In my book, I am given this proof:
$$ Cov(X,Y) = \mathbb{E}[X - \mathbb{E}X][Y - \mathbb{E}Y]$$ $$ Cov(X,Y) = \mathbb{E}[XY] - 2\mathbb{E}[X] \mathbb{E}[Y] + \mathbb{E}[X] \mathbb{E}[Y]$$
I do not see why/how $$-X \mathbb{E}Y - Y \mathbb{E}X$$ becomes $$-2\mathbb{E}[X]\mathbb{E}[Y]$$
$X$ and $Y$ are given as two random variables with finite second moments, there is no further restriction or classification upon them.
Furthermore, unrelated: Independence implies uncorrelation, but is the converse also true?
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$\begingroup$$Cov(X,Y) = \mathbb{E}[(X - \mathbb{E}X)(Y - \mathbb{E}Y)]=\mathbb{E}[XY-X\mathbb{E}(Y)-Y\mathbb{E}(X)+\mathbb{E}(X)\mathbb{E}(Y)]$. Now using linearity of expected value, you get the right result.
The converse if false, the correlation coefficient only catches linear dependance. For example, if you have $Y=X^2$ with $X\sim\mathcal{N}(0,1)$, $X$ et $Y$ are uncorrelated but obviously not independant.
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