     # A Proof of the Front-Door Adjustment Formula

We provide a proof of the the Front-Door adjustment formula using the do-calculus.

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## 1 Introduction

In (Pearl, 2009), a formula for computing the causal effect of on in the causal model of figure 1 is derived and used to motivate the definition of front-door criterion. Pearl then states, without proof, the Front-Door Adjustment Theorem (Pearl, 2009, Theorem 3.3.4). In section 3.4.3, he provides a symbolic derivation of the front door adjustment formula for the same example from the -calculus. In this short technical report, we provide a proof of Theorem 3.3.4 using the -calculus. Figure 1: A causal Bayesian network with a latent variable U.

The next section consists of the proof of the front-door adjustment formula; the theorem is restated for the reader’s convenience. The -calculus rules, the back-door criterion, the back-door adjustment formula, and the front-door criterion are in the slide set provided as an ancillary document.

## 2 Front-Door Adjustment Theorem

[Front-Door Adjustment] If a set of variables satisfies the front-door criterion relative to and if , then the causal effect of on is identifiable and is given by the formula

 P(y|^x)=∑zP(z|x)∑x′P(y|x′,z)P(x′) (1)

. By well known probability identities (for example, the Fundamental Rule and the Theorem of Total Probability),

. In Step 1, below, we show how to compute using only observed quantities. In Steps 2 and 3, we show how to compute using only observed quantities; this part of the proof is by far the hardest.

• Step 1: Compute

• in because there is no outgoing edge from in , and also by condition (ii) of the definition of the front-door criterion, all back-door paths from to are blocked.

• satisfies the applicability condition for Rule 2:

• In Rule 2, set :

 P(z|^x)=P(z|x) (2)

• Step 2:

• in because there is no incoming edge to in , and also all paths from to either by condition (ii) of the definition of the front-door criterion (blue-type paths), or because of existence of a collider node on the path (green-type paths) are blocked.

• satisfies the applicability condition for Rule 3:

• In Rule 3, set :

• because there is no outgoing edge from in , and also by condition (iii) of the definition of the front-door criterion, all back-door paths from to are blocked by .

• satisfies the applicability condition for Rule 2:

• In Rule 2, set :

•  P(y|^z)=∑xP(y|x,^z)P(x|^z)=∑xP(y|x,z)P(x) (3)

This formula is a special case of the back-door formula.

• Step 3: Compute

As already noted at the beginning of the proof,

• , as shown in Step 1 (see equation (2))

There is no rule of the -calculus that allows the elimination of the hat from , so we take a circuitous route: we first replace an observation () with an intervention () using Rule 2, and then remove an intervention variable () using Rule 3.

• because there is no outgoing edge from in , and also by condition (iii) of the definition of the front-door criterion, all back-door paths from to are blocked by .

• satisfies the applicability condition for Rule 2:

• In Rule 2, set :

• because there is no incoming edge to in , and also all paths from to are blocked either because of condition (i) of the definition of the front-door criterion (blue-type paths)[directed paths from to ], or because of the existence of a collider on the path (green-type paths) (note that the case cannot happen because there is no incoming edge to in ).

• satisfies the applicability condition for Rule 3:

• In Rule 3, set :