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In the linear regression model $Y= X\beta+u$ , when $X$ is endogenous , the OLS estimator will be biased. If we can find an instrumental variable $Z$ which is correlated with $X$, but uncorrelated with the error term, we can get the unbiased coefficient $\beta$. This requirement condition is also what I learned from my econometric classes. For example in Wooldridge book Introductory Econometrics Modern Approach (2018) chapter 15, the condition is $ cov(z, u)=0$,$cov(z, x)\neq0$.

However, when I check some material on line, it seems there is another type of requirements. The requirements are listed below:

(i) $Z$ is related to $X$,causally or not causally;

(ii) $Z$ affects the outcome variable $Y$ only through $X$ ($Z$ does not have a direct influence on $Y$ which is referred to as the exclusion restriction);

(iii) There is no confounding for the effect of $Z$ on $Y$

The same condition is used in Robins (2020) Causal Inference book chapter 16.

The key difference is in the second set condition (iii) $Z$ and $Y$ don't share common confound factors, which is not required in first set conditions. From my understanding, if we have the condition (ii), it means $Z$ can only affect $Y$ through $X$. In this case, we should have $ cov(z, u)=0$. Why do we bother to the condition (iii). If condition (iii) is violated, does it mean $Z$ could be related to $Y$ through other channel rather than $X$.

Do I miss something here?

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  • $\begingroup$ 2 and 3 are easily conflated; technically, they both translate to Cov(u,z)=0. One reason to distinguish between them is in the case of a randomly assigned instrument. Then, we know that 3 holds, but 2 may not. Further, in the case of an instrument that could be as good as random, it is easier to explore whether 3 holds (edit: actually 3 = as good as random) by controlling for potential confounders. 2 is trickier to ”test” since something that is caused by Z is not a good control variable and may introduce confounding. $\endgroup$
    – Jonathan
    Dec 21 '20 at 22:10
  • $\begingroup$ @Jonathan Is the condition (ii) enough to the conclusion Cov(u,z)=0? Or in the multiple linear model setting, when we say Cov(u,z)=0, actually we control for other control variables as well. However, condition (ii) just state the Cov(y, z |x)=0, and we didn't take account into other control variables. And that why the condition(iii) is needed, to make Z as good as random. So condition (ii) and (iii) together is the same meaning as Cov (u,z)=0 in a multiple linear regression model. $\endgroup$ Dec 21 '20 at 23:21
  • $\begingroup$ I don’t quite follow, but last sentence is correct. Both 2 and 3 are necessary for Cov(u,z)=0. $\endgroup$
    – Jonathan
    Dec 22 '20 at 17:07
  • $\begingroup$ @Jonathan I know why I have confusion here. Since I didn't formally learn the DAG setting, I misunderstood the meaning of "affect" and the " effect of confounder". I basically treat them the same thing before. Then I think condition (iii) is redundant if we have condition (ii). $\endgroup$ Dec 22 '20 at 22:10
  • $\begingroup$ @Jonathan Condition (ii) says Z only affects Y through X, so you cannot find another path which is from Z to Y. If you have that path, it means Z must be correlated with u conditional on X. In addition, condition (iii) rules out Z and u are correlated through the effect of confounder. Previously, I treat this association as the affect of Z on Y which is not through X as well (i.e. condition ii). $\endgroup$ Dec 22 '20 at 22:13

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