Order Statistic Expectation / Probability Let $Y_1<Y_2$ be order statistics from a random sample of size $2$ from a normal distribution, $\mathcal{N}(\mu,\sigma^2)$, where $\sigma^2$ is known. Show that $P(Y_1<\mu<Y_2)=\frac12$ and find $E(Y_1-Y_2)$. 
I am not exactly sure how to solve the question above. Any help would be appreciated. Thanks.

The hypothesis should be $Y_1 < Y_2$.
Now the mutual density is given by : $f_{Y_1,Y_2}(y_1,y_2) = 2 f_{X}(y_1) f_{X}(y_2)$ for any $y_1,y_2$ such that $y_1 < y_2$, and $f_{Y_1,Y_2}(y_1,y_2) = 0$ otherwise. $f_X$ is the density of $\mathcal{N}(\mu,\sigma)$.
So : $P(Y_1 < \mu <  Y_2) = \int_{D} f_{Y_1,Y_2}(y_1,y_2)dy_1dy_2$ on the domain $D=\left\{y_1 < \mu < y_2 \right\}$. If I take a pen and paper, draw this domain, I find : $f_{Y_1,Y_2}(y_1,y_2) = 1 - \int_{-\infty}^{\mu}\int_\mu^{+\infty}2 f_{X}(y_1) f_{X}(y_2)dy_1dy_2 = 1-(2F_{X}(\mu))(1-F_{X}(\mu))=\frac12$ since $F_{X}(\mu) = F_{X}(\mu)=\frac12$.
 E(Y_1-Y_2) = E(Y_1) - E(Y_2) = 0 
edit Oups, I wrote this too fast! $Y_1$ and $Y_2$ don't have the same expectation.
