A NNALES DE L ’I. H. P., SECTION B
J IANGANG Y ING
Bivariate Revuz measures and the Feynman-Kac formula
Annales de l’I. H. P., section B, tome 32, n
o2 (1996), p. 251-287
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Bivariate Revuz
measuresand the Feynman-Kac formula
Jiangang
YING*Department of Mathematics, University of California, San Diego,
La Jolla, CA 92093-0112.
Vol. 32, n° 2, 1996, p. 251-287 Probabilités et Statistiques
ABSTRACT. - In the first
part
of thepresent
paper additive andmultiplicative
functionals of aright
Markov process areinvestigated systematically
in thesetting
of weakduality, by
means of bivariate Revuzmeasures. We first
give
arepresentation
for such measures. It isproved
that additive and
multiplicative
functionals areuniquely
determinedby
theirbivariate Revuz measures and two
multiplicative
functionals are dual if andonly
if their bivariate Revuz measures are dual. In the secondpart
we prove that anysubprocess
of anearly symmetric
Markov process is alsonearly symmetric
andgive
ageneralized Feynman-Kac
formula which describes therelationship
between theircorresponding
Dirichlet forms.Key words: Markov processes, multiplicative functionals, Revuz measures, Dirichlet forms.
Part I. BIVARIATE REVUZ MEASURES 1. Introduction
In the first section we will set down all definitions and notations inforced
throughout
this paper. Under thesetting
of weakduality
and some additionalconditions,
two results ofSharpe (representation
of terminal times and* Research supported in part by NSF Grant DMS 91-01675.
1991 Mathematics Subject Classification: 60 J 45, 60 J 40, 31 C 25.
Annales de l ’Institut Henri Poincaré - Probabilités et Statistiques - 0246-0203
Vol. 32/96/02/$ 4.00/@ Gauthier-Villars
decomposition
ofMFs)
aregiven
in§2. They
are criticalthroughout
thisarticle.
Sharpe’s
canonical measure andLevy system,
which are usedto describe discontinuities of the process, are discussed in
§3. They
areemployed
togive
arepresentation
for bivariate Revuz measures in§4.
A
generalized
Revuz formula isgiven
in§5
and it isespecially
usefuland
indispensable
indealing
withmultiplicative functionals,
In§6
we useKuznetsov measures to prove that a
multiplicative
functional isuniquely
determined
by
its bivariate Revuz measure and that twomultiplicative
functionals are dual if and
only
if theirrespective
bivariate Revuz measures are dual.In the second
part
we aregoing
to prove theFeynman-Kac
formula fornearly symmetric
Markov processes(or non-symmetric
Dirichletforms)
and
general decreasing multiplicative
functionals.Let
be two
right
Markov processes on(E, ~)
with Borel sub-Markoviansemigroups ( Pt )
andrespectively,
and be in weakduality
relativeto a fixed a-finite measure m on E: for all
f, g
Ep~,
where
(., .)
denotes the innerproduct
inL2(m).
Walsh[Wa]
has shownthat under weak
duality,
(1.2) Xt_
exists in E for all te]0, ([
a.s.P"2,
and of course the dual assertion holds for
X. Assuming (1.2) identically
on H
(resp.
onH)
relieves us fromcarrying
one moreexceptional
seteverywhere.
As iscustomary,
we use theprefix
"co" or """to describe
quantities
relative toX. Though
all definitions and standard results arestated
regarding
toX,
the dual statements alsoapply
toX.
Denote
by Excq ( X)
and(Excq
or Sq if no confusion will becaused)
the cones ofq-excessive
measures andfunctions, respectively.
DEFINITION 1.3. - A real-valued process M =
(Mt :
t >0)
is called amultiplicative functional (MF) of
Xif (1 )
t ~Mt (w)
isdecreasing, right
continuous and has values in
[0, 1] for
each03C9 ~ 03A9; (2)
M isadapted, i.e., Mt E Ft for 0; (3) Mt+s(w)
=Mt(w) . for
any s, t > 0and w E f2. In
addition,
an MF M is exactprovided for
any t > 0 andAnnales de I’Institut Henri Poincaré - Probabilités et Statistiques
every sequence
tn ~~ 0,
Let
MF ( X ) (or
MF if no confusion will becaused)
be the setof all exact
multiplicative
functionals of X. For any MF M write:=
inf {t
> ~ :Mt
=~~, EM
.-{.r
E E : =1)
=1 ~,
for the lifetime and the set of
permanent points
ofM, respectively.
