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.

  • $\begingroup$ 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. $\endgroup$ – kimchi lover Feb 10 at 13:09
  • $\begingroup$ @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? $\endgroup$ – Yanior Weg Feb 10 at 13:24
  • $\begingroup$ 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? $\endgroup$ – kimchi lover Feb 10 at 14:23

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