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Pearl et al. "Causal Inference in Statistics: A Primer" (2016) p. 69 Theorem 3.4.1 provides the Front-Door Adjustment formula: $$ P(y | \text{do}(X = x)) = \sum_z P(z | x) \sum_{x'} P(y|x', z)P(x'). $$ Does it mean $$ P(y | \text{do}(X = x)) = \sum_z \left[ P(z | x) \sum_{x'} \left\{ P(y|x', z)P(x') \right\} \right] $$ or $$ P(y | \text{do}(X = x)) = \left[ \sum_z P(z | x) \right]\cdot \left[ \sum_{x'} \left\{ P(y|x', z)P(x') \right\} \right]? $$

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  • $\begingroup$ The last expression is not allowed, because $P(y|x',z)$ has $z$ in it, but that term has fallen out of scope of the $z$ summation. $\endgroup$ – Adrian Keister Apr 2 at 14:40
  • $\begingroup$ @AdrianKeister, got it, thank you. $\endgroup$ – Richard Hardy Apr 2 at 14:59
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It means $$ P(y|\text{do}(X=x)) = \sum_z \left[ P(z|x) \sum_{x'} \left\{ P(y|x',z) P(x') \right\} \right]. $$ This can be seen from equation (3.15) on p. 68 which is $$ P(y|\text{do}(X=x)) = \sum_z \sum_{x'} \left\{ P(y|x',z) P(x') P(z|x) \right\} $$ and can be rearranged to be $$ P(y|\text{do}(X=x)) = \sum_z \sum_{x'} \left\{ P(z|x) P(y|x',z) P(x') \right\}, $$ the latter ordering of conditional probabilities corresponding to the ordering in the equation at the top of this answer.

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    $\begingroup$ Right. $x'$ is just a dummy variable, and does not show up in the $P(z|x),$ so that that term can move in or out of the second sum with impunity. $\endgroup$ – Adrian Keister Apr 2 at 14:31

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