Stochastic integration with respect to fBm in the case H < 1

Một phần của tài liệu Nualart The Malliavin Calculus and Related Topics (Trang 309 - 320)

5.2 Stochastic calculus with respect to fBm

5.2.3 Stochastic integration with respect to fBm in the case H < 1

2

The extension of the previous results to the caseH < 12 is not trivial and new difficulties appear. In order to illustrate these difficulties, let us first remark that the forward integralT

0 BtdBtdefined as the limit inL2of the Riemann sums

n−1 i=0

Bti(Bti+1−Bti),

where ti = iTn, does not exists. In fact, a simple argument shows that the expectation of this sum diverges:

n i=1

E

Bti−1(Bti−Bti−1)

= 1

2 n

i=1

!t2Hi −t2Hi−1−(ti−ti−1)2H"

= 1

2T2H

1−n1−2H

→ −∞,

as ntends to infinity. Notice, however, that the expectation of symmetric Riemann sums is constant:

1 2

n i=1

E

(Bti+Bti−1)(Bti−Bti−1)

=1 2

n i=1

!t2Hi −t2Hi−1"

= T2H 2 . We recall that forH < 12 the operatorKH∗ given by (5.31) is an isometry between the Hilbert spaceHandL2([0, T]). We have the estimate :

∂K

∂t (t, s)

≤cH(1

2 −H) (t−s)H−32. (5.46) Also from (5.32) (see Exercise 5.2.3) it follow sthat

|K(t, s)| ≤C(t−s)H−12. (5.47) Consider the following seminorm on the set Eof step functions on [0, T]:

ϕ2K = T

0

ϕ2(s)(T−s)2H−1ds +

T 0

T

s |ϕ(t)−ϕ(s)|(t−s)H−32dt 2

ds.

We denote byHK the completion ofE with respect to this seminorm. The spaceHK is the class of functionsϕon [0, T] such thatϕ2K <∞, and it is continuously included inH. Ifu∈D1,2(HK), thenu∈Domδ.

Note that ifu={ut, t∈[0, T]}is a process inD1,2(HK), then there is a sequence{ϕn}of bounded simpleHK-valued processes of the form

ϕn=

n−1 j=0

Fj1(tj,tj+1], (5.48) whereFj is a smooth random variable of the form

Fj =fj(Bsj

1, ..., Bsj

m(j)),

withfj∈Cb∞(Rm(j)), and 0 =t0< t1< ... < tn=T, such that Eu−ϕn2K+E

T

0 Dru−Drϕn2Kdr−→0, as n→ ∞. (5.49) In the caseH < 12 it is more convenient to consider the symmetric inte- gral introduced by Russo and Vallois in [298]. For a processu={ut, t∈[0, T]} with integrable paths andε >0, we denote byuεtthe integral (2ε)−1t+ε

t−εusds, where we use the convention us = 0 fors /∈[0, T]. Also we put Bs =BT

fors > T andBs= 0 fors <0.

Definition 5.2.2 The symmetric integral of a process u with integrable paths with respect to the fBm is defined as the limit in probability of

(2ε)−1 T

0

us(Bs+ε−Bs−ε)ds.

asε↓0 if it exists. We denote this limit by T

0 ur◦dBr.

The following result is the counterpart of Proposition 5.2.3 in the case H < 12, for the symmetric integral.

Proposition 5.2.4 Let u={ut, t ∈[0, T]} be a stochastic process in the spaceD1,2(HK). Suppose that the trace defined as the limit in probability

TrDu:= lim

ε→0

1 2ε

T 0

#Dus,1[s−ε,s+ε]∩[0,T]

$

Hds

exists. Then the symmetric stochastic integral of uwith respect to fBm in the sense of Definition 5.2.1 exists and

T 0

ut◦dBt=δ(u) + TrDu.

In order to prove this theorem, we need the following technical result.

Lemma 5.2.1 Let u be a simple process of the form (5.48). Then uε converges to uinD1,2(HK) asε↓0.

