Actually my question arises from the definition of $E[XY]$, why is it defined as the integral of $xyf(x,y)$?
If so, what is the expectation of $xy^2$??
Thank you very much.
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$\begingroup$ It is not precisely defined in that way. The random variable $H(X,Y)$ is, well, a random variable $W$ and the expectation of $W$ is defined in the usual way, using the distribution function of $W$. However, it is a theorem that under reasonable conditions on $H$, the expectation of $W$ is the integral you gave. In elementary courses, it is often asserted without proper proof that the expectation of $H(X,Y)$ is what you wrote down. $\endgroup$– André NicolasJul 9, 2013 at 15:06
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The expected value of $g(X,Y)$ in general is
$$\iint_R dx dy \, g(x,y) f(x,y)$$
where $R$ is the region over which the random variables $X$ and $Y$ are defined to take on values. Therefore
$$E(X Y^2) = \iint_R dx dy \, x y^2 f(x,y)$$
This of course assumes that $f$ is properly normalized as a pdf.
answered Jul 9, 2013 at 14:49
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2$\begingroup$ @WayneN: please do not change the meaning of my posts through an edit. Your previous edit demonstrated that you either did not read or did not understand what I wrote. $\endgroup$ Jul 9, 2013 at 15:05
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$\begingroup$ Could I ask why the expectation is defined in this way? The formula of E[X+Y]=E[X]+E[Y] only holds true when x and y are independent. Is that correct? $\endgroup$ Jul 10, 2013 at 9:39