Let $(X,Y)$ be i.i.d $\mathcal N(0,1)$. I am trying to find $E(\max(|X|,|Y|))$ using a simple argument.
There is no trouble finding the pdf of $|X|$, nor the pdf of $Z=\max(|X|,|Y|)$.
PDF of $U=|X|$ is $f(u)=2\phi(u)=\sqrt{\frac{2}{\pi}}e^{-u^2/2}\mathbf1_{u>0}$. Let $F$ be the CDF of $U$.
As $X$ and $Y$ are i.i.d., $|X|$ and $|Y|$ are also i.i.d (both have the $\chi$ distribution with one degree of freedom). Distribution function of $Z$ is thus $\Pr(Z\le z)=(\Pr(U\le z))^2$.
So PDF of $Z$ is \begin{align}g(z)&=2F(z)f(z)\\&=2\left(\int_0^z\sqrt{\frac{2}{\pi}}e^{-u^2/2}\,\mathrm{d}u\right)\sqrt{\frac{2}{\pi}}e^{-z^2/2}\,\mathbf1_{z>0}\\&=\frac{4}{\pi}e^{-z^2/2}\int_0^ze^{-u^2/2}\,\mathrm{d}u\,\mathbf1_{z>0}\tag{1}\end{align}
The most straightforward way now is to evaluate the integral
$\displaystyle E(Z)=\frac{4}{\pi}\int_0^\infty ze^{-z^2/2}\left(\int_0^ze^{-u^2/2}\,\mathrm{d}u\right)\,\mathrm{d}z$ in this form. But I couldn't do it.
Rather, writing the above PDF in terms of $\Phi$ and $\phi$ I get
$$g(z)=2(2\Phi(z)-1)2\Phi(z)\mathbf1_{z>0}$$
And using well-known relations between $\Phi$ and $\phi$ I was able to get the right answer $E(Z)=\frac{2}{\sqrt \pi}$. But I want to reduce this last bit of computation by rather finding the expectation using a different method.
Can this expected value be found without much calculation?
I had tried setting up the integral $E(Z)=\iint\max(|x|,|y|)\phi(x)\phi(y)\,\mathrm{d}x\,\mathrm{d}y$. I also tried to find it by writing $E(Z)=\frac{1}{2}(E(|X|)+E(|Y|)+E(|\,|X|-|Y|\,|))=\sqrt{\frac{2}{\pi}}+\frac{1}{2}E(|\,|X|-|Y|\,|)$. But could not proceed in either case.
There also seems to be some connection between $E(\max(|X|,|Y|))$ and $E(\max(X,Y))$ as the latter equals $\frac{1}{\sqrt \pi}$, half of the value of the former.
If the integral $\displaystyle \int_0^\infty ze^{-z^2/2}\left(\int_0^ze^{-u^2/2}\,\mathrm{d}u\right)\,\mathrm{d}z$ can be found using some manipulation (and without resorting to $\Phi$ and $\phi$), then that also answers my question.
One way that might be more satisfying is integration by parts: $$\frac{4}{\pi}\int_0^\infty ze^{-z^2/2}\int_0^ze^{-u^2/2}du \;dz \\= -\frac{4}{\pi} e^{-z^2/2}\left.\int_0^z e^{-u^2/2}du\right|_0^\infty +\frac{4}{\pi}\int_0^\infty e^{-z^2/2} e^{-z^2/2}dz\\= 0+\frac{4}{\pi}\int_0^\infty e^{-z^2}dz \\= \frac{4}{\pi}\frac{\sqrt{\pi}}{2} = \frac{2}{\sqrt \pi}$$