LetR:==
inf{t
> 0 :Xt
E andR’ .-
0 :Xt
EClearly Esi
is also the set ofirregular points
ofSM , i. e. , EM = ~ x
EE : >
0) = 1}.
LetMF+
:={M
E MF :SM
> 0a.s.}
andMF++ . - ~ M
E MF : M does notvanish, i . e . , S~~ > ~ a.s.}.
IfEM
is
nearly optional,
M is called aright
MF. If M is exact,EM
isfinely
open and
consequently
M isright.
We define for each x E a
probability
measure onQ~
on( SZ, by
where
Moo :=
0 and are thekilling operators
on 03A9 definedby ktw( s)
=w(s)
if t > s and = Aif t
s . If M isright,
then(H, Xt, Qx)
is also aright
Markov process, called theM-subprocess
ofX and denoted
X M
or( X , M),
with state spaceClearly
thesemigroup (Qt)
and resolvent of aregiven by
and = = 0 for any x
(j.
The set ofq-excessive
functions
(resp.
excessivemeasures)
ofX ~’~
is denotedby Sq (M) (resp.
Excq(M)).
An
(Ft)-stopping
time T :03A9 ~ R+
is called a terminal time if T = t + T0 ()t identically
on{t T}.
If T is a terminaltime,
1 isan MF of X. Write
Excq(T)
andSq(T)
:= It iseasy to see that is a terminal time if M E
MF(X).
DEFINITION 1.6. - Let M e MF. A
positive, increasing, right
continuousprocess A =
(At :
t >0)
is a raw M-additivefunctional (of X ) provided
and =
At
+Mt . As
oet
a.s.for
each tand s. A raw M-additive
functional
A is an M-additivefunctional (of X ) if
A isadapted.
Vol. 32, n° 2-1996.
Let
RAF(M)
andAF(M)
denote the sets of raw M-additive functionals and M-additivefunctionals, respectively.
Write RAF forRAF(l),
theset of raw additive
functionals,
and AF forAF ( 1 ),
the set of additivefunctionals of X. For a terminal time
T, RAF(T) :=
andAF(T)
:=The energy functional L
(of X)
is defined on Exc x Sby
Refer to
[Ge]
for the basicproperties
ofL,
among which areu) =
and
L(m, U f )
=md( f )
where md is thedissipative part
of m.We now introduce the
powerful
Kuznetsov measures. Let W be the space ofpath w :
IR --~ EU ~ 0 ~
that are E-valued andright
continuous on anopen interval
] c~ ( w ) , ,~ ( w ) ~
and take the value A elsewhere. We denoteby
[A]
thepath
whichconstantly equals
to A. Let Y = EIR)
denotethe coordinate process on =
w (t) .
The shifts at : W -~ Ware definedby Ys
oYs+t.
Put and~° _ ~°.
Thenfor any m E
Exc,
there exists aunique
a-finite measureQm
on(W, ~° )
not
charging ~ [0] ~
such that iftl t2
...tn,
The measure
Qrn
is called the Kuznetsov measure of(X and)
m.Clearly
and
importantly
it istranslation-invariant;
thatis, at( Qm)
=Qm,
for each~ E IR.
Notations. - For M
e MF,
letMt
:= 1 -klt. Clearly
M EAF(M).
For any process(Zt)
andincreasing
process(At ),
we define(Z- * A)t
:=fo
and(Z * A)t
.-~’0
For anyf
Ep~
and F E
p~
x~,
definef * A
:=f ( X ) ~ A, f _ ~ A :
:=f ( X _ ) ~ A
andF * A :=
F(X_ , X ) ~ A,
wheref (X )
:=( f (Xt)), f (X-)
:_(f(X-t))
and
F(X_,X)
.-(F(Xt_, Xt)).
2.
Decomposition
and weak*duality
We will focus on a
special
class ofmultiplicative
functionals which havea nice
decomposition.
From now on we will write a. s. for anddrop
the m on Revuz measures asthey
are taken withrespect
to m if no confusion would be caused.Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
DEFINITION 2.1. -
(i) M,
N EMF (X )
arem-equivalent provided that, for
each t >0, Mt
=Nt
a.s. on~~
>t~. (ii)
Let M E MF andA,
B ERAF(M).
