Instructor's Manual for Introduction to Econometrics, 3rd edition James H. Stock, All Chapters Included

For Instructors
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Solutions to End-of-Chapter Exercises
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COMPLETE CHAPTERS
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Chapter 2
Review of Probability

2.1. (a) Probability distribution function for Y
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Outcome (number of heads) Y=0 Y=1 Y=2
Probability 0.25 0.50 0.25
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(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
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(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 ,
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and
(ui |Xi )
so that
var(Y ) = E(Y 2 ) −[E(Y )]2 = 1.50 − (1.00)2 = 0.50.
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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)
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= 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
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(b)  = E[( X −  ) ]
2 2

X X

= (0 − 0.70)2  Pr ( X = 0) + (1 − 0.70)2  Pr ( X = 1)
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= (−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




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, 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
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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:
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(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,
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W X

 V = var(20 − 7Y ) = (−7)   2Y = 49  01716 = 84084
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(c) WV = cov(3 + 6X , 20 − 7Y ) = 6  (−7) cov(X , Y ) = −42  0084 = −3528
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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
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(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
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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
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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.
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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

, 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)
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= 0  0046 +1 0954 = 0954
#(unemployed)
(b) Unemployment Rate =
#(labor force)
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= Pr (Y = 0) = 1 − Pr(Y = 1) = 1 − E(Y ) = 1 − 0954 = 0.046
(c) Calculate the conditional probabilities first:
Pr ( X = 0, Y = 0) 0037
Pr (Y = 0| X = 0) = = = 0056,
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Pr ( X = 0) 0659
Pr ( X = 0, Y = 1) 0622
Pr (Y = 1| X = 0) = = = 0944,
Pr ( X = 0)
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0659
Pr ( X = 1, Y = 0) 0009
Pr (Y = 0| X = 1) = = = 0026,
Pr ( X = 1) 0341
Pr ( X = 1, Y = 1) 0332
Pr (Y = 1| X = 1) = = = 0974
Pr ( X = 1) 0341
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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,
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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),
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Unemployment rate for college graduates = 1 − E(Y|X = 1) = 1 − 0.974 = 0.026
Unemployment rate for non-college graduates = 1 − E(Y|X = 0) = 1 − 0.944 = 0.056
(e) The probability that a randomly selected worker who is reported being unemployed is a
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college graduate is
Pr ( X = 1, Y = 0) 0009
Pr ( X = 1|Y = 0) = = = 0196
Pr (Y = 0) 0046

The probability that this worker is a non-college graduate is
Pr ( X = 0|Y = 0) = 1 − Pr ( X = 1|Y = 0) = 1 − 0196 = 0804




©2011 Pearson Education, Inc. Publishing as Addison Wesley