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The law of a stochastic integral with two independent

fractional Brownian motions

Xavier Bardina, Ciprian Tudor

To cite this version:

Xavier Bardina, Ciprian Tudor. The law of a stochastic integral with two independent fractional Brownian motions. Boletin de la Sociedad Matematica Mexicana, 2007, 13 (1), pp.XX. �hal-00130682�

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The law of a stochastic integral with two independent

fractional Brownian motions

Xavier Bardina 1 Ciprian A. Tudor 2,∗

1 Departament de Matem`atiques, Universitat Aut`onoma de Barcelona, 08193-Bellaterra, Barcelona, Spain.

Xavier.Bardina@uab.cat

2SAMOS/MATISSE, Centre d’Economie de La Sorbonne, Universit´e de Panth´eon-Sorbonne Paris 1,

90, rue de Tolbiac, 75634 Paris Cedex 13, France. tudor@univ-paris1.fr

*corresponding author

Abstract

Using the tools of the stochastic integration with respect to the fractional Brownian motion, we obtain the expression of the characteristic function of the random variable R1 0 B α sdB H s where B α and BH

are two independent fractional Brownian motions with Hurst parameters α ∈ (0, 1) and H > 1

2 respectively. The two-parameter case is also considered.

Running head: Stochastic integral with two fBms 2000 AMS Classification Numbers: 60H05, 60G15.

Key words: Stochastic integral, fractional Brownian motion, characteristic func-tion.

1

Introduction

The theory of multiple stochastic integrals with respect to Brownian motion is well-known (see for instance [9]), but in general, it is difficult to compute the law of a stochastic integral with respect to the Wiener process when the integrand is not deterministic. There are some known results in particular cases. Let us recall the context. Consider W1 and W2 two independent Brownian motions. In [6] and [19] the authors studied the law of the random variable α Z 1 0 Ws1dWs2+ β Z 1 0 Ws2dWs1.

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When α = 1 and β = 0 they showed that the characteristic function of the stochastic integral R [0,1]Ws1dWs2 is given by (1.1) Φ(t) = µ cosh2µ t 2 ¶ + sinh2µ t 2 ¶¶−12 .

In the two-parameter case in [10] (see also [12]) the authors proved that the characteristic function of the integralR

[0,1]2Ws dW1 s (here W2 1and W2denotes two independent Brownian sheets) is given by (1.2) Φ(t) = Y k≥1 cosh−12 µ 2t (2k − 1)π ¶ .

The aim of the present work is develop a similar study for the fractional Brownian motion. The recent development of the stochastic integration with respect to the fractional Brownian motion (fBm) (see for instance [14]) gives the tools for this analysis. Concretely, we will consider two independent fractional Brownian motion BH and Bα with Hurst pa-rameter α ∈ (0, 1) and H > 12, and we will find an explicit expression for the characteristic function of the stochastic integral R1

0 BsαdBsH. We mention that this kind of integrals

ap-pears in the study of stochastic wave equations with fractional noise (see [5]). Related results on the law of this integral have also been proved in [7].

2

Preliminaries: Fractional Brownian motion and Wiener

in-tegrals

Let T = [0, 1] the unit interval and let¡BH t

¢

t∈T be a fractional Brownian motion with Hurst

parameter H ∈ (0, 1). Denote by RH its covariance

RH(t, s) = E¡BH

t BsH¢ =

1 2¡t

2H+ s2H− |t − s|2H¢ .

We denote by H(H) := H the canonical space of the fractional Brownian motion BH. That is, H is the closure of the linear span of the indicator functions {1[0,t], t ∈ T }

with respect to the scalar product

h1[0,t], 1[0,s]iH= RH(t, s).

The structure of the Hilbert space H varies upon the values of the Hurst parameter. Let us recall some basic facts about this space.

• If H > 12 the elements of H may be not functions but distributions of negative order (see [15]). Therefore, it is of interest to know significant subspaces of functions contained in it.

