Continuous Time Martingales and Supermartingales

3.3 Continuous Time Martingales and Supermartingales 49

50 3 Filtrations and Martingales Important example We say that a process.Zt/t0 with values inRor inR^d has independent increments with respect to the filtration.Ft/ifZis adapted and if, for every0s<t,ZtZsis independent ofFs(for instance, a Brownian motion has independent increments with respect to its canonical filtration). IfZis a real-valued process having independent increments with respect to.Ft/, then

(i) ifZt 2L¹for everyt0, theneZtDZtEŒZtis a martingale;

(ii) ifZt 2L²for everyt0, thenYtDeZ²_t EŒeZ²_tis a martingale;

(iii) if, for some2R, we haveEŒe^Zt <1for everyt0, then XtD e^Zt

EŒe^Zt is a martingale.

Proofs of these facts are very easy. In the second case, we have for every0s<t, EŒ.eZt/²jFsDEŒ.eZsCeZteZs/²jFs

DeZ²_sC2eZsEŒeZteZsjFsCEŒ.eZteZs/²jFs DeZ²_sCEŒ.eZteZs/²

DeZ²_sCEŒeZ²_t2EŒeZseZtCEŒeZ²_s DeZ²_sCEŒeZ²_tEŒeZ²_s;

becauseEŒeZseZtDEŒeZsEŒeZtjFsDEŒeZ²_s. The desired result follows. In the third case,

EŒXtjFsD e^ZsEŒe^.ZtZs/jFs

EŒe^ZsEŒe^.ZtZs/ D e^Zs

EŒe^Zs DXs; using the fact thatEŒe^.ZtZs/jFsDEŒe^.ZtZs/by independence.

Consider the special case of Brownian motion.

Definition 3.11 A real-valued processB D .Bt/t0 is an .Ft/-Brownian motion if B is a Brownian motion and if B is adapted and has independent increments with respect to.Ft/. Similarly, a processB D .Bt/t0 with values inR^d is ad- dimensional.Ft/-Brownian motion ifBis ad-dimensional Brownian motion and if Bis adapted and has independent increments with respect to.Ft/.

Note that ifBis a (d-dimensional) Brownian motion and.Ft^B/is the (possibly completed) canonical filtration ofB, thenBis a (d-dimensional).Ft^B/-Brownian motion.

LetBbe an.Ft/-Brownian motion started from0(or from anya2R). Then it follows from the above observations that the processes

Bt; B²_t t; e^Bt22t

3.3 Continuous Time Martingales and Supermartingales 51

are martingales with continuous sample paths. The processes e^Bt22t are called exponential martingales of Brownian motion

We can also take, forf 2L².RC;B.RC/;dt/, ZtD

Z t

0 f.s/dBs:

Properties of Gaussian white noise imply thatZ has independent increments with respect to the canonical filtration ofB, and thus

Z t

0 f.s/dBs; Z ^t

0 f.s/dBs

₂

Z t

0 f.s/²ds;exp

Z t

0 f.s/dBs² 2

Z t 0 f.s/²ds

are martingales (with respect to this filtration). One can prove that these martingales have a modification with continuous sample paths – it is enough to do it for the first one, and this will follow from the more general results in Chap.5below.

Finally, ifZDNis a Poisson process with parameter(and.Ft/is the canonical filtration ofN), it is well known thatZhas independent increments, and we get that

Ntt; .Ntt/²t; exp.Ntt.e1//

are martingales. In contrast with the previous examples, these martingales do not have a modification with continuous sample paths.

Proposition 3.12 Let.Xt/t0 be an adapted process and let f W R ! RC be a convex function such that EŒf.Xt/ <1for every t0.

(i) If.Xt/t0is a martingale, then.f.Xt//t0is a submartingale.

(ii) If .Xt/t0 is a submartingale, and if in addition f is nondecreasing, then .f.Xt//t0is a submartingale.

Proof By Jensen’s inequality, we have, fors<t,

EŒf.Xt/jFsf.EŒXtjFs/f.Xs/:

In the last inequality, we need the fact thatf is nondecreasing when.Xt/is only a

submartingale. ut

Consequences If.Xt/t0is a martingale,jXtjis a submartingale and more gener- ally, for everyp1,jXtj^pis a submartingale, provided that we haveEŒjXtj^p <1 for everyt0. If.Xt/t0is a submartingale,.Xt/^C DXt_0is also a submartingale.

