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Covariance of deterministic Ito integrals

Example

Assume the Axiom of Choice and the standing hypothesis (H) of Elementary predictable Brownian integrands. For deterministic $h, k \in L^{2} [0, T]$ , $Cov (\int_{0}^{T} h d B, \int_{0}^{T} k d B) = \int_{0}^{T} h (s) k (s) d s,$ and the pair is jointly Gaussian, so $\int_{0}^{T} h d B$ and $\int_{0}^{T} k d B$ are independent exactly when $\int_{0}^{T} h k d s = 0$ . In particular deterministic integrands with disjoint supports have independent integrals.

Facts & Assumptions

Given: AC, the standing hypothesis (H), deterministic $h, k \in L^{2} [0, T]$ .

[F1]

Both integrals are centered: their laws are $N (0, \int_{0}^{T} h^{2})$ and $N (0, \int_{0}^{T} k^{2})$ . Deterministic Ito integrals are Gaussian

[F2]

The general cross identity $E [(\int_{0}^{T} h d B) (\int_{0}^{T} k d B)] = E \int_{0}^{T} h k d s$ holds for predictable $L^{2}$ integrands; for elementary (in particular deterministic step) integrands this is the polarized elementary isometry. Ito isometry and linearity in predictable L2 Cross Ito isometry

[F3]

For deterministic $h, k$ the vector of the two integrals has law $N_{2} (0, Σ)$ with $Σ_{11} = \int h^{2}$ , $Σ_{22} = \int k^{2}$ , $Σ_{12} = \int h k$ ; a multivariate normal law with diagonal covariance can be realized with independent coordinates, and its characteristic function determines the law. Multivariate normal law, including singular covariance Characteristic function of a multivariate normal law Deterministic Ito integrals are Gaussian

[F4]

AC is declared for the ambient interfaces. The Axiom of Choice

Verification

technique · direct

1.1

Since both integrals have mean $0$ by [F1], the covariance is the expectation of the product, and [F2] evaluates it as $\int_{0}^{T} h k d s$ .

F1F2given

2.1

By [F3] the pair is jointly Gaussian with covariance matrix $Σ$ whose off-diagonal entry is $\int_{0}^{T} h k d s$ ; if that entry vanishes, $Σ$ is diagonal, and the multivariate normal law with diagonal covariance is the law of a pair with independent coordinates (realization $m + Σ^{1 / 2} Z$ with independent standard normals), so the pair is independent.

F3step 1.1

3.1

Disjoint supports give $h (s) k (s) = 0$ for every $s$ , hence $\int_{0}^{T} h k d s = 0$ and independence; the degenerate cases $h = 0$ or $k = 0$ are included (a Dirac factor is independent of every variable), and AC enters only through [F4].

F3F4step 2.1given∎

Source notes

Van der Vaart, Lemma 5.22, gives the bilinear form of the isometry that computes these covariances; the independence statement is the diagonal-covariance case of the multivariate normal law.

Depends on

Used by

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Sources

Aad van der Vaart, Stochastic Integration and Differential Equations, Lemma 5.22 (standard reference, not scraped)