Extra structure on $\mathcal{V}$ so that $\mathcal{V}$ doesn't ignore the probability $\mathbb{P}$

Question

We have a probability space ($\Omega, \mathcal{F}, \mathbb{P}$), and we denote $\mathcal{V}$ the vector space of all real-valued random variables defined on $(\Omega, \mathcal{F}, \mathbb{P})$.

$\mathcal{V}$ has some nice properties. For example it's an inner-product space since : $(X,Y) \mapsto \mathbb{E}[XY]$ is an inner product. Hence it gives a geometric intuition about variance and expectation.

The main problem with this vector space $\mathcal{V}$ is that it completely ignores the probability $\mathbb{P}$. Hence if we have an other probability $\mathbb{P}'$ and we consider the vecor space $\mathcal{V}'$ of all real-valued random variables defined on $(\Omega, \mathcal{F}, \mathbb{P}')$ then we have $\mathcal{V} = \mathcal{V}'$.

That's why I am wondering if it's possible to add extra structure on $\mathcal{V}$ so that $\mathcal{V}$ doesn't ignore the probability $\mathbb{P}$ ? For example is it possible with an extra structure (let's suppose it's a function $B : \mathcal{V}^2 \to \mathbb{R}$) such that :($\pi$ is the canonical map that send a random variable to it's vector representation in $\mathcal{V}$)

$$\forall v, v' \in \mathcal{V}, B(v,v') = 0 \Leftrightarrow\forall A_1, A_2 \in \mathbb{R}, \mathbb{P}(\pi^{-1}(v) \in A_1, \pi^{-1}(v') \in A_2) = \mathbb{P}(\pi^{-1}(v) \in A_1) \mathbb{P}(\pi^{-1}(v') \in A_2)$$ ? Hence the function $B$ traduces the notion of independent vectors.

Some thoughts :

• I know that it's easy to defined a probability on $\mathcal{V}$ simply : $\mathbb{P}(v = k) = \mathbb{P}(\pi^{-1}(v) = k)$. But this is not what I want. I am really looking for an extra structure on $\mathcal{V}$ so that $\mathcal{V}$ doesn't ignore $\mathbb{P}$ but without using the canonical map $\pi$ (for me in this case we are not adding structure to $\mathcal{V}$). For example expectation can be easily traduced on $\mathcal{V}$ without using $\pi$. This extra structure on $\mathcal{V}$ transform $\mathcal{V}$ in an inner product space. I am looking for a similar result but for the probability $\mathbb{P}$.

• We know that : $X, Y$ are independent random variables $\Rightarrow \mathbb{E}[XY] = \mathbb{E}[X]\mathbb{E}[Y]$. What is nice with this is that it can be easily traduce on the inner product space $\mathcal{V}$. So maybe there is a geometric intuition in this case. We can find many answers here.

• I am sorry if my question is strange/unclear.

Thank you.