The Simple Regression Model: Introductory Econometrics: A Modern Approach (Wooldridge)
The Simple Regression Model: Introductory Econometrics: A Modern Approach (Wooldridge)
Introductory Econometrics:
A modern approach (Wooldridge)
Chapter 2
1
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
y
f(y)
E ( y | x ) = β 0 + β1 x
x1 x2
Dr. Lê Văn Chơn – FTU, 2011
2
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
y3 } u3
y2 u2 {
y1 } u1
x1 x2 x3 x4 x
Dr. Lê Văn Chơn – FTU, 2011
1 n
∑ xi ( yi − βˆ0 − βˆ1xi ) = 0
n i =1
(2.15)
3
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
∑ x ( y − [ y − βˆ x ] − βˆ x ) = 0
i =1
i i 1 1 i
n n
∑ x ( y − y ) = βˆ ∑ x ( x − x )
i =1
i i 1
i =1
i i
n n
∑ ( xi − x )( yi − y ) = βˆ1 ∑ ( xi − x ) 2
i =1 i =1
n
Provided that ∑ ( xi − x ) 2 > 0
i =1
(2.18)
n
∑ ( x − x )( yi i − y)
the estimated slope is βˆ1 = i =1
n (2.19)
∑ (x − x)
i =1
i
2
βˆ0 and βˆ1 given in (2.17) and (2.19) are called the ordinary least
squares (OLS) estimates of β 0 and β1 .
To justify this name, for any βˆ0 and βˆ1 , define a fitted value for y
given x = xi: yˆ i = βˆ0 + βˆ1 xi (2.20)
Intuitively, OLS is fitting a line through the sample points such that
the sum of squared residuals is as small as possible term
“ordinary least squares”.
Formal minimization problem:
n n
ˆ ˆ ∑ i
min uˆ 2 = ∑ ( yi − βˆ0 − βˆ1 xi ) 2 (2.22)
β 0 , β1
i =1 i =1
4
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
y3 } û3
y2 û2 {
y1 .} û1
x1 x2 x3 x4 x
Dr. Lê Văn Chơn – FTU, 2011
Once we have determined the OLS βˆ0 and βˆ1 , we have the OLS
regression line: yˆ = βˆ0 + βˆ1 x (2.23)
5
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
In most cases, every uˆi ≠ 0 , none of the data points lie on the OLS
line.
(2) The sample covariance between the regressors and the OLS
residuals is zero. n
∑ xiuˆi = 0
i =1
(3) The OLS regression line always goes through the mean of the
sample. y = βˆ + βˆ x
0 1
6
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
∑(y
i =1
i − y ) 2 is the total sum of squares (SST),
n
∑ ( yˆ
i =1
i − y ) 2 is the explained sum of squares (SSE),
n
∑ uˆ
i =1
2
i is the residual sum of squares (SSR).
Proof:
n n n
∑(y
i =1
i − y ) 2 = ∑ [( yi − yˆ i ) + ( yˆ i − y )]2 = ∑ [uˆi + ( yˆ i − y )]2
i =1 i =1
n n n
= ∑ uˆi2 + 2∑ uˆi ( yˆ i − y ) + ∑ ( yˆ i − y ) 2
i =1 i =1 i =1
n
= SSR + 2∑ uˆi ( yˆ i − y ) +SSE
i =1
n
and we know that ∑ uˆ ( yˆ
i =1
i i − y) = 0
Goodness-of-Fit
How well the OLS regression line fits the data?
7
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
Multiply the intercept and the slope in (2.26) by 1,000 (2.26) and
(2.40) have the same interpretations.
8
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
Estimating this model and the mechanics of OLS are the same:
lôg(wage) = 0.584 + 0.083educ (2.44)
wage increases by 8.3 percent for every additional year of educ.
9
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
10
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
∑ (x − x)
i =1
i
2
∑ (x − x)
i =1
i
2
∑ ( x − x )(β
i 0 + β1 xi + ui ) ∑ ( x − x )u
i i
βˆ1 = i =1
n
= β1 + i =1
SSTx
∑ (x − x)
i =1
i
2
n
Proof: E ( βˆ1 ) = β1 + E[(1 / SSTx )∑ ( xi − x )ui ]
i =1
n
= β1 + (1 / SSTx )∑ ( xi − x )E (ui ) = β1
i =1
(2.17) implies
βˆ0 = y − βˆ1 x = β 0 + β1 x + u − βˆ1 x = β 0 + ( β1 − βˆ1 ) x + u
E ( βˆ ) = β + E[(β − βˆ ) x ] = β
0 0 1 1 0
11
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
MEAP93 has data on 408 Michigan high school for the 1992-1993
school year.
mâth10 = 32.14 – 0.319lnchprg
Why? u contains such as the poverty rate of children attending
school, which affects student performance and is highly correlated
with eligibility in the lunch program.
Dr. Lê Văn Chơn – FTU, 2011
12
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
Homoskedastic case:
y
f(y|x)
E(y|x) = β0 + β1x
x1 x2 x
Dr. Lê Văn Chơn – FTU, 2011
Heteroskedastic case:
f(y|x)
E(y|x) = β0 + β1x
x1 x2 x3 x
Dr. Lê Văn Chơn – FTU, 2011
1 n 2
σ2 ∑ xi
n i =1
and Var ( βˆ0 ) = n
(2.58)
∑ (x − x)
i =1
i
2
1 n
SSTx σ2
Proof: Var ( βˆ1 ) =
SSTx2
∑ ( x − x ) Var (u ) = SST
i =1
i
2
i 2
σ2 =
SSTx
x
13
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
What we observe are the residuals, ûi . We can use the residuals to
form an estimate of the error variance.
An unbiased estimator of σ is
2
1 n 2 SSR (2.61)
σˆ =
2
∑ uˆi = n − 2
n − 2 i =1
σ
Recall that sd ( βˆ1 ) = , if we substitute σ̂ for σ 2 , then we
2
SSTx
have the standard error of βˆ1 :
σˆ σˆ
se( βˆ1 ) = =
SSTx n
∑ ( xi − x ) 2
i =1
14
CuuDuongThanCong.com https://fb.com/tailieudientucntt
15/11/2011
We still use OLS method with the corresponding first order condition
n
n
~ˆ ~ˆ ∑x y i i
∑ x (y i i −β1 xi ) = 0 β1 = i =1
n (2.66)
i =1
∑x
i =1
2
i
~ˆ
If β 0 ≠ 0 , then β1 is a biased estimator of β1.
15
CuuDuongThanCong.com https://fb.com/tailieudientucntt