www.imstat.org/aihp 2009, Vol. 45, No. 3, 876–886
DOI: 10.1214/08-AIHP194
© Association des Publications de l’Institut Henri Poincaré, 2009
Maximal Brownian motions
Jean Brossard
a, Michel Émery
band Christophe Leuridan
aaInstitut Fourier, Université Joseph Fourier et CNRS, BP 74, 38 402 Saint-Martin-d’Hères Cedex, France.
E-mails: [email protected]; [email protected]
bIRMA, Université de Strasbourg et CNRS, 7 rue René Descartes, 67 084 Strasbourg Cedex, France. E-mail: [email protected] Received 13 December 2007; accepted 29 August 2008
Abstract. LetZ=(X, Y )be a planar Brownian motion,Z the filtration it generates, andB a linear Brownian motion in the filtrationZ. One says thatB(or its filtration) is maximal if no other linearZ-Brownian motion has a filtration strictly bigger than that ofB. For instance, it is shown in [InSéminaire de Probabilités XLI265–278 (2008) Springer] thatBis maximal if there exists a linear Brownian motionC independent ofBand such that the planar Brownian motion(B, C)generates the same filtrationZ asZ. We do not know if this sufficient condition for maximality is also necessary.
We give a necessary condition forBto be maximal, and a sufficient condition which may be weaker than the existence of such aC. This sufficient condition is used to prove that the linear Brownian motion
(XdY−YdX)/|Z|, which governs the angular part ofZ, is maximal.
Résumé. SoientZ=(X, Y )un mouvement brownien plan de filtration naturelleZ, etBun mouvement brownien linéaire de la filtrationZ. On dit queBest maximal, et que la filtration naturelle deBest maximale, lorsqu’aucun autre mouvement brownien linéaire deZ n’engendre une filtration strictement plus grosse que celle deB. Il est par exemple établi dans [InSéminaire de Probabilités XLI265–278 (2008) Springer] queBest maximal dès qu’il existe dansZun mouvement brownien linéaireCindé- pendant deBet tel que le mouvement brownien plan(B, C)engendre toute la filtrationZ; nous ne savons pas si cette condition suffisante de maximalité est aussi nécessaire.
Nous donnons une conditions nécessaire de maximalité, ainsi qu’une condition suffisante peut-être plus faible que l’existence d’un telC. À l’aide de cette condition suffisante, nous démontrons que le mouvement brownien linéaire
(XdY−YdX)/|Z|, qui régit la partie angulaire deZ, est maximal.
MSC:60J65; 60G07; 60G44
Keywords:Brownian filtration; Maximal Brownian motion; Exchange property
1. Introduction
In the theory of filtered probability spaces, it is natural to study pairs of filtrations, one immersed in the other, and to attempt to understand possible different ways the smaller filtration can be immersed in the larger. A simple case is to consider various ways the filtration of a one-dimensional Brownian motion can be immersed in that of a two- dimensional Brownian motion. This study was initiated by two of us, who introduced in Section 2 of [2] the notions of complementability and maximality (Definition 1 below), and gave examples and counter-examples. The present work, which concentrates on maximality criteria, can be considered a sequel to [2].
LetZ=(Zt)t≥0 denote a planar Brownian motion withZ0=0 (all Brownian motions will be started from the origin); callZ=(Zt)t≥0the filtration generated byZ.In the sequel,Z andZ are fixed, and interest will focus on one-dimensionalZ-Brownian motions (shortly:Z-BM1) and their filtrations. IfB is aZ-BM1, its natural filtration FB is immersed inZ, that is, everyFB-martingale is also aZ-martingale. Consequently,FtB=Zt∩F∞B for every
t≥0, soFBis characterized by its endσ-fieldF∞B =σ (B). Many properties thatFBmay enjoy can equivalently be stated either in terms ofFtBfor allt, or in terms ofσ (B)only.
An example of such equivalent statements is the following double definition, borrowed from [2].
Definition 1. LetBbe aZ-BM1.
One says thatBis complementable if there exists aZ-BM1Cindependent ofB,and such that theZ-BM2(B, C) generates the filtrationZ(equivalently,theZ-BM2(B, C)generates theσ-fieldσ (Z)).
