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www.imstat.org/aihp 2011, Vol. 47, No. 2, 498–514

DOI:10.1214/10-AIHP363

© Association des Publications de l’Institut Henri Poincaré, 2011

Hiding a constant drift

Vilmos Prokaj

a

, Miklós Rásonyi

b,1

and Walter Schachermayer

c,2

aDepartment of Probability Theory and Statistics, Eötvös Loránd University, 1117 Budapest, Pázmány P. s. 1/C, Hungary, Computer and Automation Institute of the Hungarian Academy of Sciences, 1111 Budapest, Kende u. 13-17, Hungary. E-mail:[email protected] bSchool of Mathematics, University of Edinburgh, King’s Buildings, Mayfield Road, EH9 3JZ, UK. E-mail:[email protected] cFaculty of Mathematics, University of Vienna, Nordbergstrasse 15, 1090 Vienna, Austria. E-mail:[email protected]

Received 27 May 2009; revised 16 January 2010 Dedicated to Marc Yor at the occasion of his 60th birthday

Abstract. The following question is due to Marc Yor: LetB be a Brownian motion andSt=t+Bt. Can we define anFB- predictable processH such that the resulting stochastic integral(H·S)is a Brownian motion (without drift) in its own filtration, i.e. anF(H·S)-Brownian motion?

In this paper we show that by dropping the requirement ofFB-predictability ofHwe can give a positive answer to this question.

In other words, we are able to show that there is a weak solution to Yor’s question. The original question, i.e., existence of a strong solution, remains open.

Résumé. La question suivante a été posée par Marc Yor: Soit B un mouvement Brownien etSt=t+Bt. Peut-on définir un processusH qui estFB-prévisible tel que l’intégrale stochastique(H·S)soit un mouvement Brownien (sans drift) pour sa propre filtrationF(H·S)?

Dans cet article nous fournissons une réponse affirmative en relâchant la condition queH soitFB-prévisible. Autrement dit, nous montrons qu’il existe une solution faible pour cette question de Yor. La question originale (c’est à dire, l’existence d’une solution forte) reste ouverte.

MSC:Primary 60H05; 60G44; 60J65; secondary 60G05; 60H10

Keywords:Brownian motion with drift; Stochastic integral; Enlargement of filtration

1. Introduction

1.1. Main result

In this paper, we investigate the following question, due to Marc Yor: letB be a Brownian motion andSt=t+Bt. Can we define anFB-predictable processH such that the resulting stochastic integral(H ·S)is a Brownian motion (without drift) in its own filtration, i.e. anFH·S-Brownian motion? Our main result here is the following theorem.

Theorem 1. Let(Wt)t0be a real-valued standard Brownian motion on(Ω,F,P)and denote by(FtW)t0its(right continuous,saturated)natural filtration.

1On leave from the Computer and Automation Institute of the Hungarian Academy of Sciences.

2This author gratefully acknowledges financial support from the Austrian Science Fund (FWF) under Grant P19456, from the Vienna Science and Technology Fund (WWTF) under Grant MA13 and from the Christian Doppler Research Association (CDG).

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Fix μ∈R.Then there is an(FtW)t0-Brownian motion(Bt)t0as well as an(FtW)t0-predictable,{−1,+1}- valued processHsuch that the stochastic integralH·Sis a Brownian motion in its own filtration(FtH·S)t0,where St=Bt+μt.

Theorem 1 improves upon the result in a previous paper [9], where it was proved that for a given Brownian motionB, a constantμ∈Rand a thresholdδ >0 one can definet)t0and(Ht)t0both(FtB)t0-predictable such that|μμt| ≤δandβt=t

0Hs(dBs+μsds)is a Brownian motion in its own filtration. In other words any constant drift can be uniformly approximated by “strongly hidable” random drifts.

Roughly speaking, Theorem1gives a positive answer to Yor’s question, provided one replaces the filtration gen- erated by(Bt)t0 by the larger filtration(FtW)t0generated by(Wt)t0. However, Yor’s original question remains open and is left for further research.

