Is the set of differences of independent uniform random variables, independent?

Question

Consider a finite set of uniformly distributed, independent random variables $\mathbf{X} = \{X_1, X_2, \dots, X_n\}$ on the unit interval.

The absolute values of the difference between these variables form another set of random variables, $\mathbf{Y} = \{Y_{ij}, i \ne j, i,j < n\}$, where $ Y_{ij} = |X_i - X_j|$.

If each differences $Y_{ij}$ were drawn independently, the corresponding probability density function would be $f_y(y) = 2(1-y)$ for each $Y_{ij}$.

However, if I generate a realization of $\mathbf{X}$ and use it to compute $\mathbf{\hat{Y}}$, does $\hat{Y}_{ij} \in \mathbf{\hat{Y}}$ still satisfy $\hat{Y}_{ij} \sim f_y(y)$ independently?

Answer 1

No. For example, there's no reason for $|X_1 - X_2|$ and $|X_1 - X_3|$ to be independent. Indeed, you can compute $$ \text{cov}(|X_1 - X_2|, |X_1 - X_3|) = \frac{1}{180} $$

EDIT: You can get this by computing the expected values as triple integrals. It's somewhat tedious to do by hand as you'll probably want to split the domain into regions depending on the order of $x_1, x_2, x_3$. I used Maple's Statistics package:

with(Statistics):
for i from 1 to 3 do X[i]:= RandomVariable(Uniform(0,1)) od:
Covariance(abs(X[1]-X[2]),abs(X[1]-X[3]));