Uniform distribution and ordered statistics Let $X_1,....,X_{n-1}$ be $(n-1)$ random variables following a Uniform distribution. If we note $X_{(1)},..,X_{(n-1)}$ the associated ordered statistics, I would like to prove that :
$$U_i=X_{(i+1)}-X_{(i)} $$
follows a uniform distribution. I would like to know if there's a simple demonstration of that fact or if anyone has pointers.
 A: As detailed in (the simulation Bible) Devroye's Non-uniform Variate Generation (1985), Chapter V, Section 2, the vector $(S_1,\ldots,S_{n})$ is jointly uniform:

Theorem 2.1 $\quad$ If $U_{(1)}\le\cdots\le U_{(n)}$ are the Uniform $\mathcal U(0,1)$ order statistics of an $n$-sample, and the$$S_i=U_{(i)}-U_{(i-1)}\qquad(1\le i\le n+1)$$ where by convention $U_{(0)}=0$ and $U_{(n+1)}=1$, are the uniform spacings, then $(S_1,\ldots,S_{n})$ is uniformly distributed over the simplex
  $$\mathcal A_{n-1}=\left\{(x_1,\ldots,x_{n});\ 0\le x_i\,,\ \sum_{i=1}^{n} x_i\le 1\right\}$$

[The proof follows from the order statistics $U_{(1)}\le\cdots\le U_{(n)}$ being distributed as$$n!\,\prod_{i=1}^n \mathbb{I}_{(0,1)}(u_{(i)})\times\mathbb{I}_{u_{(1)}\le\cdots\le u_{(n)}}$$$n!$ being the number of permutations of $\{1,...,n\}$ and from the change of variables from the $U_{(i)}$'s to the $S_i$'s being of Jacobian determinant equal to one]
and

Theorem 2.2 $\quad$ The vector $$(S_1,\ldots,S_{n+1})$$ is distributed as
  $$(\varepsilon_0,\ldots,\varepsilon_n)\Big/\sum_{i=0}^n\varepsilon_i$$where the $\epsilon_i$'s are iid $\mathcal E(1)$.

The above is also the constructive definition of a Dirichlet $$\mathcal D\overbrace{(1,\ldots,1)}^\text{$n+1$ terms}$$ distribution and the consequence is that $S_i$ is distributed as a Beta $\mathcal B(1,n)$ random variable. Not marginaly uniform then (if conditionally so).
