On the moments of certain stochastic integrals

Statistics & Probability Letters 18 (1993) 65-72

North-Holland

30 August 1993

On the moments of certain stochastic integrals

Larry Goldstein * Department of Mathematics, University of Southern California, Los Angeles, CA, USA

Brendan McCabe Faculty of Commerce and Business Administration, University of British Columbia, Vancouver, BC, Canada

Received September 1991

Revised February 1993

Abstract: The moments of certain stochastic integrals with respect to Brownian motion are well known. In this paper the mean and

variance of some integrals involving Brownian bridges are found.

Keywords: Moments; Brownian motion; Brownian bridge; stochastic integrals; Malliavin calculus.

1. Introduction

Consider the stochastic integral /,‘W(t) dW(t) where W(t) is a Brownian motion. (See Arnold (1974, Chapters 3-5, for an elementary introduction to Brownian motion and stochastic integrals.) It is well known that E[ /;I$ dW] = 0 and that V[ j:W dWl= /$Z[W(t>12 dt = jtt dt = $. These results may readily be extended to the integrals of the form IIf dW(t) and /K(t) dW(t) where H(t) and K(t) are stochastic processes adapted to W’(t). In this case E[jH(t) dW(t)l= 0 and E[JH(t) dW(t)jK(t) dW(t)l = /E[H(t)K(t)] dt. Note it is not assumed that E[H(t)l = 0. The variance of the stochastic integral follows by setting H = K. These results, and elementary properties of Brownian motion, are used throughout. In this paper integrals of the form /W(t) dB(t) are studied where B(t) is a Brownian bridge. The relationship between W and B needs to be made precise. Two cases are studied using the relation that if U(t) is a Brownian motion then B(t) = U(t) - N(l) is a Brownian bridge on [O 11. In the first case (Section 2) U is stochastically independent of W and in the second (Section 3) U = W. Integrals of this form often arise in the study of time series, particularly in the context of random walks. Let ei be white noise and define wi = wi_i + q with w0 = 0 for convenience. Consider the statistic

z= eWi(Ei-E) i=l

Correspondence to: Larry Goldstein, Department of Mathematics, University of Southern California, 1042 W. 36th Place, DRB 155,

Los Angeles, CA 90089-1113, USA. * This work was supported in part by NSF grant DMS-90-05833.

0167-7152/93/$06.00 0 1993 - Elsevier Science Publishers B.V. All rights reserved 65

Volume 18, Number 1 STATISTICS & PROBABILITY LETIERS 30 August 1993

where E = CyC1~Jn. This is a weighted average of the deviations in the innovation process and may also be interpreted as the numerator of the correlation coefficient between the level of the process and the innovation. Let s,(t) be the partial sum process of the si, i.e., s,(t) = C~“,:)E~/T~/‘~, here u* is the variance of the white noise and (nt) means integral part of nt. A routine weak convergence argument shows

Z/Ta* = /b%(t) d&(t) - (nt>%z(l)] * i’W(t) dB(t),

thus motivating the study of such integrals. In Sections 2 and 3 a relatively elementary account of how these moments may be found is given.

However it is clear from the amount of algebra involved that these methods are almost incapable of further extension. Accordingly, in Section 4, we present a method for evaluating moments via the Malliavin calculus, a method which may prove more fruitful in similar types of problems.

2. Integrating with respect to an independent Brownian bridge

To consider integration of Brownian motion with respect to an independent Brownian bridge, let (0, gr,P) be a probability space which supports independent Brownian motions W and U adapted to an increasing family of sub-sigma algebras {9J of 9,

We wish to find the mean and variance of the stochastic integral J,‘W(t> dB(t), where B(t) = U(t) - tU(l>. The required integral may be written as

j’W( t) dU( t) - U(l)/b’( t) dt. 0 0

(2.1)

Lemma 2.1.

I = 0 and V ’ ‘W(t) dB(t) =$ 1

when B is a Brownian bridge independent of W.

Proof. It is easily seen from (2.1) that E[ jlW(t) d B(t)] = 0 as the expectation of each of the constituent terms is zero. These facts follow because U and W are independent and E[U(l)l = 0.

Using the results of the previous section combined with U(1) = 1,’ dU(t) and E[ jiW(t) d t I* = Ji/iE[W(t)W(s)] dt ds = /,‘<l - t) dt = f the variance may be calculated.