Laurent Serlet in [11] also deals with the problem of creation and deletion of drift. His approach, based on excursion theory and the notion of contour process focused on somewhat different problems.

Remark. Following the advice of the anonymous referee,we remark that in addition to the assertions of Theorem1, the following fact also holds true:the filtration(FtW)t0is weakly generated by the Brownian motion(Bt)t0,see[5], Definition6.2and Remark6.1.In other words,every(FtW)t0-martingale(Mt)t0can be represented as a stochastic integralM=c+K·B,wherec∈Ris a constant andKis a predictable process with respect to(FtW)t0.

This follows from the Itô representation theorem applied to the Brownian motion(Wt)t0in its natural filtration (FtW)t0.By Theorem1the processBis a Brownian motion in this filtration so that there is an(FtW)t0-predictable processL=(Lt)t0such thatB=L·W.ClearlyLtakes values in{−1,+1}for almost all(ω, t)Ω×R+with respect to the product ofPand the Lebesgue measure onR+.Then,obviouslyW=L·Bis also true.

Given an arbitrary martingaleM in the filtration(FtW)t0we may again apply Itô’s representation theorem to obtain a (FtW)t0-predictable processK such thatM=c+K·W with somec∈R.Then,K=KL gives the representationM=c+K·B.

1.2. Generalizations

Here is the rationale of our paper: One can easily check that for Theorem 1 we have to define H such that E(Hs|FsH·S)=0 for almost all s≥0. Our construction initially uses a larger filtration than the natural filtration of(Wt)t0. We start with a probability space on which, apart from the Brownian motionW, there is an independent random variableU uniformly distributed on]0,1[. The construction ofBt andβt=(H·S)t will be such that at each moment t the value ofHt depends on whetherU is larger or smaller than the conditional median ofU, given the sample paths)0st. This idea of construction has been already used in [9]. With this “median rule” we achieve that for allt the random variableHt is independent ofs)0st. The difficulty is of course the existence of suchH, Bandβas they are strongly related.

As a byproduct of our investigations we also get the following result which is interesting in its own right.

Theorem 2. Let t)t0 be a real-valued standard Brownian motion based on a filtered probability space (Ω,F, (Ft)t0,P).

Fixμ∈R+.Then there is an enlargement(Ft)t0of(Ft)t0such thatβis an(Ft)t0-semimartingale of the formt=dβt+νtdtwhereβis an(Ft)t0-Brownian motion and|νt| =μfor allt.

The enlargement(Ft)t0in Theorem2is an initial enlargement, that is we actually prove that there is a random variableUuniformly distributed on]0,1[, such thatFt=

s>t(Fsσ (U )). We still point out, following a suggestion of the referee, that it is also possible to define(Ft)t0such that it is the filtration generated by some Brownian motion (Wt)t0.

The statement of Theorem 2 can be generalized, by replacing the Brownian motion β with a continuous local martingale M and the constantμ by a predictable process t)t0 that can be integrated with respect to M, i.e.

t

0μ2sdMs<∞almost surely for allt≥0. We use the notationL(M)of [8] for the family of predictable processes, that can be integrated with respect to the continuous local martingale(Mt)t0.

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Theorem 3. Let(Mt)t0 be a continuous local martingale on a filtered probability space(Ω,F, (Ft)t0,P).We assume that on(Ω,F,P)there is a random variableU,uniformly distributed on]0,1[and independent fromF.

Fix an(Ft)t0-predictable,non-negative process t)t0inL(M).Then there is an enlargement(Ft)t0 of the filtration(Ft)t0such that,Mis an(Ft)t0-semimartingale of the form

dMt=dMt+νtdMt, (1)

whereMis an(Ft)t0local martingale and|νt| =μt for allt. It turns out from the proof that if

0

μ2sdMs= ∞ almost surely

then, we can do the construction in such a way thatFtFfor allt, i.e.U is not needed in this case, showing that Theorem3is indeed a generalization of Theorem2.

