# Causal Analysis in Theory and Practice

## May 4, 2008

### Alternative Proof of the Back-Door Criterion

Filed under: Back-door criterion — judea @ 6:00 pm

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].