Remark If.Xt/t0 is any martingale, Jensen’s inequality shows thatEŒjXtj^p is a nondecreasing function oftwith values inŒ0;1, for everyp1.

52 3 Filtrations and Martingales Proposition 3.13 Let.Xt/t0 be a submartingale or a supermartingale. Then, for every t> 0,

0supstEŒjXsj <1:

Proof It is enough to treat the case where.Xt/t0is a submartingale. Since.Xt/^Cis also a submartingale, we have for everys2Œ0;t,

EŒ.Xs/^CEŒ.Xt/^C:

On the other hand, sinceXis a submartingale, we also have fors2Œ0;t, EŒXsEŒX₀:

By combining these two bounds, and noting thatjxj D2x^Cx, we get sup

s2Œ0;tEŒjXsj2EŒ.Xt/^CEŒX₀ <1;

giving the desired result. ut

The next proposition will be very useful in the study of square integrable martingales.

Proposition 3.14 Let.Mt/t0be a square integrable martingale (that is, Mt 2 L² for every t0). Let0s<t and let sD t₀<t₁ < <tp Dt be a subdivision of the intervalŒs;t. Then,

E hX^p

iD1

.Mt_iMt_i1/²ˇˇˇFs

iDEŒMt²M_s²jFsDEŒ.MtMs/²jFs:

In particular, E

hX^p

iD1

.MtiMti1/²i

DEŒM²_t M_s²DEŒ.MtMs/²:

Proof For everyiD1; : : : ;p,

EŒ.MtiMti1/²jFsDEŒEŒ.MtiMti1/²jFti1jFs DE

hEŒM²ti jFti12Mti1EŒMti jFti1CM_t²_i1ˇˇˇFs

i DE

hEŒM²ti jFti1M_t²_i1ˇˇˇFs

i DEŒM_t²_i M_t²_i1 jFs

and the desired result follows by summing overi. ut

3.3 Continuous Time Martingales and Supermartingales 53 Our next goal is to study the regularity properties of sample paths of martingales and supermartingales. We first establish continuous time analogs of classical inequalities in the discrete time setting.

Proposition 3.15

(i) (Maximal inequality)Let.Xt/t0 be a supermartingale with right-continuous sample paths. Then, for every t> 0and every > 0,

P

0supstjXsj>

EŒjX₀jC2EŒjXtj:

(ii) (Doob’s inequality inL^p) Let.Xt/t0 be a martingale with right-continuous sample paths. Then, for every t> 0and every p> 1,

E h

sup

0stjXsj^pi p

p1 p

EŒjXtj^p: Note that part (ii) of the proposition is useful only ifEŒjXtj^p <1.

Proof

(i) Fixt > 0and consider a countable dense subset DofRC such that0 2 D andt 2 D. ThenD\Œ0;tis the increasing union of a sequence.Dm/m1 of finite subsetsŒ0;tof the formDm D ft^m₀;t^m₁; : : : ;t^m_mgwhere0 D t^m₀ < t^m₁ <

< t^m_m D t. For every fixedm, we can apply the discrete time maximal inequality (see AppendixA2) to the sequenceYn D Xtn^m, which is a discrete supermartingale with respect to the filtrationGnDFtn^m. We get

P

s2Dsupm

jXsj>

EŒjX₀jC2EŒjXtj:

Then, we observe that P

s2Dsupm

jXsj>

"P

sup

s2D\Œ0;t

jXsj>

whenm" 1. We have thus P

sup

s2D\Œ0;tjXsj>

EŒjX₀jC2EŒjXtj:

Finally, the right-continuity of sample paths (and the fact thatt 2 D) ensures that

sup

s2D\Œ0;t

jXsj D sup

s2Œ0;t

jXsj: (3.1)

Assertion (i) now follows.

