Tuesday, August 14, 2012

Proof IV is same as 2SLS

This is a standard proof.  Normally I am the last one to attempt proofs but I wanted to try out $ \LaTeX{} $ in blogger.

LaTeX Script:

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  % The following shows a little of the typesetting power of LaTeX
First Stage: Standard OLS \\
(1.1) X=Z\widehat{\gamma\
(1.2) Z'X=Z\widehat{\gamma\\
(1.3) Z'X=Z'Z\widehat{\gamma\\
(1.4) (Z'Z)^{-1}Z'X=(Z'Z)^{-1}Z'Z\widehat{\gamma\\
(1.5) (Z'Z)^{-1}Z'X=\widehat{\gamma\\

Second Stage, insert projection of Z on X ( \widehat{X= Z \widehat{\gamma): \\
(2.1) Y= \widehat{X\beta _{2SLS\\
(2.2) \widehat{X}' Y= \widehat{X}\widehat{X\beta _{2SLS\\
(2.3) ( \widehat{X}\widehat{X})^{-1\widehat{X}' Y=  ( \widehat{X}\widehat{X})^{-1}  \widehat{X}\widehat{X\beta _{2SLS\\
(2.4) ( \widehat{X}\widehat{X})^{-1\widehat{X}' Y=  \beta _{2SLS\\
From (1.5) we insert  \widehat{X= Z \widehat{\gamma= Z (Z'Z)^{-1}Z'X \\
and  \widehat{X}' = \widehat{\gamma}'Z = X'Z (Z'Z)^{-1}Z'. \\
(2.5) ( X'Z (Z'Z)^{-1Z'Z (Z'Z)^{-1Z'X )^{-1}  X'Z (Z'Z)^{-1Z'Y=  \beta _{2SLS\\
(2.6) ( X'Z  (Z'Z)^{-1Z'X )^{-1}  X'Z (Z'Z)^{-1Z'Y=  \beta _{2SLS\\

We will first derive the IV estimator: \\
(3.1) Y = X \beta _{IV\\
(3.2) Z'Y = Z'X \beta _{IV\\
(3.3) (Z'X)'Z'Y = (Z'X)'Z'X \beta _{IV\\
(3.4) X'ZZ'Y = X'ZZ'X \beta _{IV\\
(3.5) (X'ZZ'X)^{-1X'ZZ'Y = (X'ZZ'X)^{-1X'ZZ'X \beta _{IV\\
(3.6) (X'ZZ'X)^{-1X'ZZ'Y =  \beta _{IV\\

To show that IV is same as the 2SLS we take an addition step after (3.2). Somewhat artificially we multiply both sides by X' Z (Z'Z)^{-1}  \\
(3.3b) X' Z (Z'Z)^{-1Z'Y = X' Z (Z'Z)^{-1Z'X \beta _{IV\\
(3.4b) (X' Z (Z'Z)^{-1Z'X)^{-1X' Z (Z'Z)^{-1Z'Y \\
                              = (X' Z (Z'Z)^{-1Z'X)^{-1X' Z (Z'Z)^{-1Z'X \beta _{IV\\
(3.5d) (X' Z (Z'Z)^{-1Z'X)^{-1X' Z (Z'Z)^{-1Z'Y = \beta _{IV\\
  Which must equal (3.6) because we have done only equal operations to both sides and (3.5d) is the same as (2.6).  Thus the IV estimator is the same as the 2SLS estimator. 

1 comment:

  1. Thanks for this! Just above line (2.5) why is the last matrix Z' instead of just Z?