Stock Watson exercise and solutions

For Instructors

Solutions to End-of-Chapter Exercises

,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 |X i )
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)   E[( X   X ) 2 ]
2
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  7 E (Y )  20  7  078  1454,
E (W )  E (3  6 X )  3  6 E ( X )  3  6  070  72
(b)  W2  var (3  6 X )  62   X2  36  021  756,
 V2  var (20  7Y )  (7)2   Y2  49  01716  84084
(c)  WV  cov(3  6 X , 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 )  [ E ( X )]  0.3  0.09  0.21   0.21  0.46. To compute the skewness, use
2 2


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

, 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
# (unemployed)
(b) Unemployment Rate 
# (labor force)
 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,
Pr ( X  0) 0659
Pr ( X  0, Y  1) 0622
Pr (Y  1| X  0)    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 ( X  1, Y  1) 0332
Pr (Y  1| X  1)    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
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
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




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