Distributions Determined by Cutting a Simplex with Hyperplanes

Abstract

Suppose that $X_1, X_2, \ldots, X_n$ are random variables uniformly distributed over the simplex of points $x_1, x_2, \cdots, x_n$ such that $x_i \geqq 0$ for $i = 1, 2, \cdots, n$ and $\sum^n_1 x_i \leqq 1$. The distribution of $X = \sum^n_1 c_iX_i$ for constants $c_i$ satisfying $c_1 > c_2 > \cdots > c_n > 0$ is easily seen to be given by \begin{equation*}\begin{align*}\tag{1.1}P(X \leqq x) &= 1 - \sum^r_{j = 1} (c_j - x)^n\lbrack c_j\prod_{i \neq j} (c_j - c_i)\rbrack^{-1} \\ &= x^n\lbrack\prod^n_{i = 1} c_i\rbrack^{-1} - \sum^n_{j = r + 1} (x - c_j)^n\lbrack c_j\prod_{i \neq j} (c_i - c_j)\rbrack^ {-1}\end{align*}\end{equation*} for $0 \leqq x \leqq c_1$, where $r$ is the largest positive integer such that $x \leqq c_r$. A geometric derivation is given in Section 2 which relies on a principle of inclusion and exclusion and identifies the terms in (1.1) as volumes of various simplices. It is remarked in Section 3 that the argument of Section 2 extends in principle to give the joint distribution of a set of different linear combinations of $X_1, X_2, \cdots, X_n$. Finally, the relations to the theory of order statistics from a uniform distribution and to the theory of serial correlation are noted. A simple derivation of (1.1) will now be sketched. Suppose that $Z_1, Z_2, \cdots, Z_{n + 1}$ are $n + 1$ independent random variables each with the standard exponential density function $\exp (-z)$ for $0 \leqq z < \infty$. By a well-known argument involving the partial fraction expansion of the characteristic function (cf, Box (1954) Theorem 2.4), the density of $Z = \sum^n_1 c_jZ_j$ is found to be $\sum^n_1 (w_j/c_j) \exp (-z/c_j)$ where \begin{equation*}\tag{1.2}w_j = c^{n-1}_j\lbrack\prod_{i \neq j} (c_j - c_i)\rbrack^{-1}.\end{equation*} Now a set of random variables $X_1, X_2, \cdots X_n$ uniformly distributed over the simplex as described above may be represented as $X_j = Z_j/Y$ for $j = 1, 2, \cdots, n$ where $Y = \sum^{n+1}_1 Z_j$, and the set $X_1, X_2, \cdots, X_n$ thus created is distributed independently of $Y$. Thus $X = \sum^n_1 c_jX_j$ may be represented as \begin{equation*}\tag{1.3}X = Z/Y\end{equation*} where $X$ and $Y$ are independent. In a relation like (1.3) the marginal distributions of $Z$ and $Y$ together with the fact of independence of $X$ and $Y$ are sufficient to determine uniquely the distribution of $X$. To see this, note that $\operatorname{cf} (\log X)\cdot\operatorname{cf} (\log Y) = \operatorname{cf} (\log Z)$, where $\operatorname{cf}$ denotes the characteristic function. Thus, the characteristic function of $\log X$ is determined by those of $\log Y$ and $\log Z$. Now if $Z$ were known to have the scaled exponential density $c^{-1} \exp (-z/c)$ for $z \geqq 0$ while $Y$ had the gamma density $y^n \exp (-y)/n!$ for $y \geqq 0$, it would follow that $X$ had the scaled beta density $nc^{-1}(1 - x/c)^{n-1}$ for $0 \leqq x \leqq c$. But the density of $Z$ is, in fact, a mixture of scaled exponentials. Thus, the density of $X$ is the same mixture of scaled betas and is given by $\sum^n_1 w_jf_j(x)$, where the $w_j$'s are defined by (1.2) and \begin{equation*}\begin{align*}\tag{1.4}f_j(x) &= nc^{-1}_j (1 - x/c_j)^{n-1}\quad\text{for} 0 \leqq x \leqq c_j \\ &= 0\quad\text{otherwise}\end{align*}.\end{equation*} The result just given yields immediately the first line of (1.1). To derive the second line of (1.1), set $X_{n + 1} = 1 - \sum^n_1 X_j$ and note that $c_1 - X = c_1 X_{n + 1} + (c_1 - c_n)X_n + \cdots + (c_1 - c_2)X_2$ has a distribution of the same type as $X$. Applying the first line of (1.1) to $c_1 - X$ yields the second line of (1.1). The distribution of $\sum^{n + 1}_1d_iX_i$ is also given by (1.1), where $X_1, X_2, \cdots, X_{n + 1}$ are uniformly distributed over the simplex $x_i \geqq 0$ for $i = 1, 2, \cdots, n + 1$ and $\sum^{n + 1}_1 x_i = 1$, and where constants $d_i$ satisfy the condition \begin{equation*}\tag{1.5}d_1 > d_2 > \cdots > d_r \geqq x > d_{r + 1} > \cdots > d_{n + 1}.\end{equation*} To see this, let $c_i = d_i - d_{n + 1}$ and notice that $\sum^{n + 1}_1 d_iX_i = X + d_{n + 1}$.