Then A and Barem-equivalent provided that, for
each t >
0, At
=Bt
a. s.on ~ ~
>t ~ .
In view of
right continuity
ofMFs,
this isequivalent
to the statementthat t ~
Mt
and t ~--~Nt
are identical functions a. s. on[0, ([.
Henceequality
between MFs(resp. M-RAFs)
willalways
be understood to meanm-equivalence.
THEOREM 2.2. -
If
thefollowing
condition holds(2.2a)
everym-semipolar
set ism-polar,
then any M E
MF+
has adecomposition
where ~
E ~
x~,
0 ~1,
~ vanishes on thediagonal
Dof
E xE,
a Ep~,
A is a continuous additivefunctinal of X,
B is a Borel subsetof
E x Ewhich is
disjoint from
D andSM
=JB
.-inf~t
> 0 :(Xt- , Xt)
EB~.
The
proof
mimicks that of[Shl,
Theorem(7.1 )] .
We willjust
sketch ithere. The
Stieltjes logarithm
of Mis a
SM-additive
functional which can bedecomposed
into the sum of anatural
part
and apurely
discontinousquasi-left-continuous part.
Note that from[GS, (16,21)]
twoequivalent
statements of(2.2a)
are(2.2b)
every natural additive functional is a.s.continuous;
(2.2c)
if T is a thin natural terminaltime, ()
= 0.Hence the natural
part
is continuous and it follows from[GS (16.14)]
that.J B
withB C E
x Edisjoint
from D. Thepurely
discontinuouspart
has common discontinuities as X and must be of the formX~ )
st with
0 ~
1 since alljumps
ofslogM
are less than l.Weak
duality together
with(2.2a)
is called weak*duality.
Anexample
of weak*
duality
is apair
ofnearly symmetric
Markov processes(i.e., they satisfy
the sectorcondition)
in weakduality.
Fitzsimmons[Fi2] proved
that in this case every
semipolar
set ism-polar [Fi2 (4.13)]
andm-polar
is the same as
m-copolar [Fi2 (4.17)].
Hence these two processes are in weak*duality.
Now M E
MF~
is said to besimple
if it has thedecomposition (2.3),
inwhich case we write M E
SMF~ .
A terminal time T ESMF +
means thatT =
JB
> 0 a.s. with B e S x ? which isdisjoint
from D.Vol. 32, n° 2-1996.
3.
Levy systems
and canonical measuresA
Levy system
for X is apair (N, H),
where N is a kernel on(E, ~u)
with
N(x, {~r})
= 0 for any xE E
and H is a continuous AF of Xhaving
a bounded
1-potential,
such that for any F ep~u
x ~uvanishing
on thediagonal
and anypredictable
processY,
where
NF(x)
Refer to[Sh, §73]
for the existence ofLevy systems.
LEMMA 3.2. - Let B
E [x
~’ and T :=JB
> 0 a.s. Then(XT-, XT)
e Ba. s. on
{T
anda. s. on H.
Proof -
Define the second hitT~ :=
T o0T
+ T. The statemant(3.3)
isobviously
true on{T
=oo}.
On{T oo},
since T > 0 a.s.,T2 - T = T o 8T > O
a.s. Hence(XT-, XT)
E B a.s. on{T ~} by
thedefinition of
JB .
Then(3.3)
followsdirectly..
THEOREM 3.4. - Let
(N, H)
be aLevy system for
X and M ESMF+
with the
decomposition (2.3). Define ~
.-1B
+ andNo
.-( 1- ~ (x, y) ) N(x ,
Then(No, H) is a Levy system of the M-subprocess (X~, M).
Proof. -
We haveBy (3.2)
the discontinuouspart Md
of M takes thefollowing
formAnnales de l’Institut Henri Poincaré - Probabilités et Statistiques
Hence
and then
By definition, (No, H)
is aLevy system
of(X, M).
NWe will next introduce the canonical measure of X relative to m, which
was first used in
[Sh2].
While theLevy system
describes the discontinuities of the processcompletely,
the canonical measure does this relative to m.The canonical measure v is characterized as a a-finite measure on E x E which is carried
by
E x E - D and such that for any F Ep?
x ?vanishing
on D
It follows from
(3.1)
thatwhere
(N, H)
is aLevy system
of X. Thus(3.4) implies:
COROLLARY 2.8. - Let M E
SMF+
and denote the canonical measureof (X, M) (relative
tom).
