A NNALES DE L ’I. H. P., SECTION B
C ARL G RAHAM
The martingale problem with sticky reflection conditions, and a system of particles interacting at the boundary
Annales de l’I. H. P., section B, tome 24, n
o1 (1988), p. 45-72
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The martingale problem
with sticky reflection conditions,
and
asystem of particles interacting at the boundary
Carl GRAHAM
C.M.A.P., Ecole Polytechnique, 91128 Palaiseau Cedex
Vol. 24, n° 1, 1988, p. 45-72. Probabilités et Statistiques
ABSTRACT. - We
give
amartingale problem
with Wentzellboundary
conditions under its most natural and
general form,
without any assump- tion on the generator. We show the usualboundary sojourn-time
conditionis a consequence of the
martingale problem
as soon as the generator issufficiently non-degenerate,
and in thegeneral
casegive
a weaker condition which behaves well underlimiting procedures.
Wedevelop
thetime-change theory,
and therelationship
with somegeneralized
stochastic differentialequations.
We thengive
results on existence anduniqueness,
some of theformer
by
alimiting procedure,
and someexamples.
Weeventually
con-struct a system of
interacting particles,
with an interaction in thesojourn
term, as a limit in law.
Key words : Martingale problems, stochastic differential equations, sticky reflecting bound-
ary conditions, local times, systems of interacting particles.
RESUME. - Nous posons un
problème
demartingales
avec conditionsfrontière de Ventcel sous sa forme la
plus
naturelle etgénérale,
sans aucunehypothèse
sur legénérateur.
Nous montrons que la condition habituelle de temps desejour
a la frontière découle duproblème
demartingales
dèsClassification A.M.S. : Primary 60 H, 60 H 10, 60 K 35. Secondary : 60 B 10, 60 G 44, 60 J 50, 60 J 55.
que le
générateur
est suffisamment nondegenere,
et dans le casgeneral
donnons une condition
plus
faiblequi
se comporte bien lors de passages a la limite. Nousdéveloppons
la théorie duchangement
de temps, et la relation avec desequations
différentiellesstochastiques généralisées.
Nousdonnons alors des résultats d’existence et
d’unicité,
certains de cespremiers
par des passages a la
limite,
et desexemples.
Enfin nous construisons unsystème
departicules
eninteraction,
avec interaction dans le terme desej our,
en tant que limite en loi.Mots clés : Problemes de martingales, equations differentielles stochastiques, conditions
frontière avec reflexion collante, temps locaux, systèmes de particules en interaction.
INTRODUCTION
Our first aim was to
study
asystem
ofinteracting particles reflecting
ina domain 8 with
sticky boundary,
with an interaction in the"sojourn"
term. This was to model for
example
whathappens
in achromatography tube,
where molecules in a gaseous statepushed by
a flow of neutral gasare absorbed and released
by
aliquid
statedeposited
on thetube;
what is measured is the "mean" time takenby
a molecule to getthrough
the tube.The idea would be to construct such a system, and to
investigate
itsasymptotic
behaviour when the number ofparticles
goes toinfinity (supposing
a mean-fieldinteraction) : propagation
ofchaos,
Gaussianfluctuations, large deviations,
etc. But the construction itself is a non-classical
problem, taking place
in an"angular"
domain with an unusualboundary sojourn
condition. For thisconstruction,
we will use the results in[9] plus
alimiting procedure.
The "classical"
martingale problem
withboundary conditions,
or the"classical"
sub-martingale problem ([3], [5], [8]),
are ill-suited to thelimiting procedures
we use for the construction and for theasymptotics.
This isbecause of the presence of discontinuous terms
Ie
and either in thebody
of the(sub-)martingale problem
or in thesojourn-time
condition atthe
boundary.
Thanks to the classicalhypothesis
ofnon-degeneracy
of the generator, which enables the authors to "control" the component of the diffusion normal to theboundary,
resultsusing limiting procedures
cannevertheless be obtained : see
Proposition
20 in[3],
Theorem 3.1 in[8].
