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I am self-studying Causality: Models, Reasoning, and Inference, by Judea Pearl, and there is a question I am particularly stumped on. It reads like this:

Problem Statement: Given this fragment of a Bayesian Network, add two variables to the network, $Z$ and $W,$ such that the following three conditions hold simultaneously:

enter image description here

  1. $Z$ and $X$ are dependent given $W,$ and
  2. $Z$ and $U$ are independent given $W.$
  3. $Z$ and $W$ are ancestors of $Y$ but not of $X.$

My Work So Far: Because $Z$ and $W$ must be ancestors of $Y$ but not $X,$ there can be no arrows going into $X$ (from $Z$ or $W$). Because $U$ is an ancestor of $X,$ there can be no arrows going into $U,$ either. Likely, though, we will need arrows going into $Y.$ As I see it, there are essentially two possibilities: arrows going out of $U,$ or arrows going out of $X.$ As we need dependence on $X$ and not $U,$ I'm going to guess that we need arrows going out of $X.$ That means there are essentially four possibilities:

enter image description here

a satisfies 1 and 3, but not 2. b satisfies 2 and 3, but not 1. c satisfies 1 and 3, but not 2. And d satisfies 1 and 3, but not 2.

Do I need to look at $U?$ Or do you have other ideas?

Thanks for your time!

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    $\begingroup$ In a, if you remove the arrow from Z to Y, it will still be an ancestor of it (but not a parent) and 2 holds $\endgroup$ – CloseToC Apr 27 at 23:50
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    $\begingroup$ Are you positive this is correct? I can't see any way that an association between $X$ and $Z$ could exist without that same association existing for $U$ and $Z$. It's impossible to block the pathway between $U$ and $X$, so any association that $X$ has with its non-ancestors, $U$ will also have. $\endgroup$ – Noah Apr 28 at 1:33
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    $\begingroup$ @CloseToC: Are you sure? Conditioning on $W$ opens up the collider there, allowing the path $Z\to W\leftarrow X\leftarrow U,$ so that $Z$ and $U$ would be dependent, right? $\endgroup$ – Adrian Keister Apr 28 at 2:04
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    $\begingroup$ @Noah: Well, in my experience with Pearl's (co-authored) book Causal Inference in Statistics: A Primer, he certainly wasn't above giving trick questions for exercises. It's possible there's no answer. This problem goes with the text, but is actually a separate file. You can get it from Pearl's website bayes.cs.ucla.edu/jp_home.html and click on the Causality link at the top; you can eventually navigate to his Viewgraphs and Homeworks for Instructors link, which is where I found the homework files. $\endgroup$ – Adrian Keister Apr 28 at 2:07
  • $\begingroup$ As you have solved it, please post that as an answer (that is, in the answer box) so the Q so not linger on as unsolved. $\endgroup$ – kjetil b halvorsen Apr 29 at 22:34
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If you allow for bidirected edges, you can draw:

enter image description here

| cite | improve this answer | |
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  • $\begingroup$ Yep, spouses are allowed. Thanks very much! $\endgroup$ – Adrian Keister Apr 30 at 14:23
  • $\begingroup$ Do you mind my asking: is that a TikZ picture? If so, would you mind, please, posting your code for it? Thanks! $\endgroup$ – Adrian Keister Apr 30 at 15:17
  • $\begingroup$ Hi @AdrianKeister in this case I didn't use TikZ, I used causalfusion.net a software developed by Elias Bareinboim's team. $\endgroup$ – Carlos Cinelli Apr 30 at 16:19
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    $\begingroup$ For a short TikZ tutorial for causal graphs, check this out: dkumor.com/posts/technical/2018/08/15/causal-tikz @AdrianKeister $\endgroup$ – Carlos Cinelli Apr 30 at 16:21
  • $\begingroup$ Great, thanks for that link! $\endgroup$ – Adrian Keister Apr 30 at 16:23

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