Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From wikipedia inner product page: the expected value of product of two random variables is an inner product $\langle X,Y \rangle = \operatorname{E}(X Y)$. How it can be generalized in case of random vectors?

Or more generally for any probability measure. Let $\mathbb{P}$ be a set of all probability measures defined on $X$, and let $\mathbb{M}$ be the linear span of $\mathbb{P} - \mathbb{P}$. How an inner product can be defined on $\mathbb{M} \times \mathbb{M}$?

I've looked to the norm like $$\|P - Q\|= \sup_{f} \left| \int f \, dP - \int f \, dQ \right|$$ But it seems that this norm doesn't satisfy the parallelogram law (so $\langle x, y\rangle = \frac{1}{4}( \|x + y\|^{2} - \|x - y\|^{2})$ trick cannot be used). Is it possible to proof this?

share|cite|improve this question
I'm not sure I understand the second question or how it's a generalization of the first question. For random vectors in, say, a Hilbert space, I guess you can take the expectation of their inner product. – Qiaochu Yuan May 3 '11 at 20:38
let me clarify, in first case we assume that we have inner product defined as $\langle X, Y \rangle = \int x y \, dP_{XY}$. In second case i'm more interested possibility to define inner products based on so called integral probability metrics. – TheBug May 3 '11 at 21:10

$\mathbb M$ would be the space of signed measures on $X$ (presumably with respect to a particular $\sigma$-algebra). This is a Banach space with the total-variation norm, but not a Hilbert space, and so it doesn't have a natural inner product.

share|cite|improve this answer
Right, every Hilbert space is a Banach space because, but not every Banach space is a Hilbert space. A necessary and sufficient condition for a Banach space to be associated to an inner product (make it into a Hilbert space) is the parallelogram identity. So can you proof that TV norm doesn't satisfy the parallelogram identity. Btw, TV norm is just a special case of the norm defined in question, it depends on choise of set of $f$ functions in supremum. – TheBug May 3 '11 at 21:31
That the total variation norm doesn't satisfy the parallelogram identity is very easy (try some examples). A bit less obvious is that $\mathbb M$ is not isomorphic to a Hilbert space (except in the trivial case where $X$ consists of finitely many atoms). This can be seen from the fact that $\mathbb M$ has a closed subspace isomorphic to $\ell_1$, and $\ell_1$ is not reflexive. – Robert Israel May 5 '11 at 15:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.