Causal modeling in the presence of a latent variable

Suppose that four variables of $$X$$, $$Y$$, $$L$$, and $$C$$ have the following relationships in the form of directed acyclic graph.

$$X$$, $$Y$$, and $$C$$ are observable variables while $$L$$ is a latent (unobservable) variable. If one is interested in modeling the relationship between $$X$$ and $$Y$$ through, for example, regression, is it reasonable to include $$C$$ as a covariate and condition the $$X$$-$$Y$$ association on $$C$$? Alternatively, how about when one is interested in the association between $$C$$ and $$Y$$?

In the first scenario, C is a descendant of a mediator, so contrary to previous answers, yes it would be harmful to include if your interest is in estimating the total effect of X on Y. See DAG 12 here:

In your first scenario, you certainly can include $$C$$ on the RHS of your linear regression model, but you shouldn't need to if your DAG is correct. If your DAG is correct, then $$C$$ does not cause $$Y$$ in any way, and is not a confounder (does not set up a back door path). All of the causal effect of $$X$$ on $$Y$$ is mediated through $$L:$$ that's fine. There's nothing special you need to do, just regress $$Y$$ on $$X.$$
In your second scenario - finding the association between $$C$$ and $$Y$$ - you have a problem: $$L$$ is a confounding variable. Moreover, if you have no other variables present anywhere, then I don't see a way to estimate the causal effect of $$C$$ on $$Y.$$ $$X$$ is not an instrument, because if it were, it would have be pointing into $$C$$ and nothing else. You're not even really sure that $$C$$ has a causal effect on $$Y.$$ Technically, of course, your DAG says it doesn't have a causal effect on $$Y.$$ I'd say you're stuck, here.
• (1) It shouldn't hurt if $C$ is included in the first model. It shouldn't change anything! (2) If the paths you specify are added to the graph, then the only change to the answer to the first scenario is that it might be a bad idea to include $C$. You might get a collider bias, even though $C$ isn't a confounder. For the second scenario, there's still no causal effect of $C$ on $Y,$ but if you wanted to regress anyway, you should include $X$ because it is a backdoor path. Apr 14, 2023 at 16:46