Simultaneity

Problem

Based on the lecture Simultaneity.

  1. Show that \(x_i= \frac{β_1}{1-β_2}+ \frac{z_i}{1-β_2}+ \frac{ε_i}{1-β_2}\)
  2. Show that \(cov\left( x_i,ε_i \right)= \frac{Var\left( ε_i \right)}{1-β_2}\) .
  3. Show that \(plim b_2=β_2+\left( 1-β_2 \right) \frac{σ_ε^2}{σ_z^2+σ_ε^2}\)

Some useful covariance results:

  • \(Cov\left( X+Y+Z,W \right)=Cov\left( X,W \right)+Cov\left( Y,W \right)+Cov\left( Z,W \right)\)
  • \(Cov\left( cX,Y \right)=cCov\left( X,Y \right)\) where \(c\) is a constant
  • \(Cov\left( X,X \right)=Var\left( X \right)\)

Solution

a.

\[x_i=y_i+z_i=β_1+β_2x_i+ε_i+z_i\]

Move \(β_2x_i\) to the left,

\[x_i\left( 1-β_2 \right)=β_1+ε_i+z_i\]

Divide by \(\left( 1-β_2 \right)\) .

b.

\[cov\left( x_i,ε_i \right)=cov\left( \frac{β_1}{1-β_2}+ \frac{z_i}{1-β_2}+ \frac{ε_i}{1-β_2},ε_i \right)=\]

\[cov\left( \frac{β_1}{1-β_2},ε_i \right)+cov\left( \frac{z_i}{1-β_2},ε_i \right)+cov\left( \frac{ε_i}{1-β_2},ε_i \right)\]

  • The first term is zero since a random variable ( \(ε_i\) ) is never correlated with a constant ( \( \frac{β_1}{1-β_2}\) ).
  • The second term is zero since \(z\) is exogenous

The final term is

\[cov\left( \frac{ε_i}{1-β_2},ε_i \right)= \frac{1}{1-β_2}cov\left( ε_i,ε_i \right)= \frac{1}{1-β_2}Var\left( ε_i \right)\]

c.

\[b_2= β_2+ \frac{∑\left( x_i-\bar{x} \right)ε_i}{∑{\left( x_i-\bar{x} \right)}^2}\]

so

\[plim b_2=β_2+plim \frac{∑\left( x_i-\bar{x} \right)ε_i}{∑{\left( x_i-\bar{x} \right)}^2}= \frac{plim n^{-1}∑\left( x_i-\bar{x} \right)ε_i}{plim n^{-1}∑{\left( x_i-\bar{x} \right)}^2}\]

We have

\[plim n^{-1}∑\left( x_i-\bar{x} \right)ε_i=cov\left( x_i,ε_i \right)= \frac{Var\left( ε_i \right)}{1-β_2}= \frac{σ_ε^2}{1-β_2}\]

and

\[plim n^{-1}∑{\left( x_i-\bar{x} \right)}^2=Var\left( x_i \right)=Var\left( \frac{β_1}{1-β_2}+ \frac{z_i}{1-β_2}+ \frac{ε_i}{1-β_2} \right)=\]

\[=Var\left( \frac{z_i}{1-β_2} \right)+Var\left( \frac{ε_i}{1-β_2} \right)= \frac{σ_z^2}{{\left( 1-β_2 \right)}^2}+ \frac{σ_ε^2}{{\left( 1-β_2 \right)}^2}\]

Combining,

\[plim b_2=β_2+ \frac{ \frac{σ_ε^2}{1-β_2}}{ \frac{σ_z^2}{{\left( 1-β_2 \right)}^2}+ \frac{σ_ε^2}{{\left( 1-β_2 \right)}^2}}=β_2+\left( 1-β_2 \right) \frac{σ_ε^2}{σ_z^2+σ_ε^2}\]