TITLE: Expected time to convergence
QUESTION [24 upvotes]: Consider the following process: we place $n$ points labelled $1...n$ uniformly at random on the interval $[0,1]$. At each time step, two points $i, j$ are selected uniformly at random and $i$ updates its position to be a point chosen uniformly at random in the interval between the positions of $i$ and $j$ (so the interval $[p(i),p(j)]$ if $p(i) < p(j)$ or $[p(j),p(i)]$ otherwise, where $p(x)$ denotes the position of the point labelled $x$). 

What is the expected time until all points are within distance $\varepsilon$ of each other for some fixed $\varepsilon > 0$? 
What is the expected time until all points are either to the left or right of $\frac{1}{2}$?

Asymptotic bounds are also very interesting to me.

REPLY [2 votes]: Exact solution for $n=2$.  Consider the process when the points are at
$(a, b)$ with $0 < a < b < 1$ and a target of $\epsilon$.  It is equivalent to a process
with points $(0, 1)$ and a target of $\frac{\epsilon}{a-b}$.  With this observation, let
$f(t)$ be the expected number of steps when points are at $(0, 1)$ and the target is $\epsilon$.
The next step is terminates the process  with probability $\epsilon$.  If it doesn't terminate, the
remaining interval has a width $U(\epsilon, 1)$.
So $f$ must satisfy the following recurrence
$$f(t) = 1 + \int_t^1 f(t/x) \mathrm{d}x$$
which is satisfied by $f(t) = 1 - \ln(t).$  The initial points are both $U(0, 1)$,
so the expected number of steps with $n=2$ is given by
$$\int_0^1 (2- 2w)\left(1 - \ln\left(\frac{\epsilon}{d}\right)\right) \mathrm{d}w \\
= \frac{\epsilon^2-1}{2} + \log{\left(\frac{1}{\epsilon} \right)}$$
Which matches experiments.

For larger $n$ this gives the "trivial" lower bound of $O(-n \log{\epsilon})$
since for larger $n$, the smalest and largest points are close to $(0, 1)$ and they are
moves on average every $2/n$ steps.