would like to express my sincere appreciation to Professor V. Mandrekar
for his guidance and encouragement and to Professors Sheldon Axler, James
Hannan and Raoul LePage for reading the thesis and for their helpful com-
ments which led to the improvement of the intial draft.

I would like to thank the Department of Statistics and Probability, Michi-
gan State University and the Office of Naval Research for their financial
support.

iv

Contents

1 Introduction

2 Notations and Preliminaries

3 General Results

4 Gaussian Processes Related to Dirichlet Forms
5 Applications to Ordinary Gaussian Processes

References

20

35

52

61

Chapter 1

Introduction

The study of the Markov property for multiparameter processes was
initiated by P. Lévy[15] who conjectured that Lévy Brownian motion in odd
dimension has this Markov property. McKean [19] proved Lévy’s conjecture
and gave a precise definition of Markov property. Subsequently Pitt[24],
Kiinsch [13], Molchan[21] and Kallianpur and Mandrekar[l2] gave necessary
and sufficient conditions for a Gaussian and generalized Gaussian process to
have some type of Markov property. A systematic study of different Markov
properties is given in [17] for Gaussian processes and in [18] for the general-
ized Gaussian processes. In [7], Dynkin introduced the study of the Markov
property for Gaussian processes related to the Dirichlet space and studied
Markov property for specific Gaussian processes indexed by measures of fi-
nite energy. This was generalized in an abstract way for Gaussian processes
related to Dirichlet form of Fukushima[10] by R5chner[25].

The main techniques used by [12], [13], [21], [24] were geometric and

depended on the geometric structure of reproducing kernel Hilbert spaces. In
[12], [21] the conditions were simplified in the case the Gaussian process has a
dual process(see also Rozanov[26]). The concept of a dual prOcess originates
in [12] and [21]. In [7] and [25], the Markov property was proved by relying
heavily on probabilistic technique as the Gaussian processes considered by
them are related to Green’s functions of a symmetric Markov process.

Our purpose here is to establish general theorems, using pure geomet-
ric techniques, for Gaussian processes indexed by measures to have a Markov
property. We begin by recalling certain facts about conditional independence
and Gaussian processes from [17] and [18] in Chapter 2 and Chapter 3. As a
consequence we derive some new facts in Lemmas 2.8 - 2.11. We introduce
the concept of support (Lemma 2.2) for a linear functional on a vector space
of measures satisfying (A.1) and (A.2). Such a concept plays a role in estab-
lishing and proving our main general theorems in Chapter 3 (Theorems 3.1,
3.2). In view of Examples 2.1 the main results of [12] and [21] follow from
these general theorems.

We also need to modify the structure of the indexed sets used by
Dynkin [7] and Rbchner [25]. We demonstrate that this modification does
not affect the Markov property of the processes considered by them. However
with this modification, we can set their problems as a spacial case of Corollary
3.3. To obtain this, we need to introduce an appropriate generalization of
the concept of the dual process introduced in [12] to our setup. Finally our
results give considerable strengthening of the results of [13] and [24] as well
as generalizing the index set for the multiparameter processes. This is done

in Chapter 5.

In Chapter 4 we recall the needed concepts from the theory of Dirichlet
forms and prove a crucial analytic result (Lemm 4.7) in this setup which
allows us to relate the local property of the Dirichlet space to the condition
of Corollary 3.3.

We start the next chapter with preliminaries, notation and interrela-

tions of the concepts used throughout the thesis.

Chapter 2

Notations and Preliminaries

In this chapter, we present some concepts and results needed in the
rest of this work. We start first by introducing the reproducing kernel Hilbert

space associated with a covariance function following Aronszajn[1].

Definition 2.1 Let T be any set and C be a real valued function on T x T.

Then C is called a covariance on T if

(a) C(t,s) = C(s,t) for all s,t E T and
(b) 21.36 a,a,C(t,s) _>__ 0 for all finite subsets i ofT and {a,,s E i} of ER.

Theorem 2.1 (Aronszajn [1]) Let T be any set and C be a real valued
covariance on T. Then there exists a unique Hilbert space K (C) of functions

on T satisfying

C(-,t) E K(C) for each t E T, (2.1)
(f. C(-.t))K(0) = f(t) for each t e T and f e K(C). (2.2)

4

Here for each t, C (-, t) denotes the function of the first variable and (-, )K(c)

means the inner product in K (C)

Proof. Let RT be the real linear space of all functions on T to 32 with
coordinate-wise addition and scalar multiplication and let H be the linear
manifold in RT generated by {C(-, t),t E T}. On H define the inner product

(Lg) = 2 _ a.b..0(s.s') = gems) = 2 b.:f(s’)- (2.3)

sewer oe: s’Et'

Where f = 2,6,- a,C(',s), g = 2“,, b,:C(-,s’) with i,i’ finite subsets of T.
From the last two equalities in (2.3), we get (f, g) is independent of the
representations of f and g. From properties of C we get (f, f) 2 0 and (f, g)
is a bilinear function on H. Also f(t) = (f, C(-,t)) for each t E T and f E H
gives |f(t)|2 S (f,f)C(t,t). This implies (f,f) = 0 iff f(t) = 0 for all t.
Thus (H, (-, )) is a pre-Hilbert space. Let H be the completion of H under
norm (f, f)%, define K(C) = {f e emu) = (C(-,t),h,) for h, e 17}.
On K(C) define (f,g) = (h,,hg). Then K(C) has all the properties and is
determined uniquely by C. 0

Definition 2.2 Let T be any set. A class K (C) of functions on T forming a
Hilbert space is called the reproducing kernel Hilbert space (for short, RKHS)
of a covariance C if it satisfies (2.1) and (2.2). The above theorem gives

existence and uniqueness.

Definition 2.3 Let T be a set and (9,? , P) be a probability space, Then a

family {Xt : t E T} of real random variables is called a centered Gaussian

process if every real linear combination of finite elements of {Xt : t 6 T} is

a Gaussian random variable with mean zero.

If (X, : t E T} is a centered Gaussian process on a probability space
(0,}: P), then Cx(t,t’) = Ep(X,X;:) is a. covariance on T and K(Cx) =
(f,f(t) = Ep(X¢Y,) for a unique Y, E H(X)}, where Ep is the expectation
under P and H (X ) is the linear subspace of L2“), .77, P) generated by {X g, t E

T}. Conversely we can associate a Gaussian process with a covariance.

Lemma 2.1 Let C be a covariance on T, then there exists a Gaussian pro-
cess {Xt,t E T} defined on a suitable probability space (fl,f',P) such that
Ep(Xthl) = C(t,t’).

Proof. Let K (C) be the RKHS of C and {eh j E J } be an orthonormal
basis in K(C). Define fl = II,-fl,-,.7-' = (85.7%,!) = <8)ij where 51,- = 32,};- =
3(32) and Pj = N(O,1) forj E J. Also let {,(w) = 00,- withj E J. For
h E K(C),h = Zj(h,ej)8j define H(h) = 2,-(h,e,-)£j. Then by Parseval’s
identity we get H(h) as a Gaussian random variable for each h E K (C ) with
Xg = H(C(-,t)) we get {Xht E T} as a Gaussian process with covariance C.
Cl

Remark: The map 11 in the proof of Lemma 2.1 is an isometry between

K (C) and H (X ) We will be using this fact many times later on.
For a Gaussian process {Xht E T}, when T Q 32", we call it a
(Gaussian) random field. Let C8°(E) be the space of infinitely differentiable

functions with compact support in E, where E is a open sebset of ER". When

T = 03°(E), we call {Xht E T} generalized random field if the covariance
function C(tp, 1b) = E,(X¢X¢,) is bilinear and continuous on C3°(E) with
Schwartz topology(see [9] or [11]). We shall also be using processes indexed
by measures of bounded energy. They occurred in the works [l2],[24] and
[26]. For this we need some additional concepts.

Let E be a separable locally compact Hausdorff space. M (E) is a set
containing Radon signed measures on E with compact support. The support
of a signed Radon measure p on E is defined as the complement of the largest
open set 0 such that |p|(0) = 0, where In] is the total variation measure of
,u. We make the following assumptions on M (E)

(A.l) M (E) is a real vector space.

(A.2) M (E) has the partition of unity property, namely for any p E
M (E), if {01, 02, ..,0,.} is an open covering of the support of p (for short,
suppp) then there exist p1,p2,...,pn 6 M(E) with suppp; Q 0;,i = 1,2, ..,n
and fl=m+M2+---+#.-

(A.3) If f is a linear functional on M (E), and the support of f (for short,
supp f ) is contained in A1 U A; where A1 and A2 are two disjoint closed sub-
sets of E, then f = fl + f2, where f1, f2 are linear functionals on M (E)
with supp f,- Q A;,i = 1,2. The support of a linear functional f on M (E) is
defined as the complement of the largest open set N of E such that f (p) = 0
for all p 6 M(E) with suppp Q N.

Under the assumptions (A.l) and (A.2) the support of a linear functional
on M (E) is well defined, Actually we have the following:

Lemma 2.2 Under the assumptions (A.1 ) and (A.2) we have

(a) If f is a linear functional on M (E), then suppf = complement of
U;0.-, where the union is taken over all open set 0; such that f (p) = 0 for
all p E M(E) with suppp Q 0;.

(b) If f is a linear functional on M (E) and supp f is an empty set, then
f E 0, i.e. f(p) = 0 for all p E M(E).

(c) If f1 and f; are two linear functionals on M(E), then

surp(f1 + f2) Q (suppf1)U(suppf2)o

Proof. (a) We only need to show that Uge, 0,- Q (supp f)‘. Let p 6 M (E)
be such that suppp Q U,“ 0;. By the compactness of suppp, we may choose
finite sets 0,-1, ..., 05,. to cover suppp, using the partition of unity property
(A.2) we have p = m + + [1,, where p,- E M(E) and supppj Q 0,3,j =
1, ..,n. Then

f(#) = f(m + + Pa) = int.) = 0.

i=1

(b) Using the definition of support of a linear functional.
(c) Let A,- = suppf;,i = 1,2. Let p E M(E) with suppp Q (A; UAg)‘ =
Aim/1;. Then suppp C A? for i = 1,2. Hence f.-(p) = 0 for i = 1,2. Then
f(l‘) = f1(l‘) + f2(/‘) = 0- SO SUPPUI + f2) Q A1 UA2- 0

Lemma 2.3 Under assumptions (11.1) and (A.2), (A.3) is equivalent to the
following (A.3)’:
(A.3)’ Iff is a linear functional on M(E) and suppf Q A1 UAz where

A1 and A2 are two disjoint closed sets, then for any two disjoint open sets

01,02 with A.- Q 0;,i = 1,2, f can be decomposed into the sum of two linear
functionals f1 and f; on M(E) with suppf,- C 0;,i = 1,2.

