Missing variable

Solution

a.

\[b= \frac{∑x_iy_i}{\sum{ x_i^2 }}=β+ \frac{∑x_iε_i}{\sum{ x_i^2 }}\]

Thus,

\[E\left( b|x,z \right)=E\left( β+ \frac{∑x_iε_i}{\sum{ x_i^2 }}|x,z \right)=β+E\left( \frac{∑x_iε_i}{\sum{ x_i^2 }}|x,z \right)=\]

\[=β+ \frac{1}{\sum{ x_i^2 }}E\left( ∑x_iε_i|x,z \right)=β+ \frac{1}{\sum{ x_i^2 }}∑E\left( x_iε_i \mid x,z \right)=\]

\[=β+ \frac{1}{\sum{ x_i^2 }}∑x_iE\left( ε_i \mid x,z \right)=β+ \frac{1}{\sum{ x_i^2 }}∑x_iθz_i=β+θ \frac{∑x_iz_i}{\sum{ x_i^2 }}\]

The unconditional expectation is, by the law of iterated expectations,

\[E\left( b \right)=β+θE\left( \frac{∑x_iz_i}{\sum{ x_i^2 }} \right)\]