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Let $X_1, \ldots, X_N \sim \operatorname{Unif}[0,1]$ and consider the intervals between successive order statistics: $[0, X_{(1)}], [X_{(1)}, X_{(2)}], \ldots, [X_{(N)}, 1]$.

What is the distribution of this vector of interval lengths?

I hypothesize that it is Dirichlet$(1, \ldots, 1)$, but I cannot prove it and my literature search is taking me to obscure 1800s papers. Any thoughts?

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    $\begingroup$ This follows from the fact that the joint density of the order statistics $X_{(j)}$ is constant and the change of variable from the $x_{(j)}$ to the lengths has constant Jacobian (it's linear), so the probability is obtained by integrating a constant, which is exactly how the Dirichlet distribution with all parameters equal to $1$ works. $\endgroup$
    – Christian Remling
    Commented Feb 28 at 1:02
  • $\begingroup$ Of course, that's assuming that the $X_j$ are independent; you obviously can't say anything in general. $\endgroup$
    – Christian Remling
    Commented Feb 28 at 1:05
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    $\begingroup$ I believe your guess is right, but it's hard to see this as research-level math even if it hasn't been published. Christian Remling's comment is terse but it looks like the right idea. I think he meant the whole vector of all $N$ order statistics rather than for just one value of $j.$ $\endgroup$
    – Michael Hardy
    Commented Feb 28 at 2:30
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    $\begingroup$ Pretty sure it has been published, many times. For example, Springer's Encyclopedia of Mathematics, Dirichlet distribution explicitly says the distribution is Dirichlet. (FWIW Wikipedia also says it, but without an inline reference.) $\endgroup$
    – Jukka Kohonen
    Commented Feb 28 at 18:51
  • $\begingroup$ Did you take a look into \item {\bf Bertoin, J.:} The asymptotic behavior of fragmentation processes, J. European Math. Soc., 5 (2003) 395 - 416 and \item {\bf Bertoin, J.:} Different Aspects of a Model for Random Fragmentation Processes. ccsd-00005175, version 1, 7 Feb 2005, 26 S. $\endgroup$
    – Dieter Kadelka
    Commented Mar 5 at 15:24

2 Answers 2

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This follows from the properties of the Poisson process. Consider a Poisson process (say, of rate 1), and let its first $N+1$ points be $Y_1<Y_2<\dots<Y_{N+1}=:A$. Then $\xi_0:=Y_1$, $\xi_1:=Y_2-Y_1$, $\dots$, $\xi_N:=Y_{N+1}-Y_N$ are all i.i.d. $\exp(1)$ random variables; hence $(\xi_0/A,\xi_1/A,\dots,\xi_N/A)$ has a Dirichlet(1,1,...,1) see Wikipedia. On the other hand, by the properties of the Poisson process, $Y_1,Y_2,\dots,Y_N$, conditionally on $Y_{N+1}=A$, have i.i.d. uniform distribution on $[0,A]$ (see here).

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Suppose $1\le k_1\le k_2\le \cdots \le k_\ell \le N$ are integers. Then the probability distribution of the vector of lengths of $$ [0, X_{(k_1)}], [X_{(k_1}, X_{(k_2)}], \ldots, [X_{(k_\ell)}, 1] $$ is $\operatorname{Dirichlet}(k_1,\,k_2-k_1,\,k_3-k_2,\,\ldots, \, N-k_\ell).$

I suspect this can be proved by paraphrasing Thomas Bayes's famous posthumous paper An Essay Toward Solving a Problem in the Doctrine of Chances, published in 1761. Bayes did the case $\ell=1,$ getting what today we call a Beta distribution. The Beta distribution is a special case of the Dirichlet distribution.

The question supposed that $X_1, \ldots, X_N \sim \text{i.i.d.} \operatorname{Uniform}[0,1].$

Here I will suppose that $X_0, X_1, \ldots, X_N \sim \text{i.i.d.} \operatorname{Uniform}[0,1],$ i.e. I add one additional uniform random variable $X_0.$

For $k=1,\ldots, N$ (not $k=0,\ldots, N$) let $$ Y_k = \begin{cases} 1 & \text{if } X_{(k)}<X_0, \\ 0 & \text{if } X_{(k)}>X_0. \end{cases} \qquad \text{$(X_{(k)},$ not $X_k$)} $$ Then \begin{align} & 1/(N+1) \\[8pt] = {} & \Pr(X_0 = X_{(n+1)}) \text{ by symmetry} \\[8pt] = {} & \Pr(Y_1+\cdots+Y_N=n\mid X_0) \\[8pt] = {} & \binom Nn X_0^n (1-X_0)^{N-n} \\[8pt] \text{and } & \Pr(Y_1+\cdots + Y_N = n) \\[8pt] = {} & \operatorname E(\Pr(Y_1+\cdots + Y_N=n\mid X_0) \\[8pt] = {} & \int_0^1 \binom Nn x^n (1-x)^{N-n} \, dx \\[8pt] \text{Therefore } & \frac1{(N+1)\binom Nn} = \int_0^1 x^n (1-x)^{N-n} \, dx. \end{align} Then we have \begin{align} & \Pr(X_0\le x\mid Y_1+\cdots +Y_N = n) \\[12pt] = {} & \frac{\Pr(Y_1+\cdots+Y_N=n \mid X_0\le x) \Pr(Y_1+\cdots+Y_N=n)}{\Pr(X_0\le x)} \\[12pt] = {} & \frac{\Pr(Y_1+\cdots+Y_N=n \mid X_0\le x)\cdot 1/(N+1)} x \\[8pt] = {} & \frac1{N+1} \int_0^x \binom Nn u^n (1-u)^{N-n} \, \frac{du} x. \\[12pt] & (\text{“}du/x\text{” because } x = \Pr(X_0\le x).) \end{align}

Thus we have a Beta distribution, which is the case of the Dirichlet distribution that concerns us here.

Next: Generalize this to higher-dimensional Dirichlet distributions.

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