Proof: Letube given by the right-hand side of (5.48). Thenuis a bounded process. Hence, by the dominated convergence theorem

E T

0

(us−uεs)2(T −s)2H−1ds−→0 as ε↓0. (5.50)

Fix an index i∈ {0,1, ..., n−1}. Using thatut−us= 0 fors, t∈[ti,ti+1] we obtain

ti+1

ti

T

s |uεt−uεs−(ut−us)|(t−s)H−32dt 2

ds

≤ 2 ti+1

ti

ti+1

s |uεt−uεs|(t−s)H−32dt 2

ds

+2 ti+1

ti

T ti+1

|uεt−uεs−(ut−us)|(t−s)H−32dt 2

ds

= 2A1(i, ε) + 2A2(i, ε). (5.51)

The convergence of the expectation of the termA2(i, ε) to 0,asε↓0, follows from the dominated convergence theorem, the fact that u is a bounded process and that for a.a. 0≤s < t≤T,

|uεt−uεs−(ut−us)|(t−s)H−32 −→0 as ε↓0.

Suppose that ε < 14min0≤i≤n−1|ti+1−ti|. Then uεt−uεs = 0 if s and t belong to [ti+ 2ε, ti+1−2ε], we can make the following decomposition

E(A1(i, ε))

≤ 8 ti+2ε

ti

ti+2ε

s |uεt−uεs|(t−s)H−32dt 2

ds +8

ti+1

ti+1−2ε

ti+1

s |uεt−uεs|(t−s)H−32dt 2

ds +8

ti+2ε ti

ti+1

ti+2ε|uεt−uεs|(t−s)H−32dt 2

ds

+8

ti+1−2ε ti

ti+1

ti+1−2ε|uεt−uεs|(t−s)H−32dt 2

ds.

The first and second integrals converge to zero, due to the estimate

|uεt−uεs| ≤ c ε|t−s|.

On the other hand, the third and fourth term of the above expression converge to zero becauseuεt is bounded. Therefore we have proved that

Euưuε2K ư→0 as ε→0.

Finally, it is easy to see by the same arguments that we also have

E T

0 Dru−Druε2Kdr−→0 as ε→0.

Thus the proof is complete.

Now we are ready to prove Proposition 5.2.4.

Proof of Proposition 5.2.4: From the properties of the divergence operator, and applying Fubini’s theorem we have

(2ε)−1 T

0

us(Bs+ε−Bs−ε)ds= (2ε)−1 T

0

δ

us1[s−ε,s+ε](ã) ds +(2ε)−1

T 0

#Dus,1[s−ε,s+ε]$

Hds

= (2ε)−1 T

0

r+ε r−ε

usds

dBr

+(2ε)−1 T

0

#Dãus,1[s−ε,s+ε](ã)$

Hds

= T

0

uεrdBr+Bε.

By our hypothesis we get thatBε converges toT rDu in probability as ε↓0. In order to see thatT

0 uεrdBr converges toδ(u) inL2(Ω) asεtends to zero, we will show thatuεconverges touin the norm ofD1,2(HK). Fix δ >0. We have already noted that the definition of the space D1,2(HK) implies that there is a bounded simpleHK-valued processesϕas in (5.48) such that

Eu−ϕ2K+E T

0 ||Dru−Drϕ||2Kdr≤δ. (5.52)

Therefore, Lemma 5.2.1 implies that forεsmall enough,

Euưuε2K+E T

0 ||Dr(uưuε)||2Kdr

≤ cEu−ϕ2K+cE T

0 ||Dr(u−ϕ)||2Kdr +cEϕ−ϕε2K+cE

T

0 ||Dr(ϕ−ϕε)||2Kdr +cEϕε−uε2K+cE

T

0 ||Dr(ϕε−uε)||2Kdr

≤ 2cδ+cEϕε−uε2K+cE T

0 ||Dr(ϕε−uε)||2Kdr. (5.53)

We have

T 0

E(ϕεs−uεs)2 (T−s)2H−1ds

≤ T

0

E 1

2ε s+ε

s−ε

(ϕr−ur)dr 2

(T−s)2H−1ds

≤ T

0

E(ϕr−ur)2 1

(r+ε)∧T (r−ε)∨0

(T−s)2H−1ds

dr.

From property (i) it follows that

(2ε)−1

(r+ε)∧T (r−ε)∨0

K(T, t)2dt≤c!

