So I was tutoring an undergraduate\/master’s level applied econometrics course, and several students asked me why it is necessary to include also the exogenous control variables in both stages of 2SLS – More specifically, wouldn’t this double-count the correlation between the instrument and the controls?<\/p>\n\n\n\n
This turns out to be a fascinating question that I somehow have never thought carefully about, and seemingly also lacks documentation online (Although I am known as a horrible search engine user…). Hence I spent half an afternoon creating an illustration, which, to maximize accessibility to an audience with little matrix algebra background, is purely algebraic:<\/p>\n\n\n\n
For simplicity, consider the following regression
$$\\qquad\\qquad Y_i=\\beta_0+\\beta_1X_i+\\beta_2K_i+u_i\\,, \\qquad\\qquad(1)$$<\/p>\n\n\n\n
where $X_i$ and $u_i$ are correlated, and $K_i$ is exogenous. We will need IV(s) for $X_i$ because it’s endogenous.<\/p>\n\n\n\n
A quick reminder that the second stage of 2SLS is running the following regression<\/p>\n\n\n\n
$$Y_i=\\beta_0+\\beta_1\\hat{X}_i+\\beta_2K_i+\\nu_i\\,, $$<\/p>\n\n\n\n
where $\\hat{X}_i$ is the predicted value of $X_i$ from the first stage, whatever that might be.<\/p>\n\n\n\n
It makes sense that $K_i$ needs to be included in the second stage, as that is just running (1) using $\\hat{X}_i$ instead. In the first stage, for simplicity let’s say that we are only using one instrument, $Z_i$. So the question becomes, why does<\/p>\n\n\n\n
\\begin{equation}
\\qquad\\qquad X_i=\\pi_{0,1}+\\pi_{1,1}Z_i+\\pi_{2,1}K_i+\\xi_{1,i} \\qquad\\qquad(2)
\\end{equation}<\/p>\n\n\n\n
make more sense than<\/p>\n\n\n\n
\\begin{equation}
\\qquad\\qquad X_i=\\pi_{0,2}+\\pi_{1,2}Z_i+\\xi_{2,i}\\,? \\qquad\\qquad(3)
\\end{equation}<\/p>\n\n\n\n
Well, it’s because of the omitted variable bias<\/strong>! If $Z$ and $K$ might be correlated (The assumptions for $Z$ to be a valid IV did not prohibit $Z$ to do so!), then we could cause ourselves troubles if we didn’t account for that in the first stage.<\/p>\n\n\n\n For illustrative purposes, assume $K_i=C_i+\\theta Z_i$ where $corr(C_i,Z_i)=0$, and we are fully aware of that (i.e, actually both $C_i$ and $\\theta$ are known). Then (2) becomes<\/p>\n\n\n\n \\begin{align*} X_i&=\\pi_{0,1}+\\pi_{1,1}Z_i+\\pi_{2,1}(C_i+\\theta Z_i)+\\xi_{1,i} \\\\ &=\\pi_{0,1}+(\\pi_{1,1}+\\pi_{2,1}\\theta)Z_i+\\pi_{2,1}C_i+\\xi_{1,i}\\,, \\end{align*}<\/p>\n\n\n\n