Consider a classic confounded model: a treatment X, outcome Y, and an observed confounder Z. The graph is Z → X → Y and Z → Y. We want P(y | do(x)).
- Goal: Transform P(y | do(x)) to an observational formula.
- Apply Rule 2: We ask if we can change do(x) to see(x). Check condition: Is (Y ⊥⊥ X | Z) in G_{\underline{X}}? (Graph with arrows out of X removed). In this graph, the path X ← Z → Y is active, so they are not independent. Rule 2 fails.
- Apply Rule 3: Can we delete an action? Not applicable.
- Alternative Strategy - Conditioning: Start with P(y | do(x)) = Σ_z P(y, z | do(x)) = Σ_z P(y | do(x), z) P(z | do(x)).
- Apply Rule 2 to P(y | do(x), z): In G_{\underline{X}}, is (Y ⊥⊥ X | Z)? No, as before.
- Apply Rule 3 to P(z | do(x)): In G_{\overline{X}}, is (Z ⊥⊥ X)? Yes, because the edge from Z to X is removed. So by Rule 3, P(z | do(x)) = P(z).
- Apply Rule 2 to P(y | do(x), z) in a different graph: Consider G_{\overline{X}}. In this graph, with arrows into X removed, the only path from X to Y is X → Y, which is blocked by conditioning on nothing. However, we are conditioning on Z. In G_{\overline{X}}, Z is a non-collider on the path X → Y ← Z? No, that path doesn't exist. Actually, in G_{\overline{X}}, X and Y are d-separated given Z? Let's check: The path X → Y is blocked by conditioning on Y's child? Wait, Z is a parent of Y. The path is X → Y ← Z. This is a collider at Y? No, Y is not a collider here because both arrows point to Y. This is a 'chain' where Z influences Y. Conditioning on Z opens the path? No, conditioning on a non-collider does not open a path. In G_{\overline{X}}, the path X → Y is direct and unblocked. Conditioning on Z, a parent of Y, does not block the direct path. So X and Y are not independent given Z in G_{\overline{X}}. Therefore, Rule 2 does not apply directly to P(y | do(x), z).
The correct, simpler derivation leverages the backdoor criterion: Z satisfies it for (X, Y). The do-calculus derivation is: P(y|do(x)) = Σ_z P(y|do(x), z)P(z|do(x)) [Law of Total Probability]. By Rule 3, P(z|do(x)) = P(z) as shown. Now, in G_{\overline{X}}, the only path from X to Y is the direct edge. Is (Y ⊥⊥ X | Z) in G_{\overline{X}}? In G_{\overline{X}}, the path X → Y is still present. However, if we consider the mutually exclusive graph manipulation for Rule 2, which is G_{\overline{X}, \underline{Z}}, we remove arrows into X and arrows out of Z. In this graph, Z has no arrows out, so the path X → Y ← Z is severed. Therefore, Y is d-separated from X given Z in G_{\overline{X}, \underline{Z}}. Applying Rule 2: P(y | do(x), z) = P(y | x, z).
Final Result: P(y | do(x)) = Σ_z P(y | x, z) P(z). This is the backdoor adjustment formula, derived rigorously via do-calculus.