Independent Standard Normal Gaussian Random Variables

Question

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 $X^2 +Y^2$.

Answer 1

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.

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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.

Answer 2

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.