‘W(t) dU(t) * Cr(l)ju’W( t) dt I

+ E[U(l)L’W(t) dt]*

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Volume 18, Number 1 STATISTICS & PROBABILITY LETTERS 30 August 1993

=k1E[W(q2] dr+E[U(1)2]E[~1W(r) dt12 - 2E /‘W(t) dtE /‘W(t) dU(t). ju'dZi(t) (W

[ 0 i 0 II =joi dr+LlL’(tAS) dt dsZE[i’W(t) dt12

=ct dt-iljol(fAs) dt ds

=;-jjcds) dr-&‘tds) dt

1 = i-T

1 =- 6’ 0

3. Integrating with respect to a dependent Brownian bridge

In constrast to the last section, we integrate Brownian motion with respect to the dependent Brownian bridge obtained by setting U = W. The appropriate integral is given by

I:W(t) dW(t) - W(l)k’W(t) dt.

Applying (stochastic) integration by parts to the latter integral yields W(1) - ldt dW(t) = /t(l - t) dW(t) and so we require the mean and variance of

k’W(f) dW(t) - W(l$(l -t) dW(t) = i*W(f) dW(t) - joldW(r) f(1 -t) dW(t).

(3.1)

Lemma 3.1.

klW(t) dB(t) =+ and I I/ ‘W(t) dB(t) =A 1 where B(t) = W(t) - tW(1).

Proof. Taking expectations the first term in (3.1) is seen to be zero while the second term yields /~/~(l - t) dt ds = i. Thus E[/,‘W(t) dB(t)] = - $ in contrast with the case when B is independent of W.

The variance calculation requires E[ /,‘W(t) dW(t) - // dW(t) . /,‘<l - t> dW(t>12 from which it is seen that the expectations of the squares and cross product are needed. The first term is straight forward and E[ /,‘W(t) dW(t>12 = i. Unfortunately, there does not seem to be any easy way of evaluating the remaining two terms other than by using approximating sums. Thus to deal with the cross product term we shall interchange the order of the integration and expectation operators and evaluate

-2E ~L&A~* CA%. c(l -tk_l)AWk . i i k 1


Now, E[wi_,AWiAWjAW,] = 0 if either k or j is greater than i. Thus only those cases were both k and j are less than i are relevant. Next write Wi_ 1 as

K_l=w* +Aq+AW,+A*

i.e. isolate from Wi the contributions of the increments for Wj and W, by choosing i* = (j - 1) h(k - 1) and absorb the remainder in A*. This leaves

E( J+& + A? + AW, + A*)( AJJ$ATAW,)

to be evaluated. When the limiting operation of the integration is finally carried out only those terms of the form E(AWi2AWj2) = AtiAtj will count since the increments are independent with mean zero, i* is less than, i, j or k and integrals formed out of increments of the type AWi”, m > 2, will be zero. Thus, keeping in mind the inequalities on i, j and k the covariance term reduces to

-2 i

C C (1 - t,_,)At,Ati $ C C (1 - ti_,)AtjAti i k<i i j<i 1

= -2 C C AtkAti + C C AtjAti - C C tk_,At,Ati - C C ti_,AtjAti ( i k<i i jqi i k<i i j<i

which upon taking limits gives

-2 jolJo’dt ds - l,‘k’~ ds dt - /,l$ds dt) = -2(1- $ - $) = -1.

The final square term to be evaluated, using approximating sums, is

C CAW,A~. C x(1 - t,-I)(i - t,-I)AwrAws i j r s 1

= E I

Cdwi2 + 2x CAW;:A~ i i>j

C(1-tiJ2AK2+2C C(l-tr-1)(1 -t,-dAW,AW, i T>S 11

=E CAWi2.C(1-ti_1)*AWi2+ CAWi”2C C(l-t,_,)(l-t,_,)AW,AW, [ i i i r>s

+2x CA~~d~+pC(l-t~_~)~dw~~ i>j i

+4x CAW,AW~. C x(1 - t,_I)(l- t,_l)Aw,Aws . i>j r>s 1 As before, the only terms that are of interest in the final analysis are those with increments of the form

E(AK2AWj2) = AtiAtj. Because of the independence of the increments the second and third terms in the above expression are zero due to the inequalities on the subscripts. Thus it simplifies to

C C(l-ti_1)*AWi2AT2 +4E C C(l-ti-1)(l-tj-1)A%~A~~ 7 i j 1 [ i>j 1

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and so, eliminating the inequality on the subscripts of the second term and taking limits, we get

~1~1(l-t)2dfd~+2/01~1(1--f)(l-~)dtd~=++f=$

This results in E[ /,‘W’(t> dB(t>12 = i - 1 + i = f. The lemma follows by subtracting the mean squared. u

4. An approach using the Malliavin calculus

First and second moments of (2.1) may also be found using of the tools of the Malliavin calculus. This requires some additional background. The article of Zakai provides (1985) a nice introduction.

