1
$\begingroup$

This is a question about the definition of confounding, and/or about statistics pedagogy.

Suppose that you're doing a study to see if $X$ and $Y$ are associated, but they are not. Unbeknownst to you, another variable $Z$ is significantly associated with $Y$, but $Z$ is independent of $X$. Now, it just might happen, by chance, that in your sample $Z$ is associated with $X$. Should we say that $Z$ is a confounding variable?

I would like to say "no", because $X$ and $Z$ are not associated in reality, so that variation in $Z$ is just contributing to the noisiness of $Y$, and so if the statistics are working properly, they will take this into account.

But, you could also argue that in this sample, they are confounded.

This is an issue I'm running into while teaching an introductory stats class; the books are suprisingly vague on whether confounding needs to have an association between $X$ and $Z$. Maybe I just need different language to talk about this situation?

note: It seems like there might be different operational definitions of "confounding" out there; so please don't just cite one definition without saying why you think it's the right one.

$\endgroup$

1 Answer 1

0
$\begingroup$

The way I teach it, confounders are variables that are correlated with X and have a causal effect on Y. If Z is caused by X then Z is a mediator or intervening variable. If Z causes both X and Y then it is extraneous. Contrary to the wiki entry you reference, I view confounding as separate from extraneous. Confounding variables taint the relationship you observe between X and Y when Z is uncontrolled but X still 'causes' Y. Extraneous variables render your interpretation of the relationship between X and Y (i.e. X causes Y) spurious. Thus, confounding and extraneous are qualitatively different with the latter being more damning than the former and thus I see value in keeping them conceptually separate. In either case, omitting Z results in omitted variable bias.

You pose an interesting question in that if Z is a confounder in the sample/data on hand but is not significant, is it still a confounder? I would say no. Theoretically, I could see an argument that if Z has a relationship with X and Y in the sample (i.e. the point estimate is not zero) then it is a confounder at least in the sample and should remain in the model even if it is non significant. However, the reality is in any given sample any two variables will share some degree of covariance unless they are orthogonal by design. Thus I would say that for a variable to be a confounder it must meet the logical requirements I laid out above (i.e. correlated with X and causal effect on Y) AND those relationships must be statistically significant. If any of those criteria are not met then Z is not a confounder and does not belong in the model. Others may disagree.

$\endgroup$
1
  • $\begingroup$ "AND those relationships must be statistically significant" : The question is philisophical, but what if Z and X are uncorrelated in the population, but we don't know that, and by chance, they are correlated in the sample with, say, p=0.01? $\endgroup$
    – petrelharp
    Commented Apr 12, 2015 at 3:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.