# Computing the expectation of min and max of two iid gaussian variables

This might seem very easy but i have a question regarding the computation of $$\mathbb{E}(X_{(2)} - X_{(1)})$$ where $$X_1,X_2 \sim \mathcal{N}(\mu,\sigma^2)$$ iid and $$X_{(1)} = \min\{X_1,X_2\}$$, $$X_{(2)} = \max\{X_1,X_2\}$$. I know that the distribution of $$X_{(1)}$$ is $$-2F + F^2$$ respectively the distribution of $$X_{(2)}$$ is $$F^2$$, where $$F$$ is the original distribution function. For the computation of the expectation of the difference i used the formula $$\mathbb{E}(X) = \int \limits_{(0,\infty)} 1-F(t) \hspace{0.1cm} \mathrm{d}t - \int \limits_{(-\infty,0)} F(t) \hspace{0.1cm} \mathrm{d}t$$

and got the integral $$2 \cdot \int \limits_{\mathbb{R}} F(t) (1-F(t)) \hspace{0.1cm} \mathrm{d}t$$

Does anyone have a clue how to proceed with that ?

Just note that $$X_{(2)}-X_{(1)}=|X_1-X_2|$$
Since $$X_1,X_2$$ are independent $$N(\mu,\sigma^2)$$, we must have $$X_1-X_2\sim N(0,2\sigma^2)$$.
Therefore, using the well-known formula for the mean absolute deviation about mean, we have $$\mathbb E\left[X_{(2)}-X_{(1)}\right]=\mathbb E\left[|X_1-X_2|\right]=\sqrt{\frac{2}{\pi}}\sqrt 2\sigma=\frac{2\sigma}{\sqrt \pi}$$