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Principles Of Econometrics, 4th Edition Solution Manual

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    ’s Manual For Principles of Econometrics, Fourth Edition WILLIAM E. GRIFFITHS University of Melbourne R. CARTER HILL Louisiana State University GUAY C. LIM University of Melbourne SIMON YUNHO CHO University of Melbourne SIMONE SI-YIN WONG University of Melbourne JOHN WILEY & SONS, INC New York / Chichester / Weinheim / Brisbane / Singapore / Toronto

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    CONTENTS Solutions to Exercises in: Probability Primer 1 Chapter 2 The Simple Linear Regression Model 21 Chapter 3 Interval Estimation and Hypothesis Testing 54 Chapter 4 Prediction, Goodness of Fit and Modeling Issues 97 Chapter 5 The Multiple Regression Model 132 Chapter 6 Further Inference in the Multiple Regression Model 178 Chapter 7 Using Indicator Variables 225 Chapter 8 Heteroskedasticity 271 Chapter 9 Regression with Time Series Data: Stationary Variables 308 Chapter 10 Random Regressors and Moment Based Estimation 360 Chapter 11 Simultaneous Equations Models 387 Chapter 12 Regression with Time Series Data: Non-Stationary Variables 424 Chapter 13 Vector Error Correction and Vector Autoregressive Models 448 Chapter 14 Time-Varying Volatility and ARCH Models 472 Chapter 15 Panel Data Models 489 Chapter 16 Qualitative and Limited Dependent Variable Models 527 Appendix A Mathematical Tools 576 Appendix B Probability Concepts 586 Appendix C Review of Statistical Inference 604

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    1 PROBABILITY PRIMER Exercise Solutions

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    Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 2 EXERCISE P.1 (a) X is a random variable because attendance is not known prior to the outdoor concert. Before the concert, attendance is uncertain because the weather is uncertain. (b) Expected attendance is given by ( ) ( ) 500 0.2 1000 0.6 2000 0.2 1100 x E X x f x u  u  u ¦ (c) Expected profit is given by ( ) (5 2000) 5 ( ) 2000 5 1100 2000 3500 E Y E X E X   u  (d) The variance of profit is given by 2 var( ) var(5 2000) 5 var( ) 25 240,000 6,000,000 Y X X  u

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    Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 3 EXERCISE P.2 (a) The completed table is Y ( , ) f x y 0 1 ( ) f x Ŧ 10 0.18 0.00 0.18 X 0 0.00 0.30 0.30 10 0.07 0.45 0.52 ( ) f y 0.25 0.75 (b) ( ) ( ) 10 0.18 0 0.3 10 0.52 3.4 x E X x f x  u  u  u ¦ You should take the bet because the expected value of your winnings is positive. (c) The probability distribution of your winnings if you know she did not study is ( ,1) | 1 (1) Y f x f x y f for 10, 0,10 x  It is given in the following table X ( ,1) (1) Y f x f | 1 f x y Ŧ 10 0.00 0.75 0.0 0 0.30 0.75 0.4 10 0.45 0.75 0.6 (d) Given that she did not study, your expected winnings are | 1 | 1 10 0.0 0 0.4 10 0.6 6 x E X Y x f x y  u  u  u ¦

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    Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 4 EXERCISE P.3 Assume that total sales X are measured in millions of dollars. Then, 2 2.5,0.3 X N  , and 3 2.5 3 0.3 1.6667 1 1.6667 1 0.9522 0.0478 P X P Z P Z P Z  § · ! ! ¨ ¸ © ¹ !   

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    Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 5 EXERCISE P.4 Extending the table to include the marginal distributions for political affiliation ( PA ) and CITY yields Political Affiliation ( PA ) R I D ( ) f CITY CITY Southern 0.24 0.04 0.12 0.4 Northern 0.18 0.12 0.30 0.6 ( ) f PA 0.42 0.16 0.42 (a) ( , ) 0.18 | 0.3 ( ) 0.6 CITY f R Northern P R CITY Northern f Northern (b) Political affiliation and region of residence are not independent because, for example, ( , ) 0.18 ( ) ( ) 0.42 0.6 0.252 PA CITY f R Northern f R f Northern z u u (c) ( ) ( ) ( ) ( ) PA PA PA E PA R f R I f I D f D u  u  u 0 0.42 2 0.16 5 0.42 2.42 u  u  u (d) 2 2 ( ) 2 2 2 ( ) 2 E X E PA PA E PA E PA   where 2 2 2 2 2 2 2 ( ) ( ) ( ) 0 0.42 2 0.16 5 0.42 11.14 PA PA PA E PA R f R I f I D f D u  u  u u  u  u Thus, 2 ( ) 2 ( ) 2 2 2.42 2 11.14 27.12 E X E PA E PA  u  u

