Joint Probability Function

brd_7 · #1 $Old$ October 11th, 2008, 07:47 AM

Let X and Y have joint probability density function

fX,Y(x,y) = x+y, for 0 <=x<= 1, o<=y<=1
0, otherwise

How do i go about determining the corr(X,Y)?

Could someone outline the steps please.. Any help would be appreciated..

Laurent · #2 $Old$ October 11th, 2008, 03:29 PM

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.

mr fantastic · #3 $Old$ October 11th, 2008, 04:19 PM

Quote:

Originally Posted by Laurent $View Post$

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 ......

)

brd_7 · #4 $Old$ October 11th, 2008, 05:25 PM

Thanks so much, Ok this is what i got...

E[XY] = 1/3

E[X]^2 = (1/3 + y/2)^2 = y^2/4 + 2y/6 + 1/9

E[X^2] = 1/4 + y/3

E[Y] = 1/3 + x/2

And as a final answer.. i gained..

y^4/16 - 2y^3/24 - 13y^2/144 + 10y/216 + 10/324

Laurent · #5 $Old$ October 11th, 2008, 05:29 PM

Quote:

Originally Posted by brd_7 $View Post$

Thanks so much, Ok this is what i got...

E[X]^2 = (1/3 + y/2)^2 = y^2/4 + 2y/6 + 1/9

There's no $y$ in the answer (what is $y$ ?). In order to find the average, you integrate $y$ as well:
$E[X]=\int_0^1\int_0^1 x(x+y)dx\,dy$ . The same for $X^2$ .

brd_7 · #6 $Old$ October 12th, 2008, 04:14 AM

Oh right, that makes sense.. Im assuming you would do the same if i needed E(Y)?

Well as a final answer i got -1/11..

Thanks for all the help again.