Economics Stack Exchange is a question and answer site for professional and academic economists and analysts. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have three variables: $y$, $x_1$ and $z$.

I'm interested in finding the effect of $x_1$ on $y$. Unfortunately $x_1$ is endogenous so I use an instrument z.

I know that $z$ is correlated with $x_1$. The problem is, $z$ is correlated with another variable ($x_2$) that arguably has an effect on $y$. $z$ is unlikely to have a causal effect on $x_2$, but $x_2$ may still be endogenous for other reasons (e.g. reverse causality, etc.).

Will the following procedure give me a consistent estimate of beta-1: two stage least squares with $y$ regressed on $x_1$ and $x_2$, with $z$ as an instrument for $x_1$?

share|improve this question
Why is this just not an endogeneity issue due to omitting $x_2$ and that can be solved like you said: Include the omitted variable $x_2$ and you just happen to have an IV $z$ in hand that's correlated with (both) regressors? – Michael Apr 7 '15 at 15:10
Since nobody answered, I'm giving my two cents. If you cannot find an instrument for $x_2$ (the variable itself, and $z$ canno) , you will not be able to satisfy rank condition for identification ( $E(\mathbf{z}_i\mathbf{x}'_i)$ be full column rank is necessary to find more instruments than regressors). – An old man in the sea. May 7 '15 at 12:34

Actual availability of regressors may be an issue here, but if all four mentioned variables are available, the situation is as @Michael mentioned in a comment:

Since $X_2$ is correlated with $Y$, it should be included in the regression specification as a "control".

$$Y = \beta_0 + \beta_1X_1 + \beta_2X_2 + u$$

This is intuitive, but it also takes care of the technical aspect. Then since $X_2$ is endogenous, i.e we have ${\rm Cov}(X,u) \neq 0$, but we have an instrument that is not correlated with the error term $u$ since $X_2$ has been included in the specification, we end up with

$$Y=\beta_0 + \beta_1Z + \beta_2X_2 + u$$

for which the OLS estimator will be asymptotically consistent, since (writing $\mathbf W = [1:Z:X_2]$)

$$ \hat \beta = \beta + \left (\mathbf W'\mathbf W\right)^{-1}\mathbf W' \mathbf u$$

and its consistency will depend on the following probability limit being zero

$${\rm plim}\mathbf W' \mathbf u ={\rm plim} \left[\begin{matrix} \frac 1n\sum u_i\\ \frac 1n\sum Z_iu_i\\ \frac 1n\sum X_{2i}u_i\\ \end{matrix}\right] \rightarrow {\rm plim} \left[\begin{matrix} \frac 1n\sum E(u_i)\\ \frac 1n\sum E(Z_iu_i)\\ \frac 1n\sum E(X_{2i}u_i)\\ \end{matrix}\right]= \left[\begin{matrix} 0\\ 0\\ 0\\ \end{matrix}\right]$$

holds because both $Z$ and $X_2$ are orthogonal/uncorrelated to the error term $u$.

In other words, the fact that the regressors $X_2$ and $Z$ are correlated with each other does not create any problem (as long as of course the co-linearity between them is not near-perfect), on the contrary it more generally justifies including both in the regression specification, making it "multiple regression" -it exists exactly for such cases.

share|improve this answer
but isn't the user in the question saying that $x_2$ may be endogenous, and hence not possible to be used as an IV? – An old man in the sea. May 8 '15 at 7:52
@Anoldmaninthesea. You may be right (in which case my answer does not apply). The problem is that the OP seems to treat as equivalent the notions "x2 is correlated/has an effect on Y" with "x2 is endogenous". This is wrong of course, they do not mean the same thing. The way the OP phrased it gave me the impression that he really meant the first of the two only, in which case my answer does apply. If he knows the difference and means both for x2, (that x2 is correlated with y and correlated with some fifth variable that is included in the error term), then again my answer does not apply. – Alecos Papadopoulos May 8 '15 at 10:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.