# What justifies the multiplication step in the proof of the front-door adjustment?

$$\newcommand{\doop}{\operatorname{do}}$$ The proofs of the front-door adjustment that I've read take three steps:

1. Show $$P(M|\doop(X))$$ is identifiable
2. Show $$P(Y|\doop(M))$$ is identifiable
3. Multiply the do-free expressions for $$P(M|\doop(X))$$ and $$P(y|\doop(M))$$ to obtain $$P(Y|\doop(X))$$

where $$Y,X,M$$ meet the assumptions for the frontdoor adjustment. A graph meeting these assumptions is:

I'm sure I'm being daft here, but I don't understand what justifies simply multiplying the expressions together to get $$P(Y|\doop(X)).$$

This is like saying:

$$P(Y|\doop(X)) = P(Y|\doop(M)) \cdot P(M|\doop(X))$$

(where perhaps the assumptions for the front-door adjustment are necessary) but I don't recognize this rule in my study of causal inference.

• the law of total probability gets used.
– Ben
Commented Apr 25, 2022 at 14:37
• Thanks for the edit. I'm the worst with proper formatting! Commented Apr 25, 2022 at 15:00

$$\newcommand{\doop}{\operatorname{do}}$$It's a bit more complicated than that. As outlined in Causal Inference in Statistics: A Primer, by Pearl, Glymour, and Jewell, on p. 68, we follow this line of reasoning (variable $$Z$$ in the book changed to your $$M$$ for consistency):

First, we note that the effect of $$X$$ on $$M$$ is identifiable, since there is no backdoor path from $$X$$ to $$M.$$ Thus, we can immediately write $$P(M=m|\doop(X=x))=P(M=m|X=x).\qquad\qquad (3.12)$$ Next we note that the effect of $$M$$ on $$Y$$ is also identifiable, since the backdoor path from $$M$$ to $$Y,$$ namely $$M\leftarrow X\leftarrow U\rightarrow Y,$$ can be blocked by conditioning on $$X.$$ Thus, we can write $$P(Y=y|\doop(M=m))=\sum_x P(Y=y|M=m,X=x)\,P(X=x). \qquad(3.13)$$ Both (3.12) and (3.13) are obtained through the [backdoor, ACK] adjustment formula, the first by conditioning on the null set, and the second by adjusting for $$X.$$

We are now going to chain together the two partial effects to obtain the overall effect of $$X$$ on $$Y.$$ The reasoning goes as follows: If nature chooses to assign $$M$$ the value $$m,$$ then the probability of $$Y$$ would be $$P(Y=y|\doop(M=m)).$$ But the probability that nature would choose to do that, given that we choose to set $$X$$ at $$x,$$ is $$P(M=m|\doop(X=x)).$$ Therefore, summing over all states $$m$$ of $$M,$$ we have $$P(Y=y|\doop(X=x))=\sum_mP(Y=y|\doop(M=m))\,P(M=m|\doop(X=x))\quad (3.14)$$ The terms on the right-hand side of (3.14) were evaluated in (3.12) and (3.13), and we can substitute them to obtain a $$\doop$$-free expression for $$P(Y=y|\doop(X=x)).$$ We also distinguish between the $$x$$ that appears in (3.12) and the one that appears in (3.13), the latter of which is merely an index of summation and might as well be denoted $$x'.$$ The final expression we have is \begin{align*}&P(Y=y|\doop(X=x))=\\&\sum_m\sum_{x'}P(Y=y|M=m,X=x')\,P(X=x')\,P(M=m|X=x)\qquad (3.15)\end{align*} Equation (3.15) is known as the front-door formula.

The authors use the law of total probability in writing (3.13) and again in (3.14).

• My explanation for that would be more intuitive than rigorous: you can think of interventions (do operator) as "wiggling" nodes, and seeing how the disturbance propagates through your network. If you wiggle something upstream, such as $X,$ then it will "wiggle" anything in the direction of arrows downstream (such as $M$ and $Y.$) Does that answer your question, at least a little? Commented Apr 25, 2022 at 22:19