Expectation of the Minimum of 2 random variables Let Z = min{|X|,|Y|}, with $X , Y \sim Normal(0,1)$. 
Then, show that $E(Z) = \frac{2(\sqrt{2} - 1)}{\sqrt{\pi}}$ 
So far I got
$F_{Z}(z) = 1-[1 - F_{|X|}(z)][1-F_{|Y|}(z)]$
$F_{Z}(z) = 1-4F_{X}(-z)F_{Y}(-z)$
$F_{Z}(z) = 1-\frac{4}{2\pi}\int_{-\infty}^{-z}\int_{-\infty}^{-z}e^{-\frac{x^{2}+y^{2}}{2}}dxdy$
But I do not know where to go from here.
 A: We know that $F_{|X|}(x)=Pr(-x\leq{X}\leq{x})=F_{X}(x)-F_{X}(-x)$
Hence, $f_{|X|}(x)=f_{X}(x)+f_{X}(-x)=2f_{X}(x)$
Therefore, 
$f_{|X|}(x)=\sqrt{\frac{2}{\pi}}e^{\frac{-x^{2}}{2}}$ and similarly,
$f_{|Y|}(y)=\sqrt{\frac{2}{\pi}}e^{\frac{-y^{2}}{2}}$
Then, by symmetry of Normal curve, we can simplify to:  
$E[min(|X|,|Y|)] = \frac{2}{\pi}(\int_{x=0}^{\infty}\int_{y=0}^{x}ye^{\frac{-x^{2}}{2}}e^{\frac{-y^{2}}{2}}dydx + \int_{y=0}^{\infty}\int_{x=0}^{y}xe^{\frac{-x^{2}}{2}}e^{\frac{-y^{2}}{2}}dxdy)$
$E[min(|X|,|Y|)] = \frac{4}{\pi}\int_{x=0}^{\infty}\int_{y=0}^{x}ye^{\frac{-x^{2}}{2}}e^{\frac{-y^{2}}{2}}dydx$
$E[min(|X|,|Y|)] = \frac{4}{\pi}\int_{x=0}^{\infty}e^{\frac{-x^{2}}{2}}(1-e^{\frac{-x^{2}}{2}})dx$
$E[min(|X|,|Y|)] = \frac{4}{\pi}\int_{x=0}^{\infty}(e^{\frac{-x^{2}}{2}}-e^{-x^{2}})dx$
$E[min(|X|,|Y|)] = \frac{4}{\pi}(\frac{1}{2})\int_{0}^{\infty}(\sqrt{2\pi}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}-\sqrt{\pi}\frac{1}{\sqrt{\pi}}e^{-x^{2}})dx$
Since, we have 2 pdf's: 
$Normal(0,1) : \frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}$ 
$Normal(0,\frac{1}{2}) : \frac{1}{\sqrt{\pi}}e^{-x^{2}}$ 
Thus,
$E[min(|X|,|Y|)] = \frac{2}{\pi}(\sqrt{2\pi}-\sqrt{\pi})$
$E[min(|X|,|Y|)] = \frac{2({\sqrt{2}-1})}{\sqrt{\pi}}$
Hence, proved.
A: I have not checked your solution.  Your initial approach is not wrong either.  I have completed your first approach and came to the same result.  Attached you will find
