It can be shown that for an iid sample from a Uniform(0, 1) distribution, \begin{equation} n(1-U_{(n)}) \rightarrow exp(1) \\ n(U_{(1)}) \rightarrow exp(1) \end{equation} To see this just try finding the distribution function of the left hand side and then take the limit to infinity.
Now, it turns out we can show that they are actually asymptotically independent and it's possible to derive the joint asymptotic distribution of them. My question is how can we do this. I found a book that states this but omits the proof. I'm hoping it wasn't omitted because of it's simplicity because I can't see the answer.
Edit: Here I attempt to be fancy with characteristic functions and use a trick similar to something I've seen in a book. However, something is still wrong in this calculation but I can't figure it out. Let \begin{equation} \Phi(t_1,t_2) = \mathbb{E}\exp{(it_1 n^2 U_{(1)} + i t_2 n^2 (1-U_{(n)})}) \end{equation} Then notice the characteristic function we desire will be $\Phi(\frac{t_1}{n},\frac{t_2}{n})$. Now we expand this about 0 by differentiating under the expectation (Taylor expansion). Noting that the Uniform order statistics are actually beta distributed we can get their expectations from the formula for the beta mean. We get \begin{equation} \Phi(\frac{t_1}{n},\frac{t_2}{n}) = 1 + (i \frac{n^2}{n+1} , i \frac{n^2}{n+1}) (\frac{t_1}{n}, \frac{t_2}{n})^T + o(\|(\frac{t_1}{n}, \frac{t_2}{n})\|) \end{equation} The problem is that this goes to $1 + it_1 + it_2$ which is wrong because it should factor into the characteristic functions of two independant exponentials. Can someone figure out where I went wrong in the calculation above?