Can we just "pre-test" the backdoor criterion? I am trying to use DAGs to think more carefully about my regression models. I have a question about the "backdoor criterion", as usually seen in the DAG below (we are interested in the effect of $X$ on $Y$). One way to deal with confounding in this case is to include $Z$ in the regression, i.e.~control for it.
But if $Z$ is something that we can measure - for concreteness, in an economic context, $Z$ might be GDP, which often shows up as potential confounder in my DAGs - is it legitimate to first regress $X$ on $Z$ in order to verify that the hypothesised link $Z \to X$ actually exists? If it does, then we go ahead and include $Z$ in the model; if not, the DAG was incorrectly specified, and $Z$ is not in fact a confounder and does not need to be included.
Is this approach reasonable? What am I missing?


Indeed, given the DAG, you should only see a correlation between X and Z if there is a direct link between the two, and thus you could test for a correlation directly. These and similar tests are done by all causal discovery algorithms that automatically create the DAG from data, such as the PC algorithm.
However, from the practical perspective of a multiple regression, there are two caveats to this approach:
a) How do you perform the test? A n.s. p-value is not a proof that effect size is zero. Even if the CI is small and overlaps zero, you cannot exclude small confounding effects and thus you have to trade off a small possible bias (from not including a weak confounder) against the improved precision that you gain by having a model with fewer d.f. Plus, there is the issue of post-selection inference, i.e. you need to correct p-values for the tests performed in the model selection.
b) Second, you say that if a variable is not a confounder, it "does not need to be included". I would add to this: for reasons of bias (= causal perspective). However, there are other reasons to include a variable in a multiple regression. Most importantly, if Z is a strong predictor, i.e. if there is a strong link Z->Y, including Z in the regression can reduce the uncertainty on the estimate X->Y, even in the absence of a confounding link Z->X.
