I am trying to solve the following problem:
Two points are chosen uniformly and independently on the perimeter of a circle of radius 1. This divides the perimeter into two pieces. Determine the expected value of the length of the shorter piece.
I specifically wish to solve this problem using the theory of order statistics. I have tried to apply what I have learned in the chapter that I am reading on order statistics, and not only would I (of course) want to figure out how to solve the problem, but I am also curious to learn why my current method is not working.
So, I let $X_1$ denote the first point, and $X_2$ denote the second. These are (correct me if I'm wrong) uniformly distributed on $(0,2\pi)$ so that $F(x) = x/(2\pi)$ and $f(x) = 1/(2\pi)$. I let $X_{(1)}$ denote the smaller of the two.
I have the formula $$f_{X_{(1)}}(x) = n(1-F(x))^{n-1}f(x), $$ so in my case, that would be $$f_{X_{(1)}}(x) = 2\left(1-\frac{x}{2\pi}\right)\frac{1}{2\pi} = \frac{1}{\pi}-\frac{x}{2\pi^2}. $$
Then, I tried to compute the expected value as $$\mathrm{E}X_{(1)} = \int_0^{\pi} x\left(\frac{1}{\pi}-\frac{x}{2\pi^2}\right)dx = \frac{\pi}{3}. $$ I've chosen the bounds of the integral in this way because I'm thinking that $X_{(1)}$ must be somewhere between $0$ and $\pi$, since if it were larger than $\pi$, then it wouldn't be the smallest of the two pieces of the perimeter anymore.
The correct answer however is $\pi/2$.
Your order statistic, $X_{(1)}$, is not the length of the shorter piece. It's the smallest angle (measured from $0$).
The length of the shorter arc between the points is: $L=\begin{cases}\lvert X_1-X_2\rvert&:&\lvert X_1-X_2\rvert<\pi\\2\pi-\lvert X_1-X_2\rvert&:& \pi\leq \lvert X_1-X_2\rvert\end{cases}$
You can try to find the expectation by integration, but think about it a bit more.
What is the distribution of the angle between the points?