The instrumental variables estimator

Problem

  • Setup: same number of instruments as explanatory variables
    • Random sample \(\left( y_i,x_i,z_i \right)\) for \(i=1, \ldots ,n\) where \(x_i\) is \(k×1\) and \(z_i\) is \(k×1\)
    • A linear regression model, \(y_i=x'_iβ+ε_i\)
    • Exogeneity fails, \(E\left( ε_i \mid x_i \right)≠0\)
    • z-variables are instruments, \(E\left( ε_i|z_i \right)=0\) and \(E\left( z_ix'_i \right)=Σ_{zx'}\) is \(k×k\) and invertible

a. Show that \(E\left( {z_iε}_i \right)=E\left( z_i\left( y_i-x'_iβ \right) \right)=0\)

b. Show that the solution to

\[ \frac{1}{n}\sum_{i=1}^{n}{ z_i\left( y_i-x'_ib_{IV} \right) }=0\]

is given by

\[b_{IV}={\left( \sum_{i=1}^{n}{ z_ix'_i } \right)}^{-1}\sum_{i=1}^{n}{ z_iy_i }={\left( Z'X \right)}^{-1}Z'y\]

Solution

a. \(E\left( {z_iε}_i \right)=E\left( E\left( {z_iε}_i|z_i \right) \right)=E\left( z_iE\left( ε_i|z_i \right) \right)=E\left( z_i⋅0 \right)=0\) . The first equality by LIE

b.

\[\sum_{i=1}^{n}{ z_i\left( y_i-x'_ib_{IV} \right) }=\sum_{i=1}^{n}{ z_iy_i }-\sum_{i=1}^{n}{ z_ix'_ib_{IV} }=\sum_{i=1}^{n}{ z_iy_i }-\left( \sum_{i=1}^{n}{ z_ix'_i } \right)b_{IV}=0\]

We have

\[\left( \sum_{i=1}^{n}{ z_ix'_i } \right)b_{IV}=\sum_{i=1}^{n}{ z_iy_i }\]

or

\[b_{IV}={\left( \sum_{i=1}^{n}{ z_ix'_i } \right)}^{-1}\sum_{i=1}^{n}{ z_iy_i }\]

In matrix notation, this is \({\left( Z'X \right)}^{-1}Z'y=b_{OLS}\) .