Understanding the LRM
Problem
\(y_i\) is the height of plant \(i\) and \(x_i\) is the amount of water added to the plant. Suppose that
\[y_i=β_1+β_2x_i+ε_i\]
and that the \(x\) -variable is exogenous. Then
\[E\left( y \mid x \right)=β_1+β_2x\]
In this example,
- How do you describe \(E\left( y \mid x=5 \right)\) in words?
- How do you describe \(E\left( y \mid x \right)\) in words?
- How do you describe the result \(dE\left( y \mid x=5 \right)/dx=β_2\) in words?
- How do you describe \(dE\left( y \mid x \right)/dx=β_2\) in words?
Solution
- \(E\left( y \mid x=5 \right)\) is the expected height of a plant that receives 5 units of water.
- \(E\left( y \mid x \right)\) is the expected height of a plant that receives \(x\) units of water. It is a function of \(x\) (in this example, it is a linear function of \(x\) )
- Since \(E\left( y \mid x \right)\) is a linear function in \(x\) , \(dE\left( y \mid x=5 \right)/dx\) is the increase in the expected height of a plant if we increase the amount of water from 5 to 6 units. This is also the slope of the linear function \(β_1+β_2x\) , that is, \(β_2\) .
- Since \(E\left( y \mid x \right)\) is a linear function in \(x\) , \(dE\left( y \mid x \right)/dx\) is the increase in the expected height of a plant if we increase the amount of water by one unit (from \(x\) to \(x+1\) units). This is independent of \(x\) and equal to \(β_2\) .