TITLE: Probability distribution of the subinterval lengths from a random interval division
QUESTION [5 upvotes]: 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?

REPLY [5 votes]: 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 $f(t)$]