For our purposes, we chose to
develop
anapproach
to themartingale problem,
in its most naturalform,
which avoids the use ofIe
orlae
eitherAnnales de l’Institut Henri Poincaré - Probabilités et Statistiques
explicitly
orimplicitly through
thesojourn-time
condition at theboundary.
We show that under the classical
hypothesis
on thegenerator,
the usualsojourn-time
condition isimplied by
themartingale problem.
As we donot assume this
hypothesis,
wereplace
the usualsojourn-time
conditionby
a weaker one that behaves well under
limiting procedures.
Ourapproach
isin accordance with the
trajectorial approaches
in[1], [2], [6], [9].
It is tobe noted that the usual
sojourn-time
condition is too restrictive to get results when thegenerator
is toodegenerate.
Our paper is
organized
as follows:In part
I,
we define ourmartingale problem
andgive
differentequivalent
definitions
including
asub-martingale problem.
We show that if the non-degeneracy hypothesis holds,
the usualsojourn
condition isimplied by
themartingale problem.
In thegeneral
case, we weaken thesojourn
conditionand use this to
develop
atime-change theory.
Then we show the relation-ship
with stochastic differentialequations
and use it to get existence anduniqueness results,
some of whichinvolving limiting procedures.
We thengive examples showing
the greatergenerality obtained,
and finishby
adiscussion of
time-inhomogeneous problems.
In
part II,
we describe our system ofinteracting particles
andgive examples showing
it cannot be treatedby
usual means. Then wegive
anoriginal
construction as a limit in law.I. MARTINGALE PROBLEMS WITH BOUNDARY CONDITIONS
1. The framework
Let us consider a diffusion
reflecting
under the mostgeneral
conditions(Wentzell’s boundary conditions)
in an open subset 8 ofRd.
We suppose 8 is
given (lRd)
as follows:with for
x E a 8, I D ~ (x) ~ =1.
For such is then the normalvector n,
pointing
inwards.For and
bi
will denote real-valued bounded measurable functions onë
x[R+,
such that the matrix issymmetric
and ~0;
(ë),
definethe summations
being
from 1 to d.Let us
give
aWentzell-type boundary
condition : for 1 Yi, P real-valued bounded measurable functions such that ais
symmetric and 0, a n = o,
for(08),
defineWe say that a
probability
measure P on S2° solves themartingale problem (L, r, p) starting
at x ~ 03B8 ifP (X° = x) =1
and if there exists anincreasing, adapted
processK,
with such thatfor f ~ C2b (9)
is a continuous
(P, F0t)-martingale. Here, V f
denotes thegradient
off, ( , )
the scalarproduct
in and y,V )
is the derivativealong
yof g
sometimes denotedby a~.
r denotes(
y,D >
+A,
> and it is easy tooy
see that if P solves the
martingale problem (L, r, p),
then it will solve itfor
(L, g r, g p), where g
is any measurable function on such that for a constante, g > e > o;
it suffices tochange Kt
tot0g(Sx, s)-1 d Ks.
Note that we don’t ask for any
non-degeneracy assumption
onL,
andthat ~ n, an ~
may well vanish. In[3], [5],
the authors ask for~
n,an ~
> c >0,
and in[8] a
isuniformely elliptic.
Remark 1. - We could consider a
martingale problem
onC
(!R+, 8)
x C( I~ + , R),
the canonical processbeing
then(X, K).
This isAnnales de l’Institut Henri Poincaré - Probabilités et Statistiques
sometimes more
convenient, especially
whentaking
limits. If we check that K is anadapted
process ofX,
then the law of X will solve ourmartingale problem.
All results to ourmartingale problem
that don’tuse the
adaptation
of K to X extend to themartingale problem
onC ( f~ + , 8) x C ( ~ + , ~),
and some can be used to prove thisadaptation.
Remark 2. - We could as well take the coefficients to be
predictable
processes.
