The statistical formula for b2
Problem
This problem is quite difficult but super useful if you want to practice econometrics algebra.
a) Start from the LRM, \(y_i=β_1+β_2x_i+ε_i\) . We have
\[b_2= \frac{\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\left( y_i-\bar{y} \right) }}{\sum_{i=1}^{n}{ {\left( x_i-\bar{x} \right)}^2 }}\]
We also have the following result
\[y_i-\bar{y}=β_2\left( x_i-\bar{x} \right)+ε_i-\bar{ε}\]
Use this result to show that the terms in the sum of the numerator satisfy
\[\left( x_i-\bar{x} \right)\left( y_i-\bar{y} \right)=β_2{\left( x_i-\bar{x} \right)}^2+\left( x_i-\bar{x} \right)\left( ε_i-\bar{ε} \right)\]
b) Use a) to show that
\[\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\left( y_i-\bar{y} \right) }=β_2\sum_{i=1}^{n}{ {\left( x_i-\bar{x} \right)}^2 }+\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\left( ε_i-\bar{ε} \right) }\]
c) Show that
\[\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\left( ε_i-\bar{ε} \right) }=\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)ε_i }\]
Hint: It is not because \(\bar{ε}\) is zero (because it is not). Try instead
\[\left( x_i-\bar{x} \right)\left( ε_i-\bar{ε} \right)=\left( x_i-\bar{x} \right)ε_i-\left( x_i-\bar{x} \right)\bar{ε}\]
Now sum this, remembering that \(∑\left( x_i-\bar{x} \right)=0\) .
d) Finally show that
\[b_2=β_2+ \frac{\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)ε_i }}{\sum_{i=1}^{n}{ {\left( x_i-\bar{x} \right)}^2 }}\]
Comment: This is called the statistical formula for \(b_2\) .
Solution
a)
b) From a):
\[\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\left( y_i-\bar{y} \right) }=\sum_{i=1}^{n}{ \left( β_2{\left( x_i-\bar{x} \right)}^2+\left( x_i-\bar{x} \right)\left( ε_i-\bar{ε} \right) \right) }\]
Split the sum:
\[\sum_{i=1}^{n}{ \left( β_2{\left( x_i-\bar{x} \right)}^2+\left( x_i-\bar{x} \right)\left( ε_i-\bar{ε} \right) \right) }=\sum_{i=1}^{n}{ β_2{\left( x_i-\bar{x} \right)}^2 }+\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\left( ε_i-\bar{ε} \right) }\]
Take out \(β_2\) (constant):
\[\sum_{i=1}^{n}{ β_2{\left( x_i-\bar{x} \right)}^2 }+\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\left( ε_i-\bar{ε} \right) }=β_2\sum_{i=1}^{n}{ {\left( x_i-\bar{x} \right)}^2 }+\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\left( ε_i-\bar{ε} \right) }\]
c)
\[\left( x_i-\bar{x} \right)\left( ε_i-\bar{ε} \right)=\left( x_i-\bar{x} \right)ε_i-\left( x_i-\bar{x} \right)\bar{ε}\]
Sum:
\[\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\left( ε_i-\bar{ε} \right) }=\sum_{i=1}^{n}{ \left( \left( x_i-\bar{x} \right)ε_i-\left( x_i-\bar{x} \right)\bar{ε} \right) }\]
Split the sum:
\[\sum_{i=1}^{n}{ \left( \left( x_i-\bar{x} \right)ε_i-\left( x_i-\bar{x} \right)\bar{ε} \right) }=\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)ε_i }-\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\bar{ε} }\]
\(\bar{ε}\) is constant in the second sum, take it outside:
\[\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)ε_i }-\bar{ε}\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right) }\]
The second sum is now zero and
\[\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\left( ε_i-\bar{ε} \right) }=\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)ε_i }\]
d) Start from
\[b_2= \frac{\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\left( y_i-\bar{y} \right) }}{\sum_{i=1}^{n}{ {\left( x_i-\bar{x} \right)}^2 }}\]
Combine b) and c) and the numerator is:
\[\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)\left( y_i-\bar{y} \right) }=β_2\sum_{i=1}^{n}{ {\left( x_i-\bar{x} \right)}^2 }+\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)ε_i }\]
Plug this into the formula for \(b_2\) :
\[b_2= \frac{β_2\sum_{i=1}^{n}{ {\left( x_i-\bar{x} \right)}^2 }+\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)ε_i }}{\sum_{i=1}^{n}{ {\left( x_i-\bar{x} \right)}^2 }}=\]
(using \( \frac{a+b}{c}= \frac{a}{c}+ \frac{b}{c}\) )
\[= \frac{β_2\sum_{i=1}^{n}{ {\left( x_i-\bar{x} \right)}^2 }}{\sum_{i=1}^{n}{ {\left( x_i-\bar{x} \right)}^2 }}+ \frac{\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)ε_i }}{\sum_{i=1}^{n}{ {\left( x_i-\bar{x} \right)}^2 }}\]
The sums cancel in the first expression
\[=β_2+ \frac{\sum_{i=1}^{n}{ \left( x_i-\bar{x} \right)ε_i }}{\sum_{i=1}^{n}{ {\left( x_i-\bar{x} \right)}^2 }}\]