Determine density of $\min(X,Y)$ and $\max(X,Y)$ for independently uniform distributed variables Two independent random variables, $X$ and $Y$, are uniformly distributed on the unit interval $(-1,1)$.
Determine the density for $U=\min(X,Y)$ and for $W=\max(X,Y)$
 A: You will need to either 
1) look at the bivariate distribution of $X$ and $Y$ in order to figure out what region of the pdf for $(X,Y)$ corresponds to $U$ and $W$, or 
2) alternatively, make an algebraic argument in terms of the cdf - e.g.
$P(W\leq w)=P(X\leq w , Y\leq w)$ ...
That hint you mentioned doesn't help you any if you don't know what you're supposed to do with the standard uniforms. 
A: I think the hint given for this problem is not very helpful. Even if the joint distribution of the minimum and maximum of two independent $U(0,1)$ random variables has been solved as an example in class or in the textbook, teaching a student to rely on plugging-and-chugging from formulas instead of thinking about
the problem is very bad pedagogical practice, and even more so in this particular
case because the general result is not too difficult to derive.
If $Z = \min(X,Y)$ and $W = \max(X,Y)$, then for $w > z$,
$$\begin{align*}
F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\}\\
&= P\left[\{X \leq z, Y \leq w\} \cup \{X \leq w, Y \leq z\}\right]\\
&= P\{X \leq z, Y \leq w\} + P\{X \leq w, Y \leq z\} - P\{X \leq z, Y \leq z\}\\
&= F_{X,Y}(z, w) + F_{X, Y}(w,z) - F_{X,Y}(z,z)
\end{align*}
$$ 
while for $w < z$,
$$\begin{align*}
F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\} = P\{Z \leq w, W \leq w\}\\
&= P\{X \leq w, Y \leq w\}\\
&= F_{X,Y}(w,w).
\end{align*}
$$ 
Consequently, if $X$ and $Y$ are jointly continuous
random variables, then 
$$f_{Z,W}(z,w) = \frac{\partial^2}{\partial z \partial w}F_{Z,W}(z,w) = 
\begin{cases}
f_{X,Y}(z,w) + f_{X,Y}(w,z), & \text{if}~w > z,\\
\\
0, & \text{if}~w < z.
\end{cases}
$$
One can even think of this end result geometrically. Consider the
joint density $f_{X,Y}(x,y)$ as a solid (of volume $1$)
sitting on the $x$-$y$ plane. Slice it with a vertical cut
along the line $x=y$ and flip over the part below the line $x=y$
so that it sits on top of the part above the line $x=y$.
The resulting solid is the joint density of the minimum
and the maximum.
For example, if the solid is a rectangular parallelepiped
whose base is the square with
vertices $(1,1), (-1,1), (-1,-1), (1,-1)$, the slicing
and flipping over gives a right triangular prism of twice the height
as the parallelepiped whose base has vertices  $(1,1), (-1,1), (-1,-1)$.

If only the marginal densities are desired and not the
joint density, the solution is even easier for the case
of iid $U(-1,1)$ random variables.  For $-1 \leq z \leq 1$,
$$\begin{align}
1-F_Z(z) = P\{Z > z\} &= P\{\min(X,Y) >z\}\\
&= P\{X >z, Y > z\} = P\{X>z\}P\{Y>z\} = \left(\frac{1}{2}(1-z)\right)^2
\end{align}$$
giving, upon taking the derivative with respect to $z$ that
$$f_Z(z) = \begin{cases}\frac{1-z}{2}, &-1 \leq z \leq 1,\\0, &\text{otherwise.}
\end{cases}$$
Similarly, for $-1 \leq z \leq 1$,
$$\begin{align}
F_W(z) = P\{W \leq w\} &= P\{\max(X,Y) \leq w\}\\
&= P\{X \leq w, Y \leq w\} = P\{X\leq w\}P\{Y\leq w\} 
= \left(\frac{1}{2}(w-(-1))\right)^2
\end{align}$$
giving, upon taking the derivative with respect to $w$ that
$$f_W(w) = \begin{cases}\frac{1+w}{2}, &-1 \leq w \leq 1,\\0, &\text{otherwise.}
\end{cases}$$