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    Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 6 EXERCISE P.5 (a) The probability that the NFC wins the 12 th flip, given they have won the previous 11 flips is 0.5. Each flip is independent; so the probability of winning any flip is 0.5 irrespective of the outcomes of previous flips. (b) Because the outcomes of previous flips are independent and independent of the outcomes of future flips, the probability that the NFC will win the next two consecutive flips is 0.5 multiplied by 0.5. That is, 2 0.5 0.25 . Go Saints!

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    Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 7 EXERCISE P.6 (a) ( ) (40710 430 ) 40710 430 ( ) 40710 430 75 8460 E SALES E PRICE E PRICE    u (b) 2 2 var( ) var(40710 430 ) 430 var( ) 430 25 4,622,500 SALES PRICE PRICE   u (c) 6300 8460 6300 4622500 P SALES P Z  § · ! ! ¨ ¸ © ¹ 1.00465 1.00465 0.8425 P Z P Z !  

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    Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 8 EXERCISE P.7 After including the marginal probability distributions for both C and B , the table becomes B 0 1 2 ( ) f c C 0 0.05 0.05 0.05 0.15 1 0.05 0.20 0.15 0.40 2 0.05 0.25 0.15 0.45 ( ) f b 0.15 0.50 0.35 (a) The marginal probability distribution for C is given in the last column of the above table. (b) ( ) ( ) 0 0.15 1 0.40 2 0.45 1.3 c E C c f c u  u  u ¦ (c) > @ 2 2 2 2 2 2 var( ) ( ) ( ) 0 0.15 1 0.40 2 0.45 (1.3) 0.51 c C c f c E C  u  u  u  ¦ (d) For the two companies’ advertising strategies to be independent, the condition ( , ) ( ) ( ) C B f c b f c f b must hold for all c and b . We find that (0,0) 0.05 (0) (0) 0.15 0.15 0.0225 C B f f f z u Thus, the two companies’ advertising strategies are not independent. (e) Values for A are given by the equation 5000 1000 A B  . Its probability distribution is obtained by matching values obtained from this equation with corresponding probabilities for B . A ( ) f a 5000 0.15 6000 0.50 7000 0.35 (f) Since the relationship between A and B is an exact linear one, they are perfectly correlated. The correlation between them is 1.

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    Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 9 EXERCISE P.8 (a) X ( ) f x 1 1 6 2 1 6 3 1 6 4 1 6 5 1 6 6 1 6 (b) 1 6 4 P X 1 1 1 6 6 3 4 or 5 P X X  (c) 1 1 1 1 1 1 6 6 6 6 6 6 ( ) ( ) 1 2 3 4 5 6 3.5 x E X x f x u  u  u  u  u  u ¦ The result ( ) 3.5 E X means that if a die is rolled a very large number of times, the average of all the values shown will be 3.5; it will approach 3.5 as the number of rolls increases. (d) 2 2 2 2 2 2 2 2 1 1 1 1 1 1 6 6 6 6 6 6 ( ) 1 2 3 4 5 6 15.16667 x E X x f x u  u  u  u  u  u ¦ (e) > @ 2 2 2 var( ) ( ) 15.16667 3.5 2.91667 X E X E X   (f) The results for this part will depend on the rolls obtained by the student. Let n X denote the average value after n rolls. The values obtained by one of us and their averages are: 20 values of ^ ` 2,1,5,3,4,1,5,5,2,4,2,2,4,2,4,4,3,2,6,3 X 5 3.000 X 10 3.200 X 20 3.200 X These values are relatively close to the mean of 3.5 and are expected to become closer as the number of rolls increases.

