This question asks:

Start with a stick of length $1$. Repeatedly remove some fraction $U$ of the remaining stick, where $U$ is uniform in $[0,1]$. What is the probability that at least one of the removed pieces has length at least $\frac12$?

Eventually I managed to solve it, but the result was not very enlightening. Thus I present here a generalisation of the question where $\frac12$ is replaced with an arbitrary $0\le x\le1$. Let $f(x)$ be the probability that we eventually break off a stick of length at least $x$; it has been shown that for $\frac12\le x\le1$, $f(x)=-\ln x$.

What, then, is $f(x)$ for $0\le x\le\frac12$?

This generalisation is more difficult because now we can cut off more than one stick longer than $x$. Let's use joriki's approach to the source question: when $x\le\frac12$ either we cut off a piece longer than $x$ at the start, with probability $1-x$, or we leave a stick of length $t$ and replace $x$ with $x/t$: $$f(x)=1-x+\int_{1-x}^1f(x/t)\,dt$$ This is the same formula as in joriki's answer except that the lower bound $x$ has become $1-x$. We can make the same manipulations that follow there to get $$f(x)=1-x+x\int_x^{x/(1-x)}\frac{f(u)}{u^2}\,du$$ $$1-f(x)=x\left(1-\int_x^{x/(1-x)}\frac{f(u)}{u^2}\,du\right)$$ and eventually the functional differential equation $$x(1-x)^2f''(x)=f'\left(\frac x{1-x}\right)-f'(x)(1-x)^2$$ but how would I proceed from here?

I've determined the following approximate values for $f(x)$ with one million trials for each $x$.

.49 .712676
.48 .732251
.47 .751341
.46 .769896
.45 .787595
.44 .805675
.43 .822355
.42 .838964
.41 .854620
.40 .869991
.39 .883803
.38 .897808
.37 .910511
.36 .923427
.35 .934614
.34 .945107
.33 .954406
.32 .963026
.31 .969796
.30 .976313
.29 .981963
.28 .986094
.27 .989759
.26 .992816
.25 .995101
.24 .996743
.23 .997929
.22 .998779
.21 .999327
.20 .999609
.19 .999815
.18 .999924
.17 .999958
.16 .999990
.15 .999995
<=.14 1.000000
  • $\begingroup$ Also happy National Day to all Singaporeans. $\endgroup$ – Parcly Taxel Aug 8 '19 at 19:11

Not an answer, but a comment (not allowed to post a comment yet) :

In the setting of random permutations on $[n]$, the law of the largest cycle is known to converge (when renormalized by $n$) to Dickman law.

I presume it will be the same here, since the same stick breaking construction applies.

The distribution of the largest fragment has distribution function $F(u)= \rho (1/u)$ where $\rho$ is the Dickman function :

$$\rho(u) =1 + \sum_{k \ge 1} \frac{(-1)^k}{k!} \int \ldots \int_{I_k(u)} \frac{dy_1 ... dy_k}{y_1 ... y_k} $$

with $$ I_k(u)=\{uy_1 >1, ..., uy_k>1, y_1+...+y_k<1 \} $$

See Arratia Barbour Tavare on page 14, formula (1.35) (Chapter 1, Permutations and primes) for the derivation.

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  • $\begingroup$ Assuming $F(u)$ has the correct distribution, it would be easy to find $f(x)$. However, the formula is extremely complicated. It's impractical. $\endgroup$ – Parcly Taxel Aug 8 '19 at 19:55
  • $\begingroup$ Well it's just like that... You may also be interested in checking this short note on cycle length in random permutation inference.org.uk/itila/cycles.pdf $\endgroup$ – MO1-85-54. Aug 8 '19 at 20:15
  • $\begingroup$ The linked note has nothing to do with this question. $\endgroup$ – Parcly Taxel Aug 8 '19 at 20:24
  • $\begingroup$ Thanks for consideration. The log(2) that appears on Ex 5 is the same than for the original question; the link with stick breaking is done in Arratia Barbour Tavare though. $\endgroup$ – MO1-85-54. Aug 8 '19 at 20:30
  • $\begingroup$ It also says that the length of a cycle containing a given integer has a uniform distribution... (hence the stick breaking construction at the limit). $\endgroup$ – MO1-85-54. Aug 8 '19 at 21:05

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