Proof. That (A.3) implies (A.3)’ is obvious. To prove the converse, let
01 and 03 be two disjoint open sets of E such that 0,- 2 A;,i = 1,2. Then
f = f1+f2 with the fg’s linear functionals on M (E) and supp f.- Q 0;,i = 1, 2.
Now take another open set 0] Q 01 with 0; 2 .41. Then f = f] + f; with
811ppr Q 01 and suppf; E 02 so f1 - fl = fl - f2- By Lemma 22(6)
surp(f1 - ff) <_I(sur>pf1) U (surpf1)§ 01 and supp(f$ - f2) 9(8uppfé) U
(surpf2)§ 02. Since 01 n0. = rt. supp(f1 - fl) =supp(f$ - f2) = ¢- Then
by Lemma 1.2(b) f1 - f; = f; — f2 = 0. So suppf1 Q 0], hence

31113pr Q n 0 = A1-
A1Q0901

Similarly we can show supp f2 Q A2. D

We will give some examples in which our assumptions (A.1), (A.2)
and (A.3) are satisfied. Before we give the examples, we need the following

Lemma.

Lemma 2.4 HE is a normal space([9],p.2) and {01, ..., 0"} is an open cov-
ering of a closed set A of E then there exist open sets U1, U2, ..., U" such that
If; Q 0;,i=1,2,...,n and

UMQA

i=1
where U,- means the closure of U5.

We note that ([2],p.8) a separable locally compact Hausdorff space is

a normal space. The proof of the lemma is based on induction on n.

10

Proof. If A Q 0 where A is closed and O is open, then A and 0° are
two disjoint closed sets and hence by the normality of space there exist two
disjoint open sets U1,U2 such that A Q U1 and 0" Q U2. Hence U1 Q U; =
U; Q 0. Then U1 is the candidate, so the lemma is true for n = 1.

Assume A Q 01 U 02, A is closed and 01, 03 are open. Then An 0; Q
01. Since A1 ()0; is a closed set, there is an open set U1 such that U; Q 01
and A (I 0; Q U1. Then A = (AnU1)U(AnUf). Since Aanf is closed
and Aanf Q 02 ,then there is an open set U2 such that U; Q 02 and
AfiUlc Q U2. Then A Q U1 UU2 so the lemma is true for n = 2.

Assume that the lemma is true for n 5 1:0: 2 2). Then if

k+1 k
A Q L] 0.“ = (i=1 0;) U Ok-I-l

there exist open sets U’ and U1.“ such that

I:
U; Q U 0i)
i=1
Uk+1 Q 0H1,
and
A g U’U UH].
Now since k
U; Q U 0;
i=1
by induction there exist open sets U1, ...,Uk such that U; Q 0,,i = 1,2, ..., k
and
_ I:
U’ Q U U.-.

l

11

Then
k+1

A c_: U’UUHI Q U U,-.
Ic=l

Example 2.1 (Infinitely difierentiable functions)

Let E be an open domain in 32", M(E) = {p : dp = cpdx,cp E Cg°(E)},
where C3°(E) consists of all infinitely differentiable functions with compact
support on E. E is equiped with relative topology. Obviously M (E) is
a vector space in the sense of (A.l). If cp E C8°(E) and {01,02,...,0n}
is an open covering of suppgp, then there exist <p1,...,Lp,. E C3°(E) with
suppw; E 0,- and (p = 22;, go,- where for a function cp E CS°(E), suppcp =
closure of {x : x E E,<p(x) 75 0}. Notice that suppcp =supp(cpdx), so (A.2) is
satisfied.

To verify (A.3)’, let f be a linear functional on M (E) and supp f Q
A; UAg. Where A1 and A2 are two disjoint closed subsets of E, take two
disjoint open sets 01,02 such that A.- Q 0;,i = 1,2. Then we can take
two open sets 0; and 0; such that A; Q 05 Q 0;,i = 1,2 and C7105; is
the empty set. Take 0’ = (A1 UA2)° then {01, 0;, 03} is an open covering
of R". Then there exist (p.- E C°°(E),i = 1,2,3 such that pg 2 0, suppgo Q
0£,i = 1, 2, 3 and Zi=1¢i = 1 ([9],p.45) where C°°(E) consists of all infinitely
differentiable functions. Now for any (p 6 C8°(E), cp = Zilcpcp;

f(cp) = f(‘P‘P1) + “W?” + flees)-

Since 3UPP(W3) =supp(<p<padx) Q 03 then f(Wa) = 0- Let f.-(so) =
f(wdd = 1,2- If v E 08°(E) with surpuodx) =SUPp<p S; (Slippery

12

then (pep,- E 0,i = 1,2. So f,-(9p) = 0,i = 1,2 if supp(<pdx) =supp‘p Q
(suppcp;)°,i = 1,2. Then supp f,- Q ((suppcp;)°)° =supp<p,- Q 0;,i = 1,2. This
implies (A.3)’ holds.

Example 2.2 (Measures of finite eneryy)

Let M (E) be a vector space having the following property. If p E M (E ),
then 1 All 6 M (E) for any A E 3(E) where 8(E) is the o-field generated
by all open subsets of E and 1A}: is the measure of p restricted to A, i.e.,
lAp(B) = p(AnB) VB 6 B(E). Then M(E) satisfies (A.2) and (A.3): sup-
pose }: 6 M (E) with suppp Q UL] 0.- where 03 are open sets.By Lemma 2.4
we can find open sets Ug, (i = 1, n...) such that U.- Q 0; and suppp Q UL, Ug.
Then

I1 = (1U?.,U.-)# = IU,#+IU,nu,c#+..-+1U.uU,°n...nU¢ II '=' #1+#2+- - o+fln say,

...1
such an expression makes sense because It has compact support and hence is
a finite measure. Obviously suppp; Q U: Q 0;,i = 1,2, ...,n and p.- E M(E)
by assumption for i = 1,2, ...,n. Hence (A.2) holds.

If f is a linear functional on M (E) with supp f Q A1 UA; where A1
and A2 are two disjoint closed sets. Choose open sets 01,02 be with A.- Q
0;,i=1,2 and CIQC; = d. Let 0;; = (01U02)°, then 01U02U03 = E
and these 033 are also disjoint. Then for any p E M (E),

II = 10.11 +1021: + 10.}! E #1 + #2 + #3 say.

then

f(l‘) = f(fl1)+ f(t‘z) + “#3) ‘3 f1(l‘) + f2(l‘) + f3(#) say,

13

notice that f1, f2 and f3 are also linear on M (E ) Since supppg Q C; = 03 =
(01 U02)c Q (A1 UA2)° so f3(p) = f(p3) = 0 for all p 6 M(E). Notice that
for any p with suppp Q Cf, i = 1,2, [15:10.41 5 0 hence f.-(Iu) = 0, i = 1,2
for all p 6 M(E) with suppp Q Cf. So suppf.- Q CLi = 1,2. (A.3)’ also
holds.

From the above two examples we notice that it is easier to check (A.3)’
instead of (A.3) when we know that (A.1) and (A.2) are already satisfied.

We are interested in the Markov Property of Gaussian processes in-
dexed by M (E), here we assume that M (E) satisfies assumptions (A.1)-
(A.3). For this we need some elementary concepts of conditional indepen-
dence([16]) and related results from [17].

Let (0,]: , P) be a probability space and assume unless stated oth-
erwise that .77 is complete and all sub a-fields(algebras) contain all sets of
measure zero from .77.We mean by conditional expectation or conditional
probability the equivalence classes of random variables. Because of the above
assumptions on all sub 0 - fields, the equivalence classes for different a - fields

are the statement a.e. and consider equalities in terms of equivalence classes.

Definition 2.4 Let A,B and G be sub a-fields of f. We say that .A and B

are conditionally independent given 9 if

P(AflB|Q) = P(A|Q)P(B|g) A e .4, B e B

where P(-|g) is the conditional probability given 9. We denote this by A .11.
BIG. Ifg '2 {fl,¢}, then we say that A is independent ofB and denote this
by .A .11. B.

14

Lemma 2.5 Let A,B and G be sub a-fields of .7: such that G Q B then
A .ll. BIG if E] f IQ] = E[ f [B] for all bounded A- measurable f. In particular
ifG = {43,9} then A .11. B ifl' EIfIB] = EU] for all bounded A-measurable
f. Here EIIG] denotes the conditional expectation.

Proof. Assume E] f IG] = E[ f IB] for all bounded A-measurable f, by the
properties of conditional expectation E I f gIG] = E [E I f gIBIIG] for g bounded
B-measurable and f bounded A—measurable, EIfgIG] = EIgEIfIB]IG] =
EIfIG]E[gIG], where the last equality follows since EIfIG] = EIfIB]. To

prove the converse, observe for A E B and f bounded A—measurable

I. fdP = I. ElllflgldP = A EllllglEIfIGIdP = L EIfIgldP

Here the second equality uses conditional independence and the last equal-

ity follows from the fact that

EIIAIQIEUIQ] = EllAElflgllgl°

Using the fact that B VG = 0’{BnA; B E B, A E G} and the arguments
in the proof of Lemma2.5 where V denotes the a-field generated by both B
and G, a{. . .} denotes the generated o-field, we get.

Lemma 2.6 A .1]. BIG implies EIfIBVG] = EIfIG] for all f bounded A-

measurable.

Lemma 2.7 A .LL BIG implies the following

15

(a) For every 9‘ satisfying a g a c ova, .4 1L BIG.
(b) ifé ; ova/1 1L élg.

Proof. Lemma 2.5 and Lemma 2.6 imply A .LL (G VB)IG giving (b). To
obtain (a), by Lemma 2.5 and Lemma 2.6 EIfIG] = EIfIBVG] for all f
bounded A- measurable. Hence by Lemma 2.5, A is conditionally indepen-
dent ofBgivenGQGVB CI

Corollary 2.1 Let A,B and G be sub a-fields of .77, A .LL BIG. Then we
have the following.

(i) IfA’ Q AVG,B’ Q BVG, then A’ .11 B’Ig.

(ii) Ifg Q 9' (.3 AVBVG then A .LL BIG'.