Remark 3. - In order to
simplify
ournotations,
we take 8 to be(x e ~d, x’ > 0~
and C(x) =1- exp ( - x ~ )
or evenx~
if we don’t need the boundedness. Then forexample
a n = 0 means the first row(and
the firstcolumn)
of a isequal
to zero. Allproofs
could be carried out with ageneral C,
and tointerpret
the results weonly
need toremplace 1- exp ( - x 1 ) by
the
general ~,
allby n * an = ~ n,
an), b 1 by (
n, bB
y 1by ( n,
yB
etc.Remarks will
help
thisinterpretation.
r may then be written where A~ is a second order operatorFor
vElRd,
A d x dmatrix,
setFor a continuous
martingale M, ~ M ~
is itsquadratic
variational processas in
[5],
sometimes referred to as theincreasing
process of themartingale.
THEOREM I. 1. - The
following propositions
are allequivalent :
is
for
allf E C~0 (6)
a continuous(P, $’?)-martingale.
(b) Mt
isfor
allf E Cb (8)
a continuous(P, $’?)-martingale.
(c) Mt
isfor
allfEC2 (8)
a continuous(P, ~°)-local martingale.
is
for
all v E Rd a continuous(P, ~’°~-local martingale,
andMoreover, for
any we haveand
using
the boundednessof
thecoefficients
and E(Kt)
+ oo, we see thatthe M"
for
v E ~d and the Mffor f
with a boundedfirst
derivative areactually L2-martingales.
Proof. -
Theproof
is the same as in[3], [8]. ~ Mf ~
iseasily
established from(d)
thanks to the Ito formula. DTHEOREM I . 2. -
If
on theboundary,
either - pbl
> C >0,
oryl + p > C > 0 and all> C >
0,
then themartingale problem
isequivalent
tothe
following sub-martingale problem:
is a continuous
(P, F0t)-submartingale for
allf E Co
such thatProof. -
That themartingale problem implies
thesubmartingale
one istrivial. The other
implication
goes as follows:and on the
boundary,
for a well-chosen ~,. We can then use this instead of C in the classical
proof
of[3]
and[8].
DRemark 1. - The condition
Yl +p>C>O
and is the mostgeneral
condition considered in[3]. Generally
in[3]
and[8],
the authorsassume
Yl >C>O,
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
Remark 2. -
Naturally,
in the case of ageneral $
we should considerAn easy calculation
yields
~’~(x)
> C > 0 on a8, provided
~
n,or (
n,Y ~+p>C>0 and ~
n,THEOREM I . 3. - Assume that
Xo
E a8,
and that on aneighbourhood
weeither have all > 0 and ~ or
bi
o. Let P solve themartingale
all
problem.
Then P- a. s., V t >0, Kt>
o.Proof
with
By
the Girsanov Theorem[5],
we have aprobability
Pequivalent
_
-t
to
P,
under whichMt
=Mt + bl (XS)
o(ds -
p(XS)
dKS)
is a mar-tingale
for 0 _ t _ 1.Set T = inf
~t
>o, Kt
>0~
n 1. Thenas
X r T >_ o,
we getX1tnT = 0
andtnT0 1b
1 (Xs) 0 ds = o. Then we see that0 T
=
o,
so E(
M~T)
= 0 and~o
> o ds = o. So T = o. All this isP-a. s., and thus P-a. s. p
Remark. - We cannot have a better result. For an
example,
consider aBrownian motion B
starting
at zero,BS
=BS
if0, BS
=1 ifBS
= o.-t
Then
Bi
= 2 dBS
+ t, and K = o.2. The
sojourn
condition and thetime-change
We now
give
the fundamental results we will use when we take limits.We will first see that the usual
sojourn
conditionlae
dKS
is
implied by
themartingale problem
as soon as If thisinequality
does not
hold,
we will use as asojourn
condition the measureinequality lae s) d Ks;
a 8being closed,
thisinequality
will behaveproperly
when we takelimits,
and is trivial whenp = 0
as in[1], [2], [6].
We then
generalize
thetime-change theory, provided 11 >C>O.
Note thatin
[3], [5], [8],
the authors assume11 >C>O.
THEOREM 1. 4. -
Suppose
that P solves themartingale problem.
ThenP-a. s. , V tE IR +,
and these are all
increasing
processes.Furthermore,
and
naturally if for
x E a8,
all(x, s) ~ 0,
thenProof. -
The firstequality
of(1.5)
follows from the fact thatX: O
and thus
Xt =(Xf )+,
andusing
the Tanaka formula[12]:
We then use Theorem I .1.
