Joint density of the smallest and largest random variables among finite independent random variables with common density I am trying to show the following result.
Let $X_1, \ldots,X_n$ be independent random variables with the common density $f$ and distribution function $F$. If $X$ is the smallest and $Y$ the largest among them, the joint density of the pair $(X, Y)$ for $y>x$ is given by $$n(n-1)f(x)f(y)[F(y)-F(x)]^{n-2}$$
Some thoughts towards a partial solution
Attempt 1:
Given they all share the same density, the Joint density can be calculated as $$f_{X,Y}( x,y) = f_{Y\mid X}( y\mid x) f( x) = f_{X\mid Y}( x\mid y) f( y)$$
So we can choose $x$ in $n$ ways and fixing $x$, we can pick the maximum random variable as $C_{1}^{n-1} = (n-1)$ so this explains $n(n-1)f(x)f(y)$ part but i am unsure why we have the difference of the distribution functions of the $y$ and $x$ times $(n-2)$. i know we have $n-2$ variables to still account for and they are being integrated out. Hence we should have $(n-2)$ terms but why the difference ?
Attempt 2:
The sample space corresponding to $X_1, \ldots,X_n$ is the $n$-dimensional hypercube $\Gamma $ defined by $x_k=f$ and the probabilities equal the $n$-dimensional volume. The natural sample space with the $X_k$ as coordinate variables is the subset $\Omega$ of $\Gamma$ containing all points such that $x_1\leq \cdots \leq x_n$. The hypercube contains $n!$ congruent replicas of the set $\Omega$ and in each the ordered $n$-tuple $(X_1,\ldots,X_n)$ coincides with a fixed permutation of $X_1,\ldots, X_n$.
I am not sure i am getting anywhere with these thoughts. Any help would be much appreciated. 
 A: If you first find the joint c.d.f. of the pair, then differentiate to get the density, then you're just finding a probability and then differentiating.
$$
\begin{align}
& F_{X,Y}(x,y)=\Pr(X\le x\  \&\  Y\le y) \\[10pt]
& = \Pr(Y\le y) - \Pr(Y\le y\  \&\  X>x) \\[10pt]
& = \Pr(\text{all observations}\le y) - \Pr(y\ge \text{all observations}>x) \\[10pt]
& = (\Pr(X_1\le y))^n - (\Pr(x<X_1\le y))^n \\[10pt]
& = F(y)^n - (F(y)-F(x))^n,\qquad\text{all provided that }x\le y.
\end{align}
$$
Then apply $\dfrac{\partial^2}{\partial x\,\partial y}$.  (The first term vanishes when $\dfrac{\partial}{\partial x}$ is applied.)
A: A heuristic way of getting to the answer is to argue that for jointly continuous random variables, $f_{X,Y}(x,y)\,\Delta x\,\Delta y$ is approximately the probability that $(X,Y)$ lies in a small rectangular region of area 
$\Delta x\,\Delta y$ centered at $(x,y)$.  Thus, for $y > x$, the smallest
of $n$ random variables has value approximately $x$, the largest has
value approximately $y$, and the remaining $n-2$ have values in the
interval $(x,y)$. There are $n$ choices for the smallest, $n-1$ for the
largest, and so the probability is
$$P\{\min \approx x, \max \approx y\}
\approx f_{X,Y}(x,y)\,\Delta x\,\Delta y
\approx n(n-1)\cdot (f(x)\,\Delta x)\cdot(f_Y(y)\,\Delta y) \cdot [F(y)-F(x)]^{n-2}$$