One says thatB is maximal if for everyZ-BM1D,one has ∀t≥0FtB⊂FtD
⇒
∀t≥0FtB=FtD
(equivalently,σ (B)⊂σ (D)⇒σ (B)=σ (D)).
These two properties (to be complementable and to be maximal) can also be considered properties of the filtra- tionFB generated byB; in the sequel, we shall sometimes say that a filtration is complementable or maximal when it is generated by a complementable or maximalZ-BM1.
An interesting example is obtained by setting Ut=
t
0
XsdXs+YsdYs
√Xs2+Ys2 and Vt= t
0
XsdYs−YsdXs
√Xs2+Ys2 .
The processesUandV are independentZ-BM1 which respectively govern the radial part and angular part ofZ, but (U, V )does not generate the whole filtrationZ. Actually, as recalled in Proposition 5, it only generates the quotient ofZbySO2, that is, the filtration generated byZup to an arbitrary rotation. Yet,Uis complementable: an independent complement is exhibited in [2].
Corollary 6 of [2] states thatevery complementableZ-BM1 is maximal.Thus,U is maximal. This corollary also gives many examples of non-complementable BM1, for instanceB=
sgn(B)dB, whereBis anyZ-BM1. Indeed, the filtration generated byBis also generated by|B|, so it is strictly included in that ofB, and not maximal; by the corollary it cannot be complementable.
Conversely,is every maximalZ-BM1 complementable?We do not know. This question is already open in the sim- pler case whenBis complementable except for a germ property at time 0+(this will be made precise in Definition 3).
In that particular case, we shall give a necessary and sufficient condition for maximality (Theorem 1). Interestingly, this condition is an exchange property: it says that supremum and intersection of certainσ-fields commute.
The necessary and sufficient condition of Theorem 1 is no longer sufficient for an arbitraryZ-BM1 to be maximal, but it is still necessary (Corollary 1).
Using a variation on this condition (Proposition 4), we establish maximality for theZ-BM1V defined above.
When the present paper was first submitted, the question whetherV is complementable, a possibly stronger property, was open. We now know that the answer is positive (September 2008); the construction by one of us of an independent complement toV is rather involved and needs to be simplified before publication.
2. Notation and reminders
In a probability space(Ω,A,P), all sub-σ-fields are assumed to be(A,P)-complete. Similarly, all filtrations are right- continuous and(A,P)-complete; araw filtrationis a filtration which is not necessarily right-continuous. IfS=(St)t≥0 is any process, the filtration generated bySis denoted byFS. For 0≤a < b <∞, we define new processes as follows:
S[a,b]=(St)t∈[a,b], S]a,b]=(Sa+t−Sa)t∈[0,b−a], S[a,∞[=(St)t≥a, S]a,∞[=(Sa+t−Sa)t≥0. IfGis a filtration andS aσ-field, theexchange property
h>0
(Gt+h∨S)=Gt∨S
does not hold in general; but it does whenGandS are independent (see [5]). In plain words, independent enlarge- ments preserve right-continuity. A key ingredient in the proof of this fact is that independent enlargements preserve conditional expectations: ifT andU are two independentσ-fields, ifT isT-measurable and ifS is a sub-σ-field ofT, thenE[T|S] =E[T|S∨U].
All Brownian motions are started at the origin. A planar Brownian motionZ=(X, Y )is fixed; its natural filtration is denoted byZ=(Zt)t≥0.
Recall the previsible representation property: everyZ-martingale has the formc+
HdX+
KdY, wherecis a constant andHandKareZ-previsible.
3. A necessary condition for maximality
Definition 2. TwoZ-BM1BandCare said to be parallel ifC=
HdBfor someZ-previsible processH. SinceBandC are linear Brownian motions,H= ±1 on a(dt×dP)-full set; so one also hasB=
HdC. Paral- lelism is an equivalence relation.
Any twoZ-BM1 generating the same filtration, or more generally any twoZ-BM1B andC such that the filtra- tionFBis included inFC, are parallel. Indeed,B is aZ-martingale adapted to the sub-filtrationFC, hence also an FC-martingale; so it is a stochastic integral w.r.t.C.
