Vertical Dependency in Sequences of Categorical Random Variables

Vertical Dependency in Sequences of Categorical Random Variables

For the full text, including proofs, download the pdf here.

Graphical Interpretation and Illustrations

We may visualize the dependency structures generated by \alpha \in \mathscr{C}_{\delta} via directed dependency graphs. Each \epsilon_{n} represents a vertex in the graph, and a directed edge connects n to \alpha(n) to represent the direct dependency generated by \alpha. This section illustrates some examples and gives a graphical interpretation of the result in Theorem 2.

First-Kind Dependency

Dependency Graph for FK Dependence

For FK-dependency, \alpha(n) \equiv 1 generates the graph above. Each \epsilon_{i} depends directly on \epsilon_{1}, and thus we see no connections between any other vertices i, j where j \neq 1. There are n-1 separate subsequences of length 2 in this graph.

 

Sequential Dependency

Dependency Graph for Sequential Dependence
\alpha(n) = n-1 generates the sequential dependency structure this work has studied in detail. We can see that a path exists from any n back to 1. This  is a visual way to see the result of Theorem2,  in that if a path exists from any n back to 1, then the variables in that path must be identically distributed. Here, there is only one sequence and no subsequences.

 

A Monotonic Example

Dependency Graph for a monotonic example

Here, \alpha(n) = \lfloor \sqrt{n}\rfloor gives an example of a monotonic function in \mathscr{C}_{\delta} and the dependency structure it generates. Again, note that any n has a path to 1, and the number of subsequences is between 1 and n-1.

 

A Nonmonotonic Example

Here, \alpha(n)=\left\lfloor\frac{\sqrt{n}}{2}\left(\sin(n)\right)+\frac{n}{2}\right\rfloor illustrates a more contrived example where the function is nonmonotonic. It is neither increasing nor decreasing. The previous examples have all been nondecreasing.

 

A Prime Example

Let \alpha \in \mathscr{C}_{\delta} be defined in the following way. Let p_{m} be the mth prime number, and let \{kp_{m}\} be the set of all postive integer multiples of p_{m}. Then the set \mathscr{P}_{m} = \{kp_{m}\} \setminus \cup_{i=1}^{m-1}\{kp_{i}\} gives a disjoint partitioning of \mathbb{N}. That is, \mathbb{N} = \sqcup_{m}\mathscr{P}_{m}, and thus for any n \in \mathbb{N}, n \in \mathscr{P}_{m} for exactly one m. Now let \alpha(n) = m. Thus, the function is well-defined. We may write \alpha: \mathbb{N}_{\geq 2} \to \mathbb{N} as \alpha[\mathscr{P}_{m}] = \{m\}. As an illustration,

\begin{aligned}\alpha[\{2k\}] &= \{1\} \\\alpha[\{3k\}\setminus \{2k\}] &= \{2\} \\\alpha[\{5k\}\setminus (\{2k\}\cup \{3k\})] &= \{3\} \\&\qquad\vdots\\\alpha[\{kp_{m}\}\setminus (\cup_{i=1}^{m-1}\{kp_{i}\})] &= \{m\}\end{aligned}