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Causal graphs are an increasingly popular tool for causal inference. The underlying understanding of causality is deterministic. In the popular directed acyclic form of causal graphs, we assume that no cycles exist in causal relationships. However, I've seen variants of causal graphs which relax this assumption and allow for cyclicality.

My question is: does cyclicality really happen in nature? Is it not just the level of granularity we use to look at mechanisms that makes nature appear cyclic? Is it reasonable to say that everything could be viewed as acyclic on some level (except maybe for the quantum world where it appears that simultaneous causality really can be a thing)?

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    $\begingroup$ Because causes must precede effects, acyclic is preferred. However, there are definitely feedback loops, both in nature and in engineering. Let's say you have variables $A(t)$ and $B(t),$ and you know that $A(t)\to B(t+1).$ But then $B$ feeds back into $A$, so you might have $B(t+1)\to A(t+2).$ If you think of variables at different times as different variables, you can model the feedback loop without using a cyclic graph. You lose something in that model, of course: the relationship between $A(t)$ and $A(t+2).$ $\endgroup$
    – Adrian Keister
    Commented Jul 28, 2021 at 14:05
  • $\begingroup$ Yea, that's basically what I meant by "level of granularity". Feedback loops may appear cyclic, but we can break it down. But are there cases of "real" cyclicality (except for maybe quantum causality)? I'm thinking of things like equilibria, but I am having a hard time to find good literature on it. $\endgroup$
    – Rob G.
    Commented Jul 28, 2021 at 17:27
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    $\begingroup$ I would say no. The argument would proceed like this: If event $A$ causes event $B,$ then $A$ must precede $B.$ The time $t_a$ at which $A$ occurs must be smaller than the time $t_b$ at which $B$ occurs, for time flowing in the usual direction. If $t_a<t_b,$ it is impossible for $t_b<t_a,$ and hence impossible for $B$ to cause $A.$ $\endgroup$
    – Adrian Keister
    Commented Jul 28, 2021 at 17:30
  • $\begingroup$ That's also how I understand it. One proviso could be: in some cases systems are so complex that it is practically impossible to separate events. A slightly modified question could then be: do we gain any insights by allowing quasi-cyclicality in causal graphs? $\endgroup$
    – Rob G.
    Commented Jul 28, 2021 at 20:00
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    $\begingroup$ Ok, will do.... $\endgroup$
    – Adrian Keister
    Commented Jul 28, 2021 at 20:30

1 Answer 1

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Because causes must precede effects, acyclic is preferred. Ultimately, there can be no true cycles: if event $A$ causes event $B,$ then $A$ must precede $B.$ The time $t_a$ at which $A$ occurs must be smaller than the time $t_b$ at which $B$ occurs, for time flowing in the usual direction. If $t_a<t_b,$ it is impossible for $t_b<t_a,$ and hence impossible for $B$ to precede $A.$ Therefore, $B$ cannot cause $A.$

That said, there are definitely feedback loops, both in nature and in engineering. Let's say you have variables $A(t)$ and $B(t),$ and you know that $A(t)\to B(t+1).$ But then $B$ feeds back into $A$ at a later time, so you might have $B(t+1)\to A(t+2).$ If you think of variables at different times as different variables, you can model the feedback loop without using a cyclic graph. You lose something in that model, of course: the relationship between $A(t)$ and $A(t+2).$ You might mitigate that somewhat if you include the direct arrow $A(t)\to A(t+2).$

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    $\begingroup$ (+1) nice one Adrian ! This is such a tricky concept, and you've explained it very well indeed ! $\endgroup$
    – Robert Long
    Commented Jul 28, 2021 at 20:52
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    $\begingroup$ Thanks, Robert! Very kind of you to say so. $\endgroup$
    – Adrian Keister
    Commented Jul 28, 2021 at 20:56
  • $\begingroup$ I have a question about this: If A(t) → B(t+1) → A(t+2), and A(t) → A(t+2), doesn't A(t+2) then become a collider between A(t) and B(t+1)? $\endgroup$
    – torpedo_cantankerous_softener
    Commented May 20, 2023 at 14:23
  • $\begingroup$ @torpedo_cantankerous_softener Yes, you're quite correct: $A(t+2)$ would be a collider with those variables. Now if $A(t+2)$ is the effect you're interested in, that might not matter so much. $\endgroup$
    – Adrian Keister
    Commented May 20, 2023 at 17:57
  • $\begingroup$ And if it isn't? $\endgroup$
    – torpedo_cantankerous_softener
    Commented May 21, 2023 at 18:03

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