Proposition 1. (1)Given anyZ-BM1B,there always exists someZ-BM1independent ofB. (2) Given twoZ-BM1BandC,the following are equivalent:
(i) BandCare parallel;
(ii) there exists aZ-BM1independent ofBand independent ofC; (iii) everyZ-BM1independent ofBis also independent ofC. Proof. (1)B has the form
HdX+
KdY, whereH andK areZ-previsible, andH2+K2=1. The martingale D= −
KdX+
HdY is also aZ-BM1 sinceD, D =K2+H2=1; andDis independent ofBsinceD, B =
−KH+H K=0.
(i)⇒(iii): ifBandCare parallel,C=
HdBfor some previsibleH; so ifDis anyZ-BM1 independent ofB, one has dC, D =HdB, D =0, showing thatDis independent ofCtoo.
(iii)⇒(ii): immediate consequence of 1.
(ii)⇒(i): ifDis aZ-BM1 independent ofBand ofC, one has
dB=HBdX+KBdY, dC=HCdX+KCdY and dD=HDdX+KDdY,
where the vector(HD, KD)is (dt×dP)-a.e. orthogonal to (HB, KB)and(dt×dP)-a.e. orthogonal to(HC, KC).
Hence,(HB, KB)and(HC, KC)are(dt×dP)-a.e. parallel; so(HB, KB)=L(HC, KC)for some previsibleLde- fined(dt×dP)-a.e. ConsequentlyB=
LdC, showing parallelism.
Open questions. Given an arbitraryZ-BM1B,does there always exist aCparallel toB and maximal? Does there always exist aCparallel toBand complementable? We do not know.
Proposition 2. LetBbe aZ-BM1.
(i) Theσ-fieldsBt=
ε>0(Zε∨FtB)form a filtration(i.e.,they are right-continuous int).
(ii) The filtrationBcontainsFB and is generated by someZ-BM1B. (iii) Any suchBis parallel toB.
(iv) Applying the same procedure toB instead of B does not yield anything new:the filtrationB defined by Bt =
ε>0(Zε∨Bt)is equal toB.
Proof. (i) The raw filtrationBis right-continuous at 0 becauseBt⊂Zt and
t >0Zt is degenerate. It is also right- continuous att >0 for the following reason. Forε >0, the raw filtration(Zε∨FtB)t∈[ε,∞)is generated by theσ-field
Zεand the Brownian motionB]ε,∞[which is independent of thatσ-field; by the independent exchange property, that raw filtration is right-continuous. Consequently, for 0< ε < tone has
h>0
Bt+h⊂
h>0
Zε∨FtB+h
=Zε∨FtB,
and right-continuity ofBfollows from
h>0Bt+h⊂
ε>0(Zε∨FtB)=Bt.
(ii) The filtrationBclearly containsFB; to show thatBis generated by someZ-BM1, it suffices by Corollary 1 of [3] to verify thatBis immersed inZand is “1-Brownian after 0.”
The immersion property will be established by checking that ifR is any bounded,B∞-measurable r.v., E[R|Zt] is Bt-measurable. This is trivial for t =0 since Z0 is degenerate, so we supposet >0. For any ε∈ ]0, t], B∞ is included in Zε∨σ (B), so one can write R=f (Z[0,ε], B[0,t], B]t,∞[)for some bounded Borel functional f. Now, the first two arguments of f are Zt-measurable and the third one is independent of Zt; hence E[R|Zt] = f (Z[0,ε], B[0,t], b)w(db), wherewdenotes the law ofB]t,∞[. SoE[R|Zt]is measurable forZε∨FtB; as this holds for anyε∈ ]0, t],E[R|Zt]isBt-measurable andBis immersed inZ.
It remains to see thatBis 1-Brownian after 0, that is, given anys >0, the filtration(Bt)t≥s is generated by its initialσ-fieldBs and some 1-dimensional BM independent ofBs. Now, for allt≥s, one has
Bt=
ε∈]0,s]
Zε∨FsB∨σ B]s,t]
=
ε∈]0,s]
Zε∨FsB ∨σ B]s,t]
=Bs∨σ B]s,t]
,
where the second equality (exchanging a supremum with an intersection) is due to the processB]s,t]being independent of theσ-fieldZs, which contains all theσ-fieldsZε∨FsB. The claim is established by observing thatB]s,∞[is a BM1 independent ofZs and a fortiori ofBs. SoBis generated by someZ-BM1B.