1.3. Outline of the paper

The structure of the paper is as follows. First, we present the discussion of the discrete analogue of the problem. Then, we solve the equation formally derived from the discrete case and show that as in the discrete time case this gives a solution using the “median” strategy. It turns out that the extra randomnessU used throughout our construction is encoded in the sample paths ofβ. Using this observation we prove Theorem2. Theorem3then follows easily. Finally, we prove Theorem1completely. In theAppendixwe present minor, but useful technical results.

2. The discrete case

This section only serves as motivation for the subsequent continuous time case.

Fixμ∈Randt >0 such that|μ(t )1/2|<1. We consider a biased random walkS=(Sti)i=0on a fine grid (ti)i=0, withti=it. We suppose that the increments(Sti)i=0=(Sti+1Sti)are an i.i.d. sequence with

P

Sti= +(t )1/2

=1+μ(t )1/2

2 ,

P

Sti= −(t )1/2

=1−μ(t )1/2

2 .

The processSis a discrete analog toBt+μt, the Brownian motion with driftμ, which will be considered below, because

E(Sti)=μt, E St2

i

=t.

In addition, letUbe a uniformly distributed]0,1[-valued random variable independent ofS. The filtration(Fti)i=0is defined as the smallest one such thatUisF0-measurable and(Sti)i=0is adapted to(Fti)i=0.

We shall construct inductively a predictable,{−1,+1}-valued process(Hti)i=1such that((H·S)ti)i=0is an unbi- ased random walk (i.e., it is a martingale) in its own filtration.

To do so we construct a]0,1[-valued process(Dti(U ))i=0adapted to(Fti)i=0inductively by lettingDt0(U )=U and

Dti(U )=min

Dti(U ),

1−Dti(U ) μsign

1

2−Dti(U )

Sti, i≥0, (2)

whereDti(U )denotes the incrementDti+1(U )Dti(U ).

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The process(Hti)i=1is derived from(Dti(U ))i=0by Hti+1=sign

Dti(U )−1 2

, i≥0. (3)

Here is the idea behind this construction. At the first stepi=0, formula (3) yieldsHt1=sign(U−12), i.e., we flip a coin, independent ofS, whetherHt1 equals+1 or−1. So that obviouslyE((H·S)t1)=E(Ht1St1)=0. Now we may observe the outcome(H·S)t1=Ht1St1 which takes the value+(t )1/2or−(t )1/2. Applying Bayes’ rule, this information updates the conditional distribution ofU: conditionally on the event{Ht1St1= +(t )1/2}we obtain for the conditional distribution functionDt1(x),

Dt1(x)=

1−μ(t )1/2

x for 0≤x12,

1−μ(t )1/21

2+

1+μ(t )1/2 x12

for 12x≤1,

and an analogous expression conditionally on the event{Ht1St1= −(t )1/2}. The random variableDt1(U )equals the conditional distribution functionDt1(·)at the random pointU. Hence it makes sense to define

Ht2=sign

Dt1(U )−1 2

as the preceding argument shows thatP(Ht2=1|Gt1)=P(Ht2= −1|Gt1)=12, almost surely, whereGt1 denotes the sigma-algebra generated byHt1St1.

Now we may continue inductively and some elementary calculations show that we thus obtain the updating rule for the process(Dti(U ))i=0given by (2). We summarize these facts in the subsequent statement.

Proposition 4. Defining the process(Dti(U ))i=0and(Hti)i=1as above and denoting by(Gti)i=0the filtration gen- erated by((H·S)ti)i=0,where(H·S)ti=i

u=1HtuStu1,we have,for almost allωΩ,

Dti(U )(ω)=P(Ux|Gti)(ω), (4)

wherex=U (ω).In particular we have E(Hti|Gti−1)=0, a.s.,fori≥1,

so thatE(HtiSti1|Gti1)=0,hence(H·Sti)i=0is a martingale in its own filtration(Gti)i=0.