(a)
with v the first vector of the canonical basis-t
of and
integrate 1 ae ( X }
with respect to M 1.( X s)
dMS
is then aAnnales de I’Institut Henri Poincare - Probabilites et Statistiques
continuous
martingale
which is also of boundedvariation,
and is thereforeconstantly equal
to 0. We thus get the secondequality
in(1.5),
and(1.6) by considering
theincreasing
process of thismartingale.
DSince
L° (X 1)
isincreasing,
we have:THEOREM 1.5. - Let P solve the
martingale problem,
and seti
1
as
=(x
e0,
0x1 E) .
Then P-a.s.- 2 E t0 a11(Xs, s) 103B8~ (Xs)
ds converges,uniformely
on compact sets, tothe convergence
being
also in all theLP,
1_p
+ oo.Proof -
We use(1 . 5)
andplus
the fact that a sequence ofincreasing
continuous functionsconverging pointwise
to a continuous functionactually
convergesuniformely
on com-pact sets.
Since
the Tanaka
equation,
the baundedness of the coefficients and theexplicit
solution of the Skorokhod
equation (Lemma
III 4. 2 in[5])
show that the convergence is bounded in all the Lp and thus takesplace
in all theLp,
1p
+ ~. QTHEOREM I . 6. -
Suppose
P solves themartingale problem,
Y1 > C> o,
and that P-a. s. ,l ae
ds >_ p(X S, s)
dKs.
r
Set
At = t -
Then P-a. s. A isstrictly increasing,
lim
At
= + oo, andif
A denotes its inverse,r --~ o0
Proof. -
A isincreasing,
denoteby
A itsright-continuous
inverse:Åt=inf
Then: and sois a continuous
(F0u)-martingale,
and thus a(F+u)-martingale
where~
= FtiXf. A,_
andA,
are(~)-stopping
times.S~M
and M is
uniformely
bounded inL2
beforeAt.
We may then use theoptional sampling
theorem to get Since A does not increase betweenA t _
andA,,
C(x) = 0
for and we see thatKAt- = KAt
and thus= At.
A is then continuous and hence A isstrictly increasing.
By
theoptional sampling theorem,
E( MAn) = o,
and sinceis
bounded,
we getE
+ oo . It is dear that onand since
03B31>C>0, An0 03B31 d Ks = + ~;
so+ooP-a.s.,
andA ~ _
-f- oo .by
asimple time-change.
pr r r
Remark. - If ds =
s)
dKS,
thenAr
= ds.Moreover
by using
Theorem L4 we see thatlae ds >__
ps) d KS
as soon as = o
lae
>__1a11 _
o ps)
dKs.
Annales de l’Institut Henri Poincare - Probabilites et Statistiques
If
a=0,
Theorem I .1( d)
shows that it isenough
thatlae
ds >_ ps)
dKS (since
theincreasing
process of a mar-tingale
is indeedincreasing).
Consider also(1.8), (I. 8)’.
We can also note that as soon as
yl > o,
A isstrictly increasing
butcould have a finite limit.
From now on, we
only
considertime-homogenous problems,
thatis,
a,b,
a, y, p do notdepend
on time. We will discuss later how to get results fortime-homogenous problems by adding
time as a new coordinate.We will consider the
martingale problem
with aboundary sojourn
condition.
We will say that P is a solution to the
martingale problem (L, r, p)
with
sojourn
condition if for a certainK,
P is a solution to themartingale problem (L, r, p)
as in(1.3)
and moreover P-a. s., we have(I .12)
The
previous
remark is still ofinterest,
as is the discussion at thebeginning
of 2.Naturally,
all>_ o,
but that is all we need. We also see that the boundedness of p is notimportant.
We see that when
p = 0
as in[1], [2], [6], (1.12)
isautomatically
trueand doesn’t have to be stated.
