This random variable $Z$ seems to have the same distribution as $\min{X,Y}$ as $|X-Y|$, for $X,Y \sim \operatorname{Unif}(0,1)$.

Question

Based on these (but I hope this is self-contained):

Why does $\min(X,Y)$ and $|X-Y|$ have the same distribution when $X,Y\sim U(0,1)$?

Random points on a circle

For $X,Y \sim \operatorname{Unif}(0,1)$, $\min\{X,Y\}$ and $|X-Y|$ have the same distribution, as explained intuitively in the 1st question. Apparently, for this random variable $Z$ described in the 2nd question, $Z$ also has the same distribution as $\min\{X,Y\}$ and $|X-Y|$. $Z$ is described as

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B$. Let $Z$ be the random variable denoting the distance of the point $B$ to its neighbour in the anti-clockwise direction.

Of course the precise answer to this is that they all have pdf $f_Z(z) = 2(1-z)1_{z \in (0,1)}$. But I'm wondering here if $Z$ is actually the absolute difference or minimum or 2 iid uniform(0,1) or if $Z$ is something else altogether.

Answer 1

I believe $Z=\min\{X,Y\}$, where $X$ and $Y$ are described in David K's answer:

Let $X$ be the anticlockwise distance along the circle's circumference from $B$ to $A$. Let $Y$ be the anticlockwise distance from $B$ to $C$.

Since the three points are independently uniformly distributed along the circumference of the circle, $X$ and $Y$ are iid variables with uniform distributions on $[0,1).$