Econometric

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. σW2 = (3 – 7.2)2 x 0.3 + (9 - 7.2)2 x 0.7 = 7.56

σV2 = (20 – 14.54)2 x 0.22 + (13 – 14.54)2 x 0.78 = 8.4084

c. cov (W,V) = σWV= E [(W-μW)(V-μV)] = (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) = σWVσWσV = -3.5287.56 X 8.4084 = - 0.4425

2.14

Apply the central limit theorem, we have Y ~ N (μY, σY2) , with μY = 100 and σY2= σY2/n = 43/n

a. Pr (Y ≤ 101) = Pr (Y -10043n ≤101-10043100) = Pr (Z ≤ 1.525) ≈ Φ(1.525) = 0.9364

b. Pr (Y > 98) = 1 – Pr (Y≤ 98) = 1 – Pr (Y -10043n ≤98-10043165) = 1 – Pr (Z≤-3.9178)

≈ 1 - Φ(-3.9178) ≈ 1

c. Pr (101≤Y ≤ 103) = Pr (101-1004364≤ Y -10043n ≤103-1004364) = Pr (1.22≤Z≤3.66) ≈ Φ (3.66) – Φ(1.22) = 0.9999 – 0.8888 = 0.1111

2.23

X ~ N (0,1) and Z ~ N (0,1) => μX= μZ = 0 and σX2=σZ2= 1

X, Z independently distributed => E(Z|X) = E(Z) = 0, Cov (X,Z) = σXZ = 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. μY = E(Y) = E(X2 + Z) = E(X2) + E(Z)

E(X2) = σX2 + μX2 = 1 + 0 = 1

E(Z) = 0

=> μY = 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) = σXY + μXμY = σXY + 0 x μY = σXY

=> σXY = 0

Corr (X,Y) = σXYσXσY = 0σXσY = 0

3.10

a. nNJ = 100, sNJ = 8

=> SE (YNJ) = sNJnNJ = 8100 = 0.8

The 95% confidence interval for the mean score of all New Jersey third graders is:

μNJ = {YNJ...