Also, (1.12)
can be restatedby saying
thator that t -
t0 03C1 (Xs)
dKS
is anincreasing
process.0
THEOREM I. 7. - Let P be a solution to the
martingale problem (L, r, p)
with
sojourn condition,
and K be theincreasing
process in(I. 3).
Let p be anotherpositive
measurablefunction
ona8,
and suppose eitherp >_
p or yl >C>o. Setv V
Then A is a
strictly increasing
process which tends towardsinfinity.
Denote
by
A its inverse, X =XA,
K =KA,
P the law on gOof
X .Then P solves the
martingale problem (L, I-’, p)
withsojourn condition,
and K will be the
corresponding increasing
process.Vol. 24, n° 1-1988.
~
Proof -
That A isstrictly increasing
toinfinity
is obvioususing
eitherp~p
or Theorem1 . 6 when ’l>C>O.
We get A in much the same wayas
in
Theorem1.6 {At- ~ T} = {t ~ AT} ~ F0T
and thus is astopping time;
Abeing strictly increasing, = At.
That K increasesonly
when X is on the
boundary
is trivial from the similar property of K and X.Also,
X and K are continuous.For f ~ C2b,
is a
(P, ~ °)-martingale.
First,
let’s prove E(Kt)
+00. If then E( Kt) _
E( Kt)
+ Ify 1
> C > o,
take 03A6(x) =1- exp ( - x 1 ),
is a
martingale,
andand thus M~ is
uniformely
bounded in L2 beforeAt.
So E(M~) = 0
andby
a boundedness argument, as yi> C > o,
E(Kt)
+ oo.Now
using
Theorem I. 1 we seethat ( Mf )
isuniformely
bounded inL1 before
At.
So Mf isuniformely
bounded in L2 beforeAt,
andby
theoptional sampling theorem,
Mf(A)
is a(P,
F0(At))-martingale.
So P solves the
martingale problem (L, r, p)
with K as anincreasing
process.
Annales de l’lnstitut Henri Poincaré - Probabilités et Statistiques
and since we have the
sojourn
condition.
Let us remark that we get X and K from X and K
by taking At=t+t0 (03C1-03C1) (Xs) d Ks
and X=X(A-1)
K=K(A-1).
Hence:THEOREM 1.8. -
Suppose Y1 > C > o.
We have abijective mapping
between the setof solutions of the martingale problem (L, I-’, p)
withsojourn
condition and theone for (L, r, p),
and between thecorresponding increasing
processes K and K.
Thus the existence, or the
uniqueness, for
these twomartingale problems
are
equivalent
notions.Also, if
P is a solutionfor the martingale problem (L, r, p)
withsojourn
condition there is anunique
K such that(1.3)
holds.Proof -
Thebijection
isgiven
in Theorem 1.7 and in the remark that follows itsproof.
Ifp = o,
Theorem 1.5yields
theuniqueness
of Kusing
y 1
> C > o,
and thegeneral
case followsby
use of thebij ection.
DRemark. - We
actually
have a much better result ofuniqueness
ofK.
Indeed, using
Theorem I . 1(d)
and the fact that; M" )
is thequadratic
variational process of
Xv,
we see that is a function ofXs, 0 _
s t.Using
thetime-change
in Theorem 1.7 forp
= p+ 1 >-
p, and the reverse timechange,
we see that is a functionof
X,, 0 - s - t,
and so are both and Ina similar way, so are and In
fact, d K
is determined as a functiondepending only
on a, a,b,
y, p ofXs, 0-s_t, as soon as a (Xt) ~ 0 or 03B1 (Xt) ~ 0 or b (Xt) ~ 0 or 03B3 (Xt) ~ 0,
and is
arbitrary
whenever these fourquantities
vanishsimultaneously
onthe
boundary.
A natural choice in this case would bedK=0,
wich iscompatible
with(1.12).
It is
important
to see all this is true even if we consider themartingale
problem
on it enables us to get back to~~=c (~+, 9).
Using this,
thesojourn
conditionlae
ds >_ pd KS,
the Radon-°
Nykodym Theorem,
we first see thatand rs= lim - 1 p (Xu)dKu
and so for any~1 s
r~ >
0,
rs = r(Xu, s - ~
u _s)
with rdepending only
on a, oe,b,
y, p.Let’s summarize all this: let
(X, K)
be the coordinate process onC
((~+~ 9)
x C(f~+~ R).
