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 ?
Best Answer
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 $$\mathbb E\left[X_{(2)}-X_{(1)}\right]=\mathbb E\left[|X_1-X_2|\right]=\sqrt{\frac{2}{\pi}}\sqrt 2\sigma=\frac{2\sigma}{\sqrt \pi}$$