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Suppose I have two statements:

Statement 1: Random variable Z is the common cause for random variable X and Y (Z causes both X and Y)

Statement 2: Random variable X and Y are (conditionally) independent given Z.

What's the relationship between Statement 1 and Statement 2? Does Statement 1 imply statement 2, or does statement 2 imply statement 1, or they have other relationships?

Example: Z could be the grade level(or age) for a primary school student, X could be his height, Y could be his math ability. Thanks!

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Statement 1 is a scientific statement rather than a mathematical relationship. The idea of one random variable "causing" another doesn't have a strict mathematical definition, rather the role of the mathematician or statistician is to posit a mathematical model that captures the relationship in a way that is relevant to a particular scientific problem. For example, a mathematician might suppose that X and Y both have functional relationships with Z but not with each other, and that would capture the scientific idea of "causation".

Statement 2 is a mathematical statement, and it is one sensible and useful way to translate the first statement into mathematical terms. It is a very strong statement because it implies that Z accounts for all possible dependencies between X and Y. There cannot be any other common causes of X and Y that are not mediated by Z or correlated with Z. On the other hand, Statement 2 is more general than Statement 1 in the sense that X and Y could be conditionally independent given Z without being direct functions of Z.

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  • $\begingroup$ I don’t think your assertion that any causal model where Z is a common cause for X and Y is plausible. In particular, even within the standard model of causality, if Z’ were another common cause between X and Y, then X would not in general be conditionally independent of Y given Z $\endgroup$ Commented Oct 30, 2022 at 2:06
  • $\begingroup$ @stats_model Yes, obviously. OP's Statement 1 says "the common cause" rather than "a common cause" so I assumed that no other common causes are present. Anyway, I have deleted the last paragraph of my answer so that it doesn't depend on such a strict interpretation of OP's words. $\endgroup$ Commented Oct 30, 2022 at 3:17
  • $\begingroup$ @GordonSmyth Thank you very much! This is very helpful! $\endgroup$ Commented Nov 2, 2022 at 21:44

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