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Product identity and conditional moments for iterated Brownian integrals

Statement

Let the finite-dimensional Brownian It^o setting and the integrals Iij,n be as in Milstein updates, iterated Brownian integrals, and commutative noise. For every grid interval of length h>0, Iij,n+Iji,n=ΔWniΔWnj−δijh. In the scalar case, writing In=I11,n and ΔWn=ΔWn1, this gives In=12((ΔWn)2−h),E[In∣Ftn]=0,E[In2∣Ftn]=h22,E[ΔWnIn∣Ftn]=0. Consequently, under the commutativity condition in the Definition, the double-integral term in the general Milstein Taylor increment equals ∑i,jLiσj(Yn)Iij,n=12∑i,jLiσj(Yn)(ΔWniΔWnj−δijh).

Facts & Assumptions

Given: AC, a filtered probability space with an adapted standard m-dimensional Brownian motion whose increments after tn are independent of Ftn, and the grid interval [tn,tn+h].

[L1]

Milstein updates, iterated Brownian integrals, and commutative noise: Iij,n=∫tntn+h(Wsi−Wtni) dWsj, and the commutative condition is Liσj=Ljσi.

[F1]

Multidimensional Ito formula for Brownian-driven processes: for a Brownian It^o process, the second-order term in It^o's formula is one half of the Hessian paired with the instantaneous covariance matrix; the theorem assumes AC and the usual-filtration/integrability conventions stated in its body.

[F2]

Continuous Brownian Ito processes: Brownian increments are independent of the current sigma algebra and have their stated centered normal laws.

[F3]

Standard normal and normal laws: a standard normal has density (2π)−1/2e−x2/2; its odd moments vanish and its even moments follow by integration by parts.

[F4]

Conditional expectation given a sigma algebra: conditional expectation is the measurable integrable version characterized by its integrals on events in the conditioning sigma algebra.

[A1]

The Axiom of Choice: AC is assumed by the cited Brownian and It^o interfaces; no further choice is used here.

Proof

Given: The assumptions above.

1.1F1L1A1

Apply [F1] to f(x)=xixj on [tn,tn+h]. It gives Wtn+hiWtn+hj−WtniWtnj=∫tntn+hWsi dWsj+∫tntn+hWsj dWsi+δijh. Expand each integrand as its value at tn plus its Brownian increment; the two constant terms cancel from the two sides, leaving ΔWniΔWnj=Iij,n+Iji,n+δijh.

2.1F2F3step 1.1

Setting i=j=1 in step 1.1 yields In=((ΔWn)2−h)/2. By [F2], ΔWn is independent of Ftn and has law N(0,h). Scaling the density in [F3] and integrating by parts give E[ΔWn∣Ftn]=0, E[(ΔWn)2∣Ftn]=h, E[(ΔWn)3∣Ftn]=0, and E[(ΔWn)4∣Ftn]=3h2.

3.1F2F3F4step 2.1

Substituting the four conditional moments from step 2.1 into In=((ΔWn)2−h)/2 gives E[In∣Ftn]=(h−h)/2=0, E[In2∣Ftn]=(3h2−2h2+h2)/4=h2/2, and E[ΔWnIn∣Ftn]=(0−h⋅0)/2=0. Each equality is an equality of conditional-expectation versions by the event-integral definition [F4].

4.1L1step 1.1∎

Under commutativity, pair the summand with indices (i,j) with the summand (j,i). The coefficients agree by [L1], so the sum equals 12∑i,jLiσj(Yn)(Iij,n+Iji,n). Substituting step 1.1 gives the displayed Brownian-increment formula. The argument includes i=j, where the summand is one half of Liσi(Yn)((ΔWni)2−h).

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