Exam (elaborations) TE.ST BANK FOR INTRODUCTION TO ECONOMETRICS 3RD ED (ECON15A) Introduction to Econometrics, ISBN: 9780138009007

TEST BANK FOR INTRODUCTION TO ECONOMETRICS 3RD EDITION JAMES AND
WATSON

, For Instructors

Solutions to End-of-Chapter Exercises

,TEST BANK FOR INTRODUCTION TO ECONOMETRICS 3RD EDITION JAMES
AND WATSON




Chapter 2
Review of Probability

2.1. (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.

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)  2 = E[( X −  )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

2.3. 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