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When inferring causal effects from observational studies, one of the assumptions that's generally required is the exchangeability assumption. Suppose $A \in \{0, 1\}$ is a binary treatment, and let $Y^a$ denote the counterfactual outcome under treatment $A=a$. The exchangeability over $A$ assumption is: $$Y^a\perp\!\!\!\perp A$$

which says that $Y^a$ is independent of $A$.

My question is, why is this assumption called the "exchangeability" assumption when it's a statement about independence?

I know that exchangeable random variables have joint probability distribution does not change when the positions in the sequence in which they appear are altered. And two random variables are independent if the realization of one does not affect the probability distribution of the other. But what is the relationship between exchangeability and independence in the causal inference context?

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  • $\begingroup$ Can you please provide a source for this material. It's diffcult to explain the terminology used without seeing the context. $\endgroup$
    – Ben
    Dec 24, 2021 at 2:33

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My question is, why is this assumption called the "exchangeability" assumption when it's a statement about independence?

Exchangeability is the assumption of being able to exchange groups without changing the outcome of the study. Why? Because the relationship between treatment and outcome is not confounded. Why? Because treatment assignment is independent of everything else.

If you have people with a more severe version of the disease in one group, if you exchange the treatment group with the control group, your results will be different, so exchangeability, in this case, would have been violated.

You can also run into this assumption with a different name, depending on what source you're checking. One example is unconfoundedness. If you made treatment and outcome independent by adjusting for the appropriate variables $Z$ (check backdoor criterion, for example), you can also see this as conditional exchangeability, e.g., $Y^a \perp\!\!\!\perp A \mid Z $.

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    $\begingroup$ So exchangeability here is not referring to the probabilistic definition, i.e., that a sequence of random variables is exchangeable if permuting them in another order does not change the joint distribution, but rather that you can interchange the treated and control groups without changing the outcomes, correct? $\endgroup$
    – Adrian
    Dec 24, 2021 at 2:09
  • $\begingroup$ Yes, here it's a causal assumption. Sometimes causal assumptions are made more clear, such as "Causal Markov condition" for what sometimes is just referred as the Markov assumption. $\endgroup$ Dec 24, 2021 at 2:14

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