Proof. (i) can be deduced by using Lemma2.7 (b) twice.
(ii) can be deduced by using Lemma2.7 (a) twice. 0

Let M (E) be a set satisfying assumptions (A.1)- (A.3) and {X y, p E
M (E)} be a Gaussian centered random field indexed by M (E) We as-
sume the covariance function of {me 6 M (E)} is bilinear, i.e., C(p, u) =
E(X,,X,,) as a functional on M (E) x M (E) is bilinear. An immediate con-
sequence is that if pup; and p E M (E) such that p = mm + our: for
(11,02 6 331 then X“ = aIXu, + 02X“. For a subset S of E,S means the
closure of 5, 5° means the complement of 5, whereas 65' means the boundary

of S, that is 35 = S033. We define

F(S) = o{X,,,/1 E M(E) and suppp Q S}

16

and denote

2(5) = n F(0)

s<_:o

where the intersection is taken over all open sets 0.

We will define three different Markov Properties of {X y, p E M (E)}

Definition 2.5 (McKeanI19]) We say that {me E M (E)} has Markov
Property [(MPI) on an open subset .S' of E if for every open subset 0 of E
with 35' Q 0

F(S) JJ. F(SC)IF(0).

Definition 2.6 (Germ Field Markov Property) We say that {Ln}: 6
M (E)} has Markov Property II(MPII) on a subset 3 (not necessarily open)

ofE if
23(3) .LL 2(SE)|2(65).

Definition 2.7 We say that {X n, p E M (E)} has Markov Property [H(MPIII)
for a subset S of E if
F(§) .LL F(SE)IF(63).

Notice that MP1 is only defined for open sets, whereas MPH and
MPIII are defined for arbitrary sets. We will explore some relationships
between these Markov properties for {Xm l‘ E M (E )}

Lemma 2.8 For {me E M(E)}, F(0UO’) = F(0)VF(0’), where 0,0’
are open subsets of E.

17

Proof. We only need to prove F(0U0’) Q F(O)VF(0’). If p E M(E)
with support of It contained in OUO’, since M (E) has partition of unity
property (assumption (A.2)), p = [11 + m, here pup; E M(E), supppl Q 0
and supppg C 0’. Then X“ = X,“ + X“2 hence X“ is measurable with

respect to F(0) V F(O’). 0

Lemma 2.9 For {me E M(E)}, let S be an open subset ofE and 0 an

open set containning 05. Then
F(S) 1L FI'S‘IIFIOI in 2(3) IL 2(S°)IF(0)-

Proof. The ”if” part is easy because F(S) Q 2(S) and F(S°) Q E(S°). To

prove the converse, notice by lemma 2.8

F(.S') V F(0) = F(S u 0) 2 2(3') (2.4)

and
PG“) VF(O) = F(3‘° u 0) 2 2G). (2.5)
Then by Corollary 2.1 2(3) .LL 2(S°)|F(0) for open set 0 :_> as. 0

Lemma 2.10 For Gaussian random field {pr E M (E)} Markov Prop-
erty I on an open set 5' implies Markov property II (Germ field Markov
Property) on S.

18

Proof. By previous lemma, we know MPI for an open set S is equivalent to
2(3) iL 23(5")I1"(0)

for all open subset 0 Q 35.
We define a direct order on all open sets in terms of inclusion then for

any set A 6 2(3) and B 6 2(Sc), we have
PlAnBllel = PIA|F(0)]PIB|F(0)I

By maringale convergence theorem with net indexed o-fields(see [23],Chpter

V)
limoPIA|F(0)]P[B|F(0)] = P[A|2(6S)]P[B|E(BS)] (2.6)
and
limoP[AflB|F(0)] = P[Afl slams» (2.7)
Both limits are in L2(Q,}', P). Hence we have
PIAIIBIEWSH = PIA|2(35)IP(B|3(35)I

for all A E 2(S) and B E 2(Sc), this is Markov Property II on open set 5.
Cl

Lemma 2.11 Suppose the Gaussian randomfield {Xm p E M(E)} has Germ
Field Markov Property (MPH) on an open set .S' and also

2(33V 309°) = F(E)

. Then it also has MPI on S.

19

Proof. MPII says 2(S) .11. X(S°)I2(BS). For any open subset 0 of E with
63 Q 0
23((95) E F(0) 9 NE) = ”EVE-9°)

By (ii) of Corollary 2.1 2(3) .lL 2(S°)|F(0) which implies

F(5) .11. F(s‘)|F(0).

Relationships between MP1 and MPIII will be discussed in Chapter 4 in

case of Gaussian random fields related to Dirichlet Space.

Chapter 3

General Results

In the previous chapter, we already set up some basic notations and
lemmas. Our goal in this chapter is to explore the relationships between
the Markov property of {X m p E M (E)} and its reproducing kernel Hilbert
space. We are particularly interested in the Markov property of {X my 6
M (E)} for some classes of sets: the class of all open sets and the class of
all pro-compact open sets (a set is called pre- compact open if its closure
is compact). In the second part of this chapter, we will discuss the case in
which a dual process exists. To begin with we introduce some properties of
Gaussian spaces([l8]Appendix).

Let (0,]: , P) be a probability space,by a (centered) Gaussian space we
mean a subspace of L2(fl, .7", P) such that every finite collection of elements
of this subspace is Gaussian distributed with mean zero. We assume all

Gaussian spaces are closed unless otherwise mentioned.
Lemma 3.1 Let {X1,...,X,,} be a subset of a (centered) Gaussian space.

20

21

(XI, ...,Xn} are independent ifl' E(X,-Xj) = 0, i 74 j.
Proof. Let a = (r11, ...,n") e an, J? = (X1, ...Xn). Then

<1> for) = E(exp(iii. 55)) = exp(—-;-E(2 u;X.-)2)

Hindu.) if E(X;X,-) = 0 fori 76 j,

where (PX,(-) is the characteristic function of X.- (i = 1, 2, ..., n). Conversely,

E(X.-X,~) = EIEIXinIU(Xj)I]=EIXjEIXiI0(Xj)II
= EX,EX.-=o if #3“.

Lemma 3.2 Let H1 and H2 be subspaces of a (centered) Gaussian space H.
Then 0(H1) and 0(H2) are independent ifl' H1 1 H2.

Proof.That 0(Hl) independent of 0(H2) implies H1 1 H2 follows from
Lemma 3.1. To prove the converse, let {fin-J 6 J1} and {£2j,j 6 J2} be
orthogonal bases in H1 and H2 respectively. Then {5.3, j 6 J.- i = 1,2} has
every finite subfamily orthogonal and hence independent by Lemma 3.1. In
particular 0(H1) = a{£1,-,j 6 J1} is independent of 0(H1) = U{€2j,j 6 J2}.
Cl

Lemma 3.3 Let H, be a subspace of a Gaussian space H. Then E [YI0(H1)] =
ij H,Y for any Y E H, where Projy, is the projection operator on H,

22

Proof. Let Y = Y1 + Y2, where Y1 = ProjH,Y and Y2 = Y — Y1. Then
E(Y1Y2) = 0. Hence by Lemma 2.1, Y1 and Y; are independent. Then

EIYIO’(H1)]= Y1 + EIY2I0(H1)] = Y1 'i’ 1333 = Y1.

Lemma 3.4 Let H0, H1 and H; be subspaces of a Gaussian space H. Then
the following are equivalent:

(a) 0(H1) ..LL 0(H2)|0(Ho).

(b) H; 8H0 .L ngHo where H: = H;VH0, (i = 1,2) and erHo
means the subspace of L2(Q,f', P) generated by {n - Projyor], r) 6 H5}.

Proof. (a) implies 0(HI) JJ. 0(H;)|0(H0) by Corollary 2.1. Let X1 6 HI
and X2 6 H5. Then

E(X1X2I0(H0)) = E(X1I0(Ho))E(X2|0(H0)) = ProjHoXlProjHoX2-
Hence
E(X.-X,) = E(X.-ProjHoXj) = E(ProjHoX,-ProjHoXj) for i 761'

giving E(X1 — ProjHoX1)(Xg — Projm) = 0

Conversely, from Lemma 3.2 a(H{ 6 H0) is independent of 0(H; 6 Ho).
Now 0(HI) = a(a(H{ 6 Ho) Va(Ho)), (i = 1,2). For A.- E 0(H: 8 Ho), B.- 6
0(Ho), i = 1,2,

E(1A,n3,1A,na,Ia(Ho)) = 13,n3,E(1A,nA,Ia(Ho))

23

1' 131n33P(A1 0 A2)
= 131P(A1)132P(A2)
= E(1ArnBi[0(H0))E(1A1031I0(H0))

This gives 0(H{) .lL a(H§)|a(Ho) which gives 0(H1) .11. 0(H2)Ia(Ho). 0

Definition 3.1 Let H0,H1,H2 be subspaces of a Gaussian space. We say
that the space Ho splits H1 and H2 if 0(H1) .LL 0(H2)|0(Ho).

Lemma 3.5 Suppose H0, H1, H2 are subspaces of a Gaussian space such that
Ho Q H1 n H2. Then Ho splits H1 and H2 tflHl 0 H2 = Ho and HlJ' .L HiL
where H} are the orthogonal space of H,- in H1 V H2, (i = 1,2)

Proof. Under the assumption Ho Q H1 0 H2 and from Lemma 3.4(b) Ho
splits H1 and H 2 iff

HIVH2=(H19H0)€BHO€B(H29H0)

where G) denotes the orthogonal sum of the spaces. Then obviously this is

equivalent to H1 n H2 = Ho and H1i .L H2J'. Cl

Corollary 3.1 Let H1 and H; be two subspaces of a Gaussian space, then
”(111) JJ- 0(H2)I0(H1 0 H2) if Projmprojfle = Projrrmnoo

Lemma 3.6 Let H be a Gaussian subspace. Then L2(Q,fH,P) is generated
by M = {epr,X E H}, where f” = OIH}.

24

Proof. Clearly M Q L2(fl,fH,P). Let Y be an element in L2(Q,fg,P)
such that EYex = 0 for all X E H. then EYe‘x = 0 for all X 6 H. Since
f (t) = EYe‘X is analytic in t, if it is zero for all real t, it also vanishes for
all complex t. In particular EYeix = 0 for all X E H. Put Z 6 W if Z is
bounded and EYZ = 0. Then W contains the family {eix , X E H} which is
closed under multiplication. W is also a linear space. It contains with each
function the complex conjugate of this function and with each uniformly
bounded convergent sequence the limit of this sequence. This implies (see
P.A. Meyer(1966),Chapter1,Theorem 2) that W contains all bounded func-
tion measurable with respect to the a-algebra generated by {e‘X,X E H }
which is the same as a{H}. Y is orthogonal to all bounded f” measurable

functions, hence Y = 0. Cl

Corollary 3.2 Let {Xht E T} be a Gaussian process. Then the algebra
generated by polynomials in {X¢,t 6 T} is dense in L2(Q,0(H),P), where
H is the subspace of L2(fl,fH,P) generated by {Xht E T}.

