Linear probability model
Problem
Given:
- a random sample: \(\left( y_i,x_i \right)\) where \(y_i\) can only take values 0 or 1 and \(x_i\) is \(k×1\)
- Statistical model: \(E\left( y_i|x_i \right)=x'_iβ\) where \(β\) is a is \(k×1\) vector of unknown parameters.
- \(ε_i=y_i-E\left( y_i|x_i \right)\) such that \(y_i=x'_iβ+ε_i\)
- Show that \(P\left( y_i=1 \mid x_i \right)=x'_iβ\)
- Show that \(E\left( ε_i|x_i \right)=0\)
- Show that \(Var\left( ε_i|x_i \right)=x'_iβ\left( 1-x'_iβ \right)\)
Solution
a.
\[E\left( y_i|x_i \right)=1⋅P\left( y_i=1 \mid x_i \right)+0⋅P\left( y_i=0 \mid x_i \right)=P\left( y_i=1 \mid x_i \right)\]
and \(E\left( y_i|x_i \right)=x'_iβ\) by assumption
b.
If \(y_i=1\) then \(ε_i=1-x'_iβ\) and this has probability \(P\left( y_i=1 \mid x_i \right)=x'_iβ\) . If \(y_i=0\) then \(ε_i=-x'_iβ\) and this has probability \(P\left( y_i=0 \mid x_i \right)=1-x'_iβ\) . We have
\[E\left( ε_i|x_i \right)=\left( 1-x'_iβ \right)x'_iβ+\left( -x'_iβ \right)\left( 1-x'_iβ \right)=0\]
c.
\[Var\left( ε_i|x_i \right)=E\left( ε_i^2 \mid x_i \right)={\left( 1-x'_iβ \right)}^2x'_iβ+{\left( -x'_iβ \right)}^2\left( 1-x'_iβ \right)=\]
\[{\left( 1-x'_iβ \right)}^2x'_iβ+{\left( x'_iβ \right)}^2\left( 1-x'_iβ \right)=\left( 1-x'_iβ \right)x'_iβ\left( 1-x'_iβ+x'_iβ \right)=x'_iβ\left( 1-x'_iβ \right)\]
For the fourth equality, we have factored out \(\left( 1-x'_iβ \right)\) and \(x'_iβ\) .