TITLE: Finding the correlation coefficient of ordered statistics
QUESTION [5 upvotes]: I am working on the following problem.

Let $$X_{(1)}, \ldots ,X_{(n)}$$ be the order statistics from the uniform distribution of $[0,1]$.  Find the coefficient correlation of $X_{(1)}$ and $X_{(n)}$.

So, here are the things that I understand.
1), I know how to find the pdf for the kth order statistics and it is $nf(x)_{n-1}C_{k-1}[F(x)]^{k-1}[1-F(x)]^{n-k}$
2), As a matter of fact, $X_{(1)}$ and $X_{(n)}$ are simply the case where we have the minimum and the maximum of $X_{i}$s.
3), What we are solving for is $$\frac{Cov[X_{(1)},X_{(n)}]}{\sigma_{X1}\sigma_{Xn}}$$
and I am familiar with how to find each numbers.
So here is my question.
I encountered this problem as one of the practice actuarial exam.  Is this a typical question?  It seems unusually time consuming.
Trying to answer this question I looked at the solution and it somehow used the relationship $E[X_{(1)}]=1-E[X_{(n)}]$.  It did seem like this saves a lot of time, but I did not see how this is true and what the motivation was.
Can someone help me out?

REPLY [3 votes]: Are you allowed to use a computer during the exam? If so, it only takes about 20 seconds to solve. In particular, you are given random variable $X \sim Uniform(0,1)$ with pdf $f(x)$:

(source: tri.org.au)
Then, the joint pdf of the sample minimum $X_1$ and the sample maximum $X_n$, say $g(x_1,x_n)$, is:

(source: tri.org.au)
where I am using the OrderStat function in the mathStatica add-on to Mathematica to automate the nitty-gritties.
Then, $Corr(X_1, X_n)$ is simply:

(source: tri.org.au)
All done.
Notes

As disclosure, I should perhaps add  that I am one of the authors of the software used above.