econometric solution
Hieu Nguyen – FIN 5309 Section 1
Assignment 1
2.3
Table 2.2
X=0
X=1
Total
Y=0
0.15
0.07
0.22
Y=1
0.15
0.63
0.78
Total
0.30
0.70
1.00
With W = 3+6X and V = 20-7Y, we have:
(W|X=0)=3
(W|X=1)=9
Total
(V|Y=0)=20
0.15
0.07
0.22
(V|Y=1)=13
0.15
0.63
0.78
Total
0.30
0.70
1.00
a. E(W) = 3 x 0.3 + 9 x 0.7 = 7.2
E(V) = 20 x 0.22 + 13 x 0.78 = 14.54
b. = (3 – 7.2)2 x 0.3 + (9 - 7.2)2 x 0.7 = 7.56 = (20 – 14.54)2 x 0.22 + (13 – 14.54)2 x 0.78 = 8.4084
c. cov (W,V) = = E [(W-)(V-)] = (3 – 7.2)(20 - 14.54) x 0.15 + (3 –7.2)(13-14.54) x 0.15 + (9 – 7.2)(20 – 14.54) x 0.07 + (9 - 7.2)(13 – 14.54) x 0.63 = -3.528
corr (W,V) = = = - 0.4425
2.14
Apply the central limit theorem, we have N (, ) , with = 100 and = /n = 43/n
a. Pr ( 101) = Pr ( ) = Pr (Z 1.525) (1.525) = 0.9364
b. Pr ( > 98) = 1 – Pr ( 98) = 1 – Pr ( ) = 1 – Pr (Z-3.9178) 1 - (-3.9178) 1
c. Pr (101 103) = Pr ( ) Pr (Z3.66) (3.66) – (1.22) = 0.9999 – 0.8888 = 0.1111
2.23
X N (0,1) and Z N (0,1) => = = 0 and == 1
X, Z independently distributed => E(Z|X) = E(Z) = 0, Cov (X,Z) = = 0, E(ZX) = 0
With Y = X2 + Z, we have:
a. E(Y|X) = E(X2 + Z|X) = E(X2|X) + E(Z|X) = X2 + 0 = X2
b. = E(Y) = E(X2 + Z) = E(X2) + E(Z)
E(X2) = + = 1 + 0 = 1
E(Z) = 0 => = 1 + 0 = 1
c. E(XY) = E(X3+ZX) = E(X3) + E(ZX) = E(X3) The odd moments of N (0,1) are zero => E(X3) = 0 => E(XY) =0
d. From c) we have E(XY) = 0 We also have E(XY) = + = + 0 x =
=> = 0 Corr (X,Y) = = = 0
3.10
a. nNJ = 100, sNJ = 8
=> SE () = = = 0.8
The 95% confidence interval for the mean score of all New Jersey third graders is: = { 1.96 SE ()} = {58 1.96(0.8)} = {56.432, 59.568}
b. nIW = 200, sIW = 11
SE ( - ) = = = = 1.1158
The 90% confidence interval for the difference in mean score between Iowa and New Jersey is: - = {( - ) 1.64SE( - )} = {62 – 58 1.64 x 1.1158} = {2.17, 5.83}
c. tact = = =
Assignment 1
2.3
Table 2.2
X=0
X=1
Total
Y=0
0.15
0.07
0.22
Y=1
0.15
0.63
0.78
Total
0.30
0.70
1.00
With W = 3+6X and V = 20-7Y, we have:
(W|X=0)=3
(W|X=1)=9
Total
(V|Y=0)=20
0.15
0.07
0.22
(V|Y=1)=13
0.15
0.63
0.78
Total
0.30
0.70
1.00
a. E(W) = 3 x 0.3 + 9 x 0.7 = 7.2
E(V) = 20 x 0.22 + 13 x 0.78 = 14.54
b. = (3 – 7.2)2 x 0.3 + (9 - 7.2)2 x 0.7 = 7.56 = (20 – 14.54)2 x 0.22 + (13 – 14.54)2 x 0.78 = 8.4084
c. cov (W,V) = = E [(W-)(V-)] = (3 – 7.2)(20 - 14.54) x 0.15 + (3 –7.2)(13-14.54) x 0.15 + (9 – 7.2)(20 – 14.54) x 0.07 + (9 - 7.2)(13 – 14.54) x 0.63 = -3.528
corr (W,V) = = = - 0.4425
2.14
Apply the central limit theorem, we have N (, ) , with = 100 and = /n = 43/n
a. Pr ( 101) = Pr ( ) = Pr (Z 1.525) (1.525) = 0.9364
b. Pr ( > 98) = 1 – Pr ( 98) = 1 – Pr ( ) = 1 – Pr (Z-3.9178) 1 - (-3.9178) 1
c. Pr (101 103) = Pr ( ) Pr (Z3.66) (3.66) – (1.22) = 0.9999 – 0.8888 = 0.1111
2.23
X N (0,1) and Z N (0,1) => = = 0 and == 1
X, Z independently distributed => E(Z|X) = E(Z) = 0, Cov (X,Z) = = 0, E(ZX) = 0
With Y = X2 + Z, we have:
a. E(Y|X) = E(X2 + Z|X) = E(X2|X) + E(Z|X) = X2 + 0 = X2
b. = E(Y) = E(X2 + Z) = E(X2) + E(Z)
E(X2) = + = 1 + 0 = 1
E(Z) = 0 => = 1 + 0 = 1
c. E(XY) = E(X3+ZX) = E(X3) + E(ZX) = E(X3) The odd moments of N (0,1) are zero => E(X3) = 0 => E(XY) =0
d. From c) we have E(XY) = 0 We also have E(XY) = + = + 0 x =
=> = 0 Corr (X,Y) = = = 0
3.10
a. nNJ = 100, sNJ = 8
=> SE () = = = 0.8
The 95% confidence interval for the mean score of all New Jersey third graders is: = { 1.96 SE ()} = {58 1.96(0.8)} = {56.432, 59.568}
b. nIW = 200, sIW = 11
SE ( - ) = = = = 1.1158
The 90% confidence interval for the difference in mean score between Iowa and New Jersey is: - = {( - ) 1.64SE( - )} = {62 – 58 1.64 x 1.1158} = {2.17, 5.83}
c. tact = = =