# Should one control for non-confounders?

The standard error of the variable of interest $$x$$ can be calculated as

$$s.e.({\hat\beta_x})=\sqrt{VIF_x\frac{\sigma_\varepsilon^2}{nVar(x)}}$$

As usual, $$\sigma_\varepsilon^2=\sum_i\varepsilon_i^2$$ is the variance of the regression error and $$VIF_x$$ is the variance inflation factor of $$x$$.

If one now controls for a second variable $$z$$ (which happens to be highly significant), one inevitably reduces the residuals $$\varepsilon_i$$ what leads in turn to a reduction of $$\sigma_\varepsilon^2$$. Because $$z$$ is not a confounder, $$VIF_x$$ remains unchanged. Ultimately, $$s.e.(\hat\beta_x)$$ would go down.

If my thinking is right, then one would try to control in large datasets (say, 100k observations) for as many highly significant control variables as possible. That is because the loss in degrees of freedom is negligible and the $$p$$-value of the variable of interest goes down.

Whether to control for non-confounders seems to be quite an important question to get right in applied statistics. I am therefore wondering whether my argument is correct or whether I got something wrong?

Best wishes, Tom

• Is $x$ your main variable of interest and are you already controlling for confounders? Well, your current VIF framework is one way to think about it. The bigger issue is actually separating confounders from non-confounders, (identifying their right functional forms ...). But if you can identify them, then by all means, use them to improve precision of the $x$. Using non-confounders to improve precision is the classic application of ANCOVA, when the non-confounders are continuous. – Heteroskedastic Jim Nov 20 '18 at 14:33
• Thank you Jim for your reference to ANCOVA - never heard about this before, but certainly helps to understand the "why?" question. – Tom Pape Nov 20 '18 at 16:00