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I have an ensemble of particles labeled with $i=1,\cdots,N$. A set of measurement observables (random variables), $\{ A, B, C\}$, can be sampled on those particles. We denote $A_i$ as the observation of measurement $A$ applied on particle $i$. The ensemble satisfies the exchange symmetry that if we exchange any pairs of particles, the observable of any kind preserves its value. The symbol $\langle \cdot \rangle$ indicates the ensemble expectation value or mean value. I know that in general the covariances $\langle \Delta A_i \Delta C_j\rangle|_{i\neq j}\neq 0$ and $\langle \Delta B_i \Delta C_j\rangle|_{i\neq j}\neq 0$. The question is if the following equation valid for a product of observable and variance on different subsets of particles:

$$ \langle A_i B_i \Delta C_j\rangle|_{i\neq j} = \langle A_i B_i\rangle \langle \Delta C_j\rangle|_{i\neq j}=0? $$ I have used the fact that $\langle \Delta C\rangle=\langle C-\langle C\rangle \rangle=\langle C\rangle-\langle C\rangle=0$.

Hopefully I can get your thoughts on this simple problem. In the end, I want to show that $$ \langle (A_i B_i -\langle A_i\rangle\langle B_i\rangle)\Delta C_j\rangle|_{i\neq j}=\langle \Delta A_i\Delta B_i\rangle \langle \Delta C_j\rangle|_{i\neq j}=0. $$ Thanks!

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    $\begingroup$ So $\Delta X = X - \langle X \rangle$ in general? I don't think your equation is generally true except if the mean of $C_j$ conditional on $A_i$ and $B_i$ is the same as the unconditional mean of $C_j$, which is not true given your statement that $A_i$ and $B_i$ are each correlated with $C_j$. $\endgroup$
    – Danica
    Sep 15 '16 at 20:57
  • $\begingroup$ I understand that if the variance of $A_i$ and $B_i$ is correlated with $C_j$, they covariance is non-zero. But should the mean value of $A_i$ and $B_i$ is also correlated with the variance of $C_j$ given the exchange symmetry? I use $\Delta X= X-\langle X\rangle$ as a general statement to say that the observable is always centered around $\langle X\rangle$ independent of labeling. $\endgroup$ Sep 15 '16 at 21:15
  • $\begingroup$ I don't understand what you mean by "the mean value of $A_i$ and $B_i$ is also correlated with the variance of $C_j$." If you look at $\langle A_i B_i \Delta C_j\rangle = \langle A_i B_i C_j \rangle - \langle A_i B_i \rangle \langle C_j \rangle$, your claim is that $\langle A_i B_i C_j \rangle = \langle A_i B_i \rangle \langle C_j \rangle$. This is true when $C_j$ is independent of $A_i B_i$, but not generally true otherwise. $\endgroup$
    – Danica
    Sep 15 '16 at 21:19
  • $\begingroup$ Ok, I can have the following proof to get your conclusion: since $\langle \Delta A_i \Delta C_j\rangle=\langle A_i \Delta C_j\rangle -\langle A_i\rangle \langle \Delta C_j\rangle\neq 0$, therefore I will have $\langle A_i \Delta C_j\rangle \neq \langle A_i\rangle \langle \Delta C_j\rangle=0$ for $i\neq j$. Problem solved! I think this is rigorious now. My previous conclusion is wrong. You can write an answer for my vote if you like. Thank you for taking time explaining to me! $\endgroup$ Sep 15 '16 at 21:21
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First off, $\langle \Delta C\rangle = \sqrt{\langle C^2\rangle-\langle C\rangle^2}$ may not be zero. Secondly, if we define $D_i=A_iB_i$, then we should get

$$ \langle \Delta D_i \Delta C_j\rangle|_{i\neq j} = \langle (D_i-\langle D_i\rangle)\Delta C_j\rangle|_{i\neq j}=\left[\langle D_i\Delta C_j\rangle-\langle D_i\rangle \langle\Delta C_j\rangle\right]|_{i\neq j} $$ Hence $$ \langle A_iB_i\Delta C_j\rangle|_{i\neq j}=\langle D_i\Delta C_j\rangle|_{i\neq j}=\langle \Delta D_i\Delta C_j\rangle|_{i\neq j}-\langle D_i\rangle\langle\Delta C_j\rangle|_{i\neq j}. $$ Since either $\langle \Delta D_i\Delta C_j\rangle|_{i\neq j}$ or $\langle D_i\rangle\langle\Delta C_j\rangle|_{i\neq j}$ may not be zero, and they may not be equal in general, $\langle A_iB_i\Delta C_j\rangle|_{i\neq j}$ may not be zero.

Let's accept this as the answer. Please comment below if you find this is wrong. Thanks.

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  • $\begingroup$ Can you clean up the equations please? I just started skimming, and see saw two typos right away: missing exponent "$^2$" on the left side of the first equation, missing "$)$" in the middle of the first display equation. These typos make it harder for readers to follow your logic. $\endgroup$
    – GeoMatt22
    Sep 16 '16 at 2:49
  • $\begingroup$ The first equation is correct (without $^2$). Added the missing bracket. Thanks. $\endgroup$ Sep 25 '16 at 1:30
  • $\begingroup$ In the question you seemed to imply $\Delta C = C - \langle C\rangle$, and in any case you state "the fact that $\langle \Delta C\rangle=0$", which is consistent with that definition. In this case your first formula in the answer would be a standard deviation $\sigma_C$, where $\sigma_C^2=\langle (\Delta C)^2\rangle = \langle C^2\rangle - \langle C\rangle^2$. I could be wrong, as I only skimmed the question and answer. But this was the reasoning for my original comment. $\endgroup$
    – GeoMatt22
    Sep 25 '16 at 2:29

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