Underspecification

Problem

Based on the the lecture The linear regression model with “underspecification”, show that

  1. Show that \(E\left( ε_i \mid x_i \right)=β_3x_i^2\)
  2. Show that \(E\left( b_2 \mid x \right)=β_2+β_3 \frac{\sum{ x_i^3 }}{\sum{ x_i^2 }}\)

Solution

  1. a.

\[E\left( ε_i \mid x_i \right)=E\left( y_i-β_2x_i \mid x_i \right)=E\left( y_i \mid x_i \right)-E\left( β_2x_i \mid x_i \right)=β_2x_i+β_3x_i^2-β_2x_i=β_3x_i^2\]

  1. b. \(b_2=β_2+ \frac{∑x_iε_i}{\sum{ x_i^2 }}\) so

\[E\left( b_2 \mid x \right)=β_2+E\left( \frac{∑x_iε_i}{\sum{ x_i^2 }} \mid x \right)=β_2+ \frac{1}{\sum{ x_i^2 }}E\left( ∑x_iε_i \mid x \right)=\]

\[=β_2+ \frac{1}{\sum{ x_i^2 }}∑E\left( x_iε_i \mid x \right)=β_2+ \frac{1}{\sum{ x_i^2 }}∑x_iE\left( ε_i \mid x \right)=\]

\[=β_2+ \frac{1}{\sum{ x_i^2 }}∑x_iβ_3x_i^2=β_2+β_3 \frac{\sum{ x_i^3 }}{\sum{ x_i^2 }}\]