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    Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 10 EXERCISE P.9 (a) The area under the curve is equal to one. Recalling that the formula for the area of a triangle is half the base multiplied by the height, it is given by 1 2 2 3 3 1 u u . (b) When 1 2 x , ( ) 5 9 f x . The probability is given by the area under the triangle between 0 and 1/2. This can be calculated as 1 1 / 2 3 P X    . The latter probability is 1 1 5 5 25 2 2 2 9 36 1/ 2 3 0.69444 P X bh   u u Therefore, 25 11 36 36 (0 1/ 2) 1 1/ 2 3 1 0.30555 P X P X       (c) To compute this probability we can subtract the area under the triangle between 3/4 to 3 from the area under the triangle from 1/4 to 3. Doing so yields 1 3 4 4 1 3 1 1 1 3 2 4 2 4 4 4 1 11 11 1 9 1 2 4 18 2 4 2 121 9 144 16 5 18 1 4 3 4 3 3 2 2 0.27778 P X P X P X f f        u u  u u u u  u u  0 .2 .4 .6 .8 fx 0 1 2 3 x f(x)=2/3 - 2/9x

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    Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 11 EXERCISE P.10 (a) > @ 1 1 2 2 ( ) ( ) ( ) ( ) 2 X Y E Z E E X E Y  § ·  P  P P ¨ ¸ © ¹ (b) Assuming X and Y are independent, > @ 2 2 2 2 1 1 2 4 var( ) var var( ) var( ) ( ) 2 2 X Y Z X Y  V § ·  V  V ¨ ¸ © ¹ (c) Assuming that 2 cov( , ) 0.5 X Y V , > @ 2 2 2 2 2 1 2 1 4 var( ) var var( ) var( ) 2cov( , ) 2 3 2 0.5 4 X Y Z X Y X Y  § ·   ¨ ¸ © ¹ V V  V  u V

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    Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 12 EXERCISE P.11 Let X denote the length of life of a personal computer selected at random. The fraction of computers that fail within a given time interval is equal to the probability that X lies in that interval. (a) 1 3.4 1 1.8974 0.0289 1.6 P X P Z P Z  § ·     ¨ ¸ © ¹ (b) 4 3.4 4 0.4743 0.3176 1.6 P X P Z P Z  § · t ! ! ¨ ¸ © ¹ (c) 2 3.4 2 1.1068 0.8658 1.6 P X P Z P Z  § · t ! !  ¨ ¸ © ¹ (d) 2.5 3.4 4 3.4 2.5 4 0.7115 0.4743 1.6 1.6 P X P Z P Z   § ·        ¨ ¸ © ¹ 0.4743 0.7115 0.6824 0.2384 0.444 P Z P Z      (e) We want 0 X such that 0 0.05 P X X  . Now, ( 1.645) 0.05 P Z   , and thus a suitable 0 X is such that 0 3.4 1.645 1.64 X   Solving for 0 X yields 0 3.4 1.645 1.6 1.319 X  (which is approximately 16 months)

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    Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 13 EXERCISE P.12 (a) The probability function of X is shown below. (b) The probability that, on a given Monday, either 2, or 3, or 4 students will be absent is 4 2 ( ) (2) (3) (4) 0.26 0.34 0.22 0.82 x f x f f f     ¦ (c) The probability that, on a given Monday, more than 3 students are absent is 7 4 ( ) (4) (5) (6) (7) 0.22 0.08 0.04 0.01 0.35 x f x f f f f       ¦ (d) 7 0 ( ) . ( ) 0 0.02 1 0.03 2 0.26 3 0.34 4 0.22 x E X x f x u  u  u  u  u ¦ 5 0.08 6 0.04 7 0.01 3.16  u  u  u Based on information over many Mondays, the average number of students absent on Mondays is 3.16. (e) 2 2 var( ) [ ( )] X E X E X  7 2 2 2 2 2 2 0 2 2 2 2 ( ) 0 0.02 1 0.03 2 0.26 3 0.34 4 0.22 5 0.08 6 0.04 7 0.01 = 11.58 x E X x f x u  u  u  u  u  u  u  u ¦ 2 2 var( ) 11.58 (3.16) 1.5944 X V  2 1.2627 V V (f) ( ) (7 3) 7 ( ) 3 7 3.16 3 25.12 E Y E X E X   u  2 var( ) var(7 3) 7 var( ) 49 1.5944 78.1256 Y X X  u .00 .05 .10 .15 .20 .25 .30 .35 .40 0 1 2 3 4 5 6 7 x f(x)

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