10 iid uniform random variables on (0,1) cut it to 11 intervals
find the expectation of the longest interval
answer involves harmonic series
can anybody show how to solve this question?
This thread looked like fading into obscurity so I'll give it a bump to keep it on the radar.
For n i.i.d. uniform random variables on (0, 1) cut into n+1 intervals, the expected value for the rth shortest interval is equal to
$\displaystyle \frac{1}{n} \, \sum_{i=1}^{r} \frac{1}{n-i+1}$.
When r = n you get the expected value for the longest interval: $\displaystyle \frac{1}{n} \, \sum_{i=1}^{n} \frac{1}{n-i+1}$.
It's easy to derive this result for n = 2 and n = 3 (see, for example, Frederick Mosteller, Fifty Challenging Problems in Probability With Solutions, problem 43 (The Broken Bar). Fifty challenging problems in ... - Google Book Search)
This question is an example of the Broken-Stick Distribution, so called because it arises from a stick of unit length along which a number n of events are scattered with a uniform probability density. The events break the stick into n+1 intervals, which can be ordered in increasing size.
The Broken-Stick Distribution was popularised by Robert Macarthur (On the Relative Abundance of Bird Species, Proc. Nat. Acad. Sci. 43: 293-295 (1957). See ON THE RELATIVE ABUNDANCE OF BIRD SPECIES) who in turn cites D. E. Barton and F. N. David (Some Notes on Ordered Random Intervals, J. Royal Statistical Soc. Series B 18: 79-94 (1956). See Cookie Absent which only gives the first page, unfortunately).
Barton and David in turn cite a number of references, including:
Whitworth (1878): Choice and Chance (see A History of Probability and ... - Google Book Search ) (Amazon.com: Choice And Chance (1870): William Allen Whitworth: Books)
N. Y. Irwin (1955): A Unified Derivation of Some Well Known Frequency Distributions of Interest in Biometry and Statistics, J. Royal Statistical Soc. Series B (General) 118 (No. 4): 389-404. (See Cookie Absent which only gives the first page, unfortunately).