Recall that we assume M (E) is a vector space satisfying assumptions
(A.1)-(A.3). {me E M (E)} is a centered Gaussian process with bilinear
covariance C(p, u) = E(X,,X,,), p, V 6 M(E). Let K(C) be the reproducing
kernel Hilbert space corresponding to the covariance C(p, v) and H (X) be
the closed linear subspace of L2“), f, P) generated by {X y, p E M (E)} The
map 11 is the isometry between K (C) and H (X) with IIC(-,p) = X y. For
any subset S of E, define H (S ) the subspace of L2 generated by all X n with
suppp Q S and K (S ) is the corresponding subspace of K (C) under the map

25

II, namely K(S) = II“H(.S'). By Lemmas 3.1 and 3.5 we have the following

lemma:

Lemma 3.7 The Gaussian process {X an )1 E M (E)} has Markov Property I
on an open set S of E ifl' one of the following holds:

('7 E('71 - ProjmomrX'Iz - Projmoflz) = 0 for any ’71 6 H(S), '72 E
H (3”) and any open set 0 containing 35'.

(a) H(S u 0) n H(s'c u 0) = H(O) and

H(S u 0)i .L H(sc u 0)l

for all open set 0 containg 65, where H(S U 0)i and H(S" U 0)‘ are or-
thogonal spaces of H (S U 0) and H (5° U 0) in H (X ) respectively.

Next theorem will give the relationship between the Markov Property
I of {me 6 M (E)} for all open sets and the reproducing kernel Hilbert

space K (C ) of covariance function C(p, V).

Theorem 3.1 Let {X,,,p E M (E)} be a Gaussian process such that M (E)
satisfies assumptions (A.1)-(A.3) in Chapter 2 and covariance function C(p, v) =
E(X,,X,,) bilinear in p and V. Let K(C) be the RKHS of C(p,u). Then
{me E M(E)} has the Markov Property [for all open subsets ofE if the
following (a) and (b) hold.

(a) If fler 6 K(C) with suppf; n suppf; = 45 the" (f1,f2)K(C) = 0;
where (f1, f2)K(o) means the inner product of f1 and f; in K (C)

(b) If f 6 K (C) and f = f1+ f; where both f1 and f; are linear functionals

ofM(E) with (suppf1)n (suppfg) = 43, then f1,f2 E K(C).

26

Remark: Since each element in K ( C ) is also a linear functional on M (E),
for every f E K (C), supp f is well defined by (i) of Lemma 2.2.
Proof. We know that {X n, p 6 M (E)} has Markov Property I for all open
sets of E iff

H(S u 0) n H(§° u 0) = H(O) (3.1)
(S u 0)l .L H(s" u 0)l (3.2)

holds for all open set S and open set 0 containing 05'. We will prove that
(a) and (b) of Theorem 3.1 is equivalent to (3.1) and (3.2). We separate our
proof into two parts.

Sufficiency: Suppose (a) and (b) of Theorem 3.1 hold, we need to show
(3.1) and (3.2). To verify (3.1) it is enough to prove H(O) 2 H(S U 0) n
H (Sc U 0) this is equivalent to

 

H(0)l g H (S U 0)i v H (3‘ u 0)l (3.3)
which is the same as

K(O)‘ g K(S u 0)l v K(s‘ u 0)*. (3.4)
Let f e K(0)i, then f = II-lY for some Y e H(0)i. Then if X,. e H(O)

f(fl) = (f.C'('.#))K(c) = E(YXu) = 0-

In particular f(p) = 0 if suppp Q 0. Hence suppf Q O‘(see definiton
of supp f in chapter 2) observe the following facts: 5 n Cc = S n 0" and
3600‘ = Sic—00° because 8(Sc) Q 05' = BS Q 0. So 500“ and S'ICF‘IOc are

two disjoint closed sets, furthermore their union is 0°. By our assumption

27

(A.3) f(p) = f, (p) + f2(p) with f1,fg being linear functionals of M(E) and
supp f1 Q Sn 0‘, supp f2 Q F’n 0:. By (b) of Theorem 3.1 fl and f2 belong
to K (G) Then

f. g K((sn 0°)°)l = K(S”c u 0)i,
f2 9 K((S‘c n 0‘=)‘=)l = K(S’u 0)1 = K(S U 0)‘.

To prove (3.2), we will show the equivalence condition as follows
K(S U 0)"‘ .L K(S.c U 0)l. (3.5)

This is true if we can show that

 

K(SU 0)‘ Q span{f = f E K(C).suppf E F‘:} (3-6)

and

 

K(SCU O)J' Q span{f : f E K(C),suppf Q S}. (3.7)

But if f e K(S U 0)‘, then for every p E M(E) with suppp Q S U
0. f(#) = (f.C(rt,-))K(c~) = 0. hence surpf E (S U 0)c = (79‘U 0). =
Sc (I CC Q Sc. Similarly if f E K(S: U 0)‘L then

suppr(SCU0)°=S00°=SDOCQS.

Necessity: Suppose (3.1) and (3.2) hold. Let f1 and f; be in K (C) with
disjoint support, then there exists an open set S such that supp f1 Q S and

 

supp f2 Q Sc. Take 0 = [(suppf1)U(suppfg)]° then 0 is an open set containg
65. Since S U 0 Q (suppf2)° and Sc U 0 Q (suppf1)°, so f1 6 K(Sc U 0)1
and f2 6 K(S U 0)‘. Hence (f1,f2)K(c) = 0 by (3.5).

28

Assume f 6 K (C) and f = fl + f: where f‘ and f,» are linear functionals
of M (E) with supp f1 and supp f3 being disjoint, we choose an open set S such
that supp f1 Q S and supp f2 Q Sc. Let 0 = (supp f1U supp fg)°. By(3.1) and
(3.2) we have

H(X) = H(S u 0)* e H(ZS‘c u 0)1 e H(O)
which is the same as
K(C) .—. K(S u 0)i e K(Sc u 0)1 e K(O).

Note that suppf Q supprUsuprz = 0‘ so f E K(O)‘, hence f =
f; + f; with f{ E K(St U 0); and f; E K(S U 0)‘L, then suppfl’ Q
(3.. U 0)c = 3' n 0° = 3' 0 (suppflusuppfz) = (Smallppfr) U (gnsuppfz) =
Shauppfl quppfl. Similarly supp f; Q supp f2. Now f = fl + f2 =
fl + f5. f1 - ft = fl - f2 with supp(f1 — f{)nsupp(f£ - f2) = 4%
supp(f1 — fi) = 45 which implies f1 = f,’ and f2 = f; by virtue of (b) of
Lemma 2.2. Hence f1,f2 E K(C). U

m

0

We can also consider the Markov Property I of {X n, p E M (E)} for
all pro-compact open sets, then we have the following theorem similar to

Theorem 3.1.

Theorem 3.2 Let {me E M(E)} be the same as in Theorem 3.1 then it
has Markov Property I for all pro-compact open sets if the following (a) and
(b) hold:

(a) For any f1 and f2 6 K (C) with suppf; nsuppfg = 43 and at least one
of the suppf,-(i = 1,2) is compact, then (f1,fg)K(C) = 0

29

(b) Iff E K(C) and f = f, + f; with suppf, n suppf; = 45 and at least

one of the supp f,-(i = 1,2) is compact, where f1 and f; are linear functionals

ofM(E), then f1 and f2 belong to K(C).

Proof. All arguments in the proof of Theorem 3.1 go through, except when
one of supp f;(i = 1, 2) is compact. We can choose a pro-compact open set S

to cover the compact one and Sc to cover another. D

Now we extend the concept of dual process introduced in [12] for the
processes {me 6 M(E)}.

For the separable locally compact Hausdorff space E, we denote by
Co(E) the space of all continuous functions on E with compact support. Let
G(E) be a subset of Co(E). (G(E) need not be a subspace of Co(E)). Let
{X,, g E G(E)} be a Gaussian process defined on the same probability space
as {me E M(E)}. Then we have,

Definition 3.2 The Gaussian field {Xmg E G(E)} is called a dual process
0f{X.o# 6 M(E)} if

(i) H(X) = H(X).

(ii) E(X,,X,,) = ngdp for all g E G(E) and p E M(E), where H(X)
and H(X) are the subspaces of L2(fl,f’, P) generated by {me 6 M(E)}
and {X,,g E G(E)} respectively.

Remark: (i) we denote by g with or without sub(supper)scripts as elements

in G(E) and f with or without sub(supper)scropts as the elements in K (C)

30

(ii)For any 9 E G(E) we denote by f,(-) as f,(p) = [E gdp, p E M(E).
Since X. e H(X).f.(u) = MM.) ...e f.(-) e K(C).

For any open subset D of E, we define subspaces M (D) and M (D) of
K (C) as following:
M(D) = ep—an-{L f e K(C) suppf g D} (3.8)
M(D) = mm). 9 e G(E) suppg s D} (3.9)
where for g E G(E), suppg is defined in the usual sense, namely suppg =

closure of (e: e E E,g(e) 79 0}.
Remark: We remark that K (D) = II’1(H(D)) and is distinct from M (D)

Lemma 3.8 For any open set D, M(D) Q M(D).

Proof. Let g E G(E) such that suppG Q D then g(e) = 0 if e E ( suppg)“.

Then fo(#) = fad}! = 0 if 811mm 9 (suppy)°- That suppfo(°) (.3
(suppg)° Q D impllies f,(-) 6 M (D) 0

Definition 3.3 we say that G(E) has the partition of unity property if for
every 9 6 G(E), 01,...0n are open sets covering suppg, then g = 22;, g,-
with g; E G(E) and suppg,- Q 0;, (i = 1, ....n)
Lemma 3.9 If G(E) has the partition of unity property, then

M(Dl u D2) = Mm.) v M(D.)

for every two open sets D1 and D2.

31

Proof. Let g E G(E) with suppg Q D1 U D2, then 9 = g1 + g; with
9i 6 G(E) and suppg,- g Di (i = 1,2), hence fg(l‘) = [9100+ f92(f‘) for
,. e M(E). But f..(-) e M(Ds). I.- = 1,2), hence f.(-) e MID.) v MID.)

This gives M(Dl u 1).) = Mm.) v MD.) :1

Theorem 3.3 Let {X,,g 6 G(E)} be a dual process of {me E M(E)}.
G(E) has the partition of unity property and M (D) = M (D) for all open
sets D. Then (a) of Theorem3.1 implies (b) of Theorem3.1.

