Timeline for Front-door criteria - what does the second requirement mean?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 10, 2019 at 18:08 | comment | added | Noah | @roundsquare oh yes, one wouldn't want to adjust for P, I just wanted to note that you have to assume you are not inadvertently adjusting for P, for example, by selecting units only on one stratum of P (e.g., only participants who were not lost to follow up). | |
| Apr 10, 2019 at 18:04 | comment | added | roundsquare | Thanks! @Noah I think that what you are saying is correct, but is a different situation from using the front-door. E.g. let's say there was a node P and edges from X --> P and B --> P. In this case, we could ignore P and go ahead using the front door adjustment formula. If we did, for some reason, need to adjust for P, then we would be in a more complex situation and, I presume, would need do-calculus to figure out what to do. | |
| Apr 10, 2019 at 17:28 | vote | accept | roundsquare | ||
| Apr 10, 2019 at 12:39 | comment | added | Julian Schuessler | 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. | |
| Apr 10, 2019 at 12:06 | comment | added | roundsquare | @Jake Westfall thanks. So, in the example I gave, I could set Z to any non-empty subset of {A, B, C} right? | |
| Apr 10, 2019 at 3:36 | comment | added | Noah | 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. | |
| Apr 10, 2019 at 2:19 | comment | added | Jake Westfall | @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) | |
| Apr 9, 2019 at 22:34 | comment | added | Noah | 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. | |
| Apr 9, 2019 at 21:47 | history | answered | Jake Westfall | CC BY-SA 4.0 |