THEOREM 1 . 9. - Let P be a solution on C
( f~ +, 8)
x C( ~ +, ~) of
themartingale problem. If for
any x on theboundary,
either a(x) ~
0 ora
(x) ~
0 or b(x) ~
0 or y(x) ~ 0,
thenKt
is anunambiguous function of XS,
0 - s _ t. We mayalways
takeand K’ will be an
adapted function of
X. Then the lawof
X under P willsolve the
martingale problem
ongO,
with K’ as anincreasing
process.If
moreover thesojourn
conditionholds,
then there exists afunction
rdepending only
on a, a,b,
y, p, such thatand r
actually
is afunction of Xu,
s - ~ u _ s,for
any ~ > o.Remark. - To have
examples,
see 1.4.3. The stochastic differential
equation,
and the results on existence and
uniqueness
of solutionsWe are
going
to see that under certainassumptions
themartingale problem
isequivalent
to a stochastic differentialequation.
This will enableus to use the results on weak existence and
uniqueness
for the latter. We know that strong(trajectorial) uniqueness implies
weakuniqueness;
see[3], [5].
We also know that the
martingale problem (L, r, p)
has the samesolutions as
(L, gr, g p),
for anybounded g
such that So ifAnnales de l’lnstitut Henri Poincaré - Probabilités et Statistiques
Y 1 > C > o,
we may takey 1=1,
or any such normalization. We can remark that if the coefficients arecontinuous,
or bounded andLipschitz, dividing by
Y 1 doesn’tchange
thisproperty.
THEOREM I. 10. -
Suppose
that 6 and i are two matrices such thatoc = ii*.
Naturally,
i* n=O. Then thefollowing propositions
areequivalent:
(a)
P solves themartingale problem (L, r, p)
withsojourn
conditionstarting
at x E9,
withincreasing
process K..
(b)
There exists aprobability
space(~ t)r > o, P),
a d-dimensional Brownian motionC,
a d-dimensional continuousmartingale N,
and twocontinuous processes X and
K,
such thatKo = 0,
K isincreasing, dKt
=1~03B8(Xt) dKt, Xo
= x,Xt
E8,
and
+ Y
dKr
+ 03C4(Xt)dCKt, ( I , .17)
and P is the law
of X .
We may take
(SZ, (~ r), X )
to be an extensionof (S2°, X)
in thesense
of [5].
As soon as Y 1+ p > C > o, for
and we haveE ( exp
~,Kr)
+ oo .Proof. - (a)
==>(b):
We use Theorem 1.1(d)
and therepresentation
theorem for
martingales:
Theorem II. 7 . I’ in[5].
We reason as in[3], [5].
(b) ~ (a):
The Ito formulagives
us themartingale problem
onand Theorem 1.9
gets
usback
to Thesojourn
condition holds because
N 1 ~
is anincreasing
process.Take the
projection
of the stochastic differentialequation
on the firstaxis:
We use the classical estimates on Brownian
motion,
the boundedness of thecoefficients,
and theexplicit
solution to the Skhoro-khod
equation (Lemma
111.4.2. in[5])
to getB ~/ u /
we get
E ( exp ~, Kt)
+ oo . nRemark 1. - Thanks to the
martingale representation theorem,
we see that(perhaps
afterenlarging
theprobability space)
there is a brownianmotion B
independent
from C such thatNaturally,
ifp = 0
then r = 0 and N = B. We maygive
a definition of astrong solution for the Stochastic differential
equation, by giving
firstQ,
~ t, P, B,
C andby looking
for anadapted
continuous(X, K)
such that(1.17)
holds with Ngiven
as in(1.18).
If moreover
T=0, p=0,
we get the(strong) equations
of[1], [2], [6].
