计量经济分析（第六版）答案 Answers6.docx

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1、NYU 中STERNDepartment of EconomicsNEW YORK UNIVERSITY LEONARD N. STERN SCHOOL OF BUSINESSECONOMETRICS IFall 2007, TR, l：00- 2：20Professor William Greene Phone: 212.998.0876Office: KMC 7-78Home page:ww.stern.nyu.edu/wgreeneEmail: wgreenestern.nyu.eduURL for course web page:www.stern.nyu.edu/wgreene/Ec

2、onometrics/Econometrics.htmAssignment 6Asymptotics for Least Squares and the Classical Regression Model1. This exercise is based on the Airlines Data used in a few examples in your text. The model in this problem set islogc 二 a + piogq + y(log2q/2) + 8ilog(pf) + 521f + 53t + e.(See Table F7.1 on pag

3、e 949. The data are on the website for the text.) Linear regression of log(cost) = 1c on a constant (one), log(output) = Iq, one half the square of Iq = .5*lq2 = lq2, the log of the fuel price, Ipf, and the load factor, If, and a time trend, t, produce the following results (on the next page). Note

4、that there are three sets of results shown.a. Test the hypothesis that both the load factor and time trend coefficients are zero using an F test. Note that this requires values from both regressions below. Now, using only the first set of results, use a Wald (chi-squared) statistic, test the same hy

5、pothesis. Explain the difference in the two tests. What is the result of the test?F 二(1.577479-1.149658)/2) / 1.149658/(90-6) = 15.63. Large. Reject hypothesis.The Wald statistic is computed below. For the linear model, Wald = 2*F. In general, the chi- squared is the limiting statistic when the degr

6、ees of freedom grows large.- regr;lhs=lc;rhs=onezIq,lq2,Ipf,If,t$+ -1Ordinaryleast squares regression11Model was estimated Dec 09, 2005 at04:30:11PM|1LHS=LCMean=13.36561|1Standard deviation =1.131971|1WTS=noneNumber of observs.=90|1Model sizeParameters=6|1Degrees of freedom =84|1ResidualsSum of squa

7、res=1.149658|1Standard error of e =.1169889|1FitR-squared=.9899189|1Adjusted R-squared =.9893188|1Model testF 5,84 (prob) =1649.68 (.0000) |1DiagnosticLog likelihood=68.51107|1Restricted(b=0)=-138.3581|1Chi-sq 5 (prob)=413.74 (.0000) |1Info criter.LogAmemiya Prd. Crt.=-4.226814|1Akaike Info. Criter.

8、=-4.227012|1AutocorrelDurbin-Watson Stat.=.2169784|1Rho = core,e(-1)=.8915108|+-+4-+4-+I Variable |Coefficient|Standard Error|t-ratio|P|T|t|Mean ofX|+Constant11.3580713.6771772416.773.0000LQ.93040362.0280246533.199.0000-1.17430919LQ2.04602849.021442532.147.03471.34409282LPF.28511398.060163614.739.00

9、0012.7703594LF-1.56214282.32709516-4.776.0000*56046016T.03409548.011270133.025.00338.00000000- matr;r=0,0,0,0,1,0/0,0,0,0,0,1;aq=0/0$- matr;list; m=r*b-aq ; vm = R*varb*r * ; Wald = m* m $Matrix Mhas 2 rows and 1 columns.1+1 |-1.562142 |.03410Matrix VMhas 2 rows and 2 columns.12+1 |.106999202913D-04

10、2| -.9202913D-04.00013Matrix WALD has 1 rows and 1 columns. 1 +1 |31.25883- calc;list;fc=wald/2$ FC =.15629415347651660D+02b. The third set of regression results given corresponds to a modellogc = a + plogq + Silf + 甑 + The estimate of the scale parameter in this model is E = 1/p. Using the third se

11、t of regression results, estimate this quantity and compute an estimate of the asymptotic standard error. What is the asymptotic distribution of this estimator?- Regress;lhs=lc;rhs=one,lqzIf,t$+I OrdinaryI Model wasI LHS=LCleast squares regressionestimated Dec 09, 2005 at 04:34:43PMWTS=noneModel siz

12、eResidualsFitMeanStandard deviation Number of observs. ParametersDegrees of freedom Sum of squares Standard error of e R-squaredAdjusted R-squared13.36561 1.13197190 486 1.515892 .1327654 .9867075 .9862438Model testDiagnosticF 3,86 (prob) =2127.93 (.0000)Info criter.AutocorrelLog likelihood Restrict

