Principles Of Econometrics, 4th Edition Solution Manual

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’s ManualForPrinciples of Econometrics, Fourth EditionWILLIAM E. GRIFFITHSUniversity of MelbourneR. CARTER HILLLouisiana State UniversityGUAY C. LIMUniversity of MelbourneSIMON YUNHO CHOUniversity of MelbourneSIMONE SI-YIN WONGUniversity of MelbourneJOHN WILEY & SONS, INCNew York / Chichester / Weinheim / Brisbane / Singapore / Toronto

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CONTENTSSolutions to Exercises in:Probability Primer1Chapter 2The Simple Linear Regression Model21Chapter 3Interval Estimation and Hypothesis Testing54Chapter 4Prediction, Goodness of Fit and Modeling Issues97Chapter 5The Multiple Regression Model132Chapter 6Further Inference in the Multiple Regression Model178Chapter 7Using Indicator Variables225Chapter 8Heteroskedasticity271Chapter 9Regression with Time Series Data: Stationary Variables308Chapter 10Random Regressors and Moment Based Estimation360Chapter 11Simultaneous Equations Models387Chapter 12Regression with Time Series Data: Non-Stationary Variables424Chapter 13Vector Error Correction and Vector Autoregressive Models448Chapter 14Time-Varying Volatility and ARCH Models472Chapter 15Panel Data Models489Chapter 16Qualitative and Limited Dependent Variable Models527Appendix AMathematical Tools576Appendix BProbability Concepts586Appendix CReview of Statistical Inference604

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1PROBABILITY PRIMERExercise Solutions

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Probability Primer, Exercise Solutions,Principles of Econometrics, 4e2EXERCISE P.1(a)Xis 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()( )5000.210000.620000.21100xE Xx fxuuu¦(c)Expected profit is given by()(52000)5()20005110020003500E YEXE Xu(d)The variance of profit is given by2var()var(52000)5 var()25240,0006,000,000YXXu

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Probability Primer, Exercise Solutions,Principles of Econometrics, 4e3EXERCISE P.2(a)The completed table isY( ,)fx y01( )fxŦ100.180.000.18X00.000.300.30100.070.450.52( )fy0.250.75(b)()( )100.1800.3100.523.4xE Xx fxuuu¦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)Yfxfxyffor10, 0,10xIt is given in the following tableX( ,1)(1)Yfxf|1fxyŦ100.00 0.750.000.30 0.750.4100.45 0.750.6(d)Given that she did not study, your expected winnings are|1|1100.000.4100.66xEXYx fxyuuu¦

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Probability Primer, Exercise Solutions,Principles of Econometrics, 4e4EXERCISE P.3Assume that total salesXare measured in millions of dollars. Then,22.5,0.3XN, and32.530.31.666711.666710.95220.0478P XPZP ZP Z§·!!¨¸©¹!

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Probability Primer, Exercise Solutions,Principles of Econometrics, 4e5EXERCISE P.4Extending the table to include the marginal distributions for political affiliation (PA) andCITYyieldsPolitical Affiliation (PA)RID()fCITYCITYSouthern0.240.040.120.4Northern0.180.120.300.6()fPA0.420.160.42(a)(,)0.18|0.3()0.6CITYfR NorthernP R CITYNorthernfNorthern(b)Political affiliation and region of residence are not independent because, for example,(,)0.18()()0.420.60.252PACITYfR NorthernfRfNorthernzuu(c)()()( )()PAPAPAEPARfRIfIDfDuuu00.4220.1650.422.42uuu(d)22()222()2EXEPAPAE PAE PAwhere2222222()( )()00.4220.1650.4211.14PAPAPAEPARfRIfIDfDuuuuuuThus,2()2()222.42211.1427.12E XE PAE PAuu

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Probability Primer, Exercise Solutions,Principles of Econometrics, 4e6EXERCISE P.5(a)The probability that the NFC wins the 12thflip, given they have won the previous 11 flipsis 0.5. Each flip is independent; so the probability of winning any flip is 0.5 irrespective ofthe outcomes of previous flips.(b)Because the outcomes of previous flips are independent and independent of the outcomesof future flips, the probability that the NFC will win the next two consecutive flips is 0.5multiplied by 0.5. That is,20.50.25.Go Saints!

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Probability Primer, Exercise Solutions,Principles of Econometrics, 4e7EXERCISE P.6(a)()(40710430)40710430()40710430758460E SALESEPRICEE PRICEu(b)22var()var(40710430)430var()430254,622,500SALESPRICEPRICEu(c)6300846063004622500P SALESPZ§·!!¨¸©¹1.004651.004650.8425P ZP Z!

