Using Rotational Symmetry with Standard Normal Variables?

Question

Suppose that $X$ and $Y$ are independent standard normal random variables. Using rotational symmetry, show that $P(Y > \sqrt{3}|X|) = \frac{1}{6}$.

Hint: Recall $\arctan(\sqrt{3}) = \frac{\pi}{3}$.

I recall that rotational symmetry was used in my previous courses but not in relation to independent standard normal variables.

Answer 1

The joint PDF of $X,\,Y$ is $\frac{1}{2\pi}\exp-\frac{x^2+y^2}{2}$. Changing to polar coordinates, the PDF is $\frac{r}{2\pi}\exp-\frac{r^2}{2}$ (the extra factor comes from $dxdy=rdrd\theta$). As this is separable, $\theta$ has PDF $\frac{1}{2\pi}$ on $(-\pi,\,\pi]$. So$$P(\theta\in[0,\,\pi/3])=\int_0^{\pi/3}\frac{1}{2\pi}=\frac16.$$

Answer 2

Can you sketch a graph of the two lines $$y = \sqrt{3} x, \\ y = -\sqrt{3} x,$$ and then shade in the region in the plane that satisfies $$y > \sqrt{3}|x|?$$ What fraction of the entire plane did you shade in? Why does that relate to your question?