Generate two uniform random variables, then take the interval between them

Question

Take two independent standard uniform random variables. Then subtract the smaller from the larger. Call this random variable, $X$. I know the distribution of the smaller one, $U^{(1)}$ is $\beta(2,1)$ and that of the larger ($U^{(2)}$) is $\beta(1,2)$.

And I have a strong suspicion that the distribution of the difference is $\beta(2,2)$. But I can't seem to prove it.

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EDIT: As Henry pointed out in the comments, my intuition was wrong. The distribution of $U^{(2)}-U^{(1)}$ is actually the same as the distribution of $U^{(1)}$. This makes intuitive sense, but would be great to have some rigorous reasoning behind this.

Even though my intuition failed me once, I'm going to exercise it again and say that if you divide the $(0,1)$ interval into $n$ parts by generating $n-1$ uniform random variables, all those parts will have the same distribution. It's easy to show this for the first and last parts. But can't prove that the middle ones will all also have the same distribution.

Answer 1

Consider the following way of generating $n-1$ uniform random variables $U_1, \dots, U_{n-1} \sim U(0,1)$:

1. Generate $n$ variables $C_1, \dots, C_n$ uniformly on a circle with circumference $1$.
2. Choose one of the $\{C_1, \dots, C_n\}$ to be the $0$-point (each $C_i$ has the same probability $1/n$ of being chosen). Then unroll the circle to become the interval $[0, 1]$. The distances on the circle from the $0$-point to the other $C_i$, going clockwise, are the $U_1, \dots, U_{n-1}$.

I've tried to clarify this approach in a diagram.

Suppose that $C_4$ is chosen as the $0$-point. Then $U_1$ would be the clockwise distance on the circle from $C_4 \curvearrowright C_5$, $U_2$ would be the clockwise distance on the circle from $C_4 \curvearrowright C_6$, and so on. In this case, $U^{(1)}$ would be the clockwise distance on the circle from $C_4 \curvearrowright C_6$, $U^{(2)} - U^{(1)}$ would be the clockwise distance on the circle from $C_6 \curvearrowright C_2$, and so on.

So if you divide the $(0, 1)$ interval into $n$ parts with $n-1$ uniforms, the lengths of the parts are just the $n$ distances on the circle between adjacent $C$ points. So the lengths all have the same distribution.