(iii) Bmust be parallel toBbecause its filtration contains that ofB.
(iv) The filtrationBobtained by iterating this procedure satisfies for 0< ε≤t the inclusion Bt ⊂Zε∨Bt⊂Zε∨
Zε∨FtB
=Zε∨FtB.
Taking the intersection overεgivesBt ⊂Bt, hence equality. This extends tot=0 by right-continuity, or by observing
thatB0is degenerate.
With the notation of Proposition 2,B cannot be maximal unlessFB equals the (a priori larger) filtrationB. Thus we get the following corollary.
Corollary 1. LetBbe aZ-BM1.A necessary condition forBto be maximal is the exchange property
∀t≥0
ε>0
Zε∨FtB
=FtB (or equivalently,the equality
t >0(Zt∨σ (B))=σ (B)between the finalσ-fields of these filtrations).
4. Sufficient conditions for maximality
The following lemma was used in the proof of Corollary 6 of [2] (although it was not explicitly stated there).
Lemma 1. Given (Ω,A,P),let S, T and U be three ((A,P)-complete) sub-σ-fields of Averifying S ⊂T and S∨U=T ∨U.IfU is independent ofT,thenS=T.
Proof. It suffices to use an independent enlargement in conditional expectations so as to write, for any bounded, T-measurable r.v.T,
E[T|S] =E[T|S∨U] =E[T|T ∨U] =T .
Definition 3. LetBbe aZ-BM1.We shall say thatBis complementable after0if there exists aZ-BM1Cindependent ofBsuch that
∀t >0,∀ε∈ ]0, t] Zε∨Ft(B,C)=Zt (1)
(or equivalently,
∀ε >0 Zε∨σ (B, C)=Z∞, (2)
since the filtration(Zε∨Ft(B,C))t≥ε is immersed in(Zt)t≥ε).
Clearly, if B is complementable, it is also complementable after 0. The converse is false; for instance, if B is complementable, any Brownian motion generating the Goswami–Rao filtration associated withB(see [1]) is comple- mentable after 0, but not complementable.
Complementability after 0 is easily seen to be a property of the filtration: ifB¯ is any otherZ-BM1 generatingFB, thenB¯ is complementable after 0 if and only ifB is (with the same independent complementC). Thus, in that case we may say that the filtrationFBitself is complementable after 0.
Observe that (2) is equivalent to
ε>0(Zε∨σ (B, C))=Z∞ since Z∞ contains everything. Similarly, (1) is equivalent to
∀t >0
ε∈]0,t]
Zε∨Ft(B,C)
=Zt.
IfCin Definition 3 is not a complement toB, that is, ifσ (B, C)is strictly included inZ∞, one has
ε>0
Zε ∨σ (B, C)=Z0∨σ (B, C)=σ (B, C)=Z∞=
ε>0
Zε∨σ (B, C) ,
so the exchange property fails for theseσ-fields.
Observe also thatB and C play the same role in Definition 3: when (1) or (2) is satisfied, bothB andC are complementable after 0.
Remark 1 (Not used in the sequel). LetB andB¯ be twoZ-BM1such thatσ (B)⊂σ (B).¯ IfB is complementable after0,so is alsoB¯.Indeed,letCbe as in Definition3.Forε >0,one has
Z∞=Zε∨σ (B, C)⊂Zε∨σ (B, C)¯ ⊂Z∞,
wherefromZε∨σ (B, C)¯ =Z∞.Moreover,Cis independent ofB¯ sinceB¯ is parallel toB(Propositions1and2(iii)).
Consequently,B¯ is complementable after0.
In particular,ifB is complementable after0,so is also the filtrationBdefined in Proposition2.
Our first sufficient condition for maximality will derive from the next lemma.
Lemma 2. LetBandDbe twoZ-BM1.IfBis complementable after0andσ (B)⊂σ (D),thenFD⊂B,whereB is the filtration defined in Proposition2.
