Find $E(\max(X,Y))$ for $X$, $Y$ independent standard normal Let $X,Y$ independent random variables with $X,Y\sim \mathcal{N}(0,1)$. Let $Z=\max(X,Y)$.
I already showed that $F_Z$ of $Z$ suffices $F_Z(z)=F(z)^2$.
Now I need to find $EZ$.
Should I start like this ?
$$EZ=\int_{-\infty}^{\infty}\int_{-\infty}^\infty \max(x,y)\frac{1}{\sqrt{2\pi}}e^{-1/2x^2}\frac{1}{\sqrt{2\pi}}e^{-1/2y^2} dxdy$$
 A: To go back to $Z$ as a function of $(X,Y)$ once one has determined $F_Z$ is counterproductive. Rather, one could compute the density $f_Z$ as the derivative of $F_Z=\Phi^2$, that is, $f_Z=2\varphi\Phi$ where $\Phi$ is the standard normal CDF and its derivative $\varphi$ is the standard normal PDF, and use
$$
\mathbb E(Z)=\int zf_Z(z)\mathrm dz=\int 2z\varphi(z)\Phi(z)\mathrm dz.
$$
Since $z\varphi(z)=-\varphi'(z)$ and $\Phi'=\varphi$, an integration by parts yields
$$
\mathbb E(Z)=\int2\varphi\cdot\varphi=\frac2{\sqrt{2\pi}}\int\varphi(\sqrt2z)\mathrm dz=\frac2{\sqrt{2\pi}}\frac1{\sqrt2}=\frac1{\sqrt{\pi}}.
$$
A: While the OP asks about the case of $X$~N(0,1) and $Y$~N(0,1) independent, ... the more general problem that nests same is of course $(X,Y)$ ~ standardBivariateNormal, with joint pdf $f(x,y)$:

The general solution to $E[\text{max}(X,Y)]$ ... obtained here using the mathStatica / Mathematica combo ... is simply: 

For the special case of independence, $\rho$ = 0, and the solution simplifies to $\frac{1}{\sqrt{\pi }}$.
