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Also, why does $E(xy) = E(x)E(y)$ not hold if $x$ and $y$ are correlated? Perhaps at a more basic, intuitive level, what's the difference between $E(xy)$ and $E(x)E(y)$?

David Faux
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    Small correction. If $E(XY)=E(X)E(Y)$, then $X$ and $Y$ are uncorrelated, but not necessarily independent. – André Nicolas Dec 14 '13 at 04:35
  • Oh! Since then, their covariance is 0. Thanks! – David Faux Dec 14 '13 at 04:38
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    The answer to the question in the title is that we say that $X$ and $Y$ are uncorrelated exactly when $E[XY]$ happens to equal $E[X]E[Y]$, and if $E[XY] \neq E[X]E[Y]$ then we do not say that $X$ and $Y$ are uncorrelated random variables; we say that they are correlated random variables. – Dilip Sarwate Dec 14 '13 at 04:56

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Where $X$ and $Y$ are continuous random variables, by definition they are independent when $f_{XY}(x,y)=f_X(x)f_Y(y)$. Then we have$$\begin{align}E(XY)&=\int_{-\infty}^\infty f_{XY}(x,y)xydxdy \\ &=\int_{-\infty}^{\infty}f_X(x)f_Y(y)xydxdy \\ &=\int_{-\infty}^\infty f_X(x)xdx\int_{-\infty}^\infty f_Y(y)ydy \\ &=E(X)E(Y) \end{align}$$

The proof in the discrete case is analogous. This shows why independence of $X$ and $Y$ implies that $E[XY]=E[X]E[Y]$. The converse does not necessarily hold—that is, we can come up with examples of random variables $X,Y$ with $E[XY]=E[X]E[Y]$ but which are not independent. Whenever $E[XY]=E[X]E[Y]$, $X$ and $Y$ are uncorrelated by definition: $$\text{Cov}(X,Y) = E[XY]-E[X]E[Y]=0\implies E[XY]=E[X]E[Y]$$

crf
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  • Thank you! So you group terms into 2 integrals because y's terms are not dependent on x's integral and vice versa please? – Avv Apr 06 '22 at 22:33
  • A bit late, but yes you can split up 2D integrals when the function is separable. – APerson Jun 15 '23 at 20:56