Mean of the first order statistics in the case of i.i.d normal random variables Let $(X_1, X_2)$ denote random samples drawn from population distribution $\cal{N} (0, \sigma^2
) $. Find mean of the first order statistics, i.e., $E[X_{(1)}]$.
Now ,going by the standard approach ,the CDF of the minimum is-
$F_{X_{(1)}}(x)=1-(1-F_X(x))^2\implies f_{x_{(1)}}(x)=2(1-F(x))f(x)$
In case of normal distribution-
$f(x)=2\times\frac{1}{2\sigma^2\pi}(\int_x ^\infty e^{\frac{-x'^2}{2\sigma^2}}dx')e^{\frac{-x^2}{2\sigma^2}}$
Thus,$E[X_{(1)}]=\int_{-\infty}^{\infty}xf(x)dx$
I can't think of a way to evaluate the above integral .Is there a way ?
 A: 1st Solution. Write $f(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-x^2/2\sigma^2}$ for the p.d.f. of $\mathcal{N}(0,\sigma^2)$. Then
\begin{align*}
\mathbf{E}[X_{(1)}]
&= \mathbf{E}[\min\{X_1, X_2\}] \\
&= \iint_{\mathbb{R}^2} \min\{x_1,x_2\} f(x_1)f(x_2) \, \mathrm{d}x_1 \mathrm{d}x_2.
\end{align*}
Using the symmetry, we may write the last integral as
\begin{align*}
2 \iint_{x_1 < x_2} x_1 f(x_1)f(x_2) \, \mathrm{d}x_1 \mathrm{d}x_2
&= \frac{1}{\pi \sigma^2} \int_{-\infty}^{\infty} \int_{-\infty}^{x_2} x_1 e^{-x_1^2/2\sigma^2} e^{-x_2^2/2\sigma^2} \, \mathrm{d}x_1 \mathrm{d}x_2.
\end{align*}
Now evaluating the inner integral, the right-hand side becomes
\begin{align*}
\frac{1}{\pi} \int_{-\infty}^{\infty} \left[- e^{-x_1^2/2\sigma^2} \right]_{x_1=-\infty}^{x_1=x_2} e^{-x_2^2/2\sigma^2} \, \mathrm{d}x_2
&= - \frac{1}{\pi} \int_{-\infty}^{\infty} e^{-x_2^2/\sigma^2} \, \mathrm{d}x_2.
\end{align*}
Now computing the last integral using the gaussian integral shows that
\begin{align*}
\mathbf{E}[X_{(1)}] = - \frac{\sigma}{\sqrt{\pi}}.
\end{align*}
2nd Solution. Note that
$$ \min\{a,b\} = \frac{a+b}{2} - \frac{|a-b|}{2} $$
holds for any $a, b \in \mathbb{R}$. So
$$ \mathbf{E}[X_{(1)}]
= \mathbf{E}[\min\{X_1, X_2\}]
= \mathbf{E}\left[ \frac{X_1+X_2}{2} - \frac{|X_1 - X_2|}{2}\right]. $$
Since $\mathbf{E}[X_1] = \mathbf{E}[X_2] = 0$ and $Y := \frac{1}{2}(X_1 - X_2) \sim \mathcal{N}(0, \sigma^2/2)$, we have
$$ \mathbf{E}[X_{(1)}]
= -\mathbf{E}[|Y|]
= -2 \int_{0}^{\infty} y \cdot \frac{1}{\sqrt{\pi}\sigma} e^{-y^2/\sigma^2} \, \mathrm{d}y
= \left[ \frac{\sigma}{\sqrt{\pi}} e^{-y^2/\sigma^2} \right]_{y=0}^{y=\infty}
= - \frac{\sigma}{\sqrt{\pi}}. $$