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Define the function

(2.1) θH(s, t) = H(2H − 1)|s − t|2H−2 and let L2H(T ) be the set of function f : T → R such thatR

T

R

T |f (u)||f (v)|θ(u, v)dudv <

∞, endowed with the scalar product

(2.2) hf, giH = Z T Z T f (u)g(v)θ(u, v)dudv.

It has been proved in [15] that L2H(T ) is a strict subset of H and the scalar products h·, ·iH and h·, ·iH coincide on L2H(T ). Moreover, we have the following inclusion

(2.3) LH1(T ) ⊂ L2

H(T ) ⊂ H.

• If H < 12, then H is a set of functions; it coincides actually with the set I 1 2−H T − ¡L2(T ) ¢ where I 1 2−H

T − is the fractional integral of order 12 − H (see [8], [1], [15]). A significant

subspace of H is the set of H¨older continuous functions of order 12 − H + ε for all ε > 0,

(2.4) C12−H+ε(T ) ⊂ H ⊂ L2(T ) ⊂ L 1 H(T ). Consider EH the class of step functions of the form

(2.5) ϕ(·) =

n

X

i=1

ai1(ti,ti+1](·) n ≥ 1, ti∈ T, ai∈R.

It has been proved in [16] that EH is dense in H. For ϕ ∈ EHof the form (2.5) we define its

Wiener integral with respect to the fBm BH by

(2.6) Z 1 0 ϕ(s)dBsH := n X i=1 ai ³ BtHi+1− BHti´ The mapping ϕ →R1

0 ϕ(s)dBsH provides an isometry between EH and the first chaos of the

fBm BH and it can be extended as follows:

• If H > 12, it has been proved in [15] that EH is dense in L2H(T ) with respect to the norm

k·kH. As a consequence, the Wiener integralR01ϕ(s)dBsH can be defined in a consistent

way as limit in L2(Ω) of integrals of elementary functions for any ϕ ∈ L2H(T ). • If H < 12, then EH is dense in H (see [8], [15]) and the integral R01ϕ(s)dBHs can be

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We will need in this paper stochastic integrals of the form R

T usdBHs where u is a

stochastic process independent by BH. Using the above facts, it follows that this integral

can be defined by isometry for any u ∈ L2(Ω) × L2H(T ) if H > 12 and for any u ∈ L2(Ω; H) if H < 12.

REMARK (2.7). The integral R

T usdBHs coincides also with the Skorohod integral

introduced in [2], [1] since, by independence, the Malliavin derivative of u with respect to

BH is zero.

More generally, for H > 12, let L2H(Tn) be the set of functions f : Tn→ R such that Z Tn |f (u1, . . . , un) ||f (v1, . . . , vn) | Ã n Y i=1 θH(ui, vi) ! du1. . . dundv1. . . dvn< ∞,

endowed with the scalar product (2.8) hf, giHn = Z Tn f (u1, . . . , un) g (v1, . . . , vn) Ã n Y i=1 θH(ui, vi) ! du1. . . dundv1. . . dvn. Obviously, L2

H(Tn) is a subset of H⊗n and if f, g ∈ L2H(Tn) then we have

hf, giHn = hf, giH⊗n.

We will denote by L2s,H(Tn) the set of symmetric functions f ∈ L2H(Tn) and if f ∈ L2s,H(T2) let us introduce the (Hilbert-Schmidt) operator (see [7]) KfH : L2H(T ) → L2H(T ) given by (2.9) ¡KH f ϕ¢ (y) = Z T Z T f (x, y)ϕ(x′)θH(x, x′)dxdx′.

REMARK (2.10). Note that if f is positive and H > 12, then the operator KH f is a

positive operator. Indeed, we can write

¡KH f ϕ¢ (y) = Z T A(x′, y)ϕ(x′)dx′ where A(x′, y) =R

T f (x, y)θH(x, x′)dx is positive. Thus the eigenvalues of KfH are positive.

