I will state the question first then my work.


We have a regression model, $Y_i=\beta_0+\beta_1X_i+\mu_i$ where $Cov(\mu_i,X_i)=0$ is not guaranteed. Suppose that $Z_i$ is an instrumental variable and we have $\hat{\beta_1}$ be the two least squares estimator and we have $\hat{\beta_0}=\bar{Y_n}-\beta_1\bar{X_n}$. Show that $\widehat{Cov(\hat{\mu},Z_i)}$ is always zero even $Cov(\mu,Z_i)$ is not always $0$.

My thoughts and work:

We know that the residual is $\hat{\mu_i}=Y_i-\hat{\beta_0}-\hat{\beta_1}X_i$ so...


and thus the second term would be equal to $0$ since $B_0$ doesn't have an $i$ term. But then I'm not sure how to proceed IF this is the right way to move forward.

Step by step help would be greatly appreciated!


1 Answer 1


Given that this is a self-study question I will try to give hints to direct you towards the solution.

When you use instrumental variables via two stage least squares, for instance, you have a first stage and a second stage respectively: $$\begin{align} X_i &= \delta_0 + \delta_1 Z_i + \eta_i & \text{first stage}\newline Y_i &= \beta_0 + \beta_1 X_i + \mu_i & \text{second stage} \end{align}$$

A possible solution path would be to plug the first stage into the second stage and then use the fact that an OLS regression of this equation separates the variation in $Y_i$ into an observed part and an unobserved part. Using linear projections you can then show why this observed part (your $Z$ in this case) will be orthogonal to $\mu_i$. This is basically a purely mechanical effect from the regression (see here for an explanation), hence the estimated covariance between $Z_i$ and $\mu_i$ is going to be zero.

The next step would be to make an argument for why the population covariance $Cov(Z_i,\mu_i)\neq 0$. This basically aims at the point as to why two stage least squares could be biased. Note though that there is no test for whether $Cov(Z_i,\mu_i)=0$ in the population or not - it is always made as an assumption but it can never be tested.


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