Remark 2. -
Proposition
IV. 6 . 2 in[5]
tells us that as soon as a ispositive
definite and a Ect,
we may choose aLipschitz
continuous o such that = a. Same for a.Remark 3. - The condition a n = 0 translates into i* n =
0;
the condition n*an>0 intoTHEOREM 1.11. -
Suppose
that a =0,
that cr,b,
areuniformely Lipschitz
and
bounded,
y ECb
or y = n, y 1 > C > o. Thenfor
allpositive
p, all x E8,
there is an
unique
solution P to themartingale problem (L, r, p) starting
atX,
withsojourn
condition.If p=0,
then there is anunique
strong solution to thecorresponding
stochastic
differential equation.
Proof -
Whenp=0,
this is the result on stochastic differential equa- tions of[1], [2], [6],
where strongtrajectorial
results aregiven.
Thegeneral
case follows
by
Theorem1. 7,
thatis, by time-change.
See Theorem IV. 7 . 2in
[5]
for a direct calculation on the stochastic differentialequation.
D Remark 1. -By
theorem I. 8 there is anunique
Ksatisfying (I. 3)
and(1.12).
Remark 2. - We could
give
results underassumptions
of uniform strictellipticity
and ofcontinuity
on a. We can also diminish theassumptions
on the drifts
by
the Girsanov Theorem. We alsoonly
need to find abounded
g > C > 0
such that(L, g r, g p)
verifies theassumptions.
Fordetails,
see[3], [5], [8].
Remark 3. - We don’t suppose al l
>0,
and so canonly
use the strong results that authors get when03B3 ~ C2b
or y = n,a =0, p=0.
These are theonly equations giving uniqueness
results if we don’t takeAnnales de l’lnstitut Henri Poincaré - Probabilités et Statistiques
We now will start to use the fact that our formulation of the
martingale problem
enables us to take limits.First,
aproposition
to getcontinuity
results we will need for such
proofs.
PROPOSITION I . 12. - Let be a sequence
of real functions,
withuniformely
bounded variation on each compact set.Suppose
the sequence convergespointwise
to zero.Then
for
any T e R and anyfe
C([0, T], R), IT f (t) dun (t)
converges tozero.
Proof. - Suppose
the variations of the vn are boundedby
M on[0, T].
For E >
0,
there is an ~ such that ifLet a be a subdivision
(sj
of[0, T]
such that Max(si +
1-s~)
_ ~1, and ibe a refinement of a. Set t
( i) = {t
E si +1 ~.
Then:The first of these two terms is
equal
toI ~ (vn (st + 1) - v" (si)) ~ ]
andy
thus is less than E as soon as n is
larger
than a n(o);
the second term is lessthan E
x M = E.M
Note that if k" is a sequence of
increasing
functionsconverging pointwise to k,
thenIT
converges to because3
(kT - ko)
as soon as n islarge enough.
If we notice
that vn
needonly
converge to zero outside a countable set, theproposition
appears as ageneralization
of the classical resultrelating
weak
("narrow")
convergence on R to thepointwise
convergence of distri- bution functions wherever the limit distribution is continuous.We now get our first result
using
alimiting procedure.
THEOREM I.13. - Assume that ~,
b,
i, y are continuous andbounded, yl > C > o,
andp is a positive
measurablefunction.
Let x E6.
Then thereexists at least one solution to the
martingale problem (L, r, p)
withsojourn condition, starting
at x.If
moreover 03C3,b,
i, y areLipschitz,
and a11 > C >0,
then the solution isunique.
Proof -
The last result is Theorem IV. 7 . 2 in[5].
See also[3].
ForE > o,
put LE = L + E 0. We firstmodify
theproof
of Theorem IV. 7 . 2 in[5], by replacing
the existence of solutions for SDE withLipschitz
coef-ficients
by
the existence for SDE with continuous bounded coefficients(Theorem
IV. 2. 2 in[5]).
Since all + E >_ E, theproof
can be carriedthrough
and we see there is at least one solution PE for the
martingale problem r, p).
As we
plan
future use of Theorem1.7,
we takep = 0
atfirst,
and forsimplicity
we takeY 1 = 1 and multiply
the rest of rby - .
We will first prove
tightness
forf PE,
and then prove thanany accumulation
point
for E -~ 0 solves themartingale problem (L,. r, p).
