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I'm reading "The book of why" by Judea Pearl and although I understand qualitatively what he is saying when it comes to the bias introduced by controlling for an incorrect variable (a collider, for instance), I can't visualise the flow of association/correlation that results from such control.

What happens to the diagram below when I control for B (colored in white)? Why is there a biasing path if I control for B but not if control for C?

DAG

Why does controlling for B in the diagram here introduce bias?

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  • $\begingroup$ B is a collider (two arrow point to it). If you adjust for a collider, it opens the path. So if you just adjust for B, there exists a backdoor path X <- A -> B (now open!) <- C -> Y introducing bias. $\endgroup$ – COOLSerdash Jan 18 at 18:24
  • $\begingroup$ I’m still not getting it. I could say: if I control for C, then there is a path of association of X<-A->B<-C->Y ? $\endgroup$ – Lucidnonsense Jan 18 at 18:28
  • $\begingroup$ What stops the flow? $\endgroup$ – Lucidnonsense Jan 18 at 18:37
  • $\begingroup$ No, for non-colliders, controlling stops the flow of association. So if you only control for C, there is no backdoor path because it "stops" at C. $\endgroup$ – COOLSerdash Jan 18 at 18:41
  • $\begingroup$ But what is the rule in general? $\endgroup$ – Lucidnonsense Jan 18 at 18:43
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This is more of a comment, in response to comments to Ed Rigdon's answer:

I understand that I shouldn’t control for B because it’s a collider. I want to know how the diagram looks when a variable is controlled for. Then I would be able to see everything explicitly

A good way to do this is by drawing the conditional graph by a process called graphical moralization. The steps are very simple (this is quoted almost verbatim from Greenland and Pearl, 2017) where I have just changed the variable names to match the ones in the question:

  1. If B is a collider, join (marry) all pairs of parents of B by undirected arcs (here, a dashed line will be used).
  2. Similarly, if A is an ancestor of B and a collider, join all pairs of parents of A by undirected arcs. [obviously this is not the case here]
  3. Erase B and all arcs connecting B to other variables.

So we arrive at the following graph:

enter image description here

Note that this is not a DAG because of the presence of the dashed line. To continue using DAG theory we must retain B and use the reasoning in Noah's answer where the backdoor path is shown as $X \leftarrow A \rightarrow \fbox B \leftarrow C \rightarrow Y$

Finally I often find it instructive to do simple simulation so here I simulate data according to the original DAG and show what happens when controlling for the collider:

> set.seed(15)
> N <- 100
> A <- rnorm(N, 10, 2)
> C <- rnorm(N, 5, 1)
> B <- A + C + rnorm(N)
> X <- A + B + rnorm(N)
> Y <- X + C + rnorm(N)

> m0 <- lm(Y ~ X)
> summary(m0)

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  3.28681    0.87685   3.748 0.000301 ***
X            1.06439    0.03411  31.203  < 2e-16 ***

So we obtain good estimates for the effect of X. However:

> m1 <- lm(Y ~ X + B)
> summary(m1)

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  2.82040    0.82263   3.429 0.000892 ***
X            0.68665    0.09811   6.999 3.36e-10 ***
B            0.66931    0.16452   4.068 9.65e-05 ***

Now we have a biased estimate for X

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  • $\begingroup$ Yes! This is what I was missing! Ok so 1) we join the colliding nodes together with a bidirectional arc because we conditioned on B, 3) erase all arcs from B because B is no longer a variable. But what about (2) is it because conditioning on B will also condition only any parental relations? $\endgroup$ – Lucidnonsense Jan 19 at 9:37
  • $\begingroup$ That link is very helpful as well $\endgroup$ – Lucidnonsense Jan 19 at 9:41
  • $\begingroup$ Yes you've got it. The paper in that link is very readable and pretty much tells you all you need to know about how to use DAGs. Note that I've just updated my answer to include a simulation of the DAG in your question. $\endgroup$ – Robert Long Jan 19 at 9:43
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In the model, B = A + C, and A and C are orthogonal. If you hold B constant, you induce covariance between A and C. Say you hold B constant at 10. Then, if A is 7, C is 3. If A is 6, then C is 4. This creates a confound for the X -> Y relationship, a backdoor path from X through A through the covariance with C to Y.

C is a confound for the X -> Y relationship because X and Y are joint descendants of C. Controlling for C negates that joint dependence. You don't need to control for B because B by itself is not a confound--Y is not a descendant of B, except through X.

One of the tough parts of interpreting diagrams is looking for the paths that are not there. Those omissions can be the most important part of the diagram, but without practice they will not be the focus of our attention.

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  • $\begingroup$ I can imagine drawing a link (not a causal one, just an association) between A and C (such associations already exist due to other links) because of the controlling of B (just join them up because they both point to B). But in the other example above (hyperlink), I cant draw associations when controlling for B even though it does add bias. $\endgroup$ – Lucidnonsense Jan 18 at 18:57
  • $\begingroup$ But what is the rule? The rule that I can apply to anything. Can I backtrack along casual paths ad infinitum? $\endgroup$ – Lucidnonsense Jan 18 at 19:20
  • $\begingroup$ (+1) this is a really nice answer to whar is a very tricky problem - explaining collider bias :) $\endgroup$ – Robert Long Jan 18 at 19:38
  • $\begingroup$ @Lucidnonsense the rule is not to condition on a collider. $\endgroup$ – Robert Long Jan 18 at 20:25
  • $\begingroup$ I understand that I shouldn’t control for B because it’s a collider. I want to know how the diagram looks when a variable is controlled for. Then I would be able to see everything explicitly $\endgroup$ – Lucidnonsense Jan 18 at 21:10
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There are seven rules of association. In the first four, $R$ and $T$ are associated with each other:

$$R \rightarrow T$$ $$R \rightarrow S \rightarrow T$$ $$R \leftarrow S \rightarrow T$$ $$R \rightarrow \fbox S \leftarrow T$$

In the second three, $R$ and $T$ are not associated with each other through the path:

$$R \rightarrow \fbox S \rightarrow T$$ $$R \leftarrow \fbox S \rightarrow T$$ $$R \rightarrow S \leftarrow T$$

A box around a variable means we are conditioning on the variable.

The path $X \leftarrow A \rightarrow B \leftarrow C \rightarrow Y$, when not conditioning on $B$, is a closed path, meaning that $X$ and $Y$ are not associated with each other through this path. This is because the chain of association breaks when two arrows point to the same variable in a path (here, two arrows point to $B$). In this case, $B$ is called a collider.

Conditioning on $B$ opens the path of association. That is, $X \leftarrow A \rightarrow \fbox B \leftarrow C \rightarrow Y$ leaves open the association between $X$ and $Y$ because conditioning on a collider or a descendant of a collider opens the path of association. Other paths between $X$ and $Y$ may be open or closed, but when through this particular pathway, they are associated. This association is "noncausal" because it's an association that does not represent the causal effect of $X$ on $Y$.

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  • $\begingroup$ Great thank you! $\endgroup$ – Lucidnonsense Jan 19 at 9:27

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