Test Bank For Introduction to econometrics 3rd
Edition By James H. Stock & Mark W. Watson.
,Chapter 2
Review of Probability
2.2. We know from Table 2.2 that Pr (Y 0) 0 22, Pr (Y 1) 0 78, Pr (X 0) 0
30, Pr (X 1) 0 70. So
(a) Y E Y( ) 0 Pr (Y 0) 1 Pr (Y 1)
0 0 22 1 0 78 0 78,
X E X() 0 Pr (X 0) 1 Pr (X 1)
0 0 30 1 0 70 0 70
(b) X
2
E X[( X )2 ]
(0 0.70)2 Pr (X 0) (1 0.70)2 Pr (X 1)
( 0 70)2 0 30 0 302 0 70 0 21,
Y
2
E Y[( Y )2 ]
(0 0.78)2 Pr (Y 0) (1 0.78)2 Pr (Y 1)
( 0 78)2 0 22 0 222 0 78 0 1716
(c) XY cov (X Y, ) E X[( X )(Y Y )]
(0 0.70)(0 0.78) Pr(X 0,Y 0)
(0 0 70)(1 0 78) Pr (X 0 Y 1)
(1 0 70)(0 0 78) Pr (X 1Y 0)
(1 0 70)(1 0 78) Pr (X 1 Y 1)
( 0 70) ( 0 78) 0 15 ( 0 70) 0 22 0 15
0 30 ( 0 78) 0 07 0 30
0 22 0 63 0 084,
0 084
corr (X Y, ) XY
0 4425
X Y 0 21 0 1716
2.4. (a) E X( 3) 03 (1 p) 13 p p
(b) E X( k
) 0k (1 p) 1k p p
(c) E X( ) 0.3, and var(X) = E(X2)−[E(X)]2 = 0.3 −0.09 = 0.21. Thus =
0.21= 0.46.
var (X) E X( 2
) [E X( )]2 0.3 0.09 0.21 0.21 0.46. To compute
the skewness, use the formula from exercise 2.21:
, Solutions to End-of-Chapter Exercises 3
E X( )3 E X(3 ) 3[E X( 2 )][E X( )] 2[E X( )]3
0.3 3 0.32 2 0.33 0.084
Alternatively, E X( )3 [(1 0.3)3 0.3] [(0 0.3)3 0.7] 0.084
Thus, skewness E X( ) /3 3 0.084/0.463 0.87.
To compute the kurtosis, use the formula from exercise 2.21:
E X( )4 E X(4 ) 4[E X()][E X(3 )] 6[E X()]2 [E X( 2 )] 3[E X( )]4
0.3 4 0.32 6 0.33 3 0.34 0.0777
Alternatively, E X( )4 [(1 0.3)4 0.3] [(0 0.3)4 0.7] 0.0777
Thus, kurtosis is E X( ) /4 4 0.0777/0.464 1.76
2.3. For the two new random variables W 3 6X and V 20 7Y, we have:
(a) E V( ) E(20 7Y ) 20 7E Y( ) 20 7 0 78 14 54,
E W() E(3 6X) 3 6E X() 3 6 0 70 72
(b) W
2
var (3 6X) 62 X
2
36 0 21 7 56,
V
2
var (20 7Y ) ( 7)2 Y
2
49 0 1716 8 4084
(c) WV cov (3 6X, 20 7Y ) 6 ( 7) cov X Y( , ) 42 0 084 3 528
3 528
corr (W V, ) WV
0 4425
W V 7 56 8 4084
, 4 Stock/Watson • Introduction to Econometrics, Third Edition
2.5. Let X denote temperature in F and Y denote temperature in C. Recall that Y 0 when X 32 and Y
100 when X 212; this implies Y (100/180) (X 32) or Y 17.78 (5/9) X. Using Key
Concept 2.3, X 70oF implies that Y 17.78 (5/9) 70 21.11 C, and X 7oF
implies Y (5/9) 7 3.89 C.
2.6. The table shows that Pr (X 0,Y 0) 0 037, Pr (X 0,Y 1) 0 622,
Pr (X 1,Y 0) 0 009, Pr (X 1,Y 1) 0 332, Pr (X 0) 0 659, Pr (X 1) 0
341, Pr (Y 0) 0 046, Pr (Y 1) 0 954.
(a) E Y( ) Y 0 Pr(Y 0) 1 Pr (Y 1)
0 0 046 1 0 954 0 954
(b) Unemployment Rate #(unemployed)
#(labor force)
Pr (Y 0) 1Pr(Y 1) 1E Y( ) 10 954 0.046
(c) Calculate the conditional probabilities first:
Pr (Y 0| X 0) Pr (X 0,Y 0) 0 037
0 056, Pr (X 0)
0 659
Pr (Y 1|X 0) Pr (X 0,Y 1) 0 622 0 944,
Pr (X 0) 0 659
Pr (X 1,Y 0) 0 009
Pr (Y 0| X 1) 0 026,
Pr (X 1) 0 341
Pr (Y 1|X 1) Pr (X 1,Y 1) 0 332 0 974
Pr (X 1) 0 341
The conditional expectations are
E Y X( | 1) 0 Pr (Y 0| X 1) 1 Pr (Y 1|X 1)
0 0 026 1 0 974 0 974,
E Y X( | 0) 0 Pr (Y 0| X 0) 1 Pr (Y 1|X 0)
0 0 056 1 0 944 0 944
(d) Use the solution to part (b),
Unemployment rate for college graduates 1 E(Y|X 1) 1 0.974 0.026
Edition By James H. Stock & Mark W. Watson.
