Covariance Formula

Question

Hy everybody,

I have $f_X(x)$=$(3/14)(x^2+2)$

and $f_Y(y)$=$3/28((1/3)+y)$

and the joint $f(xy)$=$(3/56)(x^2+y)$

I have to compute the $Cov(XY)$, I proceed like this: I know they are dependent, then I apply the formula $Cov(XY)$=$E[(X-\mu_X)(Y-\mu_Y)]$

But then I compute $E[X]$ which is = $0$

So then the $Cov(XY)$ turns out to be equal to $Cov(XY)=E[XY]$

I wonder: it is always like this? I mean, if the two variables are dependent then the $Cov(XY)$ has this form, or it is just in my case??

Answer 1

$cov(X,Y)=EXY-(EX)(EY)$ so $cov(X,Y)=EXY$ iff either $EX=0$ or $EY=0$. This has nothing to do with independence. In your case it just happens that $EX=0$.