Thenv~ _ (1 -
The transform
by
amultiplicative
functional M E MF consists of twosteps
ofkilling:
a first entrance time R’ =DEcM
and amultiplicative
functional in
MF+ ( R’ ) .
Thefollowing
resultgives
us aLevy system
of theR’-subprocess
of X.THEOREM 3.9. - Let
(N, H)
be aLevy
systemof
X. Then(N’, H)
is aLévy system of the R’-subprocess of X, where N’ (x, dy)
=N(x, dy).
Vol. 32, n° 2-1996.
Proof. -
Let(Pxr : x
eE)
be theprobability
distribution of theR’-
subprocess.
For any F ep~
~ which issupported
on xEM,
apredictable
process(Zt)
and x eThe third
equality
holds sinceF ( X ~~ _ , X R~ )
= 0 due to the fine closeness of and the fact that e4. A
representation
of bivariate Revuz measuresBivariate Revuz measures were also introduced in
[Shl]
and willplay a
critical role in
coming
sections. We assume that(N, H)
is aLevy system
of X andv(dx, dy)
=N(x,
the canonical measure of X relative to m in thesequel.
DEFINITION 4.1. - Let M E MF and A E
RAF(M).
The bivariate Revuzmeasure
of
A relative to m isdefined
to beAnnales de l’Institut Henri Poincaré - Probabilités et Statistiques
The Revuz measure and
left
Revuz measureof
A relative to m aredefined
to be
As is
customary, F(x, y)
= 0 if either x = A or y = A. The existence of the limits above can be shownsimilarly
as in[FG2 (2.5)].
Note that we needto assume that
X~_
exists in(4.3)
if A doescharge (.
For anyf, g
E?,
we write
f(x)g(y)
asClearly
= butva(~ ~ 1) ~ ~A
ingeneral.
We denote the bivariate Revuz measure, Revuz measure and left Revuz measure of Mby respectively.
We also define thebivariate
potential
of A ERAF(M) by
for any F E
p~
x ? and =Ll~
for Me MF.
Denote =L!A ( 1 ~ ~ )
for A e
RAF(M), U~~
.-.),
and .- forf(x) for x ~ EM. Clearly PqM = UqM
if M ~MF+.
Recall that
[M]
denotes theStieltjes logarithm
of M EMF~~
and inthis case
~M]
E AF. Thefollowing
lemma isjust
a collection of facts about bivariate Revuz measures.They
can be checked eithereasily
orby mimicking
theproofs
of therespective
results for Revuz measuresin
[Yil].
LEMMA 4.5. - Let M E MF and A E
RAF ( lyT ) .
(4.5a) vA(F) ==1 limq qmUqAF;
(4.5b) U1F
E andvA(F)
=ifm
EDis(M);
(4.5c)
vA does notcharge
anym-bipolar
set, which is a setB ~ E
x E such that either B G E xBo for
somem-polar
setBo
or B CB1
x Efor
somem-leftpolar
setB1;
(4.5d) ~A
=vA ( ~
(g)1) if A
does notcharge (;
(4.5e) If A is natural, vA(F)
=pA ( FD ) _ for
any F Ep~
x~,
where
FD (x)
:==F(x, x).
’(4.5f) If M
eMF++,
then =Now we are
going
to prove animportant representation
theorem for bivariate Revuz measures.Vol. 32, n° 2-1996.
THEOREM 4.6. - Let M E
SMF+
with thedecomposition (2.3).
Then the bivariateRevuz
measureof
M isgiven by
where
03B4{y}
is thepoint
massat y
and v is the canonical measureof
X .Proof. - By (3.2),
if ~ :=1B
+~,
we haveDefine
Clearly Mt
=M03B4SM
= 0 and =Mdd(-Ms ) -+-Ms _ d ( - Md ) .
Then for F Ep~
x~,
since A iscontinuous,
Hence
by [FG2 (2.22)]
On the other
hand,
sinceMSM
= 0 =1,
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
Then we use the
Levy system
formula and[FG2 (2.22)]
Hence we have
That proves
(4.7)
.The
following corollary
is easy to check.COROLLARY 4.11. - Let M E
SMF+.