Proof. AsB is complementable after 0, letC be as in Definition 3. Consider anyZ-BM1D whose filtration con- tainsFB; to establishFD⊂B, it suffices to show the inclusionσ (D)⊂B∞=
t >0(Zt∨σ (B))of their terminal σ-fields.
SinceFDcontainsFB,Dis parallel toB, and also independent ofCby condition (iii) of Proposition 1; so(C, D) is aZ-BM2. Fort >0, we shall apply Lemma 1 to the threeσ-fields
S=Zt∨σ (B)=Zt∨σ B]t,∞[
, T =Zt∨σ (D)=Zt∨σ D]t,∞[
, U=σ C]t,∞[
.
First,Sis included inT becauseσ (D)containsσ (B). Second, owing to the definition ofCand to (2), Z∞=Zt∨σ (B, C)=Zt∨σ (B)∨σ
C]t,∞[
=S∨U⊂T ∨U⊂Z∞,
which impliesS∨U=T ∨U. Third,Uis independent ofT because the threeσ-fieldsZt,σ (C]t,∞[)andσ (D]t,∞[) are independent, for(C, D)is a Z-BM2. So Lemma 1 applies, yieldingS=T. Thereforeσ (D)⊂T =S=Zt∨ σ (B), and, ast >0 was arbitrary,σ (D)⊂
t >0(Zt∨σ (B)).
Proposition 3. LetB be aZ-BM1.IfB is complementable after0,then the filtrationBdefined in Proposition2is maximal.
Proof. IfFDcontainsBforDsomeZ-BM1, thenFDcontains a fortioriFBand Lemma 2 givesFD⊂B; soFD
cannot strictly containB.
Remark 2 (Not used in the sequel). DefineBto be maximal after0if forDany otherZ-BM1,one has the maximality property
ε>0
Zε∨σ (B)
⊂
ε>0
Zε∨σ (D)
⇒
ε>0
Zε∨σ (B)
=
ε>0
Zε∨σ (D)
(equivalently,with obvious notation,B⊂D⇒B=D).This property is clearly necessary forBto be maximal;it is also sufficient because,ifB⊂FD,one also hasB⊂FD⊂D,and maximality after0forcesB=D,whence B=FD.So Proposition3can be thus rephrased:complementability after0implies maximality after0.
The next maximality criterion also has two equivalent statements, in terms of the filtrationsFBandB, or of their endσ-fieldsσ (B)andB∞.
Theorem 1. Assume thatB is a Z-BM1,complementable after0. ThenB is maximal if and only if the following exchange property holds:
σ (B)=
ε>0
Zε∨σ (B)
(equivalently,the filtrationsFBandBare equal).
Proof. Corollary 1 says that the exchange property is necessary for maximality (without supposingBto be comple- mentable after 0).
Conversely, if the exchange property holds, the filtrationBfrom Proposition 2 has the same terminalσ-field as the filtrationFB, so these filtrations are equal; and Proposition 3 says thatBis maximal.
We shall now propose a variant of the sufficient condition in Theorem 1, obtained by modifying both the assumption thatBis complementable after 0 and the exchange property.
Proposition 4. LetW=(B, C)be aZ-BM2;callBthe filtration constructed fromB as in Proposition2.Assume (i) for eacht >0, σ (Z[t,∞[)⊂σ (Zt)∨σ (W]t,∞[);
(ii) for eacht >0,the conditional lawsL[Zt|FtB]andL[Zt|Bt]are equal.
ThenFB=BandBis maximal.
Observe that hypothesis (i) is stronger than supposingB to be complementable after 0, since it featuresσ (Zt) instead of the largerσ-fieldZt in the right-hand side. On the opposite, hypothesis (ii) is weaker than the exchange propertyFB=Bfrom Theorem 1.