Remark 5. A word of warning seems to be in order:it is not clear3how to definesign(0)in the formulas(2)and(3), and the above arguments do not apply to the case where this occurs.However,this is not really a problem in the present discrete time setting as this case only appears on a null set of Ω and therefore can be safely ignored by inserting the words “almost surely.”

On the other hand,in the continuous time case,this problem will become crucial and is discussed later at the end of Section3.1.

3. Continuous time

We shall now try to pass to the continuous time limit of the above random walk construction. We prove the next statement, which is nothing else but the statement of Theorem 1, written once again, for ease of the reader, but modulo the addition of an auxiliary uniform variableU.

3The anonymous referee observed that, following the tradition of P. A. Meyer, it sometimes is advantageous to define sign(0)= −1; using this convention, e.g., in the definition of local times, one thus obtains càdlàg versions of(Lat)a∈R,t0.

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Proposition 6. Let(Wt)t0be a real-valued standard Brownian motion based on(Ω,F, (Ft)t0,P),i.e.,(Wt)t0is an(Ft)t0Brownian motion.Assume that there is anF0-measurableUuniformly distributed on]0,1[.

Fixμ∈R.Then there is an(Ft)t0-Brownian motion(Bt)t0as well as an(Ft)t0-predictable,{−1,+1}-valued processHsuch that the stochastic integralβ=H·Sis a Brownian motion in its own filtration,whereSt=Bt+μt.

Moreover,the processesB,HandH·Sare adapted to(FtW,U)t0.

We split the proof into two parts. First, we look for a solution of the system of equations formally derived from the discrete time case. This means that, for a given Brownian motionBand an independent random variableU, we want to solve

dDt= −μmin

Dt, (1Dt) sign

Dt−1

2

dSt, D0=U, (5)

where

St=Bt+μt. (6)

We will also use the notation βt=

t

0

sign

Du−1 2

dSu. (7)

We shall see that the process(Dt)t0satisfying (5) cannot be adapted to the filtration(FtU,B)t0. We only can find a weak solution. In the setting of Proposition6we can derive the processesD, B, β from the given dataW andU such that (5) and (7) hold true as Itô-integrals in(Ft)t0, see Corollary8below.

Secondly, we show that if the processesB,Dandβare related to each other according to this system of equations, then they provide a solution to Yor’s question in the weak sense as formulated in Proposition6(see Lemma10below).

3.1. Heuristic description

Before turning to the proof, it is worth to have a closer look at the equations. The delicate term on the right-hand side of (5) is sign(Dt12), similarly as in Tanaka’s equation

dXt=sign(Xt)dBt. (8)

It is well known that there is no solution(Xt)t0to (8) adapted to the filtration generated by(Bt)t0; rather one has to enlarge the filtration, i.e. introduce additional sources of randomness, in order to obtain a weak solution to (8).

A similar problem appears in (5) when the process(Dt)t0hits the value 12. This happens for the first time at τ=inf

t >0:Dt=1 2

,

which is a stopping time with respect to(Ft)t0. For 0≤tτ the SDE (5) clearly has a strong solution with respect to the filtration(FtB,U)t0(hence, a fortiori, with respect to(FtW,U)t0), which is explicitly given by the formula

Dt=Uexp

μStμ2 2 t

=exp

ln(U )+μ

Bt+μ 2t

, 0≤tτ, (9)

forU∈ ]0,12[, and by the formula 1−Dt=(1U )exp

μStμ2 2 t

=exp

ln(1−U )+μ

Bt+μ 2t

, 0≤tτ, (10)

forU∈ ]12,1[.

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A unified way of writing (9) and (10) is 1− |1−2Dt| =exp

ln

1− |1−2U| +μ

Bt+1

2μt

for 0≤tτ. (11)

Using this identity we can expressτ with the help ofBandU, as τ =inf

t >0:Bt+1

2μt= −1 μln

1− |1−2U|

, (12)

soτ is a stopping time with respect to(FtB,U)t0.

We now give a heuristic and intuitive description of the present construction of a global solution to the SDE (5).