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Probability Primer, Exercise Solutions,Principles of Econometrics, 4e8EXERCISE P.7After including the marginal probability distributions for bothCandB, the table becomesB012( )fcC00.050.050.050.1510.050.200.150.4020.050.250.150.45( )fb0.150.500.35(a)The marginal probability distribution forCis given in the last column of the above table.(b)()( )00.1510.4020.451.3cE Cc f cuuu¦(c)>@222222var()( )()00.1510.4020.45(1.3)0.51cCc f cE Cuuu¦(d)For the two companies’ advertising strategies to be independent, the condition( , )( )( )CBfc bfc fbmust hold for allcandb. We find that(0,0)0.05(0)(0)0.150.150.0225CBfffzuThus, the two companies’ advertising strategies are not independent.(e)Values forAare given by the equation50001000AB. Its probability distribution isobtained by matching values obtained from this equation with corresponding probabilitiesforB.A( )fa50000.1560000.5070000.35(f)Since the relationship betweenAandBis 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, 4e9EXERCISE P.8(a)X( )fx11 621 631 641 651 661 6(b)164PX1116634 or5PXX(c)111111666666()( )1234563.5xE Xx fxuuuuuu¦The result()3.5E Xmeans that if a die is rolled a very large number of times, theaverage of all the values shown will be 3.5; it will approach 3.5 as the number of rollsincreases.(d)22222222111111666666( )12345615.16667xEXx fxuuuuuu¦(e)>@222var()()15.166673.52.91667XEXE X(f)The results for this part will depend on the rolls obtained by the student. LetnXdenote theaverage value afternrolls. The values obtained by one of us and their averages are:20 values of^`2,1,5,3,4,1,5,5,2,4,2,2,4,2,4,4,3,2,6,3X53.000X103.200X203.200XThese values are relatively close to the mean of 3.5 and are expected to become closer asthe number of rolls increases.

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Probability Primer, Exercise Solutions,Principles of Econometrics, 4e10EXERCISE P.9(a)The area under the curve is equal to one. Recalling that the formula for the area of atriangle is half the base multiplied by the height, it is given by122331uu.(b)When1 2x,( )5 9fx. The probability is given by the area under the triangle between0 and 1/2. This can be calculated as11 / 23PX. The latter probability is1155252229361/ 230.69444PXbhuuTherefore,25113636(01/ 2)11/ 2310.30555PXPX(c)To compute this probability we can subtract the area under the triangle between 3/4 to 3from the area under the triangle from 1/4 to 3. Doing so yields13441311132424441111119124182421219144165181 43 433220.27778PXPXPXffuuuuuuuu0.2.4.6.8fx0123xf(x)=2/3 - 2/9x

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Probability Primer, Exercise Solutions,Principles of Econometrics, 4e11EXERCISE P.10(a)>@1122()()()()2XYE ZEE XE Y§·P PP¨¸©¹(b)AssumingXandYare independent,>@22221124var()varvar()var()()22XYZXYV§·V V¨¸©¹(c)Assuming that2cov(,)0.5X YV,>@222221214var()varvar()var()2cov(,)2320.54XYZXYX Y§·¨¸©¹VV V uV

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Probability Primer, Exercise Solutions,Principles of Econometrics, 4e12EXERCISE P.11LetXdenote the length of life of a personal computer selected at random. The fraction ofcomputers that fail within a given time interval is equal to the probability thatXlies in thatinterval.(a)13.411.89740.02891.6P XPZP Z§· ¨¸©¹(b)43.440.47430.31761.6P XPZP Z§·t!!¨¸©¹(c)23.421.10680.86581.6P XPZP Z§·t!! ¨¸©¹(d)2.53.443.42.540.71150.47431.61.6PXPZPZ§·¨¸©¹0.47430.71150.68240.23840.444P ZP Z (e)We want0Xsuch that00.05P XX. Now,(1.645)0.05P Z, and thus a suitable0Xis such that03.41.6451.64XSolving for0Xyields03.41.6451.61.319X(which is approximately 16 months)

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Probability Primer, Exercise Solutions,Principles of Econometrics, 4e13EXERCISE P.12(a)The probability function ofXis shown below.(b)The probability that, on a given Monday, either 2, or 3, or 4 students will be absent is42( )(2)(3)(4)0.260.340.220.82xfxfff¦(c)The probability that, on a given Monday, more than 3 students are absent is74( )(4)(5)(6)(7)0.220.080.040.010.35xfxffff¦(d)70().( )00.0210.0320.2630.3440.22xE Xx fxuuuuu¦50.0860.0470.013.16uuuBased on information over many Mondays, the average number of students absent onMondays is 3.16.(e)22var()[()]XEXE X722222202222( )00.0210.0320.2630.3440.2250.0860.0470.01 = 11.58xEXx fxuuuuuuuu¦22var()11.58(3.16)1.5944XV21.2627VV(f)()(73)7()373.16325.12E YEXE Xu2var()var(73)7 var()491.594478.1256YXXu.00.05.10.15.20.25.30.35.4001234567xf(x)

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