# (yet another) length of random segment

Take a random point $Z$, i.d. in [0,1], which defines a stick.

Break the stick in two, (random i.d.).

Take the left part of the broken stick and break it again in two (i.d.)

You thus obtain three sticks.

What is the expected length of the stick where $Z$ is located?

(Prove its $5/9$. Can you come up with an intuitive reason for that?)

Generalise for n breaks of the stick in the left (i.e. always breaking the segment near the origin).

(Edit: solution was found analytically and tested on MC simulation. would be interesting to see if there is an intuitive reason though.)

// (Hope you enjoyed this series of puzzles - this was the last one!)

• And still absolutely no beginning of indication of hint about what you tried or know or think whatsoever. Sweet. – Did Jan 10 '13 at 0:52
• @did :I did actually!!! I took on board your previous comments: I mentioned that I resolved the problem, and presented you with the solution: 5/9. Further, I have confirmed it using a MC simulation! Having taken your previous comments on board.. I dont think I deserve your negative vote!.... – alexandreC Jan 10 '13 at 0:55
• @did : further.. I edited the question to include further details... (analytical + MC simulation)... :o) – alexandreC Jan 10 '13 at 1:02
• If you have a proof, why not presenting it here? (And you know that your mention of simulations is posterior to my comment.) But, as they say, something is definitely better than nothing... so, what is your proof for 5/9? – Did Jan 10 '13 at 6:15
• @did : yes, I did say it was 5/9 originally, and subsequently edited it to mention I cross checked it with MC simulation. To answer your question: I obtained 5/9 using a double integral. Could you please remove your down vote? :) Many thanks! – alexandreC Jan 10 '13 at 18:11

First, if there's only one break, we can treat $Z$ and the breakpoint as two independent uniformly distributed points which divide the stick into three parts with expected length $1/3$ each; the part in which $Z$ is located consists of two of these three parts and thus has expected length $2/3$.
We can write this as two contributions of $1/3$ each from the two cases where $Z$ is in the left part and in the right part. The contribution of $1/3$ from the right part remains the same, since the right part is not subdivided, whereas if $Z$ is in the left part then by the same reasoning the expected length of the part containing it after the second break is $2/3$ of the length of the left part, so the contribution of $1/3$ for the left part gets multiplied by $2/3$, for a total of $\frac13+\frac13\cdot\frac23=\frac59$.
Let $Z$ represent the random variable uniformly distributed on the unit interval, $Z \sim \mathcal{U}_{(0,1)}$. Let $X_2 = U$ be the location of the first break, $U\sim \mathcal{U}_{(0,1)}$. The second break occurs at $X_1 = U \cdot V$, where $V \sim \mathcal{U}_{(0,1)}$. Variable $U$, $V$ and $Z$ are independent. The length of interest is $$\ell(\omega) = \begin{cases} X_1(\omega) & Z(\omega) < X_1(\omega) \\ X_2(\omega)-X_1(\omega) & X_1(\omega) < Z(\omega) < X_2(\omega) \\ 1-X_2(\omega) & Z(\omega) > X_2(\omega) \end{cases}$$ You now have to compute $$\begin{eqnarray} \mathbb{E}\left(\ell\right) &=& \int_0^1 \int_0^1 \int_0^1 \mathrm{d} z \, \mathrm{d} u \,\mathrm{d} v \left\{ u v [ z<u v] + u (1-v) [ u v < z < u] + (1-u) [ z > u] \right\} \\ &\stackrel{\text{integrate over z}}{=}& \int_0^1 \int_0^1 \mathrm{d} u \,\mathrm{d} v \left\{ u v (u v) + u (1-v) \left( u - u v \right) + (1-u) \left(1-u \right) \right\} \end{eqnarray}$$ Splitting the latter double integral into three pieces and changing varibales $v \to 1-v$ in the second and $u\to 1-u$ in the last we get: $$\mathbb{E}\left(\ell\right) = 2 \left( \int_0^1 u^2 \mathrm{d} u \right)^2 + \int_0^1 u^2 \mathrm{d} u = \frac{1}{3} \left( 2 \cdot \frac{1}{3} + 1 \right) = \frac{5}{9}$$