Proof. If (a) of Theorem3.l holds then M (D1) J. M (D2) for every two
disjoint open sets D1 and D2. By Lemma 3.9

M(Dl u D.) = Mm.) v M(Dg) = Mar.) 69 M(Dg)
together with M(D1 U D2) = M (D1 U D2) gives

for every two disjoint open sets D1 and D2. Let f E K (C) with f = fl + f;
where f1 and f; are linear functionals of M (E) with supp flr‘lsupp f2 = ¢,
then we can choose two disjoint open sets DI and D,» such that supp f,- Q
D; (i = 1,2) and DID-D: = rt. Let D = D1 UDg, then suppf Q D1 U D2,
f 6 M(D). We can write

I = Projmmf = Projmogf + ProjM(Dg)f = ff + f5 say,

f] as an element in M (D1) is a limit of sequence f" in K (C) with supp fin Q
D1. Then fl’(p) = limnfn(p) = 0 for any p E M(E) with suppp Q Di. So

32

supp ff Q D1. Similarly supp f; Q D2. Now f = f1 + f; = f{ + f; gives
f1 "' fi = fi - f2- 31“ 3“PP(f1 - ff) Q 31 Mid SUPPUz - fi) Q 172, 80
Dr (ID—2 = 45 implies 3“PP(f1 - fi) = (fi - f2) = ¢~ Hence fi = ff 6
K(C)i=1,2. El

Theorem 3.4 Let {X,,g 6 G(E)} be a dual process of {me E M(E)},
G(E) has the partition of unity property and M (D) = M (D) for all open
sets D. Then (a) of Theorem3.2 implies (b) of Theorem3.2.

Proof. The arguments are similar to the proof of Theorem 3.3 and hence

omitted. CI

The following corollary is an immediate consequence of Theorems 3.1,

3.2, 3.3 and 3.4.

Corollary 3.3 Let {X,,g e G(E)} be a dual process of {any 5 M(E)},
G(E) has the partition of unity property and M (D) = M (D) for all open
sets D. Then ‘

(a){X,,,p E M(E)} has Markov Property [for all open sets ifl(f1, f2)K(c) 2:
0 for any f1, f2 6 K (C) with suppflnsuppfg = ()3.

(b){X,,,p E M (E)} has Markov Property I for all pre- compact open sets

if (f1, f2)K(C) = 0 for any f1, f2 6 K(C) With suppflnsuppfg = 45 and at
least one of suppf;(i = 1,2) being compact.

If we impose on G(E) the following assumption which is stroger than

the partition of unity property, then we can explore more properties about

{X,,,p 6 M(E)} and its dual process {Xmg E G(E)}.

33

Assumption 3.1 For any fixed g E G(E) there exists a positive number L9
such that for any open covers 01,...,0,. of suppg there exist g.- 6 G(E) with
suppg.- C 0;, i = 1,...,n. g = 2;, g.- and

EDI“? g L,, i = 1, .,..n.

Theorem 3.5 Let {X,,g E G(E)} be a dual process of {me 6 M(E)}
and G(E) satisfy assumption3.1, then the following are equivalent,

(0) (f1, f2)K(c~) = 0 for any f1. f2 6 K(C) with suppf: n suppf: = «t-

(b) Hf,(-) E H(D) for everyg E G(E) such that suppg Q D, where D is
open and II is the isometry map between K(C) and H(X).

(c) M(Dl) .L M(Dg) for any disjoint open sets D1 and D2 and M(D) =
M(D) for every open set D.

Proof. (a)=>(b). For open set D define Prom the projection of K (C) —->
K(D), where K(D) = II"H(D). Let g E G(E) with support ofg contained
in D, then for p E M(E) with suppp Q D.

f.(#) - Prejofo(#) = (fs(')r C(Ie. ~))K(C) - (Prejnfgt). C(It, -))K(C)-
Since C(p, ) = II‘lXu E K(D) hence
(PTOJ'DM'LCUI. '))K(C) = (fg(°),0(#,'))x(0) = foo”)-

So f, —Proij,(p) = 0 for all p E M (E) with suppp Q D. This implies that
supp(f,(’) — Proijg(')) Q Dc. We know that supp f,(-) quppg Q D. From
(a) of the theorem (f,(-) — Proijg(-),f,(-))K(c) = 0 this implies fg(-) =
Proij,(-) 5 H-1H(D)-

34

(b) =>(c) Let gl,gg 6 G(E) such that suppglr‘lsuppgg = 45 choose open
set D such that suppgl Q D and suppg; Q Dc then f,,(-) E M (D) Q M (D)
and from (b) f,,(-) E K(Dc). It is easy to check that M(D) J. K(D-c). Then

(f91(°)rfm('))K(C) = 0, namely E(X,,Xg,) = 0.

To show M (D) = M (D) for open set D. We need to use assumption 3.1.
Let f E M(D), we denote f = f - ProjM(D)f E M(D). Then there exists
fa E M(D) with suppffl Q D such that fn —r f in K(C) as n —» oo.
Denote Du = (supp fn)c ,then for any n {D,D,,} is an open cover of space
E. So for any 9 E G(E) by Assumption3.1 g = 9? + g; with g? E G(E) and
suppg? Q D,suppg; Q Du, furthermore (fgzu, f,:-)K(C) 5 L9, where L, is a

constant only depending on g. Then

(I. fg('))K(C) = lim..(fm fg('))K(C) = limo(fmfgr(-))K(0)+limn(fmfo;(°))K(0)

But (fn,f,;(-))x(c) = 0 because fn .L K(Dn) = II“H(D,,) and f,;(-) E

K(D.) by (b)- 30 (I, f9('))K(C) = limn(fm fgr(°))K(C) = limn(fo-f, fgr('))K(C)
The last equality holds because f 1 M (D) and f,,;-(-) E M (D) Then

I(fn—f,fg;'(°))K(C)I .<. IIfn-IIIK(C)IIfg;'||K(o) .<_ LgIIfn-fllxm) —» 0 as n —> 00.

so (i, f,(-))K(C) = o for all g e G(E). Since { f,(-), g e G(E)} is dense in
K(C) => f = 0 then f = Proij) f e M(D).

(c)=>(a) Let f1, f2 6 K (C) and supp flnsupp f2 = ¢- Choose open set
D such that suppfl Q D and suppf; Q Dc. Then f; e M(D) = M(D)rf2 6
mo“) -_-. Mm“). Since M(D) J. M(F), (f1, mm, = o. a

Chapter 4

Gaussian Processes Related to

Dirichlet Forms

In this chapter, we show that the probelm considered by R5chnerI25]
is a special case of Corollary 3.3. In [25], Riichner considered the Gaussian
random field induced from a Dirichlet form and prove that it has the Markov
property III for all sets iff the underlying Direchlet form has local property.
He shows that free random field studied by Nelson [22], and in some cases
the ”generalized random fields” studied by Kallianpur and MandrekarIlZ],
MolchanI21] and RozanovI26] can be handled within his framwork. First we
introduce the Dirichlet form and related potential theory. The notations and
terminologies are from the basic book by FukushimaIlO]. For details and
further information, the reader is referred to [10].

Let E be a separable locally compact Hausdorff space and m be a
positive Radon measure on E with supp(m) = E. According to [10](p.35),

35

36

a pair (11,8 ) is called a regular extended (transient) Dirichlet space with
reference measure m if the following conditions are satisfied:

(fed) 1"; is a real Hilbert space with inner product 8.

(.7792) There exists an m-integrable bounded function g, strictly positive
a.e.m. Such that .7, Q L1(E, V9) and

[luldug = flulgdm _<_ ‘/£(u,u) for every u E f,

where V denotes the measure with density g, i.e. dVg = gdm.
(17,3) .7. fl Co(E) is dense both in (12,5) and in (Co(E), II ”00), where
C0(E) denotes the set of all real continuous functions on E with compact

support and II f "00 = supp,€EIf(x)I for f E C0(E).

Let us say that funciton v is a normal contraction of a function u if
|v(x) — v(y)| S |u(x) —u(y)| and |v(x)] S |u(x)I for all x,y E E. We assume:
(11.4) Every normal contraction operates on (1",.8 ) i.e. if u E .7", and v

is a normal contraction of u then v E .7". and £(v,v) S £(u,u).

The following lemma gives slight extension of ([10] ,p.25 ).

Lemma 4.1 A regular extended (transient) Dirichlet space (far?) has the
following properties:

(i) Ifu,v E .77., then uV v,u Av,u A 1,u+,u" 6 .7... Also u,v E f. n
LOO(E, m) implies rm 6 fa.

(ii) Let {umu} Q .7], such that ufl -+ u in (fag) as n —i 00 . let <p(t) be
a realfunction such that 90(0) = 0, I<,o(t)—cp(t’)I S It—t’l for t,t’ 6 33. Then
<p(u,,),<p(u) E .7-"c and (,0(u,,) -> 50(u) weakly with respect to 8. In addition, if

(p(u) = u then the convergence is strong with respect to the norm given by E.

37

(iii) For any u E Co(E) there exist u,, 6 .77, fl Co(E), n = 1,2,..., such
that supp(u") Q {x 6 E : u(x) 9t 0} n = 1,2,..., and un converges to u

uniformly.

Proof. The arguments are similar to the proof of Theorem 1.4.2 and Lemma

1.4.2 of [10](pp.25-26). 0

Remark: For the connection between Dirichlet space on L2(E,m) and

extended Dirichlet space see ([10],p.35)

Definition 4.1 We say that (f,,£) has local property if £(u,v) = 0 for
every u,v 6 .77.3 n L2(E,m) such that supp(u - dm) and supp(u - dm) are

compact and disjoint.

Definition 4.2 A signed Radon measure p on E is a measure of bounded

energy if there exists a constant c > 0 such that

[IuIdeI S cm for every u e .7". fl Co(E).
We denote by ME all measures of bounded energy, let
M(E) = {p E M,; suppp is compact} (4.1)
MI={Iu; ”GM” #20}
M+(E) = M: n M(E)

Notice that M(E) is a vector space and also if p 6 M(E) then In}! E M(E)
for any Borel set A Q E. By Example 2.2 M (E) satisfies assumptions(A.1)—
(A.3).

38

For any A Q E we define
A. = {u : u 6 fan 2 1a.e.m on A}.

Definition 4.3 We define the 0—order capacity Capo(0) of an open set 0 Q

E as
' f.) 8 , E
Capo(0) = In 6‘0 (u u) 0 ¢ ¢
00 £0 = 43
and for an arbitrary set A Q E
Capo(A) = 3313 Capo(0) for all opene set 0 (4.2)

Lemma 4.2 The capacity defined by (4.2) is a Choquet capacity i.e.
(i) A Q B => Capo(A) Q Capo(B).
(ii) An I => Capo(U.. A.) = sun. Card/1..)-
(iii) A,, compact, A1, I => Capo(nA,,) = inf" Capo(A,,).

