How to calculate the expectation of $XY$? Suppose I am given the joint pdf of $X$, $Y$, and I am asked to find the $\operatorname{cov}(X,Y)$.
I know that $\operatorname{cov}(X,Y)=E(XY)-E(X)E(Y)$ and I know how to find $E(X)$ and $E(Y)$.
My questions are:


*

*What is the definition of $E(XY)$? Is it always equal to $$\int_{R\times R} xyf_X(x)f_Y(y)dxdy\,?$$ 
Or only if $X$, $Y$ are independent?(from the answer I have, the solution I have did not check the independence of $X$ and $Y$, and the answer $\operatorname{cov}(X,Y)$ is not zero, which proves $X$, $Y$ are not independent.)

*I remember, but not very clearly, that if the joint pdf of $X$, $Y$ $f_{X,Y}(x,y)$ can be written as $$f_{X,Y}(x,y)=g(x)h(y),$$
then $X$ and $Y$ are independent. Is it always true or need some conditions? I mean, suppose the region is not, say, $[0,1]\times[0,1]$, but, say, $0<x<1,x<y<2x$, is that saying still true?
Thank you so much!
 A: In general, for jointly continuous random variables $X$ and $Y$ with joint pdf
$f_{X,Y}(x,y)$,
$$E[g(X,Y)]=\int_{-\infty}^\infty\int_{-\infty}^\infty g(x,y)f_{X,Y}(x,y)dx dy.$$ In the special case you are
considering, this becomes
$$E[XY]=\int_{-\infty}^\infty\int_{-\infty}^\infty xyf_{X,Y}(x,y)dx dy.$$
If $X$ and $Y$ are jointly continuous random variables with joint pdf 
$f_{X,Y}(x,y)$, and $f_{X,Y}(x,y)$ factors into the product of
the marginal pdfs $f_X(x)$ and $f_Y(y)$, then $X$ and $Y$ are said
to be independent random variables.  More useful is the reverse
implication: if we assume that $X$ and $Y$ are independent continuous random
variables  with known pdfs (e.g. standard normal), then they are jointly
continuous with joint pdf $f_{X,Y}(x,y)$ equal to the product
$f_X(x)f_Y(y)$ of their individual pdfs.
Your expression $\displaystyle E[XY] = \int_{R\times R} xyf_X(x)f_Y(y)dxdy$
is incorrect in the general case, but is correct
when $X$ and $Y$ are independent continuous random variables since
$f_{X,Y}=f_X(x)f_Y(y)$ in this case. Indeed, if your expression were
correct in general, then we would have
$$E[XY] = \int_{R\times R} xyf_X(x)f_Y(y)dxdy = \int_{R} xf_X(x)dx
\int_{R} yf_Y(y)dy = E[X]E[Y]$$
so that $\text{cov}(X,Y)=E[XY]-E[X]E[Y] = 0$ for all random variables,
which is clearly not true.  So we have the following.

If $X$ and $Y$ are independent random variables, then $E[XY]=E[X]E[Y]$.

Note that this holds for all random variables, not just continuous
random variables.  Also, as you probably know, the converse is not
true: uncorrelated random variables need not be independent.
With regard to your second question,
$X$ and $Y$ are independent if you can find $g(x)$ and $h(y)$ such that the 
equality $f_{X,Y}(x,y)=g(x)h(y)$ holds at all points $(x,y)$ in the plane, not just at some points.  If the joint pdf is nonzero only for $0<x<1,x<y<2x$, then $X$ and $Y$ are dependent random variables; no need to try and see if you can express $f(x,y)$ as $g(x)h(y)$.
Finally, note that all of the above applies provided the various integrals
and expectations are defined or exist.  $E[XY]=E[X]E[Y]$ does not apply
to independent Cauchy random variables, for example, because $E[X]$ and $E[Y]$
are undefined for Cauchy random variables $X$ and $Y$.
