i got a question regarding DAGs and d-seperation within the theme Bayesian Networks.
In Task d) i am supposed to check for the following d-seperations. But i am struggeling. My other excercises provide some intel, but not enough for me to get through this task.
So i wonder, when a Chain is blocked, or not. As far as i know, there are "Head-Head", "Head-Tail" and "Tail-Tail" meeting. I am not sure if "Tail-Head" exists. Does it?
And i am also unsure, how to check if all paths are blocked.
for d-i) i have IG(Y,S|Q) So i tried to check the Chains: Y,W,S,Q and found out there is a Head-Tail meeting at Q But was it for Y,W,S? Q is instantiated, but doesnt show up here. So i might have a Head-Tail meeting at W, but as W is not instantiated, it doesn't matter, right?
So i would conclude, as not all chains are blocked and thus this d-seperation does not show up in the DAG (Figure 2) you see in the screenshot.
Can someone help me out please and explain the differences? I also dont know about ii) iii) and iv). Didn't find a good explanation in my script, nor in the internet.
For ii) i think, as the "|" is missing, we have an empty space. And as my notes go, that means, only Head-Head meetings can block some path. So for ii) we have the Chains: [Y,W] but because they are just 2 nodes, there is no H-H, H-T, or anything. I would say, this is also no d-seperation
iii) has [Y,W,X,Z] as only chain and there is a H-H meeting at X, so i would say all chains are blocked and thus we have found this d-seperation in the DAG.
iv) has [Z,X,W,S]. Here X is instantiated. And we have H-H at X. So it is blocked [Z,X,W,Q,S] Here the same. X has Head to Head so it is blocked at X, which is also instantiated, thus all Chains are blocked and we found a d-seperation. Can someone confirm? And help me out with the understanding a bit please? Thank you all :)
A trail is an undirected path between two nodes that does not contain any loops.
A trail between nodes $a$ and $z$ will be d-separated by a set of conditioning nodes, $D$, if the trail:
(Also, as you noted, if $D$ is empty, then all colliders on the trail will not be in it.)
Nodes $a$ and $z$ will be d-separated by the set, if all trails between them are d-separated by the set.
Yes. We need all trails to be d-seperated for d-seperation of the nodes. So, indeed, between $\sf Y$ and $\sf S$ there are two trails: $\sf Y{\to}W{\to}S$ and $\sf Y{\to}W{\to}Q{\to}S$. You've noted that $\sf Q$, the node in our conditioning set, is on a directed chain on the second trail but not the first. Thus the second trail is d-separated, however the first is not.
Therefore nodes $\mathsf Y$ and $\mathsf S$ are not d-separated by the set $\{\mathsf Q\}$.
ii) Indeed, when the conditioning set is empty, we need only look for colliders. On the single trail between $\sf Y$ and $\sf W$ there are no middle nodes (and so no colliders). The trail is clearly not d-separated.
Therefore nodes $\sf Y$ and $\sf W$ are not d-separated by the empty set.
Indeed. The lone trail, $\sf Y{\to}W{\to}X{\gets}Z$ contains a collider at $\sf X$, and this is not in the empty set.
Therefore the nodes $\sf Y$ and $\sf Z$ are d-separated by the empty set.
And here you go awry. $\sf X$ is a collider on both trails between $\sf Z$ and $\sf S$, however this collider is in the conditioning set.
Therefore $\sf Z$ and $\sf S$ are not d-separated by $\{\sf X\}$.
Remember:
A chain or fork node will block a trail when they are in the conditioning set.
A collider node will block the trail when it and all of its decendants are not in the conditioning set.