54 3 Filtrations and Martingales (ii) Following the same strategy as in the proof of (i), and using now Doob’s inequality inL^pfor discrete martingales (see AppendixA2), we get, for every m1,

E h

sup

s2Dm

jXsj^pi p

p1 p

EŒjXtj^p:

Now we just have to let mtend to infinity, using the monotone convergence theorem and then the identity (3.1).

u t Remark If we no longer assume that the sample paths of the supermartingaleXare right-continuous, the preceding proof shows that, for every countable dense subset DofRC, and everyt> 0,

P

sup

s2D\Œ0;t

jXsj>

1

.EŒjX₀jC2EŒjXtj/:

Letting! 1, we have in particular sup

s2D\Œ0;t

jXsj<1; a.s.

Upcrossing numbers Letf WI!Rbe a function defined on a subsetIofRC. If a< b, the upcrossing number off alongŒa;b, denoted byM_ab^f .I/, is the maximal integerk 1such that there exists a finite increasing sequences₁ < t₁ < <

sk <tkof elements ofIsuch thatf.si/ aandf.ti/ bfor everyi 2 f1; : : : ;kg (if, even forkD1, there is no such subsequence, we takeM^f_ab.I/D0, and if such a subsequence exists for everyk1, we takeM_ab^f .I/D 1). Upcrossing numbers are a convenient tool to study the regularity of functions.

In the next lemma, the notation

s##tlimf.s/ .resp. lim

s""tf.s/ / means

s#tlim;s>tf.s/ .resp. lim

s"t;s<tf.s/ /:

We say thatg W RC ! Ris càdlàg (for the French “continue à droite avec des limites à gauche”) ifgis right-continuous and has left-limits at everyt> 0.

Lemma 3.16 Let D be a countable dense subset ofRC and let f be a real function defined on D. We assume that, for every T2D,

(i) the function f is bounded on D\Œ0;T; (ii) for all rationals a and b such that a<b,

M_ab^f .D\Œ0;T/ <1:

3.3 Continuous Time Martingales and Supermartingales 55

Then, the right-limit

f.tC/WD lim

s##t;s2Df.s/ exists for every real t0, and similarly the left-limit

f.t/WD lim

s""t;s2Df.s/

exists for every real t > 0. Furthermore, the function g W RC ! Rdefined by g.t/Df.tC/is càdlàg.

We omit the proof of this analytic lemma. It is important to note that the right and left-limitsf.tC/andf.t/are defined for everyt 0(t> 0in the case off.t/) and not only fort2D.

Theorem 3.17 Let.Xt/t0 be a supermartingale, and let D be a countable dense subset ofRC.

(i) For almost every!2˝, the restriction of the function s7!Xs.!/to the set D has a right-limit

XtC.!/WD lim

s##t;s2DXs.!/ (3.2)

at every t2Œ0;1/, and a left-limit

Xt.!/WD lim

s""t;s2DXs.!/

at every t2.0;1/.

(ii) For every t2RC, XtC2L¹and

XtEŒXtCjFt;

with equality if the function t !EŒXtis right-continuous (in particular if X is a martingale). The process.XtC/t0is a supermartingale with respect to the filtration.FtC/. It is a martingale if X is a martingale.

Remark For the last assertions of (ii), we need XtC.!/ to be defined forevery

! 2 ˝ and not only outside a negligible set. As we will see in the proof, we can just takeXtC.!/D0when the limit in (3.2) does not exist.

Proof

(i) FixT 2D. By the remark following Proposition3.15, we have

s2D\supŒ0;TjXsj<1; a.s.

56 3 Filtrations and Martingales As in the proof of Proposition 3.15, we can choose a sequence .Dm/m1 of finite subsets ofD that increase to D\Œ0;T and are such that 0;T 2 Dm. Doob’s upcrossing inequality for discrete supermartingales (see AppendixA2) gives, for everya<band everym1,

EŒM_ab^X.Dm/ 1

baEŒ.XTa/: We letm! 1and get by monotone convergence

EŒMab^X.D\Œ0;T/ 1

baEŒ.XTa/ <1: We thus have

M_ab^X.Œ0;T\D/ <1; a.s.

Set ND [

T2D

n

t2D\supŒ0;TjXtj D 1o [ [

a;b2Q;a<b

fM_ab^X.D\Œ0;T/D 1g : (3.3) ThenP.N/D0by the preceding considerations. On the other hand, if! …N, the functionD3t7!Xt.!/satisfies all assumptions of Lemma3.16. Assertion (i) now follows from this lemma.

