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Let $X, Y \sim \mathcal{N}(0,1)$ be two independent normal distributions. And let $U=\max\{|X|,|Y|\}$ and $V = \min\{|X|,|Y|\}$. Find the mean of $V/U$.

I did not find this question online so far, so apologies if it was already asked.

I found the cdfs of $U$ and $V$ to be $$F_U(u) = (1+F_X(u)-F_X(-u))(1+F_Y(u)-F_Y(-u))\\F_V(v) = 1- (F_X(v)-F_X(-v))(F_Y(v)-F_Y(-v))$$ Where $F_X, F_Y$ are the cdfs of $X, Y$. We can then differentiate and find the pdfs of the two rvs. However, I do not know how to find the joint pdf $f_{U,V}$.

Alternatively, I thought of applying the Jacobian formula, however I couldn't find a bijection from $(X,Y)$ to $(U,V)$ (I don't think there is one).

I guess what I need help with is finding the joint pdf of $U$ and $V$ and, maybe, a way to find the mean without having to calculate the joint density function.

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The graph of the restriction of $V/U$ to the region $x\in[0,\infty)\ ,\ y\in[-x,x]$ is a rotation of the 3 other similar regions in $\mathbb{R}^3$ about the z-axis. Hence (and because of positivity of the integrand below) we can split the integral for the mean into 4 equal parts

$$\begin{aligned} E(V/U)&=\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{e^{-(x^2+y^2)/2}}{2 \pi }\frac{\min(|x|,|y|)}{\max(|x|,|y|)}dydx\\&= 4\int_0^\infty\int_{-x}^x\frac{e^{-(x^2+y^2)/2}}{2 \pi }\frac{|y|}{x}dydx \\&=8\int_0^\infty\int_0^x\frac{e^{-(x^2+y^2)/2}}{2 \pi }\frac{y}{x}dydx \\&=8\int_0^\infty \left.\frac{e^{-(x^2+y^2)/2}}{-2\pi x}\right|_{y=0}^{y=x} dx \\&=8\int_0^\infty \left(\frac{e^{-(x^2+x^2)/2}}{-2\pi x}-\frac{e^{-(x^2)/2}}{-2\pi x}\right) dx \\&=8\int_0^\infty \frac{e^{-x^2/2}-e^{-x^2}}{2 \pi x} dx=\log (4)/\pi \end{aligned}$$

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  • $\begingroup$ Can you please explain how you got the end result from the last integral? I've been trying to work it out, however I seem to be missing something $\endgroup$ – Andrei Crisan Jun 14 '18 at 12:45

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