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May 31, 2020 at 20:32 comment added Abdelmalek Abdesselam No need. All you have to do is say $\phi$ designates an element of the space of fields $X$ and not $Y$. Then the equation $\mathbb{E}_{\nu_C}[f_x f_y]=\mathbb{E}_{\mu_C}[\phi_x \phi_y]$ is correct. This is just a trivial consequence of the abstract change of variable theorem relating expectations for the direct image measure to expectations of the original measure.
May 31, 2020 at 19:57 comment added JustWannaKnow I must think more about all this. I'll possibly post another question on applications of these ideas. It might be better.
May 31, 2020 at 19:51 comment added Abdelmalek Abdesselam You are moving the goalposts again and replying to your question requires making corrections as a preliminary step. I suppose you mean $\phi$ is a function from $\mathbb{Z}^d\rightarrow\mathbb{R}$ and not $\mathbb{Z}^d\rightarrow\mathbb{R}^d$. So $\phi$ means an element of $Y=\mathbb{R}^{\mathbb{Z}^d}$. Then, technically $\mathbb{E}_{\mu_C}[\phi_x \phi_y]$ is not well defined because $\mu_C$ is a measure on $X$ and not $Y$.
May 31, 2020 at 19:12 comment added JustWannaKnow Let me see if I understand the application of your explanations. Suposse a field is $\phi: \mathbb{Z}^{d}\to \mathbb{R}^{d}$ so that the space of all fields is $\mathbb{R}^{\mathbb{Z}^{d}}$. For each $x \in \mathbb{Z}^{d}$ I can define random variables $f_{x}: \Omega=\mathbb{R}^{\mathbb{Z}^{d}}\to \mathbb{R}$ by $f_{x}(\phi) :=\phi(x)$. Thus, I can calculate, e.g. correlations $\mathbb{E}_{\nu_{C}}[f_{x}f_{y}] = \mathbb{E}_{\mu_{C}}[\phi(x)\phi(y)]$ both ways. That's why these are "basically the same"?
May 31, 2020 at 18:55 comment added JustWannaKnow Thank you SO much for this amazing answer! This was exactly what I was searching for!
May 31, 2020 at 18:54 vote accept JustWannaKnow
May 31, 2020 at 18:44 history edited Abdelmalek Abdesselam CC BY-SA 4.0
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May 31, 2020 at 18:35 history answered Abdelmalek Abdesselam CC BY-SA 4.0