# product of random variable and variance for an ensemble

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!

• 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 $C_j$. Sep 15 '16 at 20:57
• 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. Sep 15 '16 at 21:15
• 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. Sep 15 '16 at 21:19
• 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! Sep 15 '16 at 21:21

## 1 Answer

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.

• 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. Sep 16 '16 at 2:49
• The first equation is correct (without $^2$). Added the missing bracket. Thanks. Sep 25 '16 at 1:30
• 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. Sep 25 '16 at 2:29