Remark also that one always hasL[Zt|FtB] =L[Zt|F∞B]andL[Zt|Bt] =L[Zt|B∞]by independent enlargements;
so hypothesis (ii) in Proposition 4 amounts to
∀t >0 L
Zt|σ (B)
=L
Zt
ε>0
Zε∨σ (B)
. (3)
Proof of Proposition 4. It suffices to show thatσ (B)=B∞; this implies equality of the filtrationsFBandB, and maximality ofB follows by Theorem 1, because hypothesis (i) is stronger than complementability after 0. For an arbitrary bounded,Z∞-measurable r.v.J, we have to show
E
J
ε>0
Zε∨σ (B)
=E
J|σ (B)
. (4)
By L1-density, we may restrict ourselves to the case thatJ is measurable inσ (Z[t,∞[)for somet >0. By hypoth- esis (i),J is some measurable function ofZt and of the processesB]t,∞[ andC]t,∞[. By linearity and density, we may supposeJ=f (Zt)g(B]t,∞[)h(C]t,∞[)withf,gandhbounded and measurable. Callingwthe 1-dimensional Wiener measure, one has
E
J|Zt∨σ (B)
=f (Zt)g B]t,∞[
E h
C]t,∞[Zt∨σ (B)
=f (Zt)g
B]t,∞[ hdw sinceC]t,∞[is a BM1 independent ofZt∨σ (B]t,∞[)=Zt∨σ (B). Consequently
E
J
ε>0
Zε∨σ (B)
=
hdw g B]t,∞[
E
f (Zt)
ε>0
Zε∨σ (B) .
Using (3), the right-hand side becomes(
hdw)g(B]t,∞[)E[f (Zt)|σ (B)]; as this isσ (B)-measurable, (4) is estab-
lished.
Remark 3. The same proof also yields a slightly stronger result:Proposition4 remains true if (i)and (ii) are no longer supposed to hold for eacht >0,but only for a sequence of times tending to0.
5. Maximality of the BM1 driving the angular part ofZ
It will be convenient to considerZ=(X, Y )as the complex BMZ=X+iY. CallR the modulus ofZ. Almost surelyZdoes not come back to 0, so another complex BMW can be defined by
Wt = t
0
Rs Zs dZs.
The real and imaginary parts ofWare the BM1 given by Ut=
t
0
XsdXs+YsdYs
Rs and Vt= t
0
XsdYs−YsdXs
Rs .
The next proposition recalls some well-known facts concerningU andV (see, for instance, Proposition 3.1 and Theorem 3.4 in [4]).
Proposition 5 (Classical properties ofU andV; see [4]). (i)The linear Brownian motionsUandV are indepen- dent.
(ii) The processRis a strong solution of the stochastic differential equation dRs=dUs+ ds
2Rs
;
in particular,RandU generate the same filtration.
(iii) For0< s≤t, Zt
Rt =Zs Rs exp
i
t
s
dVr
Rr .
(iv) For eacht >0,Zt=FtW∨σ (Zt/Rt);moreover,the r.v.Zt/Rtis independent ofWand uniformly distributed on the unit circle.
It is shown in [2] that theZ-BM1U is complementable, hence also maximal. Clearly,V is not an independent complement toU, because(U, V ) does not generateZ (Proposition 5(iv)); butV (and alsoU) is complementable after 0, because ifZ has been observed during some time-interval[0, ε](however small), then observingU andV suffices to recoverZ.
We shall show thatV is maximal as an application of Proposition 4.
Theorem 2. TheZ-BM1V = XdY−YdX
R is maximal.
Proof. We shall check thatV satisfies hypotheses (i) and (ii) of Proposition 4.
Hypothesis (i) is readily verified: knowingZt and the increments ofU andV aftert, one can easily recover the path ofZaftert: the modulusR= |Z|is obtained as the solution to the stochastic differential equation
dRs=dUs+ ds 2Rs
,
and the argumentΘofZ=ReiΘis then given by dΘs=dVs/Rs.
Hypothesis (ii), or equivalently (3), says that the conditional laws of Zt given the σ-fields
ε>0(Zε∨σ (V )) andσ (V )are equal. The latter,L[Zt|V], is simply the law ofZt because Zt is independent ofV. This is a direct consequence of Proposition 5: by (ii),Rt is independent ofV, and by (iv),Zt/Rt is independent of the pair(U, V ), and a fortiori of(V , Rt). Hence the triple(V , Rt, Zt/Rt)is independent, andZt=(Zt/Rt)×Rt is independent ofV.
