6.1.1 Deriving OLS
OLS is obtained by minimizing the sum of the square
errors. This is done using the partial derivative
N
2
ˆ
min
e
ˆ ˆ  i
1 ,  2
i 1
where : eˆi  Yi  ˆ1  ˆ2 X i
  eˆi
 2 Yi  ˆ1  ˆ2 X i  0  1 
ˆ
2
1
  eˆi
 2 (Yi  ˆ1  ˆ2 X i ) X i  0  2 
ˆ
2
2
6.1.1 Deriving OLS
These can simplify to:
ˆ  ˆ X  0


Y
i 1 2 i
OR
 eˆ
i
 0  1a 

(Y  ˆ  ˆ X ) X  0

1
i
2
i
OR
 eˆ X
i
i
 0  2a 
i
6.1.1 Deriving OLS
These can be expressed in their normal equations form:
Y
i
 nˆ1  ˆ2  X i  0
 1b 
Y X
i
i
 ˆ1  X i  ˆ2  X i  0
 2b 
Notice that we have two equations with two unknowns
(β2hat and β1hat). All other components come from
our data set.
After some math, we get the OLS estimates of β1hat
and β2hat:
6.1.1 Deriving OLS
N
ˆ2 
(X
i 1
i
 X )(Yi  Y )
N
(X
i 1
N
ˆ2 
(X
i 1
N
i
(X
i 1
i
 X)
2
 X )(Yi )
i
 X )Xi
ˆ1  Y  ˆ1 X
Finally, we check the second derivative to confirm a
minimum.
6.2 OLS Estimators and Goodness of Fit
On average, OLS works well:
 The average of the estimated errors is zero
 The average of the estimated Y’s is always the
average of the observed Y’s

Proof:
Note: This comes
ei  0  1a  from our derivation

of OLS as the sum
Yi  Yi  0
of squared errors.



Y  Y
i
i
N
N