Any square integrable functional F on Wiener space may be expanded in a series of multiple Wiener-It6 integrals:

Define an operator _Y (the Malliavin derivative) by

For an integrable function h, define the directional derivative of F in the h direction by

dF( w + E@, ds) (D&(W) = ds =O,

B

and, with {hi} a complete orthonormal system in L2[0, 11, define, for two Wiener functionals F,, F2,

(DF,, DF2) = 2 (Q,~l;l)(&F2). i=l

The linear operator _P obeys a product rule and a chain rule:

_T(F,F,) = Fl_YF2 + F2L?‘Fl - 2(DF,, DF,),

2(4(F)) = @(F)WF) - @‘(F)(DF, DF)

for smooth functions 4. Lastly, as can be seen from the definition of 3,

E@?(F)) =O.

We apply the above properties of the operator _Y to compute the mean and variance when

F= i’W(t) dB(t)

for B(t) = W(t) - W(l). First, we rewrite the functional F as follows:

F(w) = I,ly dW, - W(l)lolw, dt = l,‘l,ldW, dW, - (k’dy)( klk’dW, dt)

69


and now changing the order of integration in the last integral and using that

’ //

‘dW, dt = 0 0

/‘/‘dt dW, = /l(l -s) dW, 0 S 0

we see that we have

,6( 1 - s) dW,

that is,

F=G-F,F,

where

G(w) = Lri’dW, dW,, F, = l’dW,, 0

F, = jl(l -s> dW,. 0

Lemma 4.1. For F as in (4.1),

T(F)=2F+l.

Proof. Easily,

D,F, = llh, ds, 0

D,,F2 = /'(l - s)h, ds. 0

Hence,

by Parseval’s relation. Hence

(DF,, OF;) = ;.

AppIying the product rule we have

L?(F) =L?G-FI_!TF~-F,~FI+2(DF,, DF,).

Therefore, using the definition of _Y directly to find _F’G = 2G and _YF, = F,, _5?‘F, = F2,

_!Z(F)=2G-2F,F,+1=2F+l. 0

Corollary 4.1. As EL?‘(F) = 0, we derive immediately that EF = - $. q

Lemma 4.2. For F as in (4.11,

(DF,DF)=+Wf+ I,W,d ( ’ s)‘-WIhlK ds,

(4.1)

(4.2)

and

E( DF, DF) = 5.

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Proof. We have

D,F = jl/tdWsh, dt + /‘/“s ds dW, 0 0 0 0

-(& ds)(jol (1-s) dW,) - jlolW)(~l(l -s)h, ds). (4.3)

(4.4)

Change variables in the second integral of expression (4.3):

/‘/‘dW,h, ds + l’/‘dW,h, dt = I’Wlhl dt + /‘(WI - W,)h, dt = WI j’h, dt. 0 0 0 t 0 0 0

Now, integrating by parts in the second integral that appears in expression 4.4 and integrating in the third integral, we write (4.4) as

-(l’h, ds)(cW, ds) - Wlkl(l -s)h, ds.

Combining expressions (4.3) and (4.4), we have

D,F = W$sh, ds - 0

(kk ds)(& ds),

and therefore

(DhF)2= Wf /, h, d ( ‘s s)2+ (/o(wI ds)2(cs ds)2-2WIcw, dsjolsh, d+z, ds.

Summing the above expression over a complete orthonormal set {h’} and using Parseval’s relation demonstrates the first claim of the lemma. For the next, note that

EW+ ds = I,IEW,W, ds = k’s ds = ;,

and that

= /

11 2 ,t dt+ 01(t-t2) dt=;. •I

0 /

Corollary 4.2. Var[ F] = 5.

Proof. We apply the chain rule with $J(x> = +x2. Since E[LZ’C#J(F)I = 0, We have

E[&(F)P(F) -#‘(F)(DF, DF)] =E[FP(F) -(DF, DF)] =0,

and hence

E[ F(2F + 1) - +] = 0,

using Lemma 4.2. By Corollary 4.1, E[Fl = - i, and we may solve to find EF2 = f; therefore Var[F] = A. 0

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References

Arnold, L. (19741, Stochastic Differential Equations (Wiley, New York).

Zakai, M. (1985), The Malliavin Calculus, Acta Appl. Math. 3, 175-207.

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On the moments of certain stochastic integrals

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