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Introduction to Econometrics 3rd Edition Test Bank by James H. Stock and Mark W. Watson
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Introduction to Econometrics 3rd Edition Test Bank by James H. Stock and Mark W. Watson

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  • July 8, 2022
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, For Instructors

Solutions to End-of-Chapter Exercises

,Chapter 2
Review of Probability

(a) Probability distribution function for Y
Outcome (number of heads) Y0 Y1 Y2
Probability 0.25 0.50 0.25

(b) Cumulative probability distribution function for Y

Outcome (number of heads) Y0 0Y1 1Y2 Y2
Probability 0 0.25 0.75 1.0

(c) Y = E(Y )  (0  0.25)  (1 0.50)  (2  0.25)  1.00 . F 
d
Fq, .
Using Key Concept 2.3: var(Y )  E(Y 2 ) [E(Y )]2 ,
and
(ui |Xi )
so that
var(Y )  E(Y 2 ) [E(Y )]2  1.50  (1.00)2  0.50.

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)   E[( X   )2 ]
2

X X

 (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,
 Y2  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




©2011 Pearson Education, Inc. Publishing as Addison Wesley

, Solutions to End-of-Chapter Exercises 3

(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  1 Y  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,
 XY 0084
corr (X , Y )    04425
 XY 021 01716

For the two new random variables W  3  6 X 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  72
(b)  2  var(3  6X )  62   2  36  021  756,
W X

V  var(20  7Y )  (7)   Y2  49  01716  84084
2 2


(c) WV  cov(3  6X , 20  7Y )  6  (7) cov(X , Y )  42  0084  3528
WV 3528
corr (W , V )    04425
WV 756  84084

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:
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




©2011 Pearson Education, Inc. Publishing as Addison Wesley

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