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10.6 Ordered vector

For some modeling tasks, a vector-valued random variable $X$ is required with support on ordered sequences. One example is the set of cut points in ordered logistic regression.

In constraint terms, an ordered $K$-vector $x \in \mathbb{R}^K$ satisfies

$$x_k < x_{k+1}$$

for $k \in \{ 1, \ldots, K-1 \}$.

Ordered transform

Stan’s transform follows the constraint directly. It maps an increasing vector $x \in \mathbb{R}^{K}$ to an unconstrained vector $y \in \mathbb{R}^K$ by setting

$$y_k = \left\{ \begin{array}{ll} x_1 & \mbox{if } k = 1, \mbox{ and} \\ \log \left( x_{k} - x_{k-1} \right) & \mbox{if } 1 < k \leq K. \end{array} \right.$$

Ordered inverse transform

The inverse transform for an unconstrained $y \in \mathbb{R}^K$ to an ordered sequence $x \in \mathbb{R}^K$ is defined by the recursion

$$x_k = \left\{ \begin{array}{ll} y_1 & \mbox{if } k = 1, \mbox{ and} \\ x_{k-1} + \exp(y_k) & \mbox{if } 1 < k \leq K. \end{array} \right.$$

$x_k$ can also be expressed iteratively as

$$x_k = y_1 + \sum_{k'=2}^k \exp(y_{k'}).$$

Absolute Jacobian determinant of the ordered inverse transform

The Jacobian of the inverse transform $f^{-1}$ is lower triangular, with diagonal elements for $1 \leq k \leq K$ of

$$J_{k,k} = \left\{ \begin{array}{ll} 1 & \mbox{if } k = 1, \mbox{ and} \\ \exp(y_k) & \mbox{if } 1 < k \leq K. \end{array} \right.$$

Because $J$ is triangular, the absolute Jacobian determinant is

$$\left| \, \det \, J \, \right| \ = \ \left| \, \prod_{k=1}^K J_{k,k} \, \right| \ = \ \prod_{k=2}^K \exp(y_k).$$

Putting this all together, if $p_X$ is the density of $X$, then the transformed variable $Y$ has density $p_Y$ given by

$$p_Y(y) = p_X(f^{-1}(y)) \ \prod_{k=2}^K \exp(y_k).$$