### Alternative Proof of the Back-Door Criterion

Consider a Markovian model [tex]$G$[/tex] in which [tex]$T$[/tex] stands for the set of parents of [tex]$X$[/tex]. From [tex]{em Causality}[/tex], Eq.~(3.13), we know that the causal effect of [tex]$X$[/tex] on [tex]$Y$[/tex] is given by

[tex]begin{equation} P(y|hat{x}) = sum_{t in T} P(y|x,t) P(t) %% eq 1 label{ch11-eq-a} end{equation}[/tex] (1).

Now assume some members of [tex]$T$[/tex] are unobserved, and we seek another set [tex]$Z$[/tex] of observed variables, to replace [tex]$T$[/tex] so that

[tex]begin{equation} P(y|hat{x}) = sum_{z in Z} P(y|x,Z) P(z) %% eq 2 label{ch11-eq-b} end{equation}[/tex] (2).

It is easily verified that (2) follow from (1) if [tex]$Z$[/tex] satisfies:

.

Indeed, conditioning on [tex]$Z$[/tex], ([tex]$i$[/tex]) permits us to rewrite (1) as [tex][ P(y|hat{x}) = sum_{t} P(t) sum_z P(y|z,x) P(z|t,x) ][/tex] and ([tex]$ii$[/tex]) further yields [tex]$P(z|t,x)=P(z|t)$[/tex] from which (2) follows. It is now a purely graphical exercize to prove that the back-door criterion implies ([tex]$i$[/tex]) and ([tex]$ii$[/tex]). Indeed, ([tex]$ii$[/tex]) follows directly from the fact that [tex]$Z$[/tex] consists of nondescendants of [tex]$X$[/tex], while the blockage of all back-door path by [tex]$Z$[/tex] implies , hence ([tex]$i$[/tex]). This follows from observing that any path from [tex]$Y$[/tex] to [tex]$T$[/tex] in [tex]$G$[/tex] that is unblocked by [tex]${X,Z}$[/tex] can be extended to a back-door path from [tex]$Y$[/tex] to [tex]$X$[/tex], unblocked by [tex]$Z$[/tex].