Distribution of differences between adjacent sorted uniform random variables on $[0,1]$ I saw this question on Mathematica.stackexchange, and I wonder what distribution the answer gives.
Asymmetric definition
Let $(X_1,X_2,\ldots,X_{n-1})\sim$ i.i.d. $U[0,1]$, and $(Y_1,Y_2,\ldots,Y_{n-1})$ be those values sorted in ascending order.
What is the distribution of $(Y_1,Y_2-Y_1,\ldots,Y_{n-1}-Y_{n-2},1-Y_{n-1})$?
This seems to depend on the symmetric version below. So, from the observation of the symmetry, all the components from the second to the second last will have the same distribution, with expected value $\frac{1}{n-1}$, and by symmetry the first and last has the same distribution, with expected value $\frac{1}{2(n-1)}$. I doubt this was what the original poster of the linked question was after.
Symmetric definition
Let $(X_1,X_2,\ldots,X_n)\sim$ i.i.d. $U[0,1]$, and $(Y_1,Y_2,\ldots,Y_n)$ be those values sorted in ascending order.
What is the distribution of $(1+Y_1-Y_n,Y_2-Y_1,\ldots,Y_n-Y_{n-1})$?
At least we know the expected value of each component by symmetry: $\frac{1}{n}$.
 A: I am afraid the conclusion is wrong in both cases...

(Asymmetric definition) What is the distribution of $(Y_1,Y_2-Y_1,\ldots,Y_{n-1}-Y_{n-2},1-Y_{n-1})$?

Let $Z^{(n)}=(Y_1,Y_2-Y_1,\ldots,Y_{n-1}-Y_{n-2},1-Y_{n-1})$, then $Z^{(n)}$ is almost surely in $D_n=H_n\cap(\mathbb R_+)^n$ where $H_n$ is the affine hyperplane of $\mathbb R^n$ of equation $z_1+\cdots+z_n=1$, and uniformly distributed on $D_n$. In particular, $E(Z^{(n)}_k)=1/n$ for every $1\leqslant k\leqslant n$, hence $E(Y_k)=k/n$ for every $1\leqslant k\leqslant n-1$.

(Symmetric definition) What is the distribution of $(1+Y_1-Y_n,Y_2-Y_1,\ldots,Y_{n-1}-Y_{n-2},Y_n-Y_{n-1})$?

Let $T^{(n)}=(1+Y_1-Y_n,Y_2-Y_1,\ldots,Y_{n-1}-Y_{n-2},Y_n-Y_{n-1})$, then $T^{(n)}=G(Z^{(n+1)})$ where $G:\mathbb R^{n+1}\to\mathbb R^n$ is the affine function defined by $G(z_1,\ldots,z_{n+1})=(z_1+z_{n+1},z_2,\ldots,z_n)$. This yields the distribution of $T^{(n)}$, whose support is $D_n$, and whose density is proportional to $(t_k)_{1\leqslant k\leqslant n}\mapsto t_1$ on $D_n$. In particular $E(T^{(n)}_1)=2/(n+1)$ and  $E(T^{(n)}_k)=1/(n+1)$ for every $2\leqslant k\leqslant n$.