3

The characteristic function of the double integral

Throughout this section BH and Bα will denote two independent fractional Brownian

mo-tion with parameter H and α respectively. We compute the characteristic funcmo-tion of the random variable

(3.1) S :=

Z

T

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Note that, when H > 12, the random variables S (3.1) is well-defined since obviously Bα belongs to L2(Ω) × L2

H(T ) for any α. When H < 12, if we assume that α + H > 12, then

we have Bα ∈ C12−H+ε(T ). But, in the following we will need to restrict ourselves to the situation H > 12.

We start with the following lemma which gives an approximation of the random variable S given by (3.1) when the Hurst parameter of the integrator fbm BH is bigger than one half.

LEMMA (3.2). Assume that H > 12 and α ∈ (0, 1). Denote by

(3.3) Tn=

n−1

X

i=0

Btαi³BtHi+1− BtHi´

where π : 0 = t0 < t1 < . . . < tn= 1 denotes a partition of [0, 1]. Then it holds that

Tn→ S in L2(Ω) as |π| → 0.

Proof: Using the independence of Bα and BH we can write

Btαi³BtHi+1− BtHi´= Z ti+1

ti

BtαidBsH.

To prove the lemma it is enough to prove that

n−1 X i=0 Bαti1[ti,ti+1](·) → B α · = n−1 X i=0 Bα·1[ti,ti+1](·) in L 2(Ω) × L2 H(T ) as |π| → 0.

Actually in general, to prove the convergence of a sequence of stochastic integrals of diver-gence type one needs also the converdiver-gence of the Malliavin derivatives, but in our case it is

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unnecessary due to the independence of the two fBms. We have, using formula (2.2), E ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ n−1 X i=0 ¡Bα ti− B α · ¢ 1[ti,ti+1] ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 H = n−1 X i,j=0 H(2H − 1) Z ti+1 ti Z tj+1 tj E¡Bα ti− B α s ¢³ Btαj− Brα´|r − s|2H−2drds ≤ n−1 X i,j=0 H(2H − 1) Z ti+1 ti Z tj+1 tj |ti− s|α|tj− r|α|r − s|2H−2drds ≤ H(2H − 1)|π|2α n−1 X i,j=0 Z ti+1 ti Z tj+1 tj |r − s|2H−2drds = |π|2α n−1 X i,j=0 h1[ti,ti+1], 1[tj,tj+1]iH = |π| 2α

and this goes to 0 for every α ∈ (0, 1).

We will also need to prove the following technical lemma: LEMMA (3.4). a) Assume that α > 12 and consider the function

(3.5) fH(x, y) = 1

2¡(1 − x)

2H+ (1 − y)2H− |x − y|2H¢ , x, y ∈ T = [0, 1].

Then fH ∈ L2s,α(T2).

b) Assume that H > 12 and consider the function

(3.6) fα(x, y) = 1 2¡x

+ y− |x − y|¢ , x, y ∈ T = [0, 1].

Then fα∈ L2

s,H(T2).

Proof: Let us prove first the point 2a); the point 2b) is similar. We have to show

that I := Z 1 0 Z 1 0 Z 1 0 Z 1 0 fH(x1, y1)fH(x2, y2)θα(x1, x2)θα(y1, y2)dx1dx2dy1dy2< ∞. Note that ¯ ¯fH(xi, yi) ¯ ¯ = E¡B1H − BxH i¢ ¡B H 1 − ByHi ¢ ≤ ³E¡BH 1 − BxHi ¢2´1/2³ E¡BH 1 − ByHi ¢2´1/2 = (1 − xi)H(1 − yi)H

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The integral I is therefore bounded by I ≤ (c(α))2 Z [0,1]4 (1 − x1)H(1 − y1)H(1 − x2)H(1 − y2)H|x1− x2|2α−2|y1− y2|2α−2dx1dx2dy1dy2 = µ c(α) Z 1 0 Z 1 0 (1 − x1)H(1 − x2)H|x1− x2|2α−2dx1dx2 ¶2

with c(α) = α(2α − 1). Now, using the change of variables z = x−y1−y, we get

I′ := Z 1 0 Z 1 0 (1 − x)H(1 − y)H|x − y|2α−2dydx = 2 Z 1 0 Z x 0 (1 − x)H(1 − y)H(x − y)2α−2dydx = 2 Z 1 0 (1 − x)2H+2α−1 µZ x 0 (1 − z)−H−2αz2α−2dz ¶ dx = 1 H + α Z 1 0 (1 − z)Hz2α−2dz < ∞, using that α > 12.

