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.


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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.