# A question about a generalization of covariance

Suppose, $$H$$ is a Hilbert space over $$\mathbb{R}$$. Suppose, $$X$$ and $$Y$$ are random vectors in $$H$$. Let’s define Hilbert expectation of a random vector $$X$$ in a Hilbert space $$H$$ as a vector $$v \in H$$, such that $$\forall u \in H \text{ } (E\langle X, u \rangle = \langle v, u \rangle)$$. If a Hilbert expectation exists, then it is unique, due to the fact that every Hilbert space admits an orthonormal basis. Let’s denote Hilbert expectation of a random vector $$X$$ as $$E_HX$$. Now let’s define scalar covariance of two random vectors $$X$$ and $$Y$$ as $$Cov_H(X, Y) = E\langle X, Y\rangle - \langle E_HX, E_HY \rangle$$. Is it always true, that if $$X$$ and $$Y$$ are independent, then $$Cov_H(X, Y) = 0$$?

If $$H$$ is $$l_2$$ or any its subspace, and $$X = (X_n)_{n = 1}^\infty$$ and $$Y = (Y_n)_{n = 1}^\infty$$. Then $$Cov_H(X, Y) = \Sigma_{n = 1}^{\infty} EX_nY_n - \Sigma_{n = 1}^{\infty} EX_nEY_n = \Sigma_{n = 1}^{\infty} Cov(X_n, Y_n)$$. Thus, if $$X$$ and $$Y$$ are independent, then $$Cov_H(X, Y) = 0$$. And because every separable Hilbert space is isometrically isomorphic to a subspace of $$l_2$$, the statement is proven for any separable Hilbert space.

However, I do not know, what to do in case, when $$H$$ is not separable.

• Probably yes, but your "covariance" is so at odds with the usual use of the term in even the Hilbert space $\mathbb R^2$ that I have a hard time getting my head around what you are asking. – kimchi lover Feb 10 at 13:09
• @kimchilover, I changed the name of that thing from "Hilbert covariance" to "scalar covariance" to avoid confusion with covariance operator. Is it all right now? – Yanior Weg Feb 10 at 13:24
• Your "Hilbert expectation" is usually called the weak, or Pettis, expectation. I don't know much about such things, and how to answer your questions. Have you read the Hilbert space chapter in Parasarathy's Probability measures on metric spaces? – kimchi lover Feb 10 at 14:23