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Suppose the DAG in the population is as follows:

enter image description here

We observe both $X_1$ and $X_2$.

We are interested in the effect of $X_1$ on $Y$. We want to use OLS to estimate the relationship.

Now if I take $X_2$ into account:

$Y=\beta_0 + \beta_1 X_1 + \beta_2 X_2 + \epsilon$

We have a multicollinearity problem between $X_1$ and $X_2$.

If we omit $X_2$:

$Y=\beta_0 + \beta_1 X_1 + \epsilon$

We have correlation between the independent variable and the error term.

What is the appropriate course of action?

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that's not a bug that's a feature, you just showed that when you control for x2, x1 has no effect on y, which is exactly what you wanted to know. Good job.

You also learned that x1 is highly correlated with x2 or that x1 might even be some linear transformation of x2.

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  • $\begingroup$ How do you know that x1 has no effect on x2, if "One of the features of multicollinearity is that the standard errors of the affected coefficients tend to be large. In that case, the test of the hypothesis that the coefficient is equal to zero may lead to a failure to reject a false null hypothesis of no effect of the explanator, a type II error. " (wikipedia)? $\endgroup$ May 12 at 16:21
  • $\begingroup$ x1 does not have an effect on y AFTER correcting for x2, so you cannot explain anything more in y based on x1 that you didn't already explain by x2. Standard errors are big because there is a lot of shared variance between x1 and x2, which implies that there is a little variance that is not shared, which means that there is a little information in x1 that is not already in x2 what I was talking about. You can calculate partial correlation or partial r^2 to see that $\endgroup$
    – rep_ho
    May 12 at 17:02
  • $\begingroup$ or in other words, you cannot estimate individual coefficients well due to high standard errors due to colinearity, because a lot of the variance is shared between x1 and x2. However, you can still calculate how much variance in y is explained by both of these variables together without any issues, and since variance explained by 2 variables is close to variance explained by individual variables, you infer that there is not much information in x1 that is not already in x2 $\endgroup$
    – rep_ho
    May 12 at 17:08

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