4
$\begingroup$

In this paper, I am confused about the classification of selection bias and its connection to Pearl's do-calculus.

Following notation of paper, let $E$ be a binary exposure, D be a binary outcome, $L1$ be a covariate vector, and $S$ be selection. $D^e$ refers to potential outcomes. Our goal is to target the true ATE defined as $P(D^1=1)-P(D^0=1)$

They define Type 1 Selection Bias (SB) as that which does not have internal validity but has external validity i.e.

  • $P(D=1|E=1,S=1)-P(D=1|E=0,S=1) \ne P(D^1=1|S=1)-P(D^0=1|S=1)$
  • But, $P(D^1=1|S=1)-P(D^0=1|S=1)=P(D^1=1)-P(D^0=1)$

They define Type 2 SB as that which does not have external validity but has internal validity i.e.

  • $P(D=1|E=1,S=1)-P(D=1|E=0,S=1) = P(D^1=1|S=1)-P(D^0=1|S=1)$
  • But, $P(D^1=1|S=1)-P(D^0=1|S=1) \ne P(D^1=1)-P(D^0=1)$

They provide the following DAG and explain that this is an example of Type 1 SB.

DAG

When I do Pearl's rules of do-calculus on the DAG, I'm not getting this. If I were to translate having no SB from above into sufficient action statements, I believe I'd need the following for $e=1,0$:

  • $p(d|e,s) = p(d|do(e),s)$
  • $p(d|do(e),s) = p(d|do(e))$

The first bullet is an action/observation (rule 2). For this, I would need, $D \perp E | S$ in the DAG after removing all arrows from E. So this appears to hold (since it removes S as a collider).

The second bullet is an insertion/deletion (rule 1). For this, I would need, $D \perp S|E$ in the DAG after removing all arrows into E - in this case, the DAG remains the same (an arrow would be removed if confounding were present). Regardless, this still has a backdoor path from D to L1 to S. So this appears to be violated. Hence, based on these rules, I see it as, at worst, Type 2 SB.

As they say in the paper, conditioning on L1 allows us to cure the selection bias presented by this DAG. But, this seems to address issue 2, since now we have $p(d| do(e),s,l1) = p(d|do(e),l1)$ as the above backdoor path is blocked once I also condition on $L1$. From randomization/ignorability, standardization permits identification.

Intuition-based answers aside, why is this do-calculus approach incorrect?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
2
$\begingroup$

Do-calculus is fine. The problem here is that actually your bullet 1 equation doesn't hold for this DAG, since after following the action/observation exchange rule you still have a backdoor path from the confounder $L_1$ through the given collider $S=1$ to have the fact that $D$ and $E$ are dependent given only $S$. You need to further condition on $L_1$ to fully block the backdoor.

Type 1A selection bias can be addressed, and the true causal effect recovered by measuring and adjusting for covariates that lie on the non-causal path that is opened by restricting to one (or more) level(s) of a collider via inverse probability weighting, g-computation, and sometimes stratification. For example, one can adjust for L1 in Figure 2 (d)

On the other hand, you're certainly right that the second bullet obviously violates the observation deletion rule and thus could be type 2 SB too as also mentioned in your paper and can be addressed same as for type 1A SB above.

Similarly, in Figure 2 (d), the selection S acts as both a collider and an effect measure modifier when the covariate L1 is a direct effect measure modifier. In such cases, both type 1 and type 2 selection bias can arise. Fortunately, one can eliminate both biases if the distribution of L1 in the referent population is measured and accounted for.

Improve this answer
$\endgroup$
2
  • $\begingroup$ this makes sense, but does Rule 2 remove the arrows from E, namely the arrow into a value of S=s (in this case S=1)? $\endgroup$ Commented Jan 29, 2025 at 14:36
  • 1
    $\begingroup$ @Winston I don't think so since as you rightly claimed in OP that according to rule 1 we only remove arrows coming to $E$. After conditioning on $E$ later, $S$ and $D$ are conditionally independent on the backdoor path of S<-E->D anyways, so it doesn't matter for the conclusion. Hope this completely addresses this specific post by now. $\endgroup$ Commented Jan 29, 2025 at 17:40

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.