13、ed(b=0)Chi-sq 3 (prob)LogAmemiya Prd. Crt. Akaike Info. Criter. Durbin-Watson Stat.Rho = core,e(-1)56.06677-138.3581388.85 (.0000)-3.994858-3.994917.3755908.81220464-4-4-4-I Variable | Coefficient | Standard Error |t-ratio |P|T|t | Mean of X|Constant14.5121911.1970700773.640.0000LQ.86736709.01413009

14、61.384.0000LF-1.42812721.36398158-3.924.0002T.08404797.0040523620.741.0000+-1.17430919.560460168.00000000- - calc; list ;beta = b (2) ; e = 1/beta $BETA= .86736708567543600D+00E= .11529143963553570D+01Calculator: Computed 2 scalar results- - calc; list ; vbeta = varb(2,2) ; dedb = -l/betaA2 $VBETA=

15、.19965951595452470D-03DEDB= -.13292116053234360D+01Calculator: Computed 2 scalar results- - calc; list ; ve = dedbA2 * vbeta ; sde = sqr(ve)$VE= .35275912994487820D-03SDE= . 187818830244 70100D-01The estimate of E is 1.153. The estimated asymptotic standard error is .01878.The asymptotic distributio

16、n would be normal with this mean and standard deviation.c. One common measure of economies of scale in statistical cost analysis is the measureE = l/61ogc/aiogq=-，+ /logqUsing the results from the first equation, estimate E computed at the average value of logq (not the log of the average of q). Com

17、pute the asymptotic standard error for your estimate. Explain your calculations at each step.+1 Ordinaryleast squares regression1I Model wasestimated Dec 09, 2005at 04:45:35PM|I LHS=LCMean=13.36561|1Standard deviation=1.131971|I WTS=noneNumber of observs.=90|I Model sizeParameters=6|1Degrees of free

18、dom=84|1 ResidualsSum of squares=1.149658|1Standard error of e=.1169889|I FitR-squared=.9899189|Adjusted R-squared=.9893188|1 Model testF 5,84 (prob)=1649.68 (.0000) |I DiagnosticLog likelihood=68.51107|Restricted(b=0)= -138.3581|Chi-sq 5 (prob)=413.74 (.0000) |1 Info criter.LogAmemiya Prd Crt.=-4.2

19、26814|Akaike Info. Criter.=-4.227012|I AutocorrelDurbin-Watson Stat.=.2169784|Rho = core,e(-1)=.8915108|+I Variable|Coefficient |Standard Error |t-ratio|P|T|t|Mean of X+-+-+十Constant11.3580713.6771772416.773.0000LQ.93040362.0280246533.199.0000-1.17430919LQ2.04602849.021442532.147.03471.34409282LPF.2

20、8511398.060163614.739.000012.7703594LF-1.56214282.32709516-4.776.0000.56046016T.03409548.011270133.025.00338.00000000- - calc ; beta=b(2);gamma=b(3) $- - calc ; logqb = xbr(Iq) $- matr ; vbc = varb(2:3,2:3) $- - calc ; list ; e = 1/(beta+gamma*logqb)$E=.11410940699966370D+01一- calc ; list ; dedb = -

21、e人2 ; dedg = -eA2*logqb $DEDB=13020956765814900D+01DEDG=.15290629193330670D+01Calculator: Computed 2 scalar results- - matr ; list ; ge = dedb,dedg $Matrix GEhas 1 rows and 2 columns.124-1|-1.302101.52906- - matr; ve2 = ge*vbc*ge1 $ - calc ; list;sqr(ve2)$ Result =. 16416912272909100D-01The estimate

22、 is 1.141. The estimated asymptotic standard erorr is .01642.2. Continuing part c of the previous exercise, the log quadratic cost function implies a U shaped average cost function. That in turn implies that there are economies of scale in airline services, but that they run out at some point, so th

23、at the advantage of large scale eventually disappears. In particular, I did the following: (1) Compute the means of log(pf), If, and t, (2) using the regression model estimated coefficients, compute estimated costs for a large range of values of q (3) plot fitted cost against q. The figure appears b

24、elow1.95001.65001.87501.80001.7250E stin atod A voiage C ostFunctbnAs you can see, the cost curve has a bottom, somewhere near 4. This is called the efficient scale. The efficient scale occurs where there are no economies of scale, that is, where the scale elasticity in question l.b. equals 1. The i