We motivate the construction by recalling the solution to Tanaka’s equation

dXt=sign(Xt)dBt, X0=0, (13)

where(Bt)t0is a standard Brownian motion starting atB0=0. It is well known how to construct and interpret a weak solution(Xt)t0. We summarize the construction in a heuristic way: we decompose each trajectory(Bt(ω))t0 into its running minimum(Mt(ω))t0and its positive excursions(Et(ω))t0, i.e.

Mt(ω)= inf

0stBs(ω), Et(ω)=Bt(ω)Mt(ω).

We may decompose(Et(ω))t0into its excursions. More precisely, there are sequences of random timesn)n=1and n)n=1taking values in[0,∞]such thatJσn, τnKis a.s. a sequence of disjoint intervals ofR+, whose union has full Lebesgue measure and such thatEσn=Eτn=0 whileEt >0, fort∈Kσn, τnJ. (For details we refer the reader to the classical book of Itô and McKean [4], Section 2.9, see also [6].)

Choose an i.i.d. sequencen)n=1of symmetric random signs independent of(Bt)t0and define Xt(ω)=

n=1

εn(ω)Et(ω)χJσnnK(t).

The filtration(FtX)t0generated by(Xt)t0then contains the filtration(FtB)t0generated by(Bt)t0, and(Xt)t0 as well as(Bt)t0are Brownian motions in the filtration(FtX)t0, satisfying the stochastic differential equation (13).

Summing up informally, we obtain the trajectories (Xt)t0 from the trajectories (Bt)t0 by flipping coins and pasting together the excursions of(Bt(ω))t0 multiplied with the corresponding random signsεn(ω)to obtain the trajectories(Xt(ω))t0.

This may be rephrased as follows: the processHt =

n=0εnχKσnnK(t)is predictable in(FtX)t0and we have B=H·Xas well asX=H·B, holding true with respect to this filtration.

When comparing the situation of Tanaka’s equation (13) with the present equation (5), let us start with the trivial observation that when changing the sign in Tanaka’s equation, i.e.

dXt= −sign(Xt)dBt, X0=0,

we have to replace the running minimum in the above construction by the running maximum and the positive excur- sions by the negative ones.

Now we pass to our present situation. We start with the processSt=Bt+μt where(Bt)t0is a given Brownian motion defined on((Ω,F, (Ft)t0),P). We need, as additional stochastic input, aF0-measurable random variable U, uniformly distributed on]0,1[, as well as anF0-measurable i.i.d. sequencen)n=1of random signs such thatU and(ε)n=1are independent.

We now decompose the processBt+12μt into its running maximum(Mt)t0and its negative excursions(Et)t0 from the running maximum(Mt)t0,

Mt= sup

0st

Bs+1

2μs

, Et=

Bt+1 2μt

Mt.

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Again we enumerate these excursions by random timesn, τn)n=1as above.

Let Ht=

sign U12

for 0≤τ,

εn fort > τ andt∈Kσn, τnK. (14)

Denoting by(FtB,H)t0the filtration generated by(Bt)t0and(Ht)t0we find that in this filtrationB is a Brownian motion andH is a predictable process, so that we may well define the process

βt= t

0

Hud(Bu+μu). (15)

We shall show thatt)t0is a Brownian motion in its own filtration(Ftβ)t0which is contained in(FtB,H)t0. Before doing so we interpret intuitively the construction given by (14) and (15). Let the trajectory(Bt(ω))t0as well as the random numberU (ω)be given. The trajectoryt)t0is initially given by either the trajectory(Bt+μt )t0

or−(Bt+μt )t0, depending on the sign ofU12, namely up to timeτ, which is defined in (12).

Note that at time τ the process Bt + 12μt attains for the first time the level −μ1ln(1− |1−2U|), whence, in particular,Bτ+12μτ=Mτ almost surely.

After timeτwe multiply the excursions(EtχJσnnK(t))n=1of the processBt+12μtwith the random signsn)n=1 and paste them together in order to obtain the trajectory oft)t0. This result is a variant of the construction in Tanaka’s case above: we have, fort∈Jσn, τnK, whereσnτ,

βtβσn=εn

Et+μ

2(tσn)

=εn

BtBσn+μ(tσn)

fort∈Jσn, τnK.