So, to verify hypothesis (ii), we have to check thatZt is independent of
s>0(Zs∨σ (V ))too. This property is less straightforward than the independence of Zt andV; we state it as a lemma. Strictly speaking, the elementary proof, given above, thatZt is independent ofV, was not necessary, because it is also a consequence of that lemma, for
ε>0(Zε∨σ (V ))containsσ (V ).
Lemma 3 (Key lemma). For anyt >0,the r.v.Zt is independent of theσ-field
s>0(Zs∨σ (V )).
The end of this paper is devoted to the proof of Lemma 3, which consists of five steps; the first two, Lemmas 4 and 5, are general facts for later use. Lemma 4 is certainly well known.
Lemma 4. If(Tp)p≥1is any sequence of strictly positive,identically distributed r.v.,thenlim suppTp is a.s.strictly positive.
Proof. Forδ >0, observe that
lim sup
p
Tp< δ =
∃psup
q≥p
Tq< δ
⊂lim inf
p {Tp< δ}.
By Fatou’s lemma, the probability of these events is majorized by lim infpP[Tp< δ], which equalsP[T1< δ]since theTphave the same law. SoP[lim supTp< δ] ≤P[T1< δ]; the lemma follows by lettingδgo to 0.
Lemma 5. On some probability space(Ω, A,P),letβ be aBM1and(αt)[1,2]be a continuous process independent ofβ,with diffuse law.Then the conditional law of2
1 αtdβt givenβis almost surely diffuse.A fortiori,for every real m=0,
E
exp
im 2
1
αtdβt β
<1 a.s.
Proof. Letα˜ be a process with the same law asα, independent of(β, α). The conditional law L
2 1
(αt− ˜αt)dβtα,α˜
=N
0, 2
1
(αt− ˜αt)2dt is a.s. non-degenerate. Consequently,P[2
1(αt− ˜αt)dβt=0] =0, wherefrom P
2 1
αtdβt= 2
1
˜ αtdβtβ
=0 a.s.
But, conditional onβ, the r.v.2
1αtdβt and2
1α˜tdβt are independent and have the same law; so this conditional law
must be diffuse.
For the remaining three steps of the proof, we come back to the setting of Theorem 2, withRdenoting|Z|.
Lemma 6. CallRtheσ-fieldσ ((R2k)k∈Z).Then for everyk∈Z,theσ-fieldsZ2k andσ (V , R[2k,∞[)are condition- ally independent givenσ (V )∨R.
Proof. Conditional on theσ-fieldR, the processRis still Markov (it is a succession of independent Bessel bridges);
hence theσ-fieldsF2Rk andσ (R[2k,∞[)are conditionally independent givenR.
AsV is independent ofR, theσ-fieldsF2Rk∨σ (V )andσ (R[2k,∞[)∨σ (V )are conditionally independent given R∨σ (V ).
Last, Proposition 5(iv) gives the equalityZ2k=F2Rk∨F2Vk∨σ (Z2k/R2k), withZ2k/R2k independent of(R, V ).
The result follows.
Lemma 7. The r.v.Z1/R1is independent of
s>0(Zs∨σ (V )∨R)and uniformly distributed on the unit circle.
Proof. Owing to the reverse martingale convergence theorem, it suffices to prove that for everym∈Z\ {0}, E
Z1
R1 m
Z2n∨σ (V )∨R
−→0 a.s. whenn→ −∞.
CallQnthis conditional expectation, andΔkthe angular variation ofZfrom time 2kto 2k+1: Δk=
2k+1 2k
dVs
Rs
.
As this stochastic integral can be computed in the filtration generated by the processesVt andRt1{t≥2k},Δk is mea- surable inσ (V , R[2k,∞[); so Lemma 6 ensures thatΔk is independent ofZ2k conditionally onσ (V )∨R. An easy induction shows that, forn≤ −1, the r.v.Δn, . . . , Δ−1 are conditionally independent givenZ2n∨σ (V )∨R. For n≤ −1, writing
Z1
R1 m
= Z2n
R2n m−1
k=n
eimΔk
one gets the product expansion Qn=
Z2n R2n
m−1 k=n
Ck, whereCk=E
eimΔk|Z2n∨σ (V )∨R
. (5)
Using Lemma 6 again, notice that Ck=E
eimΔk|Z2n∨σ (V )∨R
=E
eimΔk|σ (V )∨R
=E
eimΔk|σ (V , R2k, R2k+1) .
