Intuitive reason order statistics of the uniform distribution are stationary.

Question

Not sure if stationary is the right word. What I mean is that the distribution of $U_{(k)}-U_{(j)}$ is the same as that of $U_{(k-i)}-U_{(j-i)}$ and this continues to hold when $i=k$ and $U_{(0)}$ is interpreted as $0$. This can be seen mechanically, but any intuitive reason why when $n$ uniform numbers are drawn, the different in two order statistics depends only on the difference in their ranks?

Answer 1

Suppose we have a circle of length 1. Consider the following two processes:

Process 1: Choose a random point $U_0$ on the circle, and color it red. Choose $n$ more points on the circle. Cut the circle at the red point, and unfold it to make an interval, call the resulting points $U_1$ through $U_n$.

This process is equivalent to just choosing $n$ points from the interval.

Process 2: Choose $n+1$ random points on the circle. Once the points are chosen, color one of them red, and cut at that point.

Process 2 is equivalent to process 1, so also equivalent to your original problem.

In process 2, the initial choice of points divides the circle into $n+1$ intervals. Those intervals will be $U_0, U_1-U_0, \dots, U_n-U_{n-1}$. Each interval is equally likely to be each of those $n+1$ choices, since one choice corresponds to each choice of the red point. So all of those $n+1$ choices have the same distribution. The same argument works for when $k-j>1$.