In addition, the trajectories oft)t0have to be continuous; this—in conjunction with the above equation—uniquely determinest)t0pathwise.

Let us try to explain why this construction ofβindeed yields a martingale in its own filtration(Ftβ)t0. We consider the distribution function(Dt(x))ofU conditionally on(Ftβ)t0, where 0≤x≤1. Fix the random elementωΩ and suppose thatU (ω)=y∈ [0,1]. To fix ideas let us suppose thaty <12. Then, for 0≤tτ, the Bayesian updating for the conditional distribution function

Dt(x)=P

Ux|s)0st

is such that forx=y the value ofDt(y)is given by (9). Hence it is less than 12, for 0≤tτ, and hits the value 12 for the first time att=τ. According to our “median rule”Ht=sign(Dt12), this is the critical moment to change the sign ofH. The good interpretation of sign(0)in the present context is to flip a coin whether sign(0)equals+1 or

−1. If the process does “not start a negative excursion” at this moment, we have to flip a new coin at an infinitesimal time unit later and to continue to do so until “eventually a negative excursionEtχJσnnK(t)starts.” (The preceding heuristic phrase should be interpreted by visualising the process(Bt+12μt )t0as a normalized random walk on a very fine grid.) During this intervalJσn, τnKthe sign ofHt is determined by the coin flipεn (which, intuitively speaking, was done at timeσn, the “last moment before the excursionEtχ(JσnnK) started”). During this random interval the Bayesian updating rule for the conditional distribution functionDt(·)ofUevaluated aty=U (ω)yields an excursion ofDt(y)from the value 12, to the right or left, depending on the sign ofεn. Precisely at timeτnthe value ofDt(y)is back to12.

The preceding heuristics should motivate that we again use the idea of the “median rule” forH, similarly as in [9];

to do so we need, apart from the initial random inputU, also the random variables(εn)n=1. The random signεnis interpreted as a coin flipping at the random timesσn. These random timesσnfail to be stopping times in the filtration (FtB,U)t0. This is the reason why we cannot define the strategyH in such a way that it is an(FtB,U)t0-predictable process; rather we have to pass to an enlargement of the filtration and find a weak solution forH, which eventually leads to Theorems1and2.

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3.2. Proofs

We now turn to Eq. (5). First we transform it into an equation which is more tractable. So assume thatBis a Brownian motion with respect to the filtration(Ft)t0and(Dt)t0,(St)t0are(Ft)t0adapted processes satisfying (5) and (6).

Then,

dDt

1/2− |Dt−1/2|= −μsign

Dt−1 2

dSt. (16)

In order to integrate the left-hand side we considerf (Dt)where f (x)= −sign

x−1

2

ln

1− |2x−1|

forx(0,1) (17)

is the quantile function of the symmetrized exponential law with parameter 1. The functionf is continuously differ- entiable and the second derivative exists except at 12, where it has right and left limits:

f(x)= 1

2− x−1

2 1

, f(x)=sign

x−1 2

1 2−

x−1 2

2

.

The inverse off, the distribution function of the symmetrized exponential law, has the same differentiability properties asf, i.e., it is continuously differentiable and the second derivative exists except at 0, where it has right and left limits.

The discontinuities offand(f1)are harmless for the application of Itô’s formula, which, together with sign

f (x)

=sign

x−1 2

= f(x) (f(x))2, yields the next statement:

Proposition 7. Let(Xt)t0and(Dt)t0be semimartingales in the filtration(Ft)t0.Then,the differential equation dDt=min(Dt,1−Dt)dXt, with0< D0<1,

is satisfied if and only ifZt=f (Dt)is the solution of dZt=dXt+1

2sign(Zt)dXt, Z0=f (D0).

