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Vertical Dependency in Sequences of Categorical Random Variables

# Vertical Dependency in Sequences of Categorical Random Variables

### 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

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

$\alpha(n) = n-1$ 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

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 $m$th 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}