Let $\mathcal{P}_n$ be the probability simplex in $\mathbb{R}^n$, that is, the simplex whose vertices are the standard basis vectors. Then $\mathcal{P}_n$ is the set of all $n$-tuples of nonnegative real numbers whose sum equals $1$. If we assume the uniform distribution on $\mathcal{P}_n$, then clearly the expected value of a fixed variable $x_i$ equals $1/n$. But what is the probability distribution function (PDF) of $x_i$?
Intuitively, it seems that the PDF at any value $c \in [0,1]$ should equal the (normalized) volume of the $(n-1)$-dimensional "cross-section" of $\mathcal{P}_n$ determined by setting $x_i = c$, where by "normalized" I mean divided by some suitable factor so that the integral of the PDF is $1$. This cross section, I believe, can itself be viewed as a simplex in $\mathbb{R}^{n-1}$, whose vertices are the standard basis vectors scaled by $1-c$. So then perhaps my question boils down to writing down a formula for the volume of this "cross section" -- perhaps something like $(1-c)^{n-1}$ times the volume of $\mathcal{P}^{n-1}$ (which is surely well known, although not to me), and probably normalized by some suitable factor?
Is there a better approach? I very much suspect there is.
The uniform distribution on $\mathcal{P}_n$ is the Dirichlet distribution with parameters $\alpha_1 = \ldots = \alpha_n = 1$. The distribution of a component of this distribution is $\text{Beta}(1, n-1)$.