Proof. The proof is similar to that of Theorem 3.1.1 in [10]. 0

Definition 4.4 A statement depending on x E S Q E is said to hold ' quasi-
everywhere’ (for short q.e.) on S if there exists a set N Q S of zero capacity
such that the statement is true for every x E S\N.

Definition 4.5 Let EA = E U A be the one-point compactificaiton of E. A
function u defined on E is called quasi-continuous if there exists for any 6 > 0
an open set G Q E such that Capo(G) < e and "lac: is continuous. Here
"IE\G denotes the restriction of u to E\G. If we replace “IE\G by UI(EuA)\G in
the above definition then u is called quasi-continuous in the restricted sense.

Here uI(EuA)\G denotes the restriction ofu to (E U A)\G with u(A) = 0.

39

Definition 4.6 Given two functions u and v on E, v is said to be a q.e.
modification of u in the restricted sense if v is quasi-continuous in the re-

stricted sense and v = u a.e m.

Lemma 4.3 Every element u E .77., admits a q.e. modification in the re-

stricted sense denoted by it.

Proof. The proof is similar to that of Theorem 3.1.3 in [10]. C1

Using ([10],p71) we get that for any p E M; there exists a unique
element Up 6 f, such that

Isa/Ito) = / ad). Vv e r. (4.3)

Here ii denotes any quasi-continuous modification of v in the restricted sense.

Define the map

U: Mg —r .75}
Up = Up+—Up’

where p+ and p" are the positive and negative parts of p in the Jordan

decomposition. Up is called potential of p

Lemma 4.4 Let (f,,£) be a regular extended (transient) Dirichlet space,
then the linear manifolds {Up - UV: p,V 6 Mg} and {Up : p e M(E)}
are dense in (fag).

Proof. By [10](Lemma 3.3.4 and Theorem 3.3.4) we know that

f. = WfWME') - U(M3)}

40

The second part of the lemma is a consequence of the following general

result Lemma 4.5. D

For any set A Q E we define
M304) = {11; fl 6 M (E). anewl Q A}
Me(A) = {In 1‘ 6 Ms. supp/t Q A}
Lemma 4.5 Let A Q E. Then
WWW! E M£(A)} = WWW: 6 1143(4)} (4-4)

Proof. We will show that if p 6 ME (A) with suppp Q A, we can find
[I E M§(A) such that Upfl —> Up in (17,8). Let Kn be a sequence of
compact sets such that K, ‘I‘ E. Let pn = 1K..# then pn E M g (A) We will

show
Upn—rUp in (7,8) (4.5)

IIUII - Uflnllt = 5(U#. W) - 28(Ul‘rUl‘n) + 5(Upn.UIIo)
by(4.3)
g(UI‘anI‘n) = jfindfln

Where U7; is any q.e. modification of U p". FromI27](p3.2), we know that
U71; 2 0 q.e. which also implies that U}; > 0 a.e. p (also see[10]p.7l).

Hence

g(Ul‘anPn) = [K Wad” S/Wndl‘ = f"(Ul‘rUl‘nl

= [1771001. S/Wdfl=£(U#.U#)

41

Thus
IIUl‘ - Uuolli: S 2(€(U#.Uu) - €(U#.U#n))

By monotone convergence theorem

5(U#.U#o)=/l771dfln=/K deefmdp=€WmUm awn-roe.

so (4.5) holds. This completes the proof. D

For any Borel set A Q E, define a subspace of (.71, 8) as
.73“ = (u; u 6 1'}, i2 = 0 q.e. on A} (4.6)

Where q.e means quasi-everwhere, and i1 is any q.e modification of u in the
restricted sense. We denote by H64 as the orthogonal complement of .73“ in

.7}, namely
“if = fisher (4-7)

Definition 4.7 ([10],p.79) For any v 6 f; we define the spectrum of v

(denoted by S(v)) as the compelemt of the largest open set 0 such that

8(v,u) = 0 for any u 6 .7", n Co(E) with suppu Q 0. In particaular when
6 Mi, S(Up) = suppp.

Lemma 4.6 ([10],p.80) Let A be an open or closed subset of E. Then
”6‘ = WA = WlUflil‘ E M3(4)} = WWW/em 6 5425(4)}
where

W51 = sp_aTr€{v E .76, S(v) Q A} (4.8)

42

We need now the concept of ' Balayage measure ’. Let PA be the
projection on It: in (12,8) for any Borel set A. For p 6 ME, we know
from[10](section3.3) that there corrosponds a potential f = Up E .7", Let
fA = PAf then fig 6 U(Mg') and hence L; = UpA with pf E M; also
supppA Q A. Following [10], we call p" the Balayage measure (or sweeping
out) of p on A.

Lemma 4.7 Let A be a closed set and A Q D, D is open, ifu E .724 then
there exists a sequence {gn} Q .7", fl Co(E) with supp(gn) Q D and

gn —> u in (f,,8) as n —r 00 (4.9)

Proof. If u 6 £4, then by Lemma 4.1 u+,u" 6 .76 . But u: S [17] = IiIII and
ii: 5 Iu~| S IiiI so both u+ and u" are in .74. Without loss of generality, we
may assume u is nonnegtive and itself quasi-continuous.

Since .7", n Co(E) is dense in (11,8). There exists {vn} E f. n Co(E) such
that v,, -r u in (f,,8). We may assume v,, > 0, because we always can

replace v,, by v: = %v,, + §Ivnl and v: —r u in (173,8) by virtue of Lemma
4.1(ii). Let h. = v” Au = %(vn +u) - §|vn — uI, using Lemma 4.1(ii) again we
can show h,, -r u in (f;,8). Notice that h,, is bounded, ha 6 .754 and closure
of {x; x 6 E, hn(x) 75 0} is compact, we can choose mi, 6 Co(E) and w; 2 0
such that w; 2 hn q.e. and suppwf, Q D. Choose another wfi,’ E Co(E)
and wfi,’ Z 0 such that supp(wg) Q D and wfi,’ 2 w; + 1 for x Esuppwg. By
Lemma 4.1(iii) for each n we can find {wg} Q 1"; n C0(E) with w; 2 0 such
that supp(wg) Q {x; wfi,’(x) at 0} and ”w; — wglloo —r 0 as m —r 00. So for
each n we can find wn E f.nCo(E) such that urn Z 0 and w“ 2 wfi,’(x)--;- for
all x, then 11),, 2 h,, q.e.. Now for any n, select {113,} Q .7"c fl Co(E) such that

43

u; —o v,’, as m -+ coin (.77.,8), let e; = wnAug, = %(w,,—u;‘n)—%Iw,,—u;

and using Lemma4.1(ii) again we can show
e3, —> vi, in (f,,8) asm -> 00

Notice supp(eg) Q D and v; —-i u in (f,,8) as n —-> 00. So we can find

{97.} Q 1". n Co(E) such that suppg" C D and

gn—eu in(.7"¢,8) asn—eoo

Definition 4.8 Let (f¢,8) be a regular extended transient Dirichlet space
and M (E) is defined as in (4.1) . The (centered) Gaussian random field
{me E M(E)} on a probability space (9,.7, P) satisfying E(X,,X.,) =
8(U p, U V) is called (f,,8) - Gaussian field.

Remark: Rbchner considered the Gaussian field indexed by Me with covari-
ance E(X,.X.,) = 8(Up,UV).But the Markov properties of{X,,,p E M(E)}
and {X ,,, p E M,} are the same by virtue of Lemma 4.5. We point out here
that Mc may not be a vector space because the sum of two Radon measures

may not make sense.

For every 9 E .7", n Co(E) there exist p,, E M(E) and Up" —r g (by
Lemma 4.4) then {XM} is Cauchy in L2(Q,f', P). We define

A

X, = limnxooXufl (4.10)

44

for every 9 E .7". fl C0(E). Then E(X,X,,) = 8(g, Up) = fgdp = fgdp for
p E M (B)

Let G(E) = anoow). Since ancow) is dense in (1;, s), {X,,g e
G(E)} is the dual process of (f,,8) - Gaussian field {me 6 M(E)} in the

sense of Definition 3.2.
Lemma 4.8 G(E) = .7", fl Co(E) has the partition of unity property.

Proof. Suppose v E fJICo(E) with suppv Q G1UG2 (G1 and G2 are open).
Then take a pre—compact open set 0 such that suppv \Gz Q 0 Q C Q G1.
By Lemma 4.1(iii) we can choose to 6 f, flCo(E), such that w = 1 on 0 and
suppw Q G; then

v=vw+(v—vw)Evl+v2 say,

then by Lemma 4.1(i), v; E f. n C0(E) and suppv,- Q Gg, (i = 1,2). C1
Lemma 4.9 Let K(C) be the RKHS ofC(p, V) = 8(Up, UV), p,V E M(E).
Let {X,,g E .77, n Co(E)} be defined as in (4.10). Then for every open set
D, M(D) = M(D).

Remark: Recall M(D) = WK(C){fg(-); g E G(E) with suppg Q D} and
M(D) = WHO)”; f E K(C) with suppf Q D}.(cf. (3.8) and (3.9))
Proof. We only need to prove M (D) Q M (D) Let f E M (D) such that
supp f Q D, let A =supp f then there exists an element u E .7"c such that
f(p) = 8(Up,u) then 8(Up,u) = 0 for any p E M(E) with suppp Q Ac. By
Lemma 4.6

u E [s'an‘E{U,,,p E MET/hell]; = (Wo¢)1 = (Hocli = fA°

45

By Lemma 4.7 there exists {gn} Q .7", n C0(E) with suppgn Q D such that
g,, —r u in (11,8) as n —> 00. Then X,” —r Hf in Lg(fl,}', P), where 11 is
the isometry between K (C) and H (X ) = W{X,,,p 6 M (E)} such that
IIC(-,p) = X“. So f9“ —> f in K(C) where f,n(p) = fgndp. Obviously
f... e M(D) hence M(D) = M(D). a

The next theorem characterizes the relationship of the local property
of ($2, 8) (see Definition4.l) and the Markov property I of (fa, 8) - Gaussian
field {X,,,p E M(E)} for all open sets.

Theorem 4.1 (f,,8) - Gaussian field {me E M (E)} has Markov prop-
erty I for all open sets iff one of the following holds:

(a) (3,8) has local property, namely 8(u, v) = 0 for every u,v 6 .77C n
L2(E, m) such that supp(udm) and supp(vdm) are compact and disjoint.

(b) (flrf2)K(C) = 0 for f1, f2 6 K(C) With 31¢pr 0 suppf; = 43-
Proof. The equivalence of (b) and Markov property I of {X p, p E M (E)} is
the consequence of Lemma 4.8 and Lemma 4.9 and Corrollary3.3. We only
need to prove (a) and(b) are equivalent.

