Asymptotic var of IV estimator

Question

This post is asked again due to lack of answers first time around. Suppose we have a linear model $y=Q+Rx+error$, where $E(error)=0$, and $z$ is an instrument for $x$ (endogenous) where the correlation between the instrument and the error is 0 but that between the instrument and the endogenous $x$ is not zero. Moreover, $E(error$$^2$$|z)$=$\sigma^2$. Show that the asymptotic variance of ${\sqrt\ N}$*(estimator of R-true R) can be written as $\sigma^2$/($Corr(z,x)^2$*Var(x)), where estimator of R is the sample analogue of R= $(E(zx)$^-1)$E(zy)$.

Answer 1

You just gotta get your hands dirty.

Let $\mathbf x_i'=(1, x_i)$, $\mathbf X=(\mathbf x_1, \mathbf x_2, \dots, \mathbf x_n)'$, $\mathbf z_i'=(1, z_i)$ and $\mathbf Z=(\mathbf z_1, \mathbf z_2, \dots, \mathbf z_n)'$. The IV estimator is:

$$ \hat{\mathbf{R}}=(\mathbf{Z'X})^{-1}\mathbf{Z'y}. $$

The asymptotic distribution is: $$ \sqrt{n}(\hat{\mathbf{R}}-\mathbf{R}) {\buildrel d \over \longrightarrow} N(\boldsymbol{0}, \sigma^2\mathbf{Q^{-1}_{ZX}}\mathbf{Q_{ZZ}}\mathbf{Q^{-1}_{XZ}}) $$ where $n^{-1}\mathbf{Z'X}{\buildrel p \over \longrightarrow}\mathbf{Q_{ZX}}$, $n^{-1}\mathbf{Z'Z}{\buildrel p \over \longrightarrow}\mathbf{Q_{ZZ}}$ and $\mathbf{Q_{XZ}}=\mathbf{Q'_{ZX}}$. Looking at these more closely:

$$ \frac{1}{n}\mathbf{Z'X}=\frac{1}{n}\sum_{i=1}^n\mathbf{z}_i\mathbf x_i'{\buildrel p \over \longrightarrow}E(\mathbf{zx}')=\begin{pmatrix} 1 & E(x) \\ E(z) & E(xz)\end{pmatrix}=\mathbf{Q_{ZX}}\\ \frac{1}{n}\mathbf{Z'Z}=\frac{1}{n}\sum_{i=1}^n\mathbf{z}_i\mathbf z_i'{\buildrel p \over \longrightarrow}E(\mathbf{zz}')=\begin{pmatrix} 1 & E(z) \\ E(z) & E(z^2)\end{pmatrix}=\mathbf{Q_{ZZ}} $$ The whole thing together: $$ \mathbf{Q^{-1}_{ZX}}\mathbf{Q^{}_{ZZ}}\mathbf{Q^{-1}_{XZ}}=\begin{pmatrix} 1 & E(x) \\ E(z) & E(xz)\end{pmatrix}^{-1}\begin{pmatrix} 1 & E(z) \\ E(z) & E(z^2)\end{pmatrix}\begin{pmatrix} 1 & E(z) \\ E(x) & E(xz)\end{pmatrix}^{-1}\\ =\frac{1}{Cov(x,z)^2}\begin{pmatrix} E(xz) & -E(x) \\ -E(z) & 1\end{pmatrix}\begin{pmatrix} 1 & E(z) \\ E(z) & E(z^2)\end{pmatrix}\begin{pmatrix} E(xz) & -E(z) \\ -E(x) & 1\end{pmatrix}\\ =\frac{1}{Cov(x,z)^2}\begin{pmatrix}Cov(x,z) & E(xz)E(z)-E(x)E(z^2) \\ 0 & V(z)\end{pmatrix}\begin{pmatrix} E(xz) & -E(z) \\ -E(x) & 1\end{pmatrix}\\ =\frac{1}{Cov(x,z)^2}\begin{pmatrix}A & B\\ -E(x)V(z) & V(z)\end{pmatrix} $$ where $A=Cov(x,z)E(xz)-E(x)(E(xz)E(z)-E(x)E(z^2))$ and $B=E(z)(E(xz)-Cov(x,z))-E(x)E(z^2)$.

Multiplying the (2,2) element of the above matrix by $\sigma^2$ gives you the asymptotic variance of the (normalized) IV estimator of the slope coefficient. Rewrite it: $$ \sigma^2\frac{V(z)}{Cov(z,x)^2}=\sigma^2\frac{V(x)}{V(x)}\frac{V(z)}{Cov(z,x)^2}=\sigma^2\frac{1}{V(x)}\frac{1}{\left(\frac{Cov(z,x)}{\sqrt{V(x)}\sqrt{V(z)}}\right)^2}=\sigma^2\frac{1}{V(x)}\frac{1}{Corr(z,x)^2}. $$

And we are done.