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So, I've read some posts on covariance in this site and I think I have finally managed to get my head around the concept. However, there is still one thing about it that I really wish I had more intuition on, namely, when calculating the covariance between two random variables $X$ and $Y$, why do we average the product of their deviations from their respective means? In other words, why do we calculate

$$E[(X-E[X])(Y-E[Y])]$$

instead of just

$$E[XY]$$

? Wouldn't the latter calculation be enough to tell wether, on average, $X$ goes up when $Y$ goes up and vice-versa? Why would we wish to consider $(X-E[X])$ instead of just $X$? Thanks very much in advance.

daniels
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Daniels
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  • How would you propose interpreting a negative value of $E[XY]$? Consider, for instance, *perfectly positively correlated* variables $X$ and $Y$ for which $X$ is always negative and $Y$ is always positive, thereby ensuring a negative value for $E[XY]$. (For instance, $Y$ could have a uniform distribution on $(0,1)$ and $X$ could equal $Y-1$.) Note, too, that deviation from the mean is not an essential part of the *concept* of covariance. For an explanation that makes no reference to means at all, please see http://stats.stackexchange.com/a/18200. – whuber Mar 29 '17 at 17:11
  • So that you can manage this thread, please visit http://stats.stackexchange.com/help/merging-accounts to merge your accounts. – whuber Mar 29 '17 at 17:21

1 Answers1

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If $X$ and $Y$ are independent then $\mathbb{E}(XY) = \mathbb{E}(X)\mathbb{E}(Y)$. So, as a 'bilinear approximation' for 'deviating' from independence, the term $\mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y)$ seems like a good bet.

Note: much like a semi-definite bilinear form, it being zero doesn't necessarily mean the r.v.'s are independent.

sntx
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