(8) =00 Let f1.f2 6 K(C). seppfnflsurmfz = e5. Since ep—an‘EIUII; In E
M(E)} = .73, there exist u1,u2 E f. such that (f1,f2)x(c) = 8(u1,u2) and

to) = so. Ur) = [11:61}!
for all p 6 M(E), i = 1,2.
Let A; =suppf.~, i = 1,2, then 8(u.-,Up) = 0 for any p E M(E) with
suppp Q A9, i = 1,2. By Lermna 4.6

u.- e [WIUI‘W E Mid/15H]l = (WIT = fl.-

46

Where .754, = {v E .8, ii = O q.e. on A5}. Let D; be some open neighborhood
of A,- such that D1 Fl D2 = 43. By Lemma 4.7 there exist {g:,} Q .7. n Co(E)
with suppg; Q D.- and g; —+ u.- as n —» oo in (f¢,8) for i = 1,2. Then

(f1, f2)K(C) = 8(u1,u2) = lim,,8(e,‘,,ef,) = 0

because 8 (e3,, e?) = 0 for each n by the local property of (11,8).

(b) => (a) Let v1, v2 6 fan2(E, m) such that supp(vldm) and supp(vgdm)
are compact and disjoint. We need to show 8 (v1,v2) = 0. But since K (C)
and (f,,8) are isometry there exist f1, f2 6 K (C) such that 8(v1,v2) =
(f1,f2)K(c) and 8(v,-, Up) = f.-(p) forp E M(E) andi = 1,2. Let A.- =supp(v,-dm).
v,- = 0 a.e,m on A5 then u,- = 0 q.e on Af, hence v.- 6 .754, = (H345); =
(W64?)i (by Lemma 4.6). So 8(v.-,Up) = 0 for every p E M(E) with
suppp Q A? i = 1,2. Then f,-(p) = 0 for every p 6 M(E) with suppp Q Af,
hence suppf; 9 A4 5 = 1,2- SO 8(v1,v2) = (ffrlemc') = 0 ‘3

From the proof we see that (a) of Theorem 4.1 is also equivalent to
the following (b’):
(b’) (f1.f2)xIC) = 0 if fufz E K(C). 811ppr and surpfi are compact
and disjoint.

Using this, Theorem 4.1 and Corollary 3.3, we get:

Theorem 4.2 For (76,8) - Gaussian field {me E M(E)} the following
are equivalent.

(i) {Xu 6 M (E)} has Markov property I for all open sets.

(ii) {X ,, 6 M (E)} has Markov Property I for all pre-compact sets.

(iii) (f,,8) has local property.

47

From Lemma 2.11 we know that for an open set D if
2(5) v 2(D") = F(E) (4.11)

then MPI on D is equivalent to MPII on D. We shall prove (4.11) is the

case. In fact, we have:

Lemma 4.10 For (12,8) - Gaussian field {X,, E M(E)}, F(S) V F(S") =
F(E) holds for every open set S.

Proof. Let p E M(E), then p = lsp+1sep which gives X“ = X15u+X15cw
Since supp(lsep) Q S‘, X15”, is measurable F(Sc). Take Kn compact and
Kn T S then |p|(S\K,,) -+ 0. We can show 8(Umepn) —r 0, where p,, =
15\K,.Il (see the proof of Lemma 4.5). This means

U(1K”p)—>U(15p) asn—roo in (11,8)

01‘

X116.“ —+ X15” in L2“), f, P).

But XIX“, 6 17(5), so X15“ 6 f(S) and hence X” E F(S) V F(Sc). D

Corollary 4.1 (.77.,8) - Gaussian field {X ,, E M (E)} has MPH for all open
sets iff (f,,8) has local property.

In the rest of this chapter, we will prove that (.72, 8) - Gaussian field
{X n E M (E )} has MPI for all open sets is equivelent to MPIII fo all subsets
of E (see definition2.7 for MPIII). First we need a few lemmas.

48

Lemma 4.11 For (f,,8) - Gaussian field {Xu 6 M(E)}. We have

(i) If A,B are closed sets ofE and 0 open sets of E then 0 U A 2 B
implies f(O) V f(A) = f(B).

(ii) For any closed set A, F(A) = 2(A) = 002AF(0), where the inter-

section is taken over all open sets 0.

Proof. (i) is generalization of Lemma 4.10, the proof is similar to that of
Lemma 4.10.
(ii) We can choose decreasing open sets U”, n = 1,2, ..., such that Un 2

U "+1 and n°° Un = A. We will show that

n. F(Un) = F(A) (4.12)

this implies F (A) = 2(A).
Let G = finF(U,,) and f}, = F(Un). Let k E N, p1,p2, ...,pk 6 ME and
Z = III‘=1X,,... We will show that E [Z IG] is J" (A) measurable. By martingale

convergence theorem we know that
EIZIgI = limnEIZIfnl-

For any p E M (E) let p” be the Balayaged measure of p on U;(For Balayage
measure see definition before Lemma 4.7). We denote by X ”a the projection
of X“ on span{X..;u 6 M(E),suppV Q IX} and write X,,..,,.. = X” — X“...

Now Z can be written as a sum of terms

Hic=l(Xfl?)ainik=1(X#i-fl?)fi‘r airfli E {091} and 2(03' '1" fli) = k.

49

Since U p? 6 HS}: = W37: and U (p.- - p?) ..L WF, 115;, (X ,,,-,,?)I6‘ is indepen-
dent of f}. and since suppp? Q 277;, so X “I: E .75.. Then

Elzlfnl = 115:1(XV?)a‘Ele=l(Xm-V?)fiil'
Using (3.1.18) of [10]

WF.

38

w: = {1 W31" =
n=1

II
..n

n

So PWF;(U [15) -+ Pwéa(Up,-), where szy; is the projection on WE. This
means X p:- —r X ”.4 in L2(fl,f, P), where pf is the Balayage measure of p on
A. Thus 11;, (Xu?)°“ -e III‘=I(X,,.4 )°"' in probability. Since X,“_,,? —r X ”:5“?
in Lg(fl,}', P) fori = 1,2, ..., k, since {Xu:-_,,:-,i = 1,2, ..., k} are joint Gaus-
sian, by a simple argument concerning the Fourier transform of their joint
distributions, we obtain limnEIIIf-‘___,(X,,,_,,?)3‘] exists, thus limnEIZIfn] is
.77 (A) measurable because III‘=1(X,,.4)°" is .7: (A) measurable, so EIZIf'n] is
.7: (A) measurable. By Corrollary3.2, the polynomials in X y, p 6 M (E) is
dense in L2(fl,}'(X),P). So G = f(A) 0

Lemma 4.12 Let A Q E. Then the following are equivalentf for (f¢,£)
-Gaussian field {Xm p E M(E)},

(i) F(A) .LL F(FNJ‘IBA).

(ii) F(A) .LL F(A°)IF(3A).

(iii) F(A) .u F(A°)IF(6A).

Where A and A" mean the interiors of A and A‘ respectively. 6A means
the boundary of A.

50

Proof. (i)That =>(iii) is trival because F(A) g F(A) and F(A") g F(E)
(iii)=>(i) Since A U 6A = A and A" U 0A = A5. By Lermna 4.11(i)
F(A) v F(aA) = F(A), F(A‘) v F(aA) = F(Ze). Then apply Corollary
2.1(i) to get (i).
Since F(A) g F(A) g F(A) and F(Ac) g F(A°) g F(F). From (i) e

(iii) we get (i) (1) (ii) 4:) (iii). 0

Lemma 4.13 For (f,,8) - Gaussian field {Xm p E M (E)}, the following
are equivalent:

(i) F(G) .LL F(C7)IF(60) for all open subsets 0 of E.

(ii) F(A) .l.L F(F)IF(6A) for all subsets A of E.

Proof. that (ii) => (i) is trival.

(1) => (ii): For any subset A, By previous lemma it is enough to show
F(A) .11. F(A°)IF(8A). But since A is open, we have F(A) .l.L F((TAF)|F(0A).
Since at g 6.4 and 7111 (71? = 13,1704) v F(W) = F(E) 2 HM). To
apply Corollary 2.1(b), we get

11??) JJ. F(WIFIBA).
This implies F (A) JJ. F (Ac)|F (8A) and finishes the proof. D
Notice that Lemma 4.13(i) is nothing but Markov property 11 for all
open sets (see Definition2.6 and Lemma 4.11(ii)) and that Lemma 4.13(ii)

is nothing but Markov property 111 for all sets. Combining Theorem 4.2,
Corollary 4.1 and Lemma 4.13, we have the following:

51

Theorem 4.3 For (.77.,G) - Gaussian field {Xm p E M (E)}, the following
are equivalent:

(i) It has the Markov Property I for all open sets.

(ii) It has the Markov Property I for all pre-compact open sets.

(iii) It has the Markov Property H(GFMP) for all open sets.

(iv) It has the Markov Property III for all subsets.

(v) (f¢,8) has the local property.

Chapter 5

Applications to Ordinary

Gaussian Processes

The Markov Property of ordinary Gaussian stochastic processes forms
an important special case in our work. Let E be a separable locally compact
Hausdorff space and {£t,t E E} be a centered Gaussian random field. Then
the Markov property of {Eat E E} can be handled within our framework
and we deduce and extend the main results of Kiinsch[13] from our work.
We consider first the case that E is an open domain of R”(n _>_ 1) and then
consider the general case.

Let T be an open subset of 32"(n _>_ 1) and {fit 6 E} be a mean zero
Gaussian process. Let A be a subset of T,A be the closure of A in T and 0A
be the boundary of A in T. Let F(A) = a{§,,t E A}, F(Ac = o{£,,t 6 Ac}
and F(BA) = a{£,,t 6 0A}. Then we say that {fit 6 T} has the simple
Markov property on A if F (A) .11 F (A°)IF (0A). It is well known that for

52

53

n 2 2 such a definition is not reasonable because it turns out to be too
narrow and to leave out many interesting multiparameter processes. For
instant, Lévy’s mutidimensional parameter Brownian motion does not have
this property. Hence Lévy proposed in (1956) (see also McKean (1963)) the

following definition:

Definition 5.1 We say that {§,,t E E} has the Markov property on a subset
A ofT if

F(A) .l.L F(A°)|F(3A)
where for any set B Q T, F(B) = a{£t,t E B} and 2(B) = 00231370).

Here the intersection is taken over all open set 0.

The following lemma can be found in [18].