(T−r)−2α+r−2α"

.

Hence, by the dominated convergence theorem and condition (4.21) we obtain

limε↓0sup T

0

E(ϕεs−uεs)2 K(T, s)2ds

≤ T

0

E(ϕs−us)2 K(T, s)2ds ≤δ. (5.54)

On the other hand,

E T

0

T

s |ϕεt−uεt−ϕεs+uεs|(t−s)H−32dt 2

ds

≤ 1 4ε2E

T 0

ε

−ε

T s

(ϕưu)tưθư(ϕưu)sưθ(tưs)Hư32dtdθ 2

ds

= 1 4ε2E

T 0

s+ε s−ε

T+r−s

r |(ϕưu)tư(ϕưu)r|(tưr)Hư32dtdr 2

ds

≤ 1 2εE

T 0

s+ε s−ε

T+ε

r |(ϕưu)tư(ϕưu)r|(tưr)Hư32dt 2

drds

= 1 2εE

T+ε

−ε

(r+ε)∧T (r−ε)∨0

T+ε

r |ϕt−ut−ϕr+ur|(t−r)H−32dt 2

dsdr

≤E T+ε

−ε

T+ε

r |ϕt−ut−ϕr+ur|(t−r)H−32 dt 2

dr. (5.55)

By (5.54) and (5.55) we obtain

limε↓0supEϕε−uε2K≤2δ.

By a similar argument, limε↓0supE

T

0 ||Dr(ϕε−uε)||2Kdr≤2δ.

Since δ is arbitrary,uε converges tou in the norm ofD1,2(HK) asε↓ 0, and, as a consequence, T

0 uεrdBr converges in L2(Ω) to δ(u) . Thus the

proof is complete.

Consider the particular case of the process ut = F(Bt), where F is a continuously differentiable function satisfying the growth condition (5.42).

IfH > 14, the process F(Bt) the process belongs toD1,2(HK). Moreover, TrDuexists and

TrDu=H T

0

F′(Bt)t2H−1dt.

As a consequence we obtain T

0

F(Bt)0dBt= T

0

F(Bt)dBt+H T

0

F′(Bt)t2H−1dt.

(C) Itˆo’s formulas for the divergence integral in the caseH < 12

An Itˆo’s formula similar to (5.44) was proved in [9] for general Gaussian processes of Volterra-type of the formBt =t

0K(t, s)dWs, where K(t, s) is a singular kernel. In particular, the processBtcan be a fBm with Hurst parameter 14 < H < 12. Moreover, in this paper, an Itˆo’s formula for the in- definite divergence processXt=t

0usdBssimilar to (5.45) was also proved.

On the other hand, in the case of the fractional Brownian motion with Hurst parameter 14 < H < 12, an Itˆo’s formula for the indefinite symmetric integralXt=t

0usdBshas been proved in [7] assuming again 14 < H < 12. Let us explain the reason for the restriction 14 < H. In order to define the divergence integral T

0 F′(Bs)dBs, we need the process F′(Bs) to belong toL2(Ω;H). This is clearly true, providedF satisfies the growth condition (5.42), because F′(Bs) is H¨older continuous of order H−ε > 12 −H if ε <2H−12. IfH ≤ 14, one can show (see [66]) that

P(B∈ H) = 0,

and the spaceD1,2(H) is too small to contains processes of the formF′(Bt).

Following the approach of [66] we are going to extend the domain of the divergence operator to processes whose trajectories are not necessarily in the spaceH.

Using (5.31) and applying the integration by parts formula for the frac- tional calculus (A.17) we obtain for any f, g∈ H

f, gH = KH∗f, KH∗gL2([0,T])

= d2H @

s12−HDT12−−HsH−12f, s12−HDT12−−HsH−12gA

L2([0,T])

= d2H @

f, sH−12s12−HD0+12−Hs1−2HDT12−−HsH−12gA

L2([0,T]). This implies that the adjoint of the operatorKH∗ inL2([0, T]) is

KH∗,af

(s) =dHs12−HD0+12−Hs1−2HDT12−−HsH−12f.