When we consider Eq. (5), then dXt= −μsign(Dt12)dStso we have to solve dZt = −μsign(Zt)dSt+1

2μ2sign(Zt)dt

= −μsign(Zt)dBt−1

2μ2sign(Zt)dt, Z0=f (U ). (18)

The way we solve this equation depends on the initial data. In the heuristic description, when we started with a given BandU, we used that these two objects determine|Zt|, by the formula

|Zt| = |Z0| −μBt−1

2μ2t+L0t(Z)

= |Z0| −μBt−1

2μ2t+max

st

μBs+1

2μ2s− |Z0|

∨0

.

So we have to unfold this process with the help of independent random signs to obtain Z and hence D. Such an

“unfolding” result can be found in [7].

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In Proposition6we would like to findBandDsuch thatBis a Brownian motion, and (5) holds within a filtration generated by W andU. So we follow another approach here. If we set Wt =t

0sign(Zs)dBs, then (18) reads as follows

dZt= −μdWt−1

2μ2sign(Zt)dt, Z0=f (U ). (19)

Note, that the discontinuous function sign(·)now is in the drift term rather than in the diffusion term. Existence and pathwise uniqueness of the solution of (19) follows from Theorem 3.5(i) [10], Chapter IX. So, we can take(Zt)t0as the solution of (19) and then, we defineBt=t

0sign(Zs)dWs. It is clear that(Bt)t0is a Brownian motion in(Ft)t0, adapted to(FtU,W)t0. We defineSandXas before

St=Bt+μt, Xt= −μ t

0

sign(Zu)dSu. Thus, with this notation,

dZt=dXt+1

2sign(Zs)dXs.

So application of Proposition7gives the following corollary.

Corollary 8. Under the assumptions of Proposition6,there is an(Ft)t0-Brownian motion(Bt)t0and an adapted process(Dt)t0such that(5)holds true as an Itô integral in(Ft)t0.Moreover,(Bt)t0and(Dt)t0are adapted to (FtW,U)t0.

Remark 9. The SDE(19)defines a Markov process,which is ergodic ifμ=0, sometimes it is called the “bang–

bang” process.The scale functionsofZis given by the formulas(x)=sign(x)(e|x|−1),and the speed measure on the natural scale has a densityp(x)=(ss1)2.Then,the density of the invariant distribution on the original scale is proportional top(s(x))s(x)=1/s(x)=e−|x|,i.e.the symmetrized exponential distribution with parameter1.

SinceZ0has symmetrized exponential law with parameter1,the solution of(19)we used to defineDis a stationary Markov process.This means that the law ofDt is the same for allt,i.e.,Dt is uniformly distributed on]0,1[.The byproduct of the next lemma is that this is even true for the conditional law ofDt givenFtβ.In fact,item(iii)below asserts that conditional distribution function(Dˆt(x, ω))0x1of the random variableUevaluated atx=U (ω)equals Dt(ω),i.e.,Dˆt(U (ω), ω)=Dt(ω)for almost allω.This is also crucial in the construction,because it means that we indeed apply here the median strategy.The processHt=sign(Dt12)is plus or minus one if the value ofUis above or below the conditional median,respectively.

Lemma 10. Assume thatB is an (Ft)t0-Brownian motion, U is an F0 measurable random variable uniformly distributed on]0,1[,andD,S,β are(Ft)t0-adapted processes satisfying(5), (6)and(7).Then,β is a Brownian motion in its own filtration.

Moreover,the processDˆt(x)=P(Ux|Ftβ),depending on the two parametersx∈ [0,1]andt≥0has a version which:

(i) is the unique solution of the parametric family of equations dDˆt(x)= −μminDˆt(x),1− ˆDt(x)

t, Dˆ0(x)=x; (ii) is continuous in both parameters;

(iii) satisfiesDt= ˆDt(U ).

Hence the conditional law ofDt givenFtβ is uniform on]0,1[for eacht.

Proof. To prove thatβ is a Brownian motion in it own filtration(Ftβ)t0, it is enough to show that for eachT >0, t)t∈[0,T]has the correct law.

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