Angle between $(X,Y)$ and $(E(X), E(Y)) $ where X and Y are independent random variables. Suppose that $X$ and $Y$ are two independent random variables with known (different/same) probability distribution functions. Now consider the vector $(X,Y)$, I want to find the angle between $(X,Y)$ and $(E(X), E(Y))$. I am guessing that again the angle should be a random variable, but I don't know how to find its PDF. In addition, suppose that $E(X)$ and $E(Y)$ are not zero at the same time. Any help would be appreciated, Thanks.
 A: I am not sure why you are interested in the angle it makes with $(E(X),E(Y))$, because you can just find out the angle it makes with the $x$-axis and subtract the angle $(E(X),E(Y))$ makes from $x$-axis from this.
More formally:
Let $t \in [0,2\pi)$ be the angle $(E(X),E(Y))$ makes with the $x$-axis. (it equals to $\arctan(E(Y)/E(X))+2\pi k$ for some $k$ depending on which quadrant you are in).
Let $\Theta$ denote the angle $(X,Y)$ makes with the $x$-axis and let $f$ and $g$ be the density of $X$ and $Y$ respectively, then
Making a polar coordinate transformation:
Let $x=r\sin \theta $ and $y=r\cos \theta$, with $r\in[0,\infty),\theta\in[0,2\pi)$, then the Jacobian of the transformation is $r$.
$$P(\theta_1<\Theta<\theta_2)=\int^{\theta_2}_{\theta_1}\int^\infty_0f(r\sin\theta)g(r\cos\theta)r\text{d}r\text{d}\theta$$
so the density of $\Theta$ is given by 
$$h(\theta)=\int^\infty_0f(r\sin\theta)g(r\cos\theta)r\text{d}r$$.
Now the angle you want is just $u=t-\theta$, so 
$$\tilde{h}(u) = \int^\infty_0f(r\sin(t-u))g(r\cos(t-u))\text{d}r$$
Added: in fact you could have done this even if $X$ and $Y$ are not independent, just replace $f(x)g(y)$ with $f(x,y)$ where $f$ is the joint distribution.
