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I met an interesting but challenging problem in my homework:

Suppose $n-1$ independent points $x_1$, $x_2$, ..., $x_{n-1}$ are uniformly distributed on unit interval [0,1]. These $n-1$ points seperate the unit interval into $n$ pieces. Suppose the lengths of these $n$ intrvals are $v_1$, $v_2$, ..., $v_n$. What is the probablity that $v_i < a$ for $i = 1,2,...,n$ ? Here $1/n < a < 1$ is a constant.

The joint distribution of $v_1$, $v_2$, ..., $v_{n-1}$ is a uniform distribution on a $n-2$ dimention simplx with $f(v_1, v_2,...,v_{n-1}) = (n-1)!$. The simplex is $$v_i \geq 0, \; i=1,2,...,n-1$$ $$v_1 + v_2 + ... + v_{n-1} \leq 1$$ The proof of this density distribution is not easy.

I followed this idea but I found it's still hard. Maybe there are some other starting points to consider this problem.

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  • $\begingroup$ @ClementC. what do you mean? $\endgroup$ – Stupid_Guy Nov 8 '14 at 22:34
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    $\begingroup$ @Stupid_Guy Show us your thoughts. People is more eager to help if you show that you have invested some effort in the problem. Nobody likes to do some other's homework. $\endgroup$ – leonbloy Nov 8 '14 at 22:38
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    $\begingroup$ That before asking a question, you should have given it some thought and have attempted to solve it by yourself. $\endgroup$ – Clement C. Nov 8 '14 at 22:39
  • $\begingroup$ @ClementC. Sure, I will add it:-) $\endgroup$ – Stupid_Guy Nov 8 '14 at 22:49
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Here is a full analysis:

For every $a\geqslant0$ and every integer $n\geqslant1$, let $p_n(a)$ denote the probability that $n-1$ i.i.d. uniform points separate $[0,1]$ into $n$ pieces of length at most $a$, hence $p_n(a)=0$ for $a\leqslant1/n$ and $p_n(a)=1$ for $a\geqslant1$.

One can deduce $p_{n+1}$ from $p_n$ through a conditioning argument:

Throw $n$ i.i.d. uniform points in $[0,1]$. The position $x$ of the rightmost point has density $nx^{n-1}$ for $x$ and the rightmost interval has length at most $a$ if $x\geqslant1-a$. Conditionally on each $x$ in $[0,1]$, the $n-1$ other points are distributed as $n-1$ i.i.d. uniform points in $[0,x]$ hence they separate $[0,x]$ into $n$ intervals distributed as the $n$ intervals delineated by $n-1$ i.i.d. uniform points in $[0,1]$, scaled by the homothetic factor $x$. Thus, for every $a\geqslant0$,

$$p_{n+1}(a)=\int_{1-a}^1nx^{n-1}p_n(a/x)\mathrm dx.$$

One can tediously recover from this identity and from the initial condition $p_1=\mathbf 1_{[1,+\infty)}$ the easy fact that $$p_2(a)=\left\{\begin{array}{cc}0&a\leqslant\tfrac12\\2a-1&\tfrac12\leqslant a\leqslant1\\1&a\geqslant1\end{array}\right.$$ To go further, note that each function $p_n$ coincides with a specific polynomial on each interval $\left[\frac1k,\frac1{k-1}\right]$ for $k\geqslant1$, with $p_{n}=1$ on $[1,+\infty)$ and $p_{n}=0$ on $\left[0,\frac1{n}\right)$. To wit, the change of variables $$p_n(a)=(n-1)!\,a^{n-1}\,q_n(1/a),$$ yields the initial condition $q_1=\mathbf 1_{[0,1]}$ and, for every $n\geqslant1$ and every real number $t$, $$q_{n+1}(t)=\int_{t-1}^tq_n(x)\mathrm dx.$$ This concludes the proof of our final result:

For every $n\geqslant1$, $p_n(a)=(n-1)!\,a^{n-1}\,q_n(1/a)$, where $q_n$ denotes the $n$th Irwin-Hall density, that is, the density of the sum of $n$ i.i.d. random variables uniformly distributed on $[0,1]$.

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  • $\begingroup$ For some odd reason, given the fact that both $0\le p_k(a)\le1$ and $|a|<1$, your last formula reminds me of $\cos2t=2\cos^2t-1$. $\endgroup$ – Lucian Nov 9 '14 at 0:23

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