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I'm reading Causal Inference in Statistics: A Primer by Pearl, et. al. and I'm a little confused by the definition of front door. The definition in the book (Definition 3.4.1) is:

A set of variables, Z is said to satisfy the front-door criterion relative to an ordered pair of variables (X, Y) if:

  1. Z intercepts all directed paths from X to Y
  2. There is no unblocked path from X to Z
  3. All backdoor paths from Z to Y are blocked by X.

I understand the first and third requirements. But, the second criteria, I'm a little confused. Here is the situation which confused me. Let's say my causal graph looks like this:

front door criteria question

So, let's say I set Z = {B}. This satisfies criteria 1 (since the only path from X to Y goes through B. This also satisfies criteria 3 (since the only back-door path from B to Y is B <-- A <-- X <-- U --> Y and this is blocked by X.

But does it satisfy criteria 2? I think the answer is no because there is an unblocked path from X to Z namely X --> A --> B. As such, my understanding is that I need to make Z = {A, B} to satisfy all three criteria. Or am I missing something and do I need to make Z = {A, B, C}?

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Criterion 2 is written wrong. It should be:

  1. there is no unblocked back-door path from X to Z

In your diagram, there is an unblocked path from X to Z, but it is not a back-door path, which is a path between X and Z that contains an arrow into X.

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  • $\begingroup$ It wouldn't have to backdoor; if X and Z both pointed to a conditioned-upon collider, then there would be an unblocked, non-backdoor path that would indeed violate the frontdoor criterion. $\endgroup$ – Noah Apr 9 '19 at 22:34
  • $\begingroup$ @Noah Are you sure about that? The criterion I gave is a direct quote from Pearl's 2nd edition textbook (Definition 3.3.3 on p. 82) $\endgroup$ – Jake Westfall Apr 10 '19 at 2:19
  • $\begingroup$ Not sure, but the situation I described would yield the same pattern of association as an unblocked backdoor path, so that's why I thought to add it. $\endgroup$ – Noah Apr 10 '19 at 3:36
  • $\begingroup$ @Jake Westfall thanks. So, in the example I gave, I could set Z to any non-empty subset of {A, B, C} right? $\endgroup$ – roundsquare Apr 10 '19 at 12:06
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    $\begingroup$ Yes, although the estimation formula would further simplify with two or more elements from {A, B, C}, because of d-separation. @JakeWestfall 's answer is correct, see for example Pearl's "Causality" book, 2nd edition, Definition 3.3.3. $\endgroup$ – Julian Schuessler Apr 10 '19 at 12:39

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