From (5), one draws|Ck| ≤1 and|Qn| = |Cn| · · · |C−1|; to finish proving Lemma 7, it remains to see that this product tends to 0 whenn→ −∞. We shall in fact prove a stronger property, namely
lim inf
k→−∞|Ck|<1.
This stems from Lemma 4 applied to the sequence of r.v. 1− |Ck|; we only have to check that both hypotheses of this lemma are satisfied.
First, all the|Ck|are identically distributed, since, by scaling invariance ofZ, the joint law of Δk,
2−k/2V2kt
t≥0,2−k/2R2k,2−k/2R2k+1
does not depend uponk.
The other hypothesis to be checked is|C0|<1 a.s. WriteC0=E[exp(im2
1Rs−1dVs)|σ (V , R1, R2)], and observe that
E
exp
im 2
1
Rs−1dVs V =v, R1=r1, R2=r2
=E
exp
im 2
1
αtdβt β=v
,
whereαandβare as in Lemma 5, withαthe inverse of the 2-dimensional Bessel bridge fromr1tor2. Lemma 5 says that givenr1andr2, the modulus of the right-hand side is strictly less than 1 for almost allv; this proves the claim,
and completes the proof of Lemma 7.
The fifth step is the final one:
Proof of Lemma 3. By scaling, we may supposet=1; we have to establish that the r.v.
E
f (Z1)
s>0
Zs∨σ (V )
is constant, wheref is any bounded, Borel function.
By linearity and L1-density, it suffices to consider functions of the form f (z)=g(|z|)h(z/|z|), with g andh bounded. Observing thatR1isR-measurable and calling on Lemma 7, one can write
E
f (Z1)
s>0
Zs∨σ (V )∨R
=g(R1)E
h Z1
R1
s>0
Zs∨σ (V )∨R
= ¯hg(R1),
whereh¯is the constantE[h(Z1/R1)]. Therefore E
f (Z1)
s>0
Zs∨σ (V )
= ¯hE
g(R1)
s>0
Zs∨σ (V )
. (6)
UsingZs∨σ (V )=FsR∨σ (V )∨σ (Zs/Rs)and the independence ofR,V andZs/Rs, one has for 0< s≤1 E
g(R1)|Zs∨σ (V )
=E
g(R1)|FsR∨σ (V )∨σ (Zs/Rs)
=E
g(R1)|FsR
.
Letsgo to zero. By reverse martingale convergence, and recalling thatF0R+=F0U+is degenerate, one ends up with E
g(R1)
s>0
Zs∨σ (V )
=lim
s→0 s>0
E
g(R1)|FsR
=E g(R1)
;
so the right-hand side of (6) is constant. This proves Lemma 3, and Theorem 2 by the same token.
References
[1] S. Attal, K. Burdzy, M. Émery and Y. Hu. Sur quelques filtrations et transformations browniennes. InSéminaire de Probabilités XXIX56–69.
Lecture Notes in Mathematics1613. Springer, Berlin, 1995. MR1459449
[2] J. Brossard and C. Leuridan. Filtrations browniennes et compléments indépendants. InSéminaire de Probabilités XLI265–278.Lecture Notes in Mathematics1934. Springer, 2008.
[3] M. Émery. On certain almost Brownian filtrations.Ann. Inst. H. Poincaré Probab. Statist.41(2005) 285–305. MR2139021
[4] D.W. Stroock and M. Yor. Some remarkable martingales. InSéminaire de Probabilités XV 590–603.Lecture Notes in Mathematics850.
Springer, Berlin, 1980. MR0622590
[5] H. von Weizsäcker. Exchanging the order of taking suprema and countable intersections ofσ-algebras.Ann. Inst. H. Poincaré Probab. Statist.
19(1983) 91–100. MR0699981