25、nteresting quantity is the q where this occurs, which for this cost function would beUsing the regression results for the first (long) regression below, estimate q* and estimate the asymptotic standard error for your estimator. Form a confidence interval for q* What is the asymptotic distribution.-

26、- calc ; list ; escale = exp ( (1-beta)/gamma) $ ESCALE =. 4 53592238824 84 530D+01- - calc ; desdb = -escale/gamma ; desdg = -escale* (1-beta) /gamma/X2 $- - matr ; list ; ges = desdb,desdg $Matrix GES has 1 rows and 2 columns. 12+1|-98.54598-149.00432- - matr; ves = ges*vbc*ges1 $- - calc ; list;s

27、es = sqr(ves)$SES =. 5796230321684 4 540D+01- - calc ; list ; lower = escale - 1.96*ses;upper = escale + 1.96*ses $LOWER=6824 68 9042253077 0D+01UPPER= 15896533818749980D+02I Ordinary least squares regression|ILHS=LCMean=13.36561|IStandard deviation=1.131971|IWTS=noneNumber of observs.=90|IModel siz

28、eParameters=6|IDegrees of freedom=84|IResidualsSum of squares=1.149658IIStandard error of e=.1169889|IFitR-squared=.9899189|IAdjusted R-squared=.9893188|I Model test F 5,84 (prob) =1649.68 (.0000) |+I Variable|Coefficient|Standard Error|t-ratio|P|T|t|Mean ofX|+Constant11.3580713.6771772416.773.0000L

29、Q.93040362.0280246533.199.0000-1.17430919LQ2.04602849.021442532.147.03471.34409282LPF.28511398.060163614.739.000012.7703594LF-1.56214282.32709516-4.776.0000*56046016T.03409548.011270133.025.00338.00000000Estimated Asymptotic covariance matrix forthe estimated coefficients:0.458569-0.00111089-0.00078

30、2679-0.0393469-0.02486640.00721773-0.001110890.0007853810.0005366810.000142168-0.000631672-1.8712e-005-0.0007826790.0005366810.0004597821.98541e-0050.0009158073.51257e-006-0.03934690.0001421681.98541e-0050.00361966-0.00284521-0.000642815-0.0248664-0.0006316720.000915807-0.002845210.106991-9.20291e-0

31、050.00721773-1.8712e-0053.51257e-006 -0.000642815-9.20291e-0050.000127016+I Ordinary least squares regression|I Model was estimated Oct 08, 2003 at 01:23:58PM|ILHS=LCMean=13.36561|IStandard deviation=1.131971|IWTS=noneNumber of observs.=90|IModel sizeParameters=4|IDegrees of freedom=86|IResidualsSum

32、 of squares=1.577479|IStandard error of e=. 1354355|IFitR-squared=.9861674|IAdjusted R-squared=. 985684 9|I Model test F 3,86 (prob) =2043.74 (.0000) |+4-|Variable|Coefficient|Standard Error|t-ratio|P|T|tIMean ofX|Estimated Asymptotic covariance matrix for12+Constant9.13822759.2450666137.289.0000LQ.

33、92614899.0323058728.668.0000-1.17430919LQ2.05829032.024607062.369.02011.34409282LPF.41006045.0188069221.804.000012.7703594the estimated coefficients:11.06006-.00022-.00109-.0045921-.00022.00104.00073.3684100D-0431-.00109.00073.00061.8854564D-044|-.00459.3684100D-04.8854564D-04.0003534+I Ordinary lea

34、st squares regression|I Model was estimated Oct 08, 2003 at 05:23:53PM |ILHS=LCMean=13.36561|IStandard deviation=1.131971|IWTS=noneNumber of observs.=90|IModel sizeParameters=4|IDegrees of freedom=86IIResidualsSum of squares=1.568672|IStandard error of e=. 1350569|IFitR-squared=.9862447|IAdjusted R-

35、squared=.9857648|I Model test F3,86 (prob) =2055.37 (. 0000) |+4-IVariable|Coefficient |Standard Error |t-ratio|P|T|t IMean of X|+-+一 + _+Constant11.0698991.7762035714.262.0000LQ.84993303.0132353464.217.0000-1.17430919LPF.23776164.068683723.462.000812.7703594T.03218659.013004562.475.01538.000000001234+11. 60249-.00077-.05325.0096021-.00077.00018.9641626D-04-.3253667D-0431-.05325.9641626D-04.00472-.0008641.00960-.3253667D-04-.00086.00017