Lemma 5.1 The following are equivalent for a stochastic process {§t,t E E}
and a subset A of T.
(i) F(A) 1L PIA—9| U34).
(ii) For every open set 0 2 6A, F(A) .lL F(A°)IF(0).
(“*7 2(4) JJ- Z(74_°)| 23(54)-
Notice that if A is an open set then (ii) is similar to MP1 defined in
chapter 1 and (iii) is MPII(GFMP).
We assume that E(£¢£,) = R(t,s) is continuous. This is equivalent to
T —-> {£¢,t E T} is continuous in L2(Q,P). Take M(T) = {godt, cp E CS°(T)}.
We know from Example that 2.1 M (T) satisfies assumptions(A.1)-(A.3). We

associate with it the random field

X,,, = [Tswana cp e 03°(T) (5.1)

54

and get a generalized Gaussian random field {Xw (p E C8°(T)} .
Lemma 5.2 For any open set 0 Q T,
H(0;X) = H(0;£).

where
H (03X ) = WW». so 6 03" (T) with suppvJ .C_ 0}
and

H(Oifl = Witt, t6 0}-

Proof. If cp E 08°(T) with suppp Q 0 then fT ficp(t)dt as a limit of Riemann
sums belongs to H (0; 5 ), so that

H(0.X) SH(0;€)-

To prove the converse inclusion, let to E 0 and choose N so that {t E

T, It — to] < i} Q 0 for n 2 N. Let 6 > 0 and cpl be the function

-n 2
W _I e mum) It-tol <5

0 otherwise

Observe fT <p_:_(t)dt = 1, 50%, E C3°(T) and supp(cpn) Q 0, n 2 N, Then

EI/oe.(t)dt-o.l .<. EIIt—eowtwt

S suPPIr—rogi-Elét "’ €10] —’ 0 as n —* 00

Hence X,” —-> {to in probability as n -+ 00, {to E H(OiX) and H(Oié) =
H(0;X). D

From Lemma 5.1 and Lemma 5.2 we have:

55

Corollary 5.1 Let {X.,,, cp E C3°(T)} be defined as in (5.1). Then for an
any open set 0, it has MPI on 0 ifl' it has MPH on 0.

We know that the MP1 of {X,,,, cp E CS°(T)} can be characterized as
some properties of the corresponding reproducing kernel Hilbert space (see
Theorem 3.1 and Theorem 3.2). By lemma 5.2 we know that the Markov
property of {§.,t E T} for open sets is equivalent to MPI of {X,,, cp E
C3°(T)}. So we can use Theorem 3.1 and Theorem 3.2 to get similar results
for {£¢,t 6 T}. First we shall deduce some relationships between the RKHS
of {X.,,, cp E C8°(T)} and that of {§¢,t E T}.

Let K (Cx) and K (Cg) be the reproducing kernel Hilbert spaces of
{Xw so 6 CS°(T)} and {£¢,t E T} respectively, and we denote by H (X ) and
H (C) the linear spaces in Lg(fl, F, P) generated by {X,,,, (p E C8°(T)} and
{§t,t E T} respectively. Notice that since R(s,t) = E(£,£¢) is continuous,
every element in K (Cg) is also a continuous function on T. Define II"1 and

II"1 as follows

H“=H(X) —+ K(Cx).

(II'1Y)(<p) = E(YX,,,) Y€H(X).
H"‘=H(£) -r K(Ce).

(H"‘Y)(t) E(Y€¢) YE 11(6)-

We know that both If”1 and II"1 are isometric maps. Since H (X ) = H (C ),
hence J = II‘III’ is an isometric map between K (Cg) and K (Cx). We can

56

explicitly express the map J, if f E K (Cg) then

(We) = (H“(H’f))(<p)=E[(H’f)Xel
= Elms/Towns]
= fTE(H’f£:)<p(t)dt
= Inseam.

It can be easily checked that for f1, f2 and f E K (Cg)

seppr) = suppf (5.2)

and

(anszlmcx) = (fl, f2)K(C¢)- (5.3)

Where supp(J f ) is defined for a linear functional J f of Cg°(T)(see Lemma
2.2(a)) and supp f is defined as the complement in T of the largest open
set 0 such that f (t) = 0 on 0. Now we can state and prove the following
improvement of Kiinsch[l3] and Pitt[24].

Theorem 5.1 Let T be an open set of 32" and {{,,t E T} be a Gaussian
process with continuous covariance. Then it has Markov property for all

open subsets of T if the following hold.
(a) (f19f2)K(C¢) = 0 iff] and f2 6 K(Cg) such that (suppf1)n(suppf2) =

(b) Iff E K(Cg) and f = fl +f2, where f1 and f,» are continuous and
have disjoint supports. Then f1, f2 6 K (Cg).

Proof. We know that {£,,t E T} has Markov property for all Open sets
of T is equivalent to the MPI of {X,,,, (,0 E C3°(T)}, where X, is defined

57

in (5.1). Hence we only need to show that (a) and (b) of Theorem 3.1 are
equivalent to (a) and (b) of Theorem 5.1. It is easy to see that (a) of Theorem
3.1 is equivalent to (a) of Theorem 5.1 because of (5.2) and (5.3). To show
that (b) of Theorem 3.1 implies (b) of Theorem 5.1, let f E K (C5) and
f = fl + f; with f5’8 being continuous and having disjoint supports. Then
for any cp E 03°(T)

(map) = [Thea + I been 2 Me) + no.0) say,

since F1 and F2 are linear functionals on C§°(T) and suppF; =suppfg, hence
F; E K (Cx) by (b) of Theorem 3.1. Then F;(<p) = IT ffcpdt with f,-’ E
K(Ce),(i = 1,2). Now If. fgcpdt = fT flcpdt for all cp E C3°(T), (i = 1,2)
implies f.- = f,’ (i = 1,2) because f,- and f,’ are continuous. So f.- E K (C5).
Conversely if F E K(Cx) F = F1 +F2 with F1 and F2 being linear on C8°(T)
and having disjoint supports, then F (cp) = fT ftpdt with f E K (Cc). Since
supp f =suppF Q (suppFl) U (supng), define for i = 1, 2

0 otherwise

f.-(t) = I f(t) te suppe-

then we can show that f1 and f; are continuous and supp f,- quppF}, i = 1, 2,

furthermore f = fl + f2. By (b) of Theorem 5.1 f; E K (Cg). Then

F1(¢p)-/Tf1<pdt = [T fz‘Pdt - F2(<.0). 90 6 08°(T)-

Both sides are linear functionals on Cg°(T) and have disjoint supports, so
they are zero functionals by Lemma 2.2(b), hence F.-(<p) = f fggodt (i = 1,2)
andF;6K(Cx) t=1,2. Cl

58

From the above proof, we notice that the choice of M (T) is not unique.
Any M (T) which satisfies the following additional assumptions (5.a) and
(5.b) besides (A.1) - (A.3)will do the job:

(5.a) If f is a continuous function on T and for every open set 0, fT fdp =
0 for all p E M(T) with suppp Q 0. Then f = 0 on 0.

(5.b) For every open subset 0 of T, H (0;£) = H(O; X) where H (0:5) =
s—pafi {fut E 0},H(O;X) = spa—1i{X,,,p E M(T) with suppp Q 0} and
X1: = IT {tdfl-

From this we can generalize T to any separable locally compact Haus-
dorff space.

Let E be a separable locally compact Hausdorff space and {£,,t 6 E}
be a Gaussian process with continuous covariance function. Let m be a
positive Radon measure on E with supp(m) = E. Then we define

M(E) = {fine-em : n z 1,A,~ 6 3(3),..- 6 00(3)}

where Co(E) consists of all continuous functions on E with compact support.
By Example 2.2 M (E) satisfies assumptions (A.1)-(A.3). Also if f is contin-
uous and f3 fdp = 0 for all p 6 M(E) with suppp Q 0 then I}; f(pdm = 0
for all (p E Co(E) with suppgp Q 0 this implies f = 0 on 0 because f is
continuous and supp(m) = E.

Define
X, = Led). I. e M(E). (5.4)

Then we have

Lemma 5.3 for every open set 0, H(0;£) = H(0;X).

59

Proof. It is enough to show the lemm for precompact open set 0. Let
p E M (E) with suppp Q 0, then since any continuous function f can be
approximated by fn on 0 with form fn(t) = 217;", f(t,-)1A1 (t) with t.- E 0 and
A.- E B(E) i.e. fn —> f pointwise on 0. Let {u(t) = 2:2; {t,lAJt) with t.- 6 0
, {u(t) —+ {I on every t E 0 in L2(Q,F, P) . Then fE £,,(t)dp —r [E {tdp = X“
in L2(fl, F, P) because suppp Q 0. But fElfndp e H(0;£) hence X” E
H (0; £). On the other hand let to E 0 then we can choose precompact open
set 0,,,to E 0,, such that 0; Q 0 and 0:1 {to}. Let dpn = anlondm with
a,, = infil- then we can see that p,. E M (E) with supppn C 0 and

El / ode. - an 5 12/0” ao|£(t) - «tundra
= a. / least) — £(to)ldm
s sup.eo.E|€(t) - f(to)la. [0” dm
= sup..o,.EIe(t) - «to».

Since 0n are all precompact and contained in compact set 0:, also 0,, l {to}
by uniformly continuity of E [5; — {ml on 01— we have

SUPpreo.E|€(t) " {(to)| "" 0 as n "" 0°-

Hence {to E H(0;X). D

We shall have the following isometry between K (0:) and K (C x) by

J: K(Cg) _. K(Cx)
(mu) = from rem.)

60

and similar to (5.2) and (5.3), for f E K (Cg), we have

suerf) = suppf (5-5)

and (Jfr, Jf2)K(Cx) = (f1.f2)K(o¢) (5.6)

The following theorem is extension of Theorem 5.1 and the proof is almost

the same and hence omitted.

Theorem 5.2 Let E be a separable locally compact Hausdorff space,{£,,t E
E} be a Gaussian processes with continuous covariance and K (0:) be the
RKHS of its covariance. Then it has the Markov property for all open sets
if

(a) (f1,fg)x(C5) = 0 iff] and f2 6 K(Cg) with disjoint supports.

(b) Iff E K(Cg) f = fl + f;, where f1 and f2 are continuous and have
disjoint supports, then f.- E K (C5) (i = 1,2).

We also get the following theorem similar to Theorem 3.2.

Theorem 5.3 Let {£¢,t E E} be the same as in Theorem 5.2, then it has
the Markov property for all pre-compact open sets if

(a) (f1,f2)x(c,) = 0 if f1 and f; E K(Cg) with disjoint supports and one
of the supports is compact.

(b) If f E K (0:), f = fl + f; with f1 and f2 being continuous and having
disjoint supports of which one is compact. Then f.- E K (C5) (i = 1,2).

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