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I am studying Statistical Mechanics and Thermodynamics from a book that i am not sure who has written it, because of its cover is not present.

There is a section that i can not understand:

${Fj|j=1,..,N}$

$S= \sum_{j=1}^{N} F_{j}$

$<S>=< \sum_{j=1}^{N} F_{j}> = \sum_{j=1}^{N} <F_{j}>$

$\sigma^{2}_{S} =<S^{2}>-<S>^{2}$

line a:

$=\sum_{j=1}^{N}\sum_{k=1}^{N} <F_{j}F_{k}> - \sum_{j=1}^{N} <F_{j}>\sum_{k=1}^{N}<F_{k}>$

I can not understand why these terms are different

First term is:

$\sum_{i}\sum_{j}A_iA_j=(A_1A_1+A_1A_2+A_1A_3+\dots+A_1A_n)+\dots+(A_nA_1+A_nA_2+\dots+A_nA_{n-1}+\dots+A_nA_n)$

Second term is

$A_1(A_1+A_2+A_3+\dots+A_n)+\dots+A_n(A_1+A_2+\dots+A_n)$

I can see they are same.

But as you can see above it is different, how? Where am i wrong?

The question Math for Thermodynamics Basics 's answer have some clue but i could not understand exactly. Thanks.

Thanks

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1 Answer 1

up vote 3 down vote accepted

The average/expected value of a product is in general not the same as the product of expected values. (The "mean value" function is linear though: a sum of mean values is equal to the mean value of the sum.)

The product of the sum and sum of product are related by the covariance: $cov(X,Y) = <XY> - <X><Y>$, as you stated yourself.

I hope this helps.

source: http://en.wikipedia.org/wiki/Expected_value#Non-multiplicativity

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How can this be then $=\sum_{j=1}^{N}\sum_{k=1}^{N} <F_{j}F_{k}>$ $=\sum_{j=1}^{N}\sum_{k=1(k\neq j))}^{N} <F_{j}><F_{k}> +\sum_{j=1}^{N} <F_{j}^{2}>$ ? As i understood $=\sum_{j=1}^{N}\sum_{k=1}^{N} <F_{j}F_{k}>$ $=\sum_{j=1}^{N}\sum_{k=1(k\neq j))}^{N} <F_{j}F_{k}> +\sum_{j=1}^{N} <F_{j}^{2}>$ –  merveotesi Jun 18 '13 at 13:47
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What do you mean bij "this"? Your first equation is correct if $F_j$ and $F_k$ aren't related for $j\neq k$ (then the covariance equals zero), your second equation is generally true. –  BNJMNDDNN Jun 18 '13 at 14:28
    
Is following: $=\sum_{j=1}^{N}\sum_{k=1(k\neq j))}^{N} <F_{j}><F_{k}>$ equal to $=\sum_{j=1}^{N}\sum_{k=1(k\neq j))}^{N} <F_{j}F_{k}>$ because k different from j? Am i right? If it is right, why? –  merveotesi Jun 18 '13 at 15:09
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No, since $< F_j F_k > = < F_j >< F_k > + cov(F_j,F_k)$. If the covariance equals zero - which means $F_j$ and $F_k$ aren't related, they are independant - your equation is true. So you just have to determine whether your variables are dependant of each other or not. –  BNJMNDDNN Jun 18 '13 at 15:20
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