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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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Cross Ito isometry

Statement

Assume the Axiom of Choice and the standing hypothesis (H) of Elementary predictable Brownian integrands. For elementary predictable integrands H,K on [0,T] and every t∈[0,T], E[It(H) It(K)]=E∫0tHsKs ds, where the sums It(⋅) are the elementary integrals of the chosen representations Ito integral of an elementary predictable process. Both sides depend only on the (dt⊗P)-classes of H and K and are finite.

Facts & Assumptions

Given: AC, the standing hypothesis (H), a horizon T>0, elementary representations of H,K on some partitions, their common refinement, and t∈[0,T].

[F1]

On the common refinement, H+K, H−K and their scalar multiples are again elementary predictable integrands with bounded coefficients measurable at the left endpoints of that refinement; the defining sums are linear there, so It(H±K)=It(H)±It(K) identically. Ito integral of an elementary predictable process Elementary predictable Brownian integrands

[F2]

For every elementary predictable G, E[It(G)2]=E∫0tGs2 ds is finite and t↦It(G) is a continuous square-integrable martingale. Ito isometry for elementary integrands

[F3]

The pointwise identity (H+K)2−(H−K)2=4HK holds on [0,T]×Ω, and the same expansion applies to the random variables It(H)±It(K). Elementary predictable Brownian integrands

[F4]

AC is inherited from the elementary isometry and representation-independence interfaces. In particular, the proof of the elementary isometry uses conditional expectations, but its exported interface here is only the martingale and squared-isometry statement [F2]. The Axiom of Choice

Proof

technique · direct
1.1

Pass to the common refinement of the two partitions and keep the notation H,K for the refined representations; by [F1] both H+K and H−K are elementary predictable integrands on that refinement, and It(H+K)=It(H)+It(K), It(H−K)=It(H)−It(K) identically, with all quantities square-integrable.

F1F2given
1.2

Applying the elementary isometry [F2] to H+K and to H−K gives the two finite identities E[It(H+K)2]=E∫0t(H+K)2ds and E[It(H−K)2]=E∫0t(H−K)2ds.

F2given
2.1

Subtracting the second identity of step 1.2 from the first and expanding with [F3] gives E[(It(H)+It(K))2]−E[(It(H)−It(K))2]=E∫0t((H+K)2−(H−K)2)ds=4E∫0tHsKs ds, where the right-hand side is finite because ∣HK∣≤12(H2+K2) and both elementary integrands have finite energy.

F2F3step 1.2
3.1

The left-hand side of step 2.1 equals 4E[It(H)It(K)] by the algebraic expansion [F3], and 4 is invertible in R, so E[It(H)It(K)]=E∫0tHsKs ds. Representation independence follows from Elementary Ito integrals do not depend on step representation applied to H and to K; the AC bookkeeping is exactly the inherited use recorded in [F4], not an additional conditional-expectation interface asserted by this lemma.

step 2.1F3F4given∎

Source notes

Van der Vaart, Lemma 5.22, records the bilinear form of the isometry as the polarized version of the squared identity. No additional source of randomness or integrability beyond the elementary isometry is used.

Depends on

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