Then we have(a)
vsM = v;(b)
Recall the definition of
Stieltjes logarithm
of amultiplicative
functionalin
(2.4).
Now we aregoing
tocompute
its bivariate Revuz measure.PROPOSITION 4.12. - Let M e
SMF~
with thedecomposition (2.3).
Thenthe bivariate Revuz measure
of
itsStieltjes logarithm slogM
isgiven by
Proof -
Since M admits thedecomposition (2.3),
itsStieltjes logarithm slogM
admits thefollowing decomposition
Vol. 32, n° 2-1996.
Hence for any F e
p~
x~, using
theLevy system formula, (3.2)
and[FG2 (2.22)]
we findThat
completes
theProof..
Combining (4.7), (4.11 a)
and(4.13),
we can see that if M eSMF+,
thenBut
(4.15)
holds even without theassumption
that M has asimple decomposition. Actually
we have a moregeneral
formula. Recall a useful formula[Yil (3.12)].
If M EMF+,
m E Dis and u eS,
thenTHEOREM 4.17. -
(i)
LetM,
N E MF withEN
C Thenwhere is the bivariate Revuz measure
of
~l relative to m* =under the
N-subprocess of
X .(ii)
Let M EMF. ThenProof - (i) Clearly EMN
= For any F epE
x~,
Annales de l ’lnstitut Henri Poincaré - Probabilités et Statistiques
But
where denotes the
probability
measurescorresponding
to theN-subprocess
of X. Thus if we write(M_ * N)t
:=M- * N,
thenWe claim that the
following identity
holdswhere := Px(N)
~0 f(Xt)d(-Mt)
= Pxf~0 f (Xc)Ned(-Mt).
In fact for F E
pE
~ and x E with o0Since M e
MF- (X, N),
we can use(4.1 f )
and find(ii)
Let N = Then M =MN, EN
= andb~ (i)
we havevM*
= NvM* ~ vsM .
butFollowing (4.18)
andusing
the fact thatslogM
EAF(SM),
it is easy to check thatThen
using (4.16) again,
we haveVol. 32, n° 2-1996.
That
completes
theproof.
Our last result of this section describes therelationship
between the left Revuz measure~M
and leftmarginal
measureof vM.
PROPOSITION 4.19. - Assume that
X ~ _
exists in E a. s. andM~
=h 0. Let M EMF+
and 03BA be thekilling
measureof
X relative to m(the left
Revuzmeasure
of ~)
Then A M
= 01)
+ ~.Proof. -
Forany f
Ep~
we haveThe second term is the left Revuz measure of
(J AF(M),
which is
equal
to r~by
anargument
similar to that used to prove(4.17). N
5. A
generalized
Revuz formulaThe Revuz formula was first
given
in[Rel]
understrong duality
and laterproved by
in[GS]
and[GG]
under weakduality.
Wequote
thefollowing
form of Revuz formula from
[GS (9.9)]
and[GG (A8)].
Let A ERAF(X)
which may
charge
the life time(.
Assume that A has a a-finite left Revuz measure, which is defined as(5.1) Hypothesis. X~ _
exists in E a.s. >0 ~ .
Under
(5.1)
it holds that For anyf , g E p~, using (5.1)
we havewhere
( f * A) t :
-Jo f(Xs)dAs.
Note thatf * A
nevercharges (
and thus(5.2)
holds without(5.1).
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
In this section we will deduce a similar formula for an
M-additive
functional. Let M EMF(X)
and A ERAF(M) having
a a-finiteleft Revuz measure relative to
1EM .
m EExc(M),
andM
the dualmultiplicative
functional ofM; namely, M
EMF(X)
such that(X, M)
and
(X , M)
are in weakduality
withrespect
to m. Write m* :=mlEM
and let
(Qt)
and(resp., (Qt)
and( Tl q ) )
be the transitionsemigroup
and resolvent of
(X, M) (resp., (X , M)).
Here A maycharge (.
Now we will
give
the somepreliminary
lemmas and thegeneralized
Revuz formula. The
proofs
wepresent
here areanalogous
to those in[GS].
LEMMA 5.3. -
Let ~
E7Z+
and bea-finite
measure on E. Thenfor
each t >0,
Proof -
The directcomputation gives
That
completes
theproof..
