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?


I'm going to assume this is self-study, and make some suggestions rather than giving an explicit answer. The proof is in two steps:

  1. Find the joint distribution function of $U_{(1)}, U_{(n)}$;
  2. Show that in the limit as $n \to \infty$ the joint CDF factors.

There is an alternative: you could find the joint characteristic function (or moment generating function, if you prefer) and again show that the limiting characteristic function factors.

  • $\begingroup$ +1. The joint distribution is an easy consequence of the results at Joint PDF of functions of order statistics. The convergence is pretty fast, too, as attested by examining the bivariate and marginal distributions in this R simulation for $n=30$: n <- 30; sim <- c(0,n) + c(n,-n) * apply(matrix(runif(n*1e4), n), 2, range). $\endgroup$
    – whuber
    Aug 9 '14 at 20:32
  • $\begingroup$ When I take the joint distribution (CDF) of them to infinity I'm getting zero. I probably calculated it wrong? $\endgroup$
    – Stacy
    Aug 9 '14 at 21:04
  • $\begingroup$ The question asks for the joint distribution of $nU_{[1]}$ and $n(1-U_{[n]})$. The support of this distribution is $\{(x,y)|0\le x\le n-y, 0 \le y \le n\}$. The limiting support is $[0,\infty)\times[0,\infty)$. There is nothing degenerate about the limiting distribution. $\endgroup$
    – whuber
    Aug 10 '14 at 0:26
  • $\begingroup$ I tried being fancy with characteristic functions (see edit above). Something is wrong in the calculation it should factor :( $\endgroup$
    – Stacy
    Aug 10 '14 at 0:57

Assuming $U_1,U_2,\ldots,U_n$ are i.i.d uniform on $(0,1)$.

For any (real) $u,v>0$, we have $u+v<n$ for $n$ sufficiently large, so that

\begin{align} P\left[nU_{(1)}>u\,,\,n(1-U_{(n)})>v\right]&=P\left[\frac un<U_{(1)},U_{(n)}<1-\frac vn\right] \\&=P\left[\frac un<U_1,\ldots,U_n< 1-\frac vn\right] \\&=\left\{P\left[\frac un<U_1<1-\frac vn\right]\right\}^n \\&=\left(1-\frac {u+v}n\right)^n \\&\stackrel{n\to\infty} \longrightarrow e^{-u}\cdot e^{-v} \end{align}

This shows $nU_{(1)}$ and $n(1-U_{(n)})$ are asymptotically independent, each with limiting distribution a standard exponential.


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