One has
![{\rm Corr}(X,Y)=\frac{{\rm Cov}(X,Y)}{\sqrt{{\rm Var}(X){\rm Var}(Y)}}=\frac{E[XY]-E[X]E[Y]}{\sqrt{(E[X^2]-E[X]^2)(E[Y^2]-E[Y]^2)}}](http://www.mathhelpforum.com/math-help/latex2/img/ed2f4d03b5f802571a4299d54d2f6021-1.gif "{\rm Corr}(X,Y)=\frac{{\rm Cov}(X,Y)}{\sqrt{{\rm Var}(X){\rm Var}(Y)}}=\frac{E[XY]-E[X]E[Y]}{\sqrt{(E[X^2]-E[X]^2)(E[Y^2]-E[Y]^2)}}")
, right? So what you need to compute is
![E[XY]](http://www.mathhelpforum.com/math-help/latex2/img/6e440be6519e9782522cae0d403d758f-1.gif "E[XY]")
,
![E[X]](http://www.mathhelpforum.com/math-help/latex2/img/f564e7c6618bc2c0c99eae5a9376fbaf-1.gif "E[X]")
,
![E[Y]](http://www.mathhelpforum.com/math-help/latex2/img/a7536de6a5a3e3d205f9bbe8598f2e41-1.gif "E[Y]")
,
![E[X^2]](http://www.mathhelpforum.com/math-help/latex2/img/4d1d522fcf91e6ad6f6948a5e295ff59-1.gif "E[X^2]")
and
![E[Y^2]](http://www.mathhelpforum.com/math-help/latex2/img/8a23c1b4166a611aa981e5f879fa62eb-1.gif "E[Y^2]")
. In fact, the pdf of
is symmetric in
, so that
and
have same distribution and the formula reduces to:
![{\rm Corr}(X,Y)=\frac{E[XY]-E[X]^2}{E[X^2]-E[X]^2}](http://www.mathhelpforum.com/math-help/latex2/img/6656230d0c56b5f73ca794f9dfc9c3ba-1.gif "{\rm Corr}(X,Y)=\frac{E[XY]-E[X]^2}{E[X^2]-E[X]^2}")
.
To compute
, just integrate
times the pdf of
(over the square
).
You can do the same with the other ones (integrating
and
), but you may find it quicker to first determine the pdf of
. For that, you just have to integrate the pdf of
with respect to the variable
, keeping
fixed.