THEOREM 5.5. -
Suppose
that a.s. t ~--~f (Xt_)
isleft
continuouson 0, oo ~.
Then under(S.1 )
Vol. 32, n° 2-1996.
By
the lemmaabove,
theproof
of this theorem isexactly
the same asthe
proof
of[GS (8.7)].
We won’trepeat
it here.THEOREM 5.7. - Under
(5.1 ),
thefollowing Revuz formula holds,
Proof -
Since isa-finite,
it suffices to prove(5.8)
when(E)
oo. First notice thatnothing
will bechanged
if wereplace
m* in the
rightside
of the above formulaby
m. Letf
eSq(M)
and bebounded. Then t ~--~
f(Xt-)
is left continuous on a.s. LetCt(x)
= Thus =Ct
+QtCs
andBy
theprevious
theorem and the facte-qtQtf
ESq(M),
we haveTherefore
(Qt f’)
is theright
derivative of( f, Ct)
andwhere N := Thus t ~--~
(,f, Ct)
isabsolutely
continuous and it follows fromCo
= 0 thatFor any bounded continuous h > 0 which
supported by qVqh
---~ hboundedly.
Hence(5.8)
holds for suchh..
Finally
we willpresent
two useful consequences. Let.~(~c, v)
:=v)
for ic eS
and v E S. Since L isjust
the Dirichlet form of X in some sense, wethereby
obtain some idea on transformations of Dirichlet forms.Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
COROLLARY 5.9. -
( 1 )
Thefollowing generalized Revuz formula
holdswithout
(5. I )
(2) If
M E MF andSM
> 0 a.s., thenProof. -
Part(1)
is immmediate from(5.7)
as we indicated in(5.2). (2)
Since icm
eExc(X)
for ic ev~’, using [Yi (2.5)]
and(5.5),
we haveThe reason that we can
apply (5.6)
here is that doesnot
charge (..
6.
Uniqueness
and dualmultiplicative
functionals Thefollowing
lemma is an m-a.e. version of[BG (IV.2.12)].
LEMMA 6.1. - Let
AI, A 2
E AF.Suppose that, for
somefixed
q >0,
UqA1 1
00 a. e. m, and thatfor
any bound and continuousf
Ep~,
=
U2f
a.e. m. ThenA1
andA2
arem-equivalent.
Proof. -
It is clear that a.e. m above can bereplaced by
q.e.,namely,
except
a set ofm-capacity
zero. It follows from theproof
of[BG, IV(2.12)]
that = a.e. m for
p >
q.Since q
---~ isdecreasing
and continuous for any x E
E,
there exists N E ~ withm(N)
= 0 suchthat oo and hold for any
p > q
andx
¢
N. Thus it follows thatA1
andA2
arem-equivalent.
First we have the
following uniqueness
result.PROPOSITION 6.2. -
Suppose
thatA2
E AF which area-integrable.
~4~
andA2
arem-equivalent if
andonly if
their bivariateRevuz
measuresVol. 32, n° 2-1996.
are the same
Proof. -
Theonly
ifpart
is trivial. Now assume that vAl = vA2. Sincethey
area-finite,
there exist a sequence of Borel functionsbn E ~
X ~such that
1)
0bn
1 for any n;2) bn ~ 1;
3)
Bothbn .
vAi andbn .
03BDA2 are finite.Let
Thus
From
(6.2),
it follows that = a.e. m for anyf
e Onthe other
hand, by
the Revuzformula,
Thus oo a.e. m. Therefore and are
m-equivalent
andby passing
tolimit, Al and A2
arem-equivalent..
Let
M,
N E MF andS’~,T, Sl~
the lifetimes ofM, N, respectively.
If Mand N are
m-equivalent,
then P~ a.s. =SN,
0EN
ism-polar
and
03BDmM*
= X where m* := Thefollowing
result is apartial
answer to its inverseproblem.
THEOREM 6.3. - Assume that
M,
N ESMF+.
