Mean of the first order statistics in the case of i.i.d normal random variables

Question

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 ?

Answer 1

1^st 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*}

2^nd 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

$$ \mathbf{E}[X_{(1)}] = -\mathbf{E}[|Y|] = -2 \int_{0}^{\infty} y \cdot \frac{1}{\sqrt{\pi}\sigma} e^{-y^2/\sigma^2} \, \mathrm{d}y = \left[ \frac{\sigma}{\sqrt{\pi}} e^{-y^2/\sigma^2} \right]_{y=0}^{y=\infty} = - \frac{\sigma}{\sqrt{\pi}}. $$