We state now our main result. The point b) allows to consider the situation when the Hurst parameter of the integrand α is less than 12.

THEOREM (3.7). a). Let α > 12 and H > 12. Then the characteristic function of the random variable S given by (3.1) is

E¡eitS¢ =Y i≥1 µ 1 1 + t2µ i ¶12

where (µi)i≥1 are the eigenvalues of the operator KfαH given by (2.9 ) where fH is defined

by (3.5).

b). Assume that H > 12 and α ∈ (0, 1). Then the characteristic function of S (3.1) is E¡eitS¢ =Y i≥1 µ 1 1 + t2ξ i ¶12 − →

where (ξi)i≥1 are the eigenvalues of the operator KfHα given by (2.9) and fα is defined by

(3.6).

REMARK (3.8). If α = 12, then the operator Kα

fH must be replaced by (3.9) µ K 1 2 fHϕ ¶ (y) = Z 1 0 fH(x, y)ϕ(x)dx.

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Proof of Theorem (3.7). We prove first a). By Lemma (3.2) we have

E¡eitS¢ = lim n→∞E¡e

itTn¢

where Tn is given by (3.3) with ti = ni, for every i = 0, . . . , n − 1. Let us compute the

characteristic function of the random variable Tn.

We will use the following fact: If X, Y are two independent random variables, then E (Φ(X, Y )/X) = ϕ(X)

where ϕ(x) = E(Φ(x, Y )). Let us put (3.10) X =³B0α, Bα1 n , . . . , Bαn−1 n ´ and Y =³BH1 n − B0H, . . . , BHn n − B H n−1 n ´ . Therefore, we obtain ϕ(x) = E³eitPn−1k=0xkYk ´ = e−t 2 2x TAHx

where the matrix AH =³AHk,l´

k,l=0,...,n−1 is given by AHk,l = E³BHk+1 n − BHk n ´ ³ BHl+1 n − BHl n ´ = 1 2n2H ¡|k − l + 1| 2H+ |k − l − 1|2H− 2|k − l|2H¢ . We will obtain

E¡eitTn¢ = E(e−t 2 2Sn) where Sn := n−1 X k,l=0 AHk,lBαk n Bαl n = n−1 X k,l=1 AHk,lBαk n Bαl n = n−1 X k,l=1 AHk,l Ãk−1 X k′=0 µ Bαk′+1 n − Bαk′ n ¶! Ãl−1 X l′=0 µ Bαl′+1 n − Bαl′ n ¶! = n−2 X k′,l=0 µ Bαk′+1 n − Bαk′ n ¶ µ Bαl′+1 n − Bαl′ n ¶ n−1 X l=l′+1 n−1 X k=k′+1 AHk,l.

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We calculate first n−1 X l=l′+1 n−1 X k=k′+1 AHk,l = 1 2n2H n−1 X l=l′+1 " n−1 X k=k′+1 ¡|k − l + 1|2H + |k − l − 1|2H − 2|k − l|2H¢ # = 1 2n2H n−1 X l=l′+1 " n−1 X k=k′+1 ¡|k − l + 1|2H − |k − l|2H¢ − n−1 X k=k′+1 ¡|k − l|2H− |k − l − 1|2H¢ # = 1 2n2H n−1 X l=l′+1 £|n − l|2H− |k+ 1 − l|2H− |n − 1 − l|2H+ |k− l|2H¤ = 1 2n2H " n−1 X l=l′+1 ¡|l − k′|2H− |l − 1 − k|2H¢ − n−1 X l=l′+1 ¡|l + 1 − n|2H− |l − n|2H¢ # = 1 2n2H £(n − k ′− 1)2H+ (n − l− 1)2H− |l− k|2H¤ = fHµ k ′+ 1 n , l′+ 1 n ¶

where the function fH is given by (3.5). By combining the above calculations we get

Sn= n−1 X k,l=0 fHµ k + 1 n , l + 1 n ¶ ³ Bαk+1 n − Bαk n ´ ³ Bαl+1 n − Bαl n ´ .

