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Take three random points, iid in [0,1].

The first two points devide the interval in three.

What is the expected length of the segment where the third point is lying?

Generalise for n.

alexandreC
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  • "The first two points" means the two smaller values? Or just two values $x_1 x_2$ from the set $x_1 x_2 x_3$? – leonbloy Jan 09 '13 at 19:04
  • Related: http://math.stackexchange.com/questions/272927/expected-length-of-arc-in-a-randomly-divided-circle – leonbloy Jan 09 '13 at 19:04
  • @leonboy: means the two values x1 and x2 from the set. no sorting is involved. – alexandreC Jan 09 '13 at 19:06
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    Could you start showing what you tried to solve the questions you post on the site as if you were dumping undigested problems here with no indication that you have put any thought into them? – Did Jan 09 '13 at 19:07
  • @did: ok, thank you for suggestion :) – alexandreC Jan 09 '13 at 19:09
  • @alexandreC : Then this is practically equivalent to the previous related question... – leonbloy Jan 09 '13 at 19:18
  • @leonbloy: yes, I think it is equivalent indeed, but I was not entirely sure - thats why I posted it. – alexandreC Jan 09 '13 at 19:20

2 Answers2

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Join $0$ and $1$ to make a circle with circumference $1$; place a black dot at the join. Then place the next $n$ points independently at random, marking the first $n-1$ with blue dots and the last with a red dot. Rotate the circle by a random amount. You now have $n+1$ dots at independent random positions, dividing the circle into $n+1$ arcs, and you want to know the expected combined length of the arcs flanking the red dot. The arcs are all equivalent by symmetry, so each has expected length $1/(n+1)$, and the answer is $2/(n+1)$.

mjqxxxx
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  • nice. Can you cross check your result with a double integral for n = 2?... – alexandreC Jan 09 '13 at 19:00
  • Could you explain transition from expected length of $1/(n+1)$ to the answer $2/(n+1)$ please? – Tautvis Feb 22 '18 at 17:04
  • @Tautvis: there are two arcs adjacent to the red dot; each has expected length $1/(n+1)$, so their union has expected length $2/(n+1)$. – mjqxxxx Feb 25 '18 at 06:31
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A full description of the distribution (i.e., not just the expected value) is of interest in classic distributed hash table systems (e.g., BitTorrent).

Cecile Penland "The distribution of Random Segment Lengths" has the answer:

Segmenting a line of length L to N segments yields segment length distribution $P_{L,N}(d)=(N-1)\frac{(L-d)^{N-2}}{L^{N-1}}$

Interestingly, taking N and L to infinite yields:

$P_{L,N}(d) \rightarrow \frac{N}{L} exp ( - \frac{N}{L}d)$. I.e., the exponential distribution with the natural expected value.

Ran Wolff
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