Set H2 = (KH∗)−1

KH∗,a−1

(L2([0, T])). Denote by SH the space of smooth and cylindrical random variables of the form

F =f(B(φ1), . . . , B(φn)), (5.56) where n≥1,f ∈Cb∞(Rn) (f and all its partial derivatives are bounded), andφi∈ H2.

Definition 5.2.3 Let u = {ut, t ∈ [0, T]} be a measurable process such that

E T

0

u2tdt

<∞.

We say that u ∈ Dom∗δ if there exists a random variable δ(u) ∈ L2(Ω) such that for all F ∈ SH we have

R

E(utKH∗,aKH∗DtF)dt=E(δ(u)F).

This extended domain of the divergence operator satisfies the following elementary properties:

1. Domδ⊂Dom∗δ, andδ restricted to Domδcoincides with the diver- gence operator.

2. Ifu∈Dom∗δthenE(u) belongs toH.

3. Ifuis a deterministic process, thenu∈Dom∗δif and only ifu∈ H. This extended domain of the divergence operator leads to the following version of Itˆo’s formula for the divergence process, established by Cheridito and Nualart in [66].

Theorem 5.2.2 Suppose thatF is a function of classC2(R)satisfying the growth condition (5.42). Then for allt∈[0, T], the process{F′(Bs)1[0,t](s)} belongs toDom∗δand we have

F(Bt) =F(0) + t

0

F′(Bs)dBs+H t

0

F′′(Bs)s2H−1ds. (5.57) Proof: Notice thatF′(Bs)1[0,t](s)∈L2([0, T]×Ω) and

F(Bt)−F(0)−H t

0

F′′(Bs)s2H−1ds∈L2(Ω). Hence, it suffices to show that for anyF ∈ SH

E(F′(Bs)1[0,t](s), DsFH

=E

F(Bt)−F(0)−H t

0

F′′(Bs)s2H−1ds

F

. (5.58) TakeF =Hn(B(ϕ)), whereϕ∈ H2andHnis thenth-Hermite polynomial.

We have

DtF =Hn−1(B(ϕ))ϕt Hence, (5.58) can be written as

E

Hn−1(B(ϕ))F′(Bs)1[0,t](s), ϕsH

= E((F(Bt)−F(0)−H t

0

F′′(Bs)s2H−1ds)Hn(B(ϕ)))

Using (5.33) we obtain e2H

t 0

E(F′(Bs)Hn−1(B(ϕ))) (D

1 2−H

+ D

1 2−H

− ϕ)(s)ds

= E((F(BtH)−F(0)−H t

0

F′′(Bs)s2H−1ds)Hn(B(ϕ))). (5.59) In order to show (5.59) we will replaceF by

Fk(x) =k 1

−1

F(x−y)ε(ky)dy,

where ε is a nonnegative smooth function supported by [−1,1] such that 1

−1ε(y)dy= 1.

We will make use of the following equalities:

E(F(Bt)Hn(B(ϕ))) = 1

n!E(F(Bt)δn(ϕ⊗n))

= 1

n!EDn(F(Bt)), ϕ⊗nH⊗n

= 1

n!E(F(n)(Bt))1(0,t], ϕnH. Let p(σ, y) := (2πσ)−12exp

−y2σ2

. Note that ∂p∂σ = 12∂∂y2p2. For all n ≥ 0 ands∈(0, t],

d

dsE(F(n)(Bs)) = d ds

R

p(s2H, y)F(n)(y)dy

=

R

∂p

∂σ(s2H, y)2Hs2H−1F(n)(y)dy

= Hs2H−1

R

∂2p

∂y2(s2H, y)F(n)(y)dy

= Hs2H−1

R

p(s2H, y)F(n+2)(y)dy

= Hs2H−1E(F(n+2)(Bs)). (5.60) Forn= 0 the left hand side of (5.59) is zero. On the other hand it follows from (5.60) that

E(F(Bt))−F(0)−H t

0

E(F′′(Bs))s2H−1ds= 0 This shows that (5.59) is valid forn= 0.

Fixn≥1. Equation (5.60) implies that for alls∈(0, t], d

ds

E(F(n)(Bs))1(0,s], ϕnH

= Hs2H−1E(F(n+2)(Bs))1(0,s], ϕnH

+e2HnE(F(n)(Bs))

×1(0,s], ϕnH−1(D+12−HD−12−Hϕ)(s).

