# Intuition behind conditioning Y on X in the front-door adjustment formula

I am trying to understand the practicalities of Pearl's front-door criteria to estimate causal effects under an unobserved unconfounder. First, to give context to the question we have a graph that looks like this: Where $$U$$ is unobserved and, hence, no back-door adjustment formula can be used.

We can infer the causal effects on $$Y$$ of an intervention $$do(X=x)$$ by using the front-door adjustment formula Pearl (2009 p.81-83):

$$P(y | \textit{do}(X = x)) = \sum_z P(z | x) \sum_{x'} P(y|x', z)P(x').$$

My question is: What is the intuition of conditioning $$Y$$ on $$x'$$ ($$P(y|x', z)$$) if $$Y$$ is conditionally independent of $$X$$ given $$Z$$? Does that mean, in practical terms, that the way to estimate the front-door adjustment is by using the following:

$$P(y | \textit{do}(X = x)) = \sum_z P(z | x) \sum_{x'} P(y|z)P(x').$$

• Or does it mean that I have to build an alternative estimator using both z and x as predictors? Feb 5, 2020 at 15:10
• If you haven't read it already, I would recommend Pearl's book The Book of Why, pages 224 to 231. He explains the intuition behind the front-door adjustment formula. Feb 11, 2020 at 18:37

Y is independent of X given Z, but Z isn't independent of X. Which you're manipulating (or, well, simulating the manipulation, anyway).

An intervention on the value of X changes the probabilities of Zs you're going to observe. That's why we need the first part, summing/integrating over the post-intervention distribution $$p(z|x)$$.

However, both X and Y are influenced by the unobserved variable 'in the wild'. By intervening on X, we eliminate U's influence on X - but not on Y. It's still there, and we still need to account for it.

The $$p(x')$$ is the pre-intervention probability on values of X, so you cannot use your final equation. The term is actually derived using the independence assumptions to substitute for a $$\sum_Up(y|z,u)p(u)$$ term. Since U is not easily measurable while X' can be, we're inferring U from pre-intervention X.

$$P(y | \textit{do}(X = x)) = \sum_z P(z | x) \sum_{x'} P(y|x', z)P(x').$$
$$P(y | \textit{do}(X = x)) = \sum_z P(z | x) \sum_{x'} P(y|z)P(x')$$
because $$P(y|x', z) \neq P(y|z)$$ in the pre-intervention distribution given the presence of U, the confounder.