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I'm referring to the back-door adjustment and front-door adjustment here:

Back-door adjustment:The archetypal epidemiological problem in statistics is to adjust for the effect of a measured confounder. The back-door criterion of Pearl generalizes this idea.

Front-door adjustment: If some variables are unobserved then we may need to resort to other methods for identifying the causal effect.

The page also comes with precise mathematical definitions for the above two terms.

How do you make a layman understand the difference between back-door and front-door adjustment, based on the above mathematical definitions?

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  • $\begingroup$ Did my answer help? $\endgroup$ Jul 26, 2018 at 7:38
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    $\begingroup$ @JulianSchuessler, sorry to say I read through a few times but still couldn't quite understand it. Maybe some diagrams will help? $\endgroup$
    – Graviton
    Jul 26, 2018 at 7:45
  • $\begingroup$ I've updated my answer and inserted relevant diagrams. $\endgroup$ Jul 30, 2018 at 13:04
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    $\begingroup$ the link in OP is not working for me. $\endgroup$ Sep 9, 2020 at 6:32
  • $\begingroup$ Try this wayback machine link web.archive.org/web/20170910225539/https://www.stats.ox.ac.uk/… $\endgroup$ Mar 12, 2021 at 15:35

1 Answer 1

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Let's say you are interested in the causal effect of $D$ on $Y$. The following statement are not quite precise but I think convey the intuition behind the two approaches:

Back-door adjustment: Determine which other variables $X$ (age, gender) drive both $D$ (a drug) and $Y$ (health). Then, find units with the same values for $X$ (same age, same gender), but different values for $D$, and compute the difference in $Y$. If there is a difference in $Y$ between these units, it should be due to $D$, and not due to anything else.

The relevant causal graph looks like this:

enter image description here

Front-door adjustment: This means that you need to understand precisely the mechanism by which $D$ (let's now say it's smoking) affects $Y$ (lung cancer). Let's say it all flows through variable $M$ (tar in lungs): $D$ (smoking) affects $M$ (tar), and $M$ (tar) affects $Y$; there is no direct effect. Then, to find the effect of $D$ on $Y$, compute the effect of smoking on tar, and then the effect of tar on cancer - possibly through backdoor adjustment - and multiply the effect of $D$ on $M$ with the effect of $M$ on $Y$.

The relevant causal graph looks like this (where $U$ is not observed):

enter image description here

Here, front-door adjustment works because there is no open back-door path from $D$ to $M$. The path $D \leftarrow U \rightarrow Y \leftarrow M$ is blocked. This is because the arrows "collide" in $Y$. So the $D \rightarrow M$ effect is identified.

Similarly, the $M \rightarrow Y$ effect is identified because the only back-door path from $M$ to $Y$ runs over $D$, so you can adjust for it using the back-door strategy.

In sum, you can identify the "submechanisms", and there is no direct effect, so you can piece together the submechanisms to estimate the overall effect. This will not work if $U$ infuences $M$, because then identifying the submechanisms does not work.

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  • $\begingroup$ Then, to find the effect of $D$ on $Y$, compute the effect of smoking on tar, and then the effect of tar on cancer - possibly through backdoor adjustment - and multiply the effect of $D$ on $M$ with the effect of $M$ on $Y$-- how can this actually be done? All I can see is that we have no way to measure the effect of $M$ on $Y$, we have no way to isolate the $D$ from $M$. $\endgroup$
    – Graviton
    Aug 1, 2018 at 10:22
  • $\begingroup$ BTW, with the diagram, the difference between front and back door adjustment is a lot clearer. I'm going to accept your answer ( despite that there are some parts that I'm still not quite clear of) $\endgroup$
    – Graviton
    Aug 1, 2018 at 10:22
  • $\begingroup$ I've updated my answer. $\endgroup$ Aug 5, 2018 at 15:56
  • $\begingroup$ If I'm using a standard OLS regression for front-door adjustment, the regression model would be $E(Y) = \beta_0 + \beta_1M+\beta_2D$ is that correct? $\endgroup$
    – RobertF
    Sep 16, 2020 at 14:32
  • $\begingroup$ You would take the coefficient $\beta_1$ from that regression and multiply it with $\delta_1$ from the regression $M = \delta_0 + \delta_1D + \epsilon$. $\endgroup$ Sep 17, 2020 at 15:21

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