University Cornell Department of Economics Econ 620 Spring 2007 Instructor: Professor Kiefer TA: Jae-ho Yun ([e-mail])
Problem Set # 1: Solution Key
1. b
1
X (xi x)( + xi + x2 + "i ) x)yi 0 1 2 i = (xi x)2 (xi x)2 X (xi x)x2 X (xi x) X (xi x)xi i + 2 = + 1 0 (xi x)2 (xi x)2 (xi x)2 X (xi x)" + (xi x)2 X (xi x)x2 X (xi x)" i = + 1+ 2 (xi x)2 (xi x)2 = Note: X (xi (xi X (xi x) = 0; x)2 (xi X (xi x)xi =1 x)2
X (xi
Take the conditional expectation; E( b 1 jxi ) = + x)x2 i x)2
1
2
(xi
Therefore, b 1 is biased. V ar( b 1 jxi )
X (xi Note: E( (xi
x)" j xi ) = 0 x)2
Next, the conditional variance is;
= E[( b 1 E( b 1 jxi ))2 j xi ] X (xi x)" = E[( )2 j xi ] (xi x)2 (cross product terms are all zero)
2
=
P (xi 1
x)2
2. i) The correlation between y and z is 1 if a>0, -1 if a<0 and 0 if a=0. ii) The support of the joint distribution of y and z can be represented by the following set; f(x; y) : z = ay; y 2 Rg (It is a straight line) iii) The correlation between y and x ( x = y 2 ) is zero, since Cov(y; y 2 ) = E(y y 2 ) E(y)E(y 2 ) = E(y 3 ) = 0 iv) The support is f(x; y) : x = y 2 ; y 2 Rg v) In the above problem, clearly, there are dependence between x and y. However, our linear relationship measure, correlation does not capture this relationship. 3. (a) OLS is BLUE if "i are homoskedastic with mean zero. 1 1 ( 1) + ( 1) = 0 for i=1,2 2 2 1 1 ( 1)2 + ( 1)2 = 1 for i=1,2 E("2 ) = i 2 2 : Cov("i ; "j ) = E("i "j ) E("i )E("j ) = 0 for i 6= j E("i ) =
Mean: Variance: Covarince
Hence, our OLS is BLUE. (b) b P P xy xi (xi + "i ) P i2i = P 2 = = xi xi 1 "1 + 2 "2 = 1+ 12 + 22 P xi "i + P 2 xi
(c) The alternative estimator
We have the following exact distribution of b Prob. :25 :25 :25 :25 b 2=5 4=5 6=5 8=5 P P y (xi + "i ) P i = P = xi xi "1 + "2 = 1+ 1+2 = P "i +P xi
is clearly unbiased. Hence, the exact distribution of is Prob. :25 :50 1=3 1
:25 5=3
2
(d) The exact variance of the two estimators: "1 + 2"2 1 1 )= (V ("1 ) + 4V ("2 )) = 5 25 5 "1 + "2 1 2 = V( ) = (V ("1 ) + V ("2 )) = 3 9 9
V( Hence, V (
V (b)
= V(
)
4. (a) Recall that
) > V ( ); b which is consistent with b being BLUE. b X (xi x)(yi y) (xi x)2
=
and
X
(xi
Hence,
x)(yi X (xi = =
= y bx X y) = xi yi nxy X x)2 = x2 nx2 i b
(b) R2 is de...ned as the ratio of the explained sum of squares(ESS) to total sum of squares(TSS). P 2 b 2 P(xi x)2 2 xi P =b P 2 (yi y)2 yi b (n
b
b
30 = 0:5 60 440 0:5 22
220 = 15 22
R2 =
nx2 60 = 0:52 8900 22 ny 2
202
= 0:15
(c) By the normality assumption, we know that
2
Moreover,
N( ; P (xi 2)S 2
2 2
x)2
)
(n
2)
P 2 where S2 = n 1 2 ei : We can also show that b and S 2 are independent each other. Then, 3
r
r
P
b
2 x)2 2)S 2 2
(xi
=
(n
(n 2)
pP
b
S (xi x)2
t(n
2)
We want to reject the null hypothesis if b
t=
under the null hypothesis. On the other hand, 1 n 1 n 2 4:25 2 X X 1 n 2 X
2
pP
S (xi x)2
> t0:975 (20)
S2
= = =
e2 = i
(yi
2 (yi + b 2 + b x2 i
2b yi + 2b b xi
b
b xi )2
2 b xi yi )2
Hence, the test statistics is given by 0:5 t= q 0
4:25 60
= 1:8787
Since t0:975 (20) = 2:086;we cannot reject the null hypothesis. 5. i)
b1
0 0 (X1 M2 X1 ) 1 X1 M2 y 0 0 (X1 M2 X1 ) 1 X1 M2 (X1 1 + X2 2 + ") 0 1 0 0 0 (X1 M2 X1 ) X1 M2 X1 1 + (X1 M2 X1 ) 1 X1 M2 X2 0 0 +(X1 M2 X1 ) 1 X1 M2 " 0 1 0 = X 1 M2 " 1 + 0 + (X1 M2 X1 )
= = =
2
Take expectation.
E(b1 )
= = = =
1
0 0 + E[(X1 M2 X1 ) 1 X1 M2 "] 0 1 0 X1 M2 E[&q...