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So I have a fairly basic setup, with $X\rightarrow Y$. However, I've run across a potential third variable, Z, that is probably correlated with both X and Y. However, the causality for Z is unusual for an omitted variable bias. All the examples of omitted variables I've read have the causality $Z\rightarrow X$ and $Z\rightarrow Y$. In my model, however, the causality is reversed: $X\rightarrow Z$ and $Y\rightarrow Z$. Do I still need to include Z in my regression because it is correlated with both variables, or can I squeak my way out of it because causality runs in the other direction?

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If you regress $y_i = \alpha + \beta X_i + \epsilon_i$ instead of the long regression $y_i = \alpha + \beta X_i + \delta Z_i + \epsilon_i$, the omitted variable bias formula is $$\widehat{\beta} = \beta + \delta \frac{Cov(X_i,Z_i)}{Var(X_i)} $$

For a proof of this see here. This shows you that the bias relates to the covariances. For this point it matters little whether $X$ causes $Z$ or vice versa. Given that $Cov(X_i,Z_i)\neq 0$ and $\delta \neq 0$ (since $Z$ is related to $Y$, again the way of causation does not make a difference), your short regression will suffer from omitted variable bias.

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To address the comments it should also be mentioned that the long regression is not without problems either. Including a variable in a regression that in itself is an outcome that is determined by the dependent variable leads to the problem of "bad controls" (see section 3.2.3 in Angrist and Pischke, p. 47). So whether you use the long- or the short-regression you will get biased estimates in either case. The typical solution would be to omit $Z_i$ from the model (I assume it is unobserved which is the reason for it to be an omitted variable in the first place) and to use an instrumental variable for $X_i$.

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    $\begingroup$ But typically, one wouldn't want to control for an intermediate variable, eg $X \rightarrow Z \rightarrow Y$. Your algebra still holds of course, but controlling for $Z$ would be removing some of $X$'s bona-fide influence on $Y$. Can you address that point in your answer? $\endgroup$
    – Andrew M
    Jan 27, 2015 at 2:16
  • $\begingroup$ @AndrewM I hope the extended discussion addresses your concern. $\endgroup$
    – Andy
    Jan 27, 2015 at 6:28

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