Let $X_1$ and $X_2$ be i.i.d. normal can we find the distribution of $(U_1,U_2)=(\max(X_1,X_2), \min(X_1,X_2))$ Let $X_1$ and $X_2$ be i.i.d. normal. 
The question is: 
Can we find the joint distribution of  a pair 
\begin{align}(U_1,U_2)=(\max(X_1,X_2), \min(X_1,X_2)).
\end{align} 
What I did 
Note that this is just the ordering of $X_1$ and $X_2$. In other words,
If $X_1>X_2$ then $(U_1,U_2)=(X_1,X_2)$ and if $X_2<X_1$ then $(U_1,U_2)=(X_2,X_1)$.
Then for every set $A$ we have that 
\begin{align}
P( (U_1,U_2) \in A)&= P( (U_1,U_2) \in A \mid  X_1>X_2)  P( X_1>X_2)+ P( (U_1,U_2) \in A \mid  X_1\le X_2)  P( X_1 \le X_2)\\
&= \frac{1}{2} P( (U_1,U_2) \in A \mid  X_1>X_2)+  \frac{1}{2} P( (U_1,U_2) \in A \mid  X_1 \le X_2)\\
&=P( (X_1,X_2) \in A \mid  X_1>X_2)+  \frac{1}{2} P( (X_2,X_1) \in A \mid  X_1 \le X_2)
\end{align}
Can somehow finish this?
 A: The density of this distribution is the same as for $X_1$ and $X_2$, just multiplied by $2$ and restricted to $U_1\gt U_2$. That's true for arbitrary i.i.d. variables, not necessarily normally distributed.
A: Suppose $X_{(1)}=\min(X_1,X_2)$ and $X_{(2)}=\max(X_1,X_2)$. Let $f$ and $F$ denote the density and distribution function of the (normal) population.
Then for $x<y$, the distribution function of $(X_{(1)},X_{(2)})$ is 
\begin{align}F(x,y)&=\Pr(X_{(1)}\le x,X_{(2)}\le y)\\&=\Pr(X_{(2)}\le y)-\Pr(X_{(1)}>x,X_{(2)}\le y)\\&=\Pr(X_1,X_2\le y)-\Pr(x<X_1,X_2\le y)\\&=(F(y))^2-\left(F(y)-F(x)\right)^2\end{align}
The second equality follows from the fact that $\Pr(A\cap B^c)=\Pr(A)-\Pr(A\cap B)$.
So the pdf of $(X_{(1)},X_{(2)})$ is \begin{align}f(x,y)&=\frac{\partial^2}{\partial x\partial y}F(x,y)\\&=\frac{\partial}{\partial x}\left(2F(y)f(y)-2(F(y)-F(x))f(y)\right)\mathbf1_{x<y}\\&=2f(x)f(y)\mathbf1_{x<y}\end{align}
This method is easily generalised to find the distribution function and hence the density of $(\min (X_1,\cdots,X_n),\max (X_1,\cdots,X_n))$ for any continuous population when your sample size is $n$, say. Here of course $n=2$.
