probability - Density of order statistics

I need help with order statistics:

Given a sample $X_1, \ldots, X_n$, $X_i \sim U_{0,1}$, i.e. the $X_i$ are uniformly distributed on $[0,1]$, determine the following for the corresponding order statistics:

a) the density of $X_{(k)}$

b) the joint density of $X_{(1)}, X_{(n)}$

c) the density of the range $R:=X_{(n)} - X_{(1)}$

d) the limit distribution for $2n(1-R)$ with $n \rightarrow \infty$.

Here is my idea for the first one:

a) For the density of an order statistic we've shown:

$$f_{X_{(k)}}(t) = \binom{n}{k} k F_X(t)^{k-1}(1-F_X(t))^{n-k}f_X(t)$$
Given the fact that $X_i \sim U_{0,1}$, the density is pretty easy to determine, i.e.

$$f_{X_{(k)}}(t) = \binom{n}{k} k t^{k-1}(1-t)^{n-k} \mathbb{1}_{[0,1]}$$

b) For b), I think I can use the following formula:

$$f_{(i),(j)} = \dfrac{n!f(x_i)f(x_j)(F(X_i))^{i-1}(F(x_j)-F(x_i))^{j-1-i}(1-F(x_j))^{n-j}}{(i-1)!(j-1-i)!(n-j)!}$$ to get

$$f_{(1),(n)} (x_1,x_n) = (n-1)n(x_n-x_1)^{n-2}$$
Is that correct?

c) My idea was to use the transformation rule for densities, so

$$\begin{pmatrix} x \\ y \end{pmatrix} = \phi ( z,u) = \begin{pmatrix} z-u \\ u \end{pmatrix}$$

$$\begin{pmatrix} z \\ u \end{pmatrix} = \phi^{-1}(x,y) = \begin{pmatrix} x+u \\ y \end{pmatrix}$$

with $J_{\phi^{-1}}(x,y) = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1$

Then $f_R = f_{(n)}(\phi^{-1}(x,y))f_{(1)}(\phi^{-1}(x,y))\cdot1 = \cdots$ - how do I proceed now?

d) Here I don't know how to start...

Thank you for the help!

Answer

For b) as @air commented, independance is false. I would compute the joint CDF $F_{X_{(1)},X_{(n)}} (x,y) := \Bbb P [ X_{(1)} \leq x \cap X_{(n)} \leq y]$, and use the following lemma :

If $F_{X,Y}$ is twice continuously differentiable,then $(X,Y)$ has density $f_{X,Y} = \frac {\partial ^2f}{\partial x \partial y}F_{X,Y}$. (The result holds with weaker conditions, using the more general form of the differenciation under integral theorem)

The rest follows, your idea of using change of variables is good.