2SLS and Control Functions (identity of estimator)

Question

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 $\hat b = \hat \beta_{2SLS}$.

Answer 1

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 $$ \widehat{\delta}=(Z'P_{X}Z)^{-1}Z'P_{X}y $$