Smaller over larger, for normals

Idea. The standard normal distribution in the plane is rotation invariant, so the angle of $(|Z_2|,|Z_1|)$ is uniform on $[0,\frac{\pi }{2}]$. The ratio of the two values is the tangent of that angle, and the expectation becomes an elementary integral.

Step 1: the ratio. Let $\Theta$ be the angle of the point $(|Z_2|,|Z_1|)$ with the horizontal axis. It is uniform on $[0,\frac{\pi }{2}]$, and $\frac{|Z_1|}{|Z_2|}=\tan \Theta$.

Step 2: the smaller over the larger. $S=\min (\tan \Theta ,\cot \Theta )$. By symmetry about $\frac{\pi }{4}$, $S$ has the distribution of $\tan \Theta$ with $\Theta$ uniform on $[0,\frac{\pi }{4}]$. So $$\begin{array}{rl}\mathbb{E}[S]& =\frac{4}{\pi }{\int }_0^{\pi /4}\tan \theta d\theta \\ & =\frac{4}{\pi }[-\ln \cos \theta {]}_0^{\pi /4}=\frac{4}{\pi }\cdot \frac{\ln 2}{2}.\end{array}$$

Step 3. $$\mathbb{E}[S]=\frac{2\ln 2}{\pi }\approx \mathrm{0.441.}$$

Compare. For two uniforms on $[0,1]$ the same quantity is $\frac{1}{2}$. And the larger over the smaller has infinite expectation here: the ratio of two independent normals is a Cauchy variable, whose tails are too heavy.