I am working on a problem where I have a "target" variable $Z$ that I know for sure is influenced by a "predictor" $Y$. I also have a second predictor $X$ that is correlated with $Y$ (about -.3), and I would like to know whether is has an influence on $Z$

If I fit a linear model for $Z$ on $X$, I get a very significatively non-zero coefficient for the slope, but of course, this tells nothing, as any variable that is correlated with $Y$ would do so.

Next thing I tried was to fit a linear model for $Z$ on both $X$ and $Y$. The coefficients for both variables is significant, but, since the variables are correlated, I am not sure about the power of the model to "distinguish" between the effects of each of the predictors.

So I finally went for a third approach: I will build the model in two steps: First, I will fit a model for $Z$ on $Y$ to eliminate its effects. Then, I'll take its residuals $\epsilon_i$ and fit a second linear model for $\epsilon$ on $X$

Now my question is: Is this a valid method to achieve what I actually want to achieve? Also, should I add more complex terms in the "first-step" model, to eliminate other non-linear effects of $Y$ on $Z$?

NOTE: When I say "significant", I don't mean one of those $0.04$ $p$-values that fill mediocre research papers. I mean something more in the roum of $10^{-10}$, as sample and effect sizes are both quite big

  • $\begingroup$ Would you please post the data, or a link to the data? $\endgroup$ Jun 28, 2019 at 12:59
  • $\begingroup$ Unfortunatelly, the data belongs to my company, not myself $\endgroup$
    – David
    Jun 28, 2019 at 23:04
  • $\begingroup$ As far as I can see your third approach is the same as your second one. Have you actually tried it? $\endgroup$
    – mdewey
    Aug 4, 2020 at 10:16
  • $\begingroup$ None of this modeling is capable of assessing any amount of "influence:" all it can do is measure associations. When explanatory variables are correlated and you don't have any other variables to (conditionally) remove those correlations, it isn't possible to decompose their associations with the response variable, anyway. You have to think of them as working together. $\endgroup$
    – whuber
    Aug 11, 2020 at 20:32


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