# Splitting intervals with cut-off

I would like determine the cumulative distribution function (cdf) of the following random-variable X:

Suppose we have the following process: The unit interval is split into two pieces at a point $u$, chosen uniformly at random in $[0,1]$. Now we obtain two intervals of length $r$ and $1-r$.

Any of the two intervals that has length $\ell > x_c>0$ will again be split into two by a uniform random number $u'$ in $[0,1]$, i.e. into $\ell u'$ and $\ell(1-u')$. We iterate this process until there is no fragment with length larger than $x_c$ remaining.

Now if we denote the the set of final interval lengths by $L$. Then we define $X$ to be the length of a randomly chosen interval length in $L$.

My numerical observation indicates that $X$ is uniform in $[0,x_c]$, i.e.

$$F(x) \equiv \mathbb{P}[X\leq x] = \frac{x}{x_c} \mathbb{I}{\{x\leq x_c\}}$$

Unfortunately I can't prove it.

In a next step I would be interested to loose the condition of $u$ being uniform. And then also generalize to splittings into $m>2$ pieces.

You are correct (assuming $0 \lt x_c \lt 1$): the length of each final piece is uniformly distributed on $[0,x_c]$ .
Any final piece has been created by splitting a piece longer than $x_c$ into two pieces, each with a uniformly distributed length. But to be a final piece it must be shorter than $x_c$. So its distribution conditioned on being less than $x_c$ of a stick's length is uniform in this range.
• the distribution conditioned on being less than $x_c$ of a stick's length may depend on the length of its parent
• Thanks. I understand that when the picked fragment has only been fragmented once we obtain the following contribution to $F(x)$ from one-step fragments: $\mathbb{P}[u\leq x\vert u\leq x_c] = \mathbb{P}[u\leq x, u\leq x_c] / \mathbb{P}[u\leq x_c]=\mathbb{P}[u\leq x]/x_c = x/x_c$. But what about fragments obtained from multiple splitting steps? – antarcticfox Sep 26 '14 at 9:42