What is the expected value of $\min\{|X|,|Y|\}/\max\{|X|,|Y|\}$ assuming $X$ and $Y$ are independent? So I need to compute $$E\left[\frac{\min\{|X|,|Y|\}}{\max\{|X|,|Y|\}}\right]$$ given $X,Y \sim$ Normal$(0,1)$ and independent. 
What I am having trouble seeing is 


*

*whether $\min\{|X|,|Y|\}$ and $\max\{|X|,|Y|\}$ are independent. If so I can factor the expectation into $$E\left[\min\{|X|,|Y|\}\right]\cdot E\left[\frac{1}{\max\{|X|,|Y|\}}\right]$$

*how I should find the expected value of $1/\max\{|X|,|Y|\}$.

 A: Here is a (small) hint:  for any two nonzero real numbers $a, b$, we must have $$\frac{\min\{a,b\}}{\max\{a,b\}} = \begin{cases}\frac{a}{b}, & a < b, \\ 1, & a = b, \\ \frac{b}{a}, & a > b. \end{cases}$$

Addendum.  Since an answer was provided, I suppose it is only natural to furnish an outline of a proof.  In light of the above hint, let $$g(x,y) = \frac{\min\{x,y\}}{\max\{x,y\}}.$$  Note that $$F_{|X|}(x) = \Pr[-x \le X \le x] = F_X(x) - F_X(-x),$$ consequently $$f_{|X|}(x) = f_X(x) + f_X(-x) = 2f_X(x) = \sqrt{\frac{2}{\pi}} e^{-x^2/2}, \quad x \ge 0,$$ and similarly for $|Y|$.  We then have $$\begin{align*} \operatorname{E}[g(|X|,|Y|)] &= \int_{x=0}^\infty \int_{y=0}^\infty g(x,y) f_{|X|,|Y|}(x,y) \, dy \, dx \\ &= \frac{2}{\pi} \Biggl( \int_{x=0}^\infty \int_{y=0}^x \frac{y}{x} e^{-x^2/2} e^{-y^2/2} \, dy \, dx + \int_{y=0}^\infty \int_{x=0}^y \frac{x}{y} e^{-x^2/2} e^{-y^2/2} \, dx \, dy \Biggr) \\ &= \frac{4}{\pi} \int_{x=0}^\infty \frac{e^{-x^2/2}}{x} \int_{y=0}^x ye^{-y^2/2} \, dy \, dx \\ &= \frac{4}{\pi} \int_{x=0}^\infty \frac{e^{-x^2/2}(1-e^{-x^2/2})}{x} \, dx. \end{align*}$$  This much is quite straightforward.  The evaluation of this integral may be accomplished by introducing a parameter $a$ as follows:  define $$h(a,x) = \frac{e^{-ax^2/2}(1-e^{-x^2/2})}{x},$$ so that $$\frac{\partial h}{\partial a} = -\frac{x}{2} e^{-ax^2/2}(1-e^{-x^2/2}).$$  Then $$\int_{x=0}^\infty \frac{\partial h}{\partial a} \, dx = -\frac{1}{2a(a+1)}, \quad a > 0,$$ hence $$H(a) = \int_{x=0}^\infty h(a,x) \, dx = -\int \frac{da}{2a(a+1)} = \frac{1}{2} \log(1 + a^{-1}) + C,$$ where we require $H(\infty) = 0$, thus $C = 0$.  Therefore, the expectation is $$\operatorname{E}[g(|X|,|Y|)] = \frac{4}{\pi} H(1) = \frac{\log 4}{\pi},$$ as claimed.
A: Approach 1:  Joint pdf
If $X$ and $Y$ are standard Normal and independent, then the joint pdf of $(X,Y)$, say $f(x,y)$ is:

(source: tri.org.au) 
Then, the desired expectation is:

(source: tri.org.au) 

Approach 2: Order statistics
Let $X$ be half-Normal with pdf $f(x)$:

(source: tri.org.au) 
Then, the joint pdf of the $1^{st}$ and $2^{nd}$ order statistics, $(X_1,X_2)$, in a sample of size 2, denoted, say $g(x_1, x_2)$, is:

(source: tri.org.au) 
Then, the desired expectation is:   

(source: tri.org.au) 

Notes


*

*The Expect and OrderStat function used above to help automate are from the mathStatica package for Mathematica. 

*As disclosure, I should add that I am one of the authors.
