Sigma-separation question in cyclic causal graph – understanding sigma-separation

Question

Main Question

In https://arxiv.org/pdf/1807.03024.pdf, a generalization of d-separation in DAGs is introduced, called $\sigma$-separation for cyclic graphs.

1. I am wondering how $v_1 \perp v_6$ using the $\sigma$-separation criterion?
2. Based on that answer, why is $v_1 \not\perp v_6 | v_3, v_5$

Background

The following graph is given as Figure 3:

and they then provide the following table comparing the d-separation statements and the $\sigma$-separation statements:

Background details on a path to be $\sigma$-connected (i.e. similar to d-connected)

Answer 1

$ \newcommand{\pperp}{\perp\kern-5pt\perp} \newcommand{\npperp}{\not\perp\kern-8pt\perp} $ First, please note that there are different definitions of a path out there and in the definition of $Z$-$\sigma$-open paths, they use the following definition for a path (unfortunately, they don't give this definition in the paper, at least I didn't find it):

A path as a sequence of consecutive edges in the graph, without any restrictions on the types or orientations of the edges involved.

In particular, this means that a path can also go through the same node or edge multiple times.

Your first question:
You cannot find a path from $v_1$ to $v_6$ that does not contain a collider: A path starting with $v_1\to v_2\to v_5$ contains $v_2$ as a collider, so we have to go via $v_1\to v_2\to v_3\to v_4$. But however we continue this path to $v_6$, there will always be a collider in this path. Thus, all paths from $v_1$ to $v_6$ contain colliders, thus $v_1 \pperp v_6$.

Your second question:
Using the notation $Z=\{v_3, v_5\}$, we need to find a path from $v_1$ to $v_6$ that is $Z$-$\sigma$-open. Once we have that, we have shown that $v_1\npperp v_6 | Z$.

Now, the path $p = v_1\to v_2\to v_3\to v_4\to v_5\leftarrow v_4\leftrightarrow v_6$ is a $Z$-$\sigma$-open path, because:

• $p$ is $Z$-$\sigma$-open at $v_1$ because $v_1\not\in Z$ (point 1. in the definition),
• $p$ is $Z$-$\sigma$-open at $v_2$ because of (c) in the definition,
• $p$ is $Z$-$\sigma$-open at $v_3$ because of (c) in the definition (this is different from d-separation!),
• $p$ is $Z$-$\sigma$-open at $v_4$ because of (c) in the definition,
• $p$ is $Z$-$\sigma$-open at $v_5$ because of (a) in the definition,
• $p$ is $Z$-$\sigma$-open at $v_4$ because of (b) in the definition, and
• $p$ is $Z$-$\sigma$-open at $v_6$ because $v_6\not\in Z$ (point 1. in the definition).

In a nutshell:
We have a $\sigma$-open path through the conditioned node $v_3$ because at $v_3$ the path "stays inside a loop" and the path is open at the conditioned node $v_5$ simply because it is a collider.