1. ## Covariance question

Hi,
I just want you expert to confirm something for me.

Suppose you have two random variables X and Y. Further suppose that fXY(x,y) = g(x) where x and y are bounded, fXY(x,y)= 0 elsewhere. That is fXY(x,y) is just a function of x (or y). Does it follow that Cov (X,Y) = 0????

2. ## Re: Covariance question

If $f_{XY}(x,y) = f_X(x)$ then $f_Y(y)=1$

$E[XY] = \displaystyle{\int_{-\infty}^\infty\int_{-\infty}^\infty} ~x y f(x,y)~dx~dy =$

$\displaystyle{\int_{-\infty}^\infty y \int_{-\infty}^\infty} ~x f(x)~dx~dy =$

$\displaystyle{\int_{-\infty}^\infty y f_Y(y) \int_{-\infty}^\infty} ~x f(x)~dx~dy =$

$\left(\displaystyle{\int_{-\infty}^\infty} y f_Y(y)~dy \right) \left(\displaystyle{\int_{-\infty}^\infty} ~x f(x)~dx\right) = E[y]E[x]$

$E[XY] -E[x]E[y]=0$

3. ## Re: Covariance question

Originally Posted by romsek
If $f_{XY}(x,y) = f_X(x)$ then $f_Y(y)=1$

$E[XY] = \displaystyle{\int_{-\infty}^\infty\int_{-\infty}^\infty} ~x y f(x,y)~dx~dy =$

$\displaystyle{\int_{-\infty}^\infty y \int_{-\infty}^\infty} ~x f(x)~dx~dy =$

$\displaystyle{\int_{-\infty}^\infty y f_Y(y) \int_{-\infty}^\infty} ~x f(x)~dx~dy =$

$\left(\displaystyle{\int_{-\infty}^\infty} y f_Y(y)~dy \right) \left(\displaystyle{\int_{-\infty}^\infty} ~x f(x)~dx\right) =$

$E[y]E[x]$

$E[XY] = E[x]E[y]=0$
This is exactly what I did. I think that the last line should be E[XY] - E[X][Y] = 0 and not E[XY] = E[X][Y] = 0. Am I correct about that?

4. ## Re: Covariance question

Originally Posted by JaguarXJS
This is exactly what I did. I think that the last line should be E[XY] - E[X][Y] = 0 and not E[XY] = E[X][Y] = 0. Am I correct about that?
yes