Then M and N arem-equivalent if
andonly if
=Proof -
It suffices to check thesufficiency. Clearly
M and N have thedecompositions
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
where
1) SM = JK
>0, SN
=JL
>0,
and both K and L aredisjoint
fromD;
2)
Ep~
x 1 and 03A8 1everywhere;
3) a, b
ep~
andA, B
E AFbeing
continuous.Then we know that
Hence vM = vN
implies
thatThus we have
Since
everywhere, v(KC
0L~)
=0,
orequivalently v ( K ~ L )
= 0.Choose a sequence of Borel subsets
(En )
of E x Esatisfying:
1) En I E x E - D;
2) v(En)
oo for any n;3) En C ~ (x, ~) : d(x, y)
>n ~,
where d is a metric on Ecompatible
with the
given topology.
Existence of such sequence
~En ~
is easy to check. LetClearly v{(K
nEn )
~(L
n = 0. Defineand
where and
,5‘N, n
are thekth-iterates
ofSM,n
andSN,n .
Then
Kn, Ln
E AF andFrom the Lemma
(6.2),
we know that for anyf, g
Ep~,
Vol. 32, n° 2-1996.
On the other
hand, by
the Revuzformula,
Hence
U1Kn
1 oo a.e. m. Therfore Kn and Ln arem-equivalent.
ButIt follows that P"2 a.s.
SM,n
=SN,n
and thus Pm a.s.SM
=SN.
Now we have
slogM, slogN
EAF(S),
where S : :==8M
=SN.
Butby (4.15)
It follows from
(6.2)
thatslogM
andslogN
arem-equivalent.
Hence Mand N are
m-equivalent.
Before our next
result,
we would like to recall somebackground.
Twomultiplicative
functionals M EMF(X)
andM
EMF(X)
are said to bedual if their
corresponding subprocesses
are dual withrespect
to m. In[Shl], Sharpe proved
that if X andX
are instrong duality
relative tom and M e
MF(X)
andA~
eMF(X),
then M and~V
are dual if andonly if, 1)
thecorresponding
exactregularizations
ofSM
andSN
are dualterminal
times; 2)
the bivariate Revuz measures of theStieltjes logarithms
of M and
N
aredual;
that isHere we are
going
to prove some similar results in weakduality by
meansof Kuznetsov measures.
Let
Q
be the Kuznetsov measure of X and m. SZ is identified as asubspace
ofW, i. e. ,
Shift
operators
and truncated shiftoperators
are defined asClearly Btlo
is the shiftoperator
of X andB(~a t~) C S2.
We also definea reversal
operator
A : W asAnnales de l’lnstitut Henri Poincaré - Probabilités et Statistiques
for any s E R
(write
:= 03BBw and =Yt(w))
and the backward shiftoperator ât
on Wnaturally
asWe have
Hence We also have A :== and
/3
:= -cx.Similarly, SZ
is identified asor
equivalently,
Let
Q
be the Kuznetsov measureof X
and m onW, i.e.,
fort1 t2
...tn,
It is known that
Q
is the reversal ofQ, i.e.,
In fact for any
tl t2
...tn,
Let A E
RAF(X)
which induces aHRM,
denotedby A*,
of Yby
the formula
Vol. 32, n° 2-1996.
Precisely,
A* is carriedby a, ,~~
andNow define an RM on W
LEMMA 6.8. -
A*
is a backward HRMof
Y in the senseProof. -
Checkdirectly by definition,
That
completes
theproof..
DEFINITION 6.9. -
1 )
Let A ERAF(X)
andA
ERAFeK). They
are saidto be dual
m-equivalent
to whereA*
isdefined
in(6. 7). 2)
Twomeasures 1 and 2 on E E are said to be dual if 1 (dx, dy) =
2(dy, dx).
Remark. - The construction of the dual
multiplicative
functionals is similar. Forexample,
a terminal time T of X induces astationary
terminaltime T* of Y
by
Define the dual of T*
by
Then T is the dual terminal time of T
(See [Mi2]). Clearly
the birth timea and the death
time $
are dual.THEOREM 6.10. -
1 )
Let A EAF (X )
andA
EAF(X)
which are a-integrable.
Then A andA
are dualif
andonly if
vA andÂ
are dual.2)
Furthermoreif
A andA
arenatural,
thenthey
are dualif
andonly if
PA =
pA.
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
Proof. - 1)
First assume A and~4
are dual. We canidentify ~4
here with the dual constructed above. Nowby [Ge (8.21)],
for any F Ep~
x?,
where §
EpR
= 1 andF(x, ~)
.-F(y, x).
Conversely,
assume that vA andvA
are dual. LetA’
:=A* ~~
Thenwe have
From
(6.2)
it follows that~4 and A’
arem-equivalent.
