# Inconsistent assumptions about instrumental variable

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?

• 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. Dec 21 '20 at 22:10
• @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. Dec 21 '20 at 23:21
• I don’t quite follow, but last sentence is correct. Both 2 and 3 are necessary for Cov(u,z)=0. Dec 22 '20 at 17:07
• @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). Dec 22 '20 at 22:10
• @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). Dec 22 '20 at 22:13