Econ0701-2280 (A/B)
Introductory Econometrics
Tutorial Problem Set 3 Answer Key
September 28, 2019
1
Question 1(Properties of BLUE estimator)
For the simple linear regression model y = 0 + 1 +u; we can obtain the OLS
and
respectively.
estimater b and b , where b and b are estimating
0
1
0
1
0
1
Please prove:
1. b 1 =
Pn
i=1 xi (yi y)
Pn
i=1 xi (xi x)
and so c1 =
Proof. We …rst derive c1 .
Pn
(x x)(yi y)
i=1
Pn i
:
2
i=1 (xi x)
The simple linear regression model
y=
0
+
1x
+ u;
suggests
u=y
0
1 x:
We assume E (u) = 0 and E (ujx) = 0:1
The assumption E (u) = 0 suggests
E (y
1
As long as the intercept
assume E (u) = 0:
0
1 x)
0
=0
(1)
is included in the equation, it is without generality to
1
and the assumption E (ujx) = 0 implies E (ux) = 0, which further suggests
E (x (y
1 x))
0
= 0:
(2)
Given a sample data, we choose b 0 and b 1 to solve the sample counter-
part of (1) and (2), i.e.,
n
1X
yi
n
b
i=1
n
1X
xi yi
n
b
i=1
Equation (3) can be rewritten as
y
b =y
0
Plugging in (5) into (4),
xi yi
b x
1
y
i=1
n
X
xi (yi
i=1
y) + b 1
n
X
b xi
1
0
b xi
1
xi (x
Xn
i=1
i=1
(xi
xi (yi
y) =
x) (yi
y) =
=
(3)
= 0:
(4)
b x:
1
=
0
=
0
i=1
b
Xn
Xi=1
n
Xni=1
i=1
(5)
()
xi )
Moreover,
Xn
= 0;
b + b x;
0
1
=
()
n
X
b xi
1
0
Pn
xi (xi
Pi=1
n
i=1 xi (yi
=
1
Xn
y
xi yi
x
xi yi
2nxy + nxy =
2
Xi=1
n
i=1
xi =
Xn
xi yi
yi
y
i=1
X
n
x)
:
y)
xi yi
Xi=1
n
i=1
xi +
xi yi
nyx
Xn
i=1
nyx
xy
That is
Xn
i=1
Likewise,
Xn
i=1
X
n
xi (xi
x) =
(xi
x)2 =
i=1
xi (yi
y) =
Xn
Xi=1
n
i=1
x2i
x2i
Xn
i=1
(xi
x) (yi
y) :
nx2
Xn
Xn
2x
xi + nx2 =
x2i
i=1
i=1
Therefore,
Pn
b = Pni=1 xi (yi
1
i=1 xi (xi
y)
=
x)
3
Pn
x) (yi
i=1 (xi
Pn
x)2
i=1 (xi
y)
:
nx2 :
2. E b 1 =
1.
Proof.
b
1
1
=
=
=
E b1
1
Pn
Pn
i 1 (xi
0
Pn
x) (
i 1 (xi
Pn
+ Pi n 1
=
Pn
0
i 1 (xi
Pn
x) +
(xi
Pn
i 1 E ((xi
Pn
i 1 (xi
1
Pn
x)
i 1 xi (xi
Pn
2
x)
i 1 (xi
1
+
Pn
i 1 (xi
:
x) ui
2
x)
x) ui )
2
x)
4
!
=
=
E
Pn
Pn
i 1 (xi
x) ui
i 1 (xi
x)2
Pn
x) E (ui )
i 1 (xi
Pn
x)2
i 1 (xi
x) ui
1
1
x)2
x)2
i 1 (xi
Pn
i 1 (xi
+ ui )
x) ui
x) ui
i 1 (xi
Pn
1 xi
2
x)
i 1 (xi
= E
+
i 1 (xi
Pn
1
1
x)2
i 1 (xi
=
=
x) yi
i 1 (xi
Pn
= 0:
3. V ar b 1 =
2
SSTx ,
where SSTx =
Proof.
b
1
1
V ar b 1
=
Pn
i 1 (xi
Pn
i 1 (xi
= V ar b 1
x) ui
x)2
1
Pn
i=1 (xi
x)2 .
:
= V ar
Pn
i 1 (xi
Pn
i 1 (xi
x) ui
x)2
!
2
=
SSTX
4. Cov b 0 ; b 1 < 0 when x > 0:
Proof.
Cov b 0 ; b 1
= Cov y
= 0
b x; b = Cov y; b
1
1
1
xCov b ; b =
5
xV ar b =
Cov b 1 x; b 1
2
x
SSTX
< 0:
:
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Question 2 (Special case of the OLS estimator)
For the simple linear regression model y = 0 + 1 x + u, we can obtain
the OLS estimator b 0 and b 1 , where b 0 and b 1 are estimating 0 and 1
respectively. Under the case of b = 0, please judge the validity of the
1
following statements and explain.
1. b 0 = 0:
Answer:
b
2.
1
0
= 0:
=
y
b x=y
1
=) b 0 = 0 i¤ y = 0:
Answer: This is untrue in general because the
1
is the population parameter
which value is unobservable. We don’t know the true value of
1.
3 SSR = 0:
Answer:
SSR =
n
X
i=1
(yi
2
ybi ) =
n
X
b
yi
i=1
0
b xi
1
2
=
n
X
(yi
y)2 = SST:
i=1
The statement SSR = 0 is true only if SST = 0, which means there
is no variation on the value of y. All data points are on the horizontal
line.
4 R2 = 0:
Answer:
Since SSR = SST :
SSE
SST SSR
SST SST
=
=
= 0:
R2 =
SST
SST
SST
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This statement is correct. Grephically speaking, b 1 implies a horizon-
tal …tted line. The independent variable x gives no explanatory power
to the dependent variable y.
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