Expectation of max of $2$ symmetric random variables

Question

Let :

• $X_1,X_2$ independent, with same law
• $ Var(X_1)= \sigma^2$, $E(X_1)=0$
• $G$ is their cumulative function
• they have a density $g$
• Let $X= \max(X_1,X_2)$ with cumulative function $F$
• We assume that $X_1$ (and $X_2$) have a symmetric law : $\forall z ~~g(z)=g(-z)$

1. We want to show that $E(X^2)= \sigma^2$
2. In the particular case of $X_1$ and $X_2$ follow $N(0, \sigma^2)$ that $EX= \frac{\sigma}{ \sqrt{ \pi}} $

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My attempt :

$ \begin{align*} B &=E(X) \\ &= \int_{- \infty}^{\infty} 2t g(t) G(t) dt \\ \int_{- \infty}^{0} 2t g(t) G(t) dt &= \int_{ \infty}^{0} 2 u g(-u) G(-u) du \\ &= \int_{ \infty}^{0} 2 u g(u) G(-u) du \\ &=\int_{ \infty}^{0} 2 u g(u) (1-G(u) ) du \\ &=\int_{0}^{ \infty} 2 u g(u) (G(u)-1 ) du \\ B &=E(X) \\ &= \int_{0}^{ \infty} 2 u g(u) (G(u)-1 ) du + \int_{0}^{ \infty} 2 u g(u) G(u) du \\ &= \int_{0}^{ \infty} 2u g(u) [ 2 G(u) -1] du \\ \end{align*} $

2. The question is done. Can we use a different method ?

$X=\max(X_1,X_2)= \frac{ X_1 + X_2 + |X_1-X_2|}{2}$, if the law is symmetric

and we take the expectation $ E(X)= \frac{ |X_1-X_2|}{2}= \frac{ E|Z|}{2}$ with $Z\sim N(0, 2 \sigma^2)$ or $E(X)= E(|U|)$ with $U\sim N(0,1) \times \sigma \sqrt{2}$ and $E|U|=\sqrt{ \frac{2}{\pi}}$

Answer 1

Try using $X^2=\frac{X_1^2+X_2^2+(X_1-X_2)^2+2X_1|X_1-X_2|+2X_2|X_1-X_2|+2X_1X_2}{4}$, and using the fact that $\mathbb E[X_1 X_2]=0$, you get that $\mathbb E[X^2] = \sigma^2+\frac{\mathbb E[(X_1+X_2) |X_1-X_2|]}{2}$, by symmetry. Now to show that $\mathbb E[(X_1+X_2) |X_1-X_2|]=0$, observe that $X_1 |X_1-X_2|$ and $-X_1 |X_1-X_2|$ have the same distribution by your assumptions, they also have the same distribution as $X_2|X_1-X_2|$ by i.i.dness, so that $\mathbb E[X_2|X_1-X_2|]=\mathbb E[-X_1|X_1-X_2|]$ and so you get $\mathbb E[X^2]=\sigma^2$.