Let us denote by (µi)i≥1the eigenvalues of the operator KfαH and by (gi)i≥1the correspond-ing eigenfunctions. Then, uscorrespond-ing Lemma (3.4), we can write

fH(x, y) =X

i≥1

µigi(x)gi(y)

with the vectors (gi)i≥1 orthogonal in L2s,α(T ) and the µi are square-summable.

The sum Sn becomes

Sn = n−1 X k,l=0   X i≥1 µigi( k + 1 n )gi( l + 1 n )   ³ Bαk+1 n − Bαk n ´ ³ Bαl+1 n − Bαl n ´ = X i≥1 µi Ãn−1 X k=0 gi( k + 1 n ) ³ Bαk+1 n − Bαk n ´ !2 . Since α > 12 and gi∈ L2s,α(T ) it follows from [15] that

n−1 X k=0 gi( k + 1 n ) ³ Bαk+1 n − Bαk n ´|π|→0 −→ Z 1 0 gi(x)dBα(x) in L2(Ω)

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and therefore we have that Snn→∞−→ X i≥1 µiHi2 in L1(Ω) where ³Hi = R1 0 gi(x)dBα(x), i ≥ 1 ´

are independent, standard normal random variables. As a consequence, since the eigenvalues are positive (see Remark 2.10)

E(eitT) = E  exp  −t 2 2 X i≥1 µiHi2     = Y i≥1 E µ exp µ −t 2 2µiH 2 i ¶¶ = Y i≥1 µ 1 1 + t2µ i ¶12 .

Let us discuss now the point b). We follow the lines of a) by interchanging the roles of X and Y in (3.10). We obtain that E(eitS) = limn→∞E

µ e−t 2 2Sn ¶ where Sn = n−1 X k,l=0 E³Bαk n Bαl n ´ ³ BHk+1 n − BHk n ´ ³ BHl+1 n − BHl n ´ = n−1 X k,l=0 fαµ k n, l n ¶ ³ BHk+1 n − BHk n ´ ³ BHl+1 n − BHl n ´

where fα is given by (3.6). Now we use Lemma (3.4) b) and we proceed as in the proof of the point a).

REMARK (3.11). As a final comment, let us note that the points a). and b). of the above theorem agree if α and H are bigger than 12. In fact it can be shown that in this case KfHα and KfαH have the same eigenvalues and in this case their characteristic functions coincide term by term. Indeed, let us suppose that λ 6= 0 is an eigenvalue for KfHα. Then there exists a non identically zero function ϕα,H ∈ L2H(T ) such that

(KfHαϕα,H)(y) = λϕα,H(y) or H(2H − 1) Z 1 0 Z 1 0 1 2¡x 2α+ y− |x − y|¢ ϕ α,H(x′)|x − x′|2H−2dxdx′= λϕα,H(y).

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Let us denote by

χα,H(y) = ϕH,α(1 − y).

It is easy to check that χα,H ∈ L2α(T ) and by using the change of variables 1 − x = u and

1 − x′ = v we obtain

(KfαHχα,H)(y) = λχα,H(y)

which implies that λ is also an eigenvalue for KfαH.

