# is endogeneity a problem in the first-stage regression in a two-stage least square regression?

We have this model:

$$D=\alpha_1+\gamma_1 Z + \epsilon_1$$

$$Y=\alpha_2+\gamma_2 D + \epsilon_2$$

Notations are as usual: Y is the outcome, D is the treatment, Z is the instrument

Two conditions for instrument variable to work in the literature: relevance (there is a correlation between Z and D) and exogeneity (no correlation between Z and error term $$\epsilon_2$$)

However, it seems no one cares about endogeneity in the first stage regression, which is the correlation between Z and $$\epsilon_1$$. Why is that the case? I have never seen any applied micro paper addressing this.

Yes it is a problem. The first stage itself has to satisfy the same assumptions that standard OLS would and $$cov(Z,\epsilon_1)\neq 0$$ would violate them (see A Guide to Modern Econometrics by Verbeek).

Furthermore, actually the two conditions you mention are not enough. The instrument should also not be 'weak', that is the first stage should have $$F$$-statistics above $$10$$ (as a rule of thumb). Also the instrument should have an effect on $$Y$$ only through $$D$$ - the so called exclusion restriction (see Mostly Harmless Econometrics by Angrist and Pischke).

• relevance is equal to weak instrument. relevance mean the the correlation between Z and D must be significantly larger than 0. exogeneity condition is similar to exclusion restriction. they are just different names for the same thing – FARRAF Sep 27 '20 at 22:35
• @FARRAF no it’s not equal. Even with non-zero correlation, and individual betas being significant using t test, the F-statistics can be <10 – 1muflon1 Sep 27 '20 at 22:36

I'm going to assume that by "conditions for instrumental variables to work" you mean "instrumental variables is consistent." However, there are other properties to consider, like performance in small samples, etc. $$\newcommand{\Cov}{\text{Cov}}$$

In this simple case, the IV estimator in sample of size $$n$$ is $$\hat \gamma_2 = \frac{\sum_{i=1}^n (Z_i - \bar Z)(Y_i - \bar Y)}{\sum_{i=1}^n (Z_i - \bar Z)(D_i - \bar D)}.$$ For the instrumental variables estimator to be consistent, we only need $$\text{Cov}(Z, \epsilon_2) = 0 \quad \text{ (instrument exogeneity)} \tag{1}$$ and $$\text{Cov}(Z, D) \neq 0. \quad \text{ (instrument relevance)} \tag{2}$$ As long as these conditions are satisfied, it doesn't matter if $$Z$$ is correlated with $$\epsilon_1$$. The IV estimator will be consistent. To see this, let's analyze the model in terms of populations, \begin{align} \Cov(Z,Y) &= \gamma_2 \Cov(Z, D) + \Cov(Z, \epsilon_2), \end{align} and solve $$\gamma_2 = \frac{\Cov(Z, Y)}{\Cov(Z, D)} - \frac{\Cov(Z, \epsilon_2)}{\Cov(Z, D)}.$$ If conditions (1) and (2) hold, then $$\frac{\Cov(Z, \epsilon_2)}{\Cov(Z, D)} = 0$$ and the IV estimate
$$\gamma_2 = \frac{\Cov(Z, Y)}{\Cov(Z, D)}$$ is well-defined. The estimate is consistent since the probability limit is the population analog, $$\hat \gamma_2 \overset{p}{\rightarrow} \frac{\Cov(Z, Y)}{\Cov(Z, D)} = \gamma_2.$$

Note that weak instruments become a problem as soon as we relax the assumption that $$\text{Cov}(Z, \epsilon_2) = 0$$ exactly, as you might want to do when analyzing the properties of the estimator in small samples.

So, are there any consequences if $$Z$$ is correlated with $$\epsilon_1$$? Well, it means that a regression of $$D$$ onto $$Z$$ will give a biased and inconsistent estimate of $$\gamma_1$$. However, the usual assumption underlying IV is that you don't really care about estimating $$\gamma_1$$. You want to consistently estimate $$\gamma_2$$, which you can do as long as assumptions (1) and (2) hold, as shown above.