It follows that

E(F(n)(Bt))1(0,s], ϕnH = H t

0

E(F(n+2)(Bs))1(0,s], ϕnHs2H−1ds +e2Hn

t 0

E(F(n)(Bs))

×1(0,s], ϕnH−1(D+12−HD−12−Hϕ)(s), (5.61) (5.61) is equivalent to (5.59) because

E(F(n)(Bt))1(0,s], ϕnH =n!E(F(Bt)Hn(B(ϕ))), E(F(n)(Bs))1(0,s], ϕnH−1= (n−1) !E(F′(Bs)H(n−1)(B(ϕ))), and

E(F(n+2)(Bs))1(0,s], ϕnH=n!E(F′′(Bs)Hn(B(ϕ))).

This completes the proof (5.59) for the functionFk. Finally it suffices to

letktend to infinity.

(D)Local time and Tanaka’s formula for fBm

Berman proved in [22] that that fractional Brownian motionB ={Bt, t≥ 0}has a local timelat continuous in (a, t)∈R×[0,∞) which satisfies the occupation formula

t 0

g(Bs)ds=

R

g(a)ltada (5.62) for every continuous and bounded functiongonR. Moreover,lat is increas- ing in the time variable. Set

Lat = 2H t

0

s2H−1la(ds).

It follows from (5.62) that 2H

t 0

g(Bs)s2H−1ds=

R

g(a)Latda.

This means thata→Lat is the density of the occupation measure à(C) = 2H

t 0

1C(Bs)s2H−1ds,

whereC is a Borel subset ofR. Furthermore, the continuity property oflat implies thatLat is continuous in (a, t)∈R×[0,∞).

As an extension of the Itˆo’s formula (5.57), the following result has been proved in [66]:

Theorem 5.2.3 Let 0< t <∞ anda∈R. Then 1{Bs>a}1[0,t](s)∈Dom∗δ , and

(Bt−a)+= (−a)++ t

0

1{Bs>a}dBs+1

2Lat. (5.63) This result can be considered as a version of Tanaka’s formula for the fBm. In [69] it is proved that forH > 13, the process1{Bs>a}1[0,t](s) belongs to Domδand (5.63) holds.

The local time lat has H¨older continuous paths of order δ < 1−H in time, and of order γ < 1−H2H in the space variable, provided H ≥ 13(see Table 2 in [117]). Moreover, lta is absolutely continuous ina if H < 13, it is continuously differentiable if H < 15, and its smoothness in the space variable increases whenH decreases.

In a recent paper, Eddahbi, Lacayo, Sol´e, Tudor and Vives [88] have proved that lat ∈ Dα,2 for all α < 12H−H. That means, the regularity of the local timelat in the sense of Malliavin calculus is the same order as its H¨older continuity in the space variable. This result follows from the Wiener chaos expansion (see [69]):

lat = ∞ n=0

t 0

s−nHp(s2H, a)Hn(as−H)In

1[o,s] ⊗n ds.

In fact, the series ∞ n=0

(1 +n)αE ( t

0

s−nHp(s2H, a)Hn(as−H)In

1[o,s] ⊗n ds

2)

= ∞ n=0

(1 +n)αn!

t 0

t 0

(sr)−nHp(s2H, a)p(r2H, a)Hn(as−H)Hn(ar−H)

×RH(r, s)ndrds

is equivalent to ∞ n=1

n−12+α t

0

t 0

RH(u, v)(uv)−nH−1dudv

= ∞ n=0

n−12+α 1

0

RH(1, z)z−nH−1dz.

Then, the result follows from the estimate

1

0

RH(1, z)z−nH−1dz

≤Cn−2H1 .

Exercises

5.2.1 Show the tranfer principle stated in Propositions 5.2.1 and 5.2.2.

5.2.2 Show the inequality (5.34).

5.2.3 Show the estimate (5.47).

5.2.4 Deduce the Wiener chaos expansion of the local timelta.

Một phần của tài liệu Nualart The Malliavin Calculus and Related Topics (Trang 309 - 320)

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