4

The two-parameter case

In this section, we will briefly discuss the case of the fractional Brownian sheet. Let us denote by¡Bα1,α2 s,t ¢ s,t∈T and ³ BH1,H2 s,t ´

s,t∈T two independent fractional Brownian sheets. We recall

that a fractional Brownian sheet ³BH1,H2

s,t

´

s,t∈T with Hurst parameters H1, H2 ∈ (0, 1) is a

centered Gaussian process starting from 0 with covariance given by E³BH1,H2

s,t BHu,v1,H2

´

= RH1(s, u)RH2(t, v), s, t, u, v ∈ T,

where RHi is the covariance of the one-parameter fBm with Hurst index H

i (i = 1, 2). We

refer to [4] or [3] for the basic properties and [17], [18] or [11] for elements of the stochastic calculus with respect to this process. We only point here the following facts:

• the canonical Hilbert space H(H1, H2) of the Gaussian process BH1,H2 is defined as the

closure of the linear vector space generated by the indicator functions {1[0,s]×[0,t], s, t ∈

T } with respect to the scalar product

h1[0,s]×[0,t], 1[0,u]×[0,v]iH(H1,H2) = RH1(s, u)RH2(t, v).

• if H1 or H2 is bigger than 12, then the elements of H(H1, H2) maybe not functions

but distributions. In this case it is convenient to work with the following subspace of H(H1, H2)

L2H1,H2¡T2¢ := L2

H1(T ) ⊗ L

2 H2(T )

which is a space of functions (and which plays the role played by L2H(T ) in the one-parameter case). Therefore Wiener integrals with respect to BH1,H2 can be naturally defined for integrands in L2H1,H2¡T2¢.

We prove here the following result.

THEOREM (4.1). a). Assume that Hi > 12 and αi > 12, i = 1, 2. Then the

characteristic function of the random variable

(4.2) A := Z T Z T Bα1,α2 u,v dBu,vH1,H2

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is given by (4.3) E¡eitA¢ = Y i,j≥1 µ 1 1 + t2µ i,1µj,2 ¶12

where (µi,1)i are the eigenvalues of the operator KfαH11 given by (2.9), (µj,2)j are the

eigen-values of Kα2

fH2 and fH1, fH2 are defined by (3.5). b). If Hi > 12 and αi ∈ (0, 1), then E¡eitA¢ = Y i,j≥1 µ 1 1 + t2ξ i,1ξj,2 ¶12

where for j = 1, 2, (ξi,j)i are the eigenvalues of the operator KfHαjj, where fαj is defined by (3.6).

Proof: We prove only the first part because the second point is similar. Denote by

An:= n−1 X k,l=0 Bα1,α2 k n, l n BH1,H2(∆ k,l) where BH1,H2(∆ k,l) = BHk+11,H2 n , l+1 n − BH1,H2 k n, l+1 n − BH1,H2 k+1 n , l n + BH1,H2 k n, l n

As in Lemma (3.2), we can prove that An→ A when n → ∞ in L2(Ω) for αi> 12, Hi > 12,

i = 1, 2. We obtain, using the methods used in the proof of Lemma (3.2) (see also [10]) that E(eitA) = lim

n→∞E¡e itSn¢ with Sn= n−1 X k,l=0 n−1 X k′,l=0 fH1µ k + 1 n , k′+ 1 n ¶ fH2µ l + 1 n , l′+ 1 n ¶ Bα1,α2(∆ k,l)Bα1,α2(∆k′,l′).

By Lemma (3.4) a) we get that fHi ∈ L2

s,αi(T ) (i = 1, 2) and thus fHi = P

kµk,igk,i where

(gk,i)k≥1 are the eigenvectors of KfαHii (i = 1, 2).

Sn= Y i,j≥1 µi,1µj,2   n−1 X k,l=0 gi,1( k + 1 n )gj,2( l + 1 n )B α1,α2(∆ k,l)   2 .

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Since gi,1 ∈ L2α1(T ) for every i ≥ 1 and gj,2 ∈ L

2

α2(T ) for every j ≥ 1, we have that gi,1⊗ gj,2∈ L2α1,α2(T

2) and it is not difficult to see that n−1 X k,l=0 gi( k + 1 n )gj( l + 1 n )B α1,α2(∆ k,l) →n→∞ Z T Z T

gi(x)gj(y)dBx,yα1,α2 := Hi,j

and the random variables Hi,j are mutually independent and N (0, 1) distributed. The

conclusion follows easily.

Acknowledgements

The first author was partially supported by DGES Grants 01345 and BFM2003-00261.

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