Therefore A and~4
are dual.
2)
follows from1) directly,
due to the fact that bivariate Revuzmeasures of natural additive functionals are carried
by
thediagonal
D..THEOREM 6.11. - Let M E
MF++(X)
andM
EMF++ (X ).
Then thefollowing
statements areequivalent.
(1)
M andM
aredual;
(2) [M]
and~M~
aredual;
(3)
vM andvM
are dual. -Proo, f : -
Since z/M = and = theequivalence
of(ii)
and(iii)
follows from(6.10).
Now we aregoing
to prove theequivalence
of(i)
and(iii).
First we assume that M andM
are dual.Using
the Revuzformulas
(5.2)
and(5.10),
we find for any boundedf , g
ep~,
Let .-
^Yl f )
and .-f ~ dx)
for anyfixed
f
e Then1Û1
= Pick h ~ ~strictly positive
suchthat
m(h)
oo. Then =Yl f )
_(h, (h,
oo; thatis, 1Û1
and are a-finite. Thisimplies
that =by [Ge (2.12)]
or =dx).
Vol. 32, n° 2-1996.
On the other
hand,
let03BD1(dy)
.-dy)
and .-g)
for any
fixed g
eClearly 03BD1V1
=and, using
the Revuz formula(6.10) again,
= =g) _ h)
~; thatis, 03BD1V1
and are a-finite. Hence vi = v2 ;i. e. ,
vM andvM
are dual.Conversely,
if vM andvM
aredual,
let N EMF(X)
denote the dualof
M.
Then we haveHence vM = vN .
By (6.2)
M and N arem-equivalent.
Therefore M andM
aredual..
PROPOSITION 6.12. -
1 )
Let S eSMF+(X) and S
eSMF +(.X) be
terminal times. Then ,S’ and
,S’
are dualif
andonly if vs
andvs
are dual.2)
The canonical measures v and v
of
X andX
are dual.Proof. - 1)
First let S andS
are dual. There exist Borel subsets such thatLet
/t tJ
Since
Gn
=Gn,
andJCn
are dual[Ge3].
Setand
Obviously, Sn
and9n
are also dual terminal times. Letwhere are the
corresponding kth-iterate. Clearly
An andA’~
aredual.
By (6.10)
Annales de l’lnstitut Henri Poincaré - Probabilités et Statistiques
Since n is
arbitrary
and vs andVs
do notcharge D,
it follows that vs andvs
are dual.Conversely
if vs andVs
aredual,
let.K : _ ~ ( x, ~ ) : K ~
andS’ - ~K .
Then ,5’ and9’
are dual and we haveHence Vs
=vs, . By (6.3) ,S’ _ ,S’’
and. S and9
are dual.2) Using
the notationabove,
sinceJGn
andJCn
aredual,
their bivariate Revuz measures aredual;
thatis, 1Gn - v
and1Gn - v
are dual. Thus v andv are
dual..
COROLLARY 6.13. - Let M e
SMF+(X)
andM
ESMF+ (X ).
Then Mand M are dual
if
andonly if
v~ and are dual.Proof -
Sinceduality
of M andM implies duality
ofSM
and(see [De]),
thenecessity
is a direct consequence of(6.11)
and(6.12).
Assumenow that vM and
vM
aredual;
thatis,
where B and 03A6
(resp. B
and)
are theparameters
associated with thedecomposition
of M(resp. M)
as in(2.3).
Since v and v aredual,
we findthat vsM and are dual and that vslogM and
slog
are dual. Thus Mand M are dual
by (6.11 )
and(6.12). N
ACKNOWLEDGEMENTS
Part of the paper is based on the author’s Ph.D. dissertation. The author would like to thank Professors R.K.
Getoor,
Ph.D.advisor,
andP.J. Fitzsimmons for their
inspiring suggestions
andgreat helps.
Part II. THE FEYNMAN-KAC FORMULA 1. Introduction
Though
the work onsymmetric
Markov processes has beencompleted
in
[Yi2],
the similar result fornon-symmetric
processes is notjust
trivialgeneralization
because the methodemployed
in[Fi]
or[Yi2]
can not beused in
non-symmetric
case. Given anearly symmetric
process,i.e.,
oneVol. 32, n° 2-1996.