,Chapter 2
Review of Probability
2.2. We know from Table 2.2 that Pr (Y 0) 0 22, Pr (Y 1) 0 78, Pr (X 0) 0
30, Pr (X 1) 0 70. So
(a) Y E Y( ) 0 Pr (Y 0) 1 Pr (Y 1)
0 0 22 1 0 78 0 78,
X E X() 0 Pr (X 0) 1 Pr (X 1)
0 0 30 1 0 70 0 70
(b) X
2
E X[( X )2 ]
(0 0.70)2 Pr (X 0) (1 0.70)2 Pr (X 1)
( 0 70)2 0 30 0 302 0 70 0 21,
Y
2
E Y[( Y )2 ]
(0 0.78)2 Pr (Y 0) (1 0.78)2 Pr (Y 1)
( 0 78)2 0 22 0 222 0 78 0 1716
(c) XY cov (X Y, ) E X[( X )(Y Y )]
(0 0.70)(0 0.78) Pr(X 0,Y 0)
(0 0 70)(1 0 78) Pr (X 0 Y 1)
(1 0 70)(0 0 78) Pr (X 1Y 0)
(1 0 70)(1 0 78) Pr (X 1 Y 1)
( 0 70) ( 0 78) 0 15 ( 0 70) 0 22 0 15
0 30 ( 0 78) 0 07 0 30
0 22 0 63 0 084,
0 084
corr (X Y, ) XY
0 4425
X Y 0 21 0 1716
2.4. (a) E X( 3) 03 (1 p) 13 p p
(b) E X( k
) 0k (1 p) 1k p p
(c) E X( ) 0.3, and var(X) = E(X2)−[E(X)]2 = 0.3 −0.09 = 0.21. Thus =
0.21= 0.46.
var (X) E X( 2
) [E X( )]2 0.3 0.09 0.21 0.21 0.46. To compute
the skewness, use the formula from exercise 2.21:
, Solutions to End-of-Chapter Exercises 3
E X( )3 E X(3 ) 3[E X( 2 )][E X( )] 2[E X( )]3
0.3 3 0.32 2 0.33 0.084
Alternatively, E X( )3 [(1 0.3)3 0.3] [(0 0.3)3 0.7] 0.084
Thus, skewness E X( ) /3 3 0.084/0.463 0.87.
To compute the kurtosis, use the formula from exercise 2.21:
E X( )4 E X(4 ) 4[E X()][E X(3 )] 6[E X()]2 [E X( 2 )] 3[E X( )]4
0.3 4 0.32 6 0.33 3 0.34 0.0777
Alternatively, E X( )4 [(1 0.3)4 0.3] [(0 0.3)4 0.7] 0.0777
Thus, kurtosis is E X( ) /4 4 0.0777/0.464 1.76
2.3. For the two new random variables W 3 6X and V 20 7Y, we have:
(a) E V( ) E(20 7Y ) 20 7E Y( ) 20 7 0 78 14 54,
E W() E(3 6X) 3 6E X() 3 6 0 70 72
(b) W
2
var (3 6X) 62 X
2
36 0 21 7 56,
V
2
var (20 7Y ) ( 7)2 Y
2
49 0 1716 8 4084
(c) WV cov (3 6X, 20 7Y ) 6 ( 7) cov X Y( , ) 42 0 084 3 528
3 528
corr (W V, ) WV
0 4425
W V 7 56 8 4084
, 4 Stock/Watson • Introduction to Econometrics, Third Edition
2.5. Let X denote temperature in F and Y denote temperature in C. Recall that Y 0 when X 32 and Y
100 when X 212; this implies Y (100/180) (X 32) or Y 17.78 (5/9) X. Using Key
Concept 2.3, X 70oF implies that Y 17.78 (5/9) 70 21.11 C, and X 7oF
implies Y (5/9) 7 3.89 C.
2.6. The table shows that Pr (X 0,Y 0) 0 037, Pr (X 0,Y 1) 0 622,
Pr (X 1,Y 0) 0 009, Pr (X 1,Y 1) 0 332, Pr (X 0) 0 659, Pr (X 1) 0
341, Pr (Y 0) 0 046, Pr (Y 1) 0 954.
(a) E Y( ) Y 0 Pr(Y 0) 1 Pr (Y 1)
0 0 046 1 0 954 0 954
(b) Unemployment Rate #(unemployed)
#(labor force)
Pr (Y 0) 1Pr(Y 1) 1E Y( ) 10 954 0.046
(c) Calculate the conditional probabilities first:
Pr (Y 0| X 0) Pr (X 0,Y 0) 0 037
0 056, Pr (X 0)
0 659
Pr (Y 1|X 0) Pr (X 0,Y 1) 0 622 0 944,
Pr (X 0) 0 659
Pr (X 1,Y 0) 0 009
Pr (Y 0| X 1) 0 026,
Pr (X 1) 0 341
Pr (Y 1|X 1) Pr (X 1,Y 1) 0 332 0 974
Pr (X 1) 0 341
The conditional expectations are
E Y X( | 1) 0 Pr (Y 0| X 1) 1 Pr (Y 1|X 1)
0 0 026 1 0 974 0 974,
E Y X( | 0) 0 Pr (Y 0| X 0) 1 Pr (Y 1|X 0)
0 0 056 1 0 944 0 944
(d) Use the solution to part (b),
Unemployment rate for college graduates 1 E(Y|X 1) 1 0.974 0.026