(ii) To defineXtC.!/for every!2˝and not only on˝nN, we set

XtC.!/D ( lim

s##t;s2DXs.!/if the limit exists 0 otherwise.

With this definition,XtCisFtC-measurable.

Fixt 0and choose a sequence.tn/n0inDsuch thattndecreases strictly totasn! 1. Then, by construction, we have a.s.

XtCD lim

n!1Xtn:

Set Yk D Xt_k for every integer k 0. Then Y is a backward super- martingale with respect to the (backward) discrete filtrationHk D Ft_k (see Appendix A2). From Proposition 3.13, we have sup_k0EŒjYkj < 1. The convergence theorem for backward supermartingales (see AppendixA2) then implies that the sequenceXt_nconverges toXtCinL¹. In particular,XtC 2L¹.

Thanks to the L¹-convergence, we can pass to the limit n ! 1 in the inequalityXtEŒXt_n jFt, and we get

XtEŒXt_CjFt

3.3 Continuous Time Martingales and Supermartingales 57 (we use the fact that the conditional expectation is continuous for theL¹-norm, and it is important to realize that an a.s. convergence would not be sufficient to warrant this passage to the limit). Furthermore, thanks again to the L¹- convergence, we haveEŒXtC D limEŒXtn. Thus, if the functions !EŒXs is right-continuous, we must haveEŒXt D EŒXtC D EŒEŒXt_C j Ft, and the inequalityXtEŒXt_C jFtthen forcesXtDEŒXt_C jFt.

We already noticed thatXtC isFtC-measurable. Lets < tand let.sn/n0

be a sequence inDthat decreases strictly tos. We may assume thatsn tnfor everyn. Then as previouslyXsn converges toXsCinL¹, and thus, ifA2 FsC, which impliesA2Fsnfor everyn, we have

EŒXsC1AD lim

n!1EŒXsn1A lim

n!1EŒXtn1ADEŒXtC1ADEŒEŒXtC jFsC1A: Since this inequality holds for everyA2FsC, and sinceXsCandEŒXtC jFsC are bothFsC-measurable, it follows that XsC EŒXtC j FsC . Finally, if X is a martingale, inequalities can be replaced by equalities in the previous considerations.

u t Theorem 3.18 Assume that the filtration .Ft/is right-continuous and complete.

Let XD .Xt/t0be a supermartingale, such that the function t ! EŒXtis right- continuous. Then X has a modification with càdlàg sample paths, which is also an .Ft/-supermartingale.

Proof LetDbe a countable dense subset ofRC as in Theorem3.17. LetN be the negligible set defined in (3.3). We set, for everyt0,

Yt.!/D

XtC.!/if!…N

0 if!2N:

Lemma3.16then shows that the sample paths ofYare càdlàg.

The random variable XtC is FtC-measurable, and thus Ft-measurable since the filtration is right-continuous. As the negligible set N belongs to F1, the completeness of the filtration ensures thatYt isFt-measurable. By Theorem3.17 (ii), we have for everyt0,

XtDEŒXtC jFtDXtC DYt; a:s:

becauseXtCisFt-measurable. Consequently,Yis a modification ofX. The process Y is adapted to the filtration.Ft/. SinceY is a modification ofX the inequality EŒXtjFsXs, for0s<t, implies that the same inequality holds forY. ut Remarks

(i) Let us comment on the assumptions of the theorem. A simple example shows that our assumption that the filtration is right-continuous is necessary. Take

58 3 Filtrations and Martingales

˝ D f1; 1g, with the probability measurePdefined byP.f1g/DP.f1g/D 1=2. Let"be the random variable ".!/ D !, and let the process.Xt/t0 be defined byXt D 0if0 t 1, andXt D "ift> 1. Then it is easy to verify thatX is a martingale with respect to its canonical filtration.Ft^X/(which is complete since there are no nonempty negligible sets!). On the other hand, no modification ofXcan be right-continuous attD1. This does not contradict the theorem since the filtration is not right-continuous (F₁^X_C 6DF₁^X).

(ii) Similarly, to show that the right-continuity of the mapping t ! EŒXt is needed, we can just takeXtD f.t/, wheref is any nonincreasing deterministic function. If f is not right-continuous, no modification of X can have right- continuous sample paths.