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mr fantastic · #3 October 11th, 2008, 03:19 PM

Quote:

Originally Posted by Laurent

One has ${\rm Corr}(X,Y)=\frac{{\rm Cov}(X,Y)}{\sqrt{{\rm Var}(X){\rm Var}(Y)}}=\frac{E[XY]-E[X]E[Y]}{\sqrt{(E[X^2]-E[X]^2)(E[Y^2]-E[Y]^2)}}$ , right? So what you need to compute is $E[XY]$ , $E[X]$ , $E[Y]$ , $E[X^2]$ and $E[Y^2]$ . In fact, the pdf of $(X,Y)$ is symmetric in $x,y$ , so that $X$ and $Y$ have same distribution and the formula reduces to: ${\rm Corr}(X,Y)=\frac{E[XY]-E[X]^2}{E[X^2]-E[X]^2}$ .
To compute $E[XY]$ , just integrate $xy$ times the pdf of $(X,Y)$ (over the square $0\leq x,y\leq 1$ ).
You can do the same with the other ones (integrating $x$ and $x^2$ ), but you may find it quicker to first determine the pdf of $X$ . For that, you just have to integrate the pdf of $(X,Y)$ with respect to the variable $y$ , keeping $x$ fixed.

I was just about to reply when I saw yours. I misread the question as find Cov(X, Y) ..... So it's lucky you replied first (although my reply would have got the OP half way ......

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