# 2SLS and Control Functions (identity of estimator)

I would like to show the identity of the 2 Stage Least Squares estimator and the control function estimator.

Assume a linear regression model

$$y = X\beta + u$$ where $$X =[X_1 \ X_2]$$ is $$n \times k$$ and where $$X_2$$ is endogenous ($$\mathbb E[x_{2i}u_i]\not =0$$). Let $$Z$$ be a $$n \times l$$ matrix $$l\geq k$$ of instruments (that includes $$X_1$$ which is assumed to exogenous $$\mathbb E[x_{1i}u_i]=0$$) and

$$X = Z\Gamma + V$$

where $$\Gamma:=\mathbb E[z_iz_i^\top]^{-1}\mathbb E[z_i y_i]$$ such that by construction $$\mathbb E[z_iv_i^\top]=0$$

The 2SLS Estimator

The two stage least squares estimator is defined as $$\hat \beta_{2SLS}:=(\hat X^\top X)^{-1}(\hat X^\top y)$$ where $$\hat X = Z(Z^\top Z)^{-1} Z X = P_Z X$$.

The control function approach regress $$X$$ on $$Z$$ to get residuals $$\hat V$$

$$X = Z\hat \Gamma + \hat V$$ and then includes these residuals in a regresssion of $$y$$ on $$X$$ and $$\hat V$$ to get the estimator

$$\begin{bmatrix}\hat b \\ \hat \rho \end{bmatrix} = \begin{bmatrix} X^\top X & X^\top \hat V \\ \hat V^\top X & \hat V^\top \hat V\end{bmatrix}^{-1} \begin{bmatrix} X^\top y \\ \hat V^\top y\end{bmatrix}$$

where the result I am looking for then is that $$\hat b = \hat \beta_{2SLS}$$.

When we regress the regressors $$Z$$ on the instruments $$X$$, we 1. get residuals $$\widetilde{Z}:=M_{X}Z=Z-X(X'X)^{-1}X'Z$$.
We then, 2., regress $$y$$ on $$Z$$ and $$\widetilde{Z}$$, $$\begin{equation} y=Z\widehat{\delta}+\widetilde{Z}\widehat{\theta}+\widehat{u} \end{equation}$$ Recall that the Frisch-Waugh-Lovell theorem states that we can obtain subvectors of coefficients on variables of "interest" of a long regression by regressing the residuals of a regression of the dependent variable on the remaining ("non-interesting'') explanatory variables on the residuals of a regression of the submatrix of interest on the remaining variables.
We use FWL in step 2 to show that $$\begin{eqnarray*} \widehat{\delta}&=&(Z'M_{\widetilde{Z}}Z)^{-1}Z'M_{\widetilde{Z}}y\\ &=&(Z'(I-P_{\widetilde{Z}})Z)^{-1}Z'(I-P_{\widetilde{Z}})y \end{eqnarray*}$$ Now, $$\begin{eqnarray*} P_{\widetilde{Z}}&=&M_{X}Z(Z'M_{X}'M_{X}Z)^{-1}Z'M_{X}\\ &=&M_{X}Z(Z'M_{X}Z)^{-1}Z'M_{X} \end{eqnarray*}$$ so that $$(I-P_{\widetilde{Z}})Z=Z-M_{X}Z(Z'M_{X}Z)^{-1}Z'M_{X}Z=Z-M_{X}Z=P_{X}Z$$ such that $$\widehat{\delta}=(Z'P_{X}Z)^{-1}Z'P_{X}y$$