# 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.

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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 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 $$P\{\min \approx x, \max \approx y\} \approx f_{X,Y}(x,y)\Delta x\Delta y \approx n(n-1)\cdot (f(x)\Delta x)\cdot(f_Y(y)\Delta y) \cdot [F(y)-F(x)]^{n-2}$$