That will prove the theorem
by using
Theorem I. 7.Under
P,
we havewith and
Ko = 0
and X and K continuous. Theexplicit
solution to the Shorokhod
problem yields:
The coefficients are
uniformely
bounded forE 1,
and(1.20) plus
theclassical results on Brownian motion
yield
that for anyTe!R+,
there isCT
such that for E 1 and we have(the eighth
powercoming
fromCK).
Theorem 1.4.3 in
[5]
thengives
thetightness
for the laws of(X, K)
under the
P£,
0 E 1.Annales de l’lnstitut Henri Poincaré - Probabilités et Statistiques
Take E" - 0 such that the laws of the
couple
converges to poo onand
L",
P"corresponding
to En,0 _ sl _ ...
..., gp continuous
bounded,
and setthen we have
Since the coefficients are
continuous,
we seeusing Proposition
1.12 andwhat follows that
(X, K)
-~ is continuous.Using (1.21)
tos
get the needed uniform
integrability,
we obtain at the limitwhere
naturally Mf
is likeMf, n
with L instead of L". SoMf
is a(P°°, F0t)-martingale,
where(F0t)
is the natural filtration onIRd)
xC(IR+, R).
Nowif g
is continuouspositive
with compact support ine,
and K increases
only
when X is on 89. Since 9 isclosed, Similarily,
K isincreasing,
andKo=0;
E°°
(Kr)
+ oo comes from(1 . 21). Using
Theorem1 . 9,
ordirectly (1 . 20),
we see that K is
actually
anadapted
process ofX,
and thus if poo is theprojection
of poo onMf
is a(Poo, ~ °)-martingale.
So poo is a solutionto the
martingale problem (L, r, 0).
Timechange gives
thegeneral
case. D
Up
to now, werequire Y1 > C > 0
as most authors do. In[5], Chapter IV,
at the end of Section7,
the authors argue that thisassumption implies
that there is reflection at the
boundary
and is toorestrictive;
instead we should suppose1’1 + p > C > 0,
which would allowsojourn
without reflec- tion. Thenthey
construct a process under thatassumption (and
thanks to the Poisson
point
process of excursions and ajump-type
stochas-tic differential
equation
on theboundary. Naturally, they
don’t getunique-
ness this way, and their construction is
quite
intricate. See[5],
IV(7. 13)
and
following,
and the referencesgiven
there.We will now
give
ageneralization
of Theorem I . 13 when y 1+ p > C > 0 by
alimiting
method. Thisenglobes
the result in[5]. Naturally,
y 1 andp >__ 0.
A similar idea appears in[3], Proposition 20,
when p > C> 0,
all >c>O.
THEOREM I.14. - Assume that ~,
b,
i, y are continuous andbounded,
p is apositive
measurablefunction,
and yl + p > C > 0. Let x E 8.Then there exists at least one solution to the
martingale problem (L, r, p)
with
sojourn condition, starting
at x.Proof. -
ForE > 0,
T is like r with y 1replaced by
y 1 + E. We useTheorem 1 . 13 to get existence for the
martingale problem (L, hE, p).
Wewill follow the
proof
of Theorem I13,
except we don’t take yl =1 andp = 0.
For we will get instead of(I . 21), using
and since
and
using y 1-f- p > C > 0
and an easyconvexity
argument,Now if p is
continuous,
we can finish ourproof
as for TheoremI. 13,
and since 10 isclosed,
theinequality lae
ds >_ pdKs
carries to thelimit. We need the
continuity
of p forthis,
as well as to take limits in themartingale problem.
Now if p is notcontinuous,
take Thenp
iscontinuous,
and so(L, r, p)
has a solution. Now sincep >_ p,
we can usetime-change
as in Theorem 1.7(the
notations are inreverse but
coherent)
to get existence for(L, r, p).
We use Theorem I. 9 to get back to aRemark. - Assume and p is continuous. Then the law of
lae
ds under theconverging subsequence
ofP~,
convergesweakly
to itslaw under poo. To see
this, just
use Theorem 1.4 and thecontinuity
ofJo
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques