Independent Standard Normal Gaussian Random Variables I am having trouble making sense of this. I know what independence of random variables means. Suppose $X$ and $Y$ are independent, standard normal (Gaussian) random variables. Then, it is supposed to be that $X^2 + Y^2$ and $\frac X Y$ are also independent random variables. 
I just cannot intuitively make sense of this. Where does that fact that they are standard normal Gaussian come into play? Also, if we know that $\frac X Y \leq k$ for some fixed $k$, then it seems obvious that it should affect the probability distribution of $X^2 +Y^2$. 
 A: Consider the random point $(X,Y)$ in $\mathbb{R}^2$. The ratio $X/Y$ tells us what angle the segment from $(0,0)$ to $(X,Y)$ makes with the $x$-axis, while $X^2+Y^2$ tells us how far $(X,Y)$ is from $(0,0)$. 
The distribution of $(X,Y)$ is symmetric under rotations, so the distribution of the angle $\Theta$ is uniform and independent of the radius $R=\sqrt{X^2+Y^2}$. 
This intuitive explanation can be made more rigorous by converting to polar coordinates.  

The joint density of two independent standard normals $(X,Y)$ is 
$$f(x,y)={1\over 2\pi} \exp(-(x^2+y^2)/2).$$ Converting to polar coordinates 
we get the  joint density of $(R,\Theta)$ as
$$g(r,\theta)={r\over 2\pi}\exp(-r^2/2)={1\over 2\pi}\cdot r\exp(-r^2/2).$$
This product form of $g(r,\theta)$ shows that $\Theta$  and $R$ are independent.   
A: Additional remark to Byron's answer: It can actually be shown that an (isotropic) Normal distribution is the only rotationally symmetric distribution for which $X$ and $Y$ as well as $X/Y$ and $X^2 + Y^2$ are independent. 
See e.g. 

Ali, M. M. (1980). Characterization of the Normal Distribution Among
  the Continuous Symmetric Spherical Class. Journal of the Royal
  Statistical Society B, 42(2), 162-164.

