Probability distribution of the subinterval lengths from a random interval division

Question

For $a \in \mathbb{R}_{+}$ and $n \in \mathbb{N}_{+}$ draw $n-1$ points $X_1, \ldots, X_{n-1}$ independently, uniformly at random from the interval $I = [0, a]$. These points partition $I$ into $n$ disjoint subintervals $I_1 \, \dot\cup \ldots \dot\cup \, I_n = I$. Let $Y_i := |I_i| \in [0, a]$ denote the interval length of the $i$'th subinterval.

How are the $Y_i$'s distributed? I am especially interested in the expectation and the variance.

Here are my thoughts about this: I suppose that all $Y_i$'s are identically distributed, i.e., $Y_i \sim Y$ for some random variable $Y$. Further, I suppose that the expectation of $Y$ is $\mathbb{E}(Y) = \frac{a}{n}$. However, I have no clue about the variance, and I can neither prove my conjectures, nor find an answer on the internet.

Can you help me on this?

Answer 1

That $\forall i, Y_i \sim Y$ is evident from the symmetry observed if you consider the interval wrapped around as a circle, where $n$ random points are chosen to break the circumference into $n$ random arcs. The length of any arc is obviously identically distributed.

To find the mean, observe that $\sum {Y_i} = a$
Take expectations on both sides, and use $\mathbb{E}(Y_i) = \mathbb{E}(Y)$ to prove your conjecture.

To find the distribution of $Y$ and other properties such as variance, perhaps the easiest way is to focus on $Y_n$. Let $f(t)$ and $F(t)$ denote the distribution function of $Y$ and its cumulative. Then we have,

$\mathbb{P}(Y_n \ge t) = \mathbb{P} (\max X_i \lt a-t)$
$\quad \quad \quad \quad = \mathbb{P}(\forall i, X_i < a-t)$
$\quad \quad \quad \quad = \prod_{i} \dfrac{a-t}{a}$
$\quad \quad \quad \quad = \left( \dfrac{a-t}{a} \right)^{n-1}$

$F(t) = \mathbb{P}(Y_n \lt t) = 1 - \mathbb{P}(Y_n \ge t)$
$\quad= 1 - \left( \dfrac{a-t}{a} \right)^{n-1}$

Differentiating, $f(t) = \dfrac{n-1}{a} \left( \dfrac{a-t}{a} \right)^{n-2}$

Hope that helps! [and thanks for the correction in $f(t)$]

Answer 2

Assume without loss of generality that $a=1$ and define $I_i=[Z_{i-1},Z_i]$ hence $Z_0=0$, $Z_n=1$ and $(Z_i)_{1\leqslant i\leqslant n-1}$ is the ordered sample $(X_i)_{1\leqslant i\leqslant n-1}$.

For $x\lt y$ in $(0,1)$, $[Z_{i-1}\leqslant x,y\leqslant Z_i]$ means that $(x,y)\subseteq I_i$ which means that $i-1$ points from the sample $(X_i)_{1\leqslant i\leqslant n-1}$ are in $(0,x)$ and the $n-i$ others are in $(y,1)$. This happens with probability $$ \mathbb P(Z_{i-1}\leqslant x,y\leqslant Z_i)=\binom{n-1}{i-1}x^{i-1}(1-y)^{n-i}, $$ hence $\mathbb P(Z_{i-1}\in\mathrm dx,Z_i\in\mathrm dy)=f_i(x,y)\mathrm dx\mathrm dy$, with $$ f_i(x,y)=\binom{n-1}{i-1}(i-1)(n-i)x^{i-2}(1-y)^{n-i-1}\mathbf 1_{0\lt x\lt y\lt1}. $$ Projecting this on $Y_i=Z_i-Z_{i-1}$ yields $\mathbb P(Y_{i}\in\mathrm dz)=g_i(z)\mathrm dz$, with $$ g_i(z)=\int_0^{1-z}f_i(x,x+z)\mathrm dx. $$ The change of variables $x=(1-z)t$ and the binomial identity $$ \int_0^1t^{i-2}(1-t)^{n-i-1}\mathrm dt=\mathrm{B}(i-1,n-i)=\frac{(i-2)!(n-i-1)!}{(n-2)!}, $$ yield $$ g_i(z)=(n-1)(1-z)^{n-2}\mathbf 1_{0\lt z\lt 1}. $$ It follows that the density of $1-Y_i$ is $(n-1)z^{n-2}\mathbf 1_{0\lt z\lt 1}$, hence, for every $a\gt1-n$, $$ \mathbb E((1-Y_i)^a)=\frac{n-1}{a+n-1}. $$ In particular, $$ \mathbb E(Y_i)=1-\mathbb E(1-Y_i)=\frac1n, $$ and $$ \mathbb E(Y_i^2)=\mathbb E((1-Y_i)^2)-2\mathbb E(1-Y_i)+1=\frac2{n(n+1)}, $$ hence $$ \mathrm{var}(Y_i)=\frac{n-1}{n^2(n+1)}. $$

Answer 3

Here is a partial answer that should get you started.

Let $m=n-1$. The values $X_i$ are distinct almost surely. Then, sort the resulting sequence and denote by $Z_i$ the $i$-th element in the sorted sequence. Thus, $Z_1$ is the smallest among the $X_i$ and $Z_m$ is the largest.

We have $Y_k=Z_{k}-Z_{k-1}$ for all $k$ (where for convenience, we put $Z_0=0$ and $Z_{m+1}=b$).

Let us compute the distribution of the $Z_k$. We have

$$ P(Z_k \leq t)=\int_{0}^{t} I_1(x_k)I_2(x_k)dx_k, \tag{1} $$

where

$$ I_1(x_k)=\int_{0}^{x_k}\int_{0}^{x_{k-1}} \ldots \int_{0}^{x_2}dx_1dx_2 \ldots dx_{k-1}=\frac{x_k^{k-1}}{(k-1)!} \tag{2} $$

and

$$ I_2(x_k)=\int_{x_k}^{a}\int_{x_{k+1}}^{a} \ldots \int_{x_{n-1}}^{a}dx_ndx_{n-1} \ldots dx_{k+1}= \frac{(a-x_k)^{n-k}}{(n-k)!} \tag{3} $$

We deduce that $Z_k$ has the the following density :

$$ d_k(t)=\frac{t^{k-1}}{(k-1)!} \times \frac{(a-t)^{n-k}}{(n-k)!} \tag{4} $$

So

$$ E(Z_k)=\int_{0}^a \frac{t^{k}}{(k-1)!} \times \frac{(a-t)^{n-k}}{(n-k)!} dt=\frac{k}{m+1}a \tag{5} $$

and

$$ E(Z_k^2)=\int_{0}^a \frac{t^{k+1}}{(k-1)!} \times \frac{(a-t)^{n-k}}{(n-k)!} dt=\frac{6k}{(m+1)(m+2)}a^2 \tag{6} $$

So $E(Y_k)=E(Z_k)-E(Z_{k-1})=\frac{a}{m+1}=\frac{a}{n}$ as you rightly conjectured.