第十章习题答案计量经济学

1、第十章习题讲解做单位根检验有个潜规则先从最复杂的模型开始如果最复杂的模型说非平稳那就是非平稳如果最复杂的模型说平稳 那么要用简单的模型检验 如果所有模型都说是平稳的那就是平稳的10.1答：（1）做利润和红利的连线图PFT利润图红利图从图中可以初步判断，禾I润和红利都是非平稳的。（2）首先对利润进行单位根检验，选择有截距项和时间趋势（trend andintercept ）,结果如下Null ypothesis: R has a un it rootExoge nous: Con sta ntLag Len gth: 0 (Automatic based on SIC, MAXLAG=11)t-

2、StatisticAugme nted Dickey-Fuller test statistic-1.408419Test critical values:1% level-3.5073945% level-2.89510910% level-2.584738Null Hypothesis: PFT has a unit rootExogenous: Constant, Linear TrendLag Length: 0 (Automatic based on SIC, MAXLAG=11)t-StatisticProb.*Augmented Dickey-Fuller test statis

3、tic-1.7970790.6978Test critical values:1% level-4.0669815% level-3.46229210% level-3.157475*MacKinnon (1996) one-sided p-values.可以看到ADF值大于所有临界值且P值很大，所以不能拒绝原假设，即利润 是非平稳的。然后对红利进行单位根检验，选择有截距项和时间趋势，结果如下Null Hypothesis: I has a un it rootExoge no us: Con sta ntLag Len gth: 3 (Automatic based on SIC, MAXL

4、AG=11)t-StatisticProb.*Augme nted Dickey-Fuller test statistic1.3859460.9989Test critical values:1% level-3.5102595% level-2.89634610% level-2.585396Null Hypothesis: BNU has a unit rootExogenous: Constant, Linear TrendLag Length: 1 (Automatic based on SIC, MAXLAG=11)t-StatisticProb.*Augmented Dickey

5、-Fuller test statistic-2.8935590.1698Test critical values:1% level-4.0682905% level-3.46291210% level-3.157836*MacKinnon (1996) one-sided p-values.可以看到ADF值大于所有临界值且P值很大，所以不能拒绝原假设，即红利 也是非平稳的。(3)首先对GD进行平稳性检验，选择只带截距项，结果如下5000Null Hypothesis: GDP has a unit rootExogenous: ConstantLag Length: 2 (Automatic

6、 based on SIC, MAXLAG=11)t-StatisticProb.*Augmented Dickey-Fuller test statistic-0.5111750.8829Test critical values:1% level-3.5092815% level-2.89592410% level-2.585172可以看到ADF值大于所有临界值且P值很大，所以不能拒绝原假设，即 GD是 非平稳的。之后对GDPS行一阶差分，并检验其平稳性，结果如下Null Hypothesis: D(GDP) has a unit rootExogenous: Constant, Linea

7、r TrendLag Length: 1 (Automatic based on SIC, MAXLAG=11)t-StatisticProb.*Augmented Dickey-Fuller test statistic-10.036000.0000Test critical values:1% level-4.0696315% level-3.46354710% level-3.158207可以看出GD经过一阶差分后是平稳的，因此GDP 1(1)同理可得，PDI I (1)且 PCE I (1)。10.2(1)对固定厂房设备投资进行平稳性检验，首先先看图SHEB从图中我们可以初步判断，固定

8、厂房设备投资可能是非平稳的。之后进行ADF 检验，检验结果如下Null Hypothesis: SHEB has a unit rootExogenous: ConstantLag Length: 3 (Automatic based on SIC, MAXLAG=4)t-StatisticProb.*-0.1244720.9324-3.857386Augmented Dickey-Fuller test statisticTest critical values:1% level5% level10% level-3.040391 -2.660551可以看到ADF值大于所有临界值且P值很大，

9、所以不能拒绝原假设，即固定厂 房设备投资是非平稳的。（2）对销售量进行平稳性检验，首先先看图XS从图中我们可以初步判断，销售额可能是非平稳的。之后进行 AD检验，检 验结果如下Null Hypothesis: X has a unit rootExoge no us: Con sta ntLag Len gth: 0 (Automatic based on SIC, MAXLAG=4)t-StatisticProb.*Augme nted Dickey-Fuller test statistic-0.7430460.8145Test critical values:1% level-3.788

10、0305% level-3.01236310% level-2.646119Null Hypothesis: XS has a unit rootExogenous: Constant, Linear TrendLag Length: 0 (Automatic based on SIC, MAXLAG=4)t-StatisticProb.*Augmented Dickey-Fuller test statistic-1.8862380.6258Test critical values:1% level-4.4678955% level-3.64496310% level-3.261452可以看

11、到adF值大于所有临界值且p值很大，所以不能拒绝原假设，即销售额 是非平稳的。10.3答：(1)回归结果是虚假的！回归结果如下图Dependent Variable: BNUMethod: Least SquaresDate: 06/03/10 Time: 20:21Sample: 1980Q1 2001Q4Included observations: 88CoefficientStd. Errort-StatisticProb.PFT0.6282190.05286611.883120.0000C-13.026447.371237-1.7671980.0807R-squared0.621493

12、Mean dependent var69.24205Adjusted R-squared0.617092S.D. dependent var38.36748S.E. of regression23.74163Akaike info criterion9.194802Sum squared resid48475.19Schwarz criterion9.251105Log likelihood-402.5713Hannan-Quinn criter.9.217485F-statistic141.2085Durbin-Watson stat0.083355Prob(F-statistic)0.00

13、0000运用经验法则可以看出R2 = 0.621493 >D.W=0.083355所以此回归为伪回归(2)首先对利润进行一阶差分，选择带时间趋势和截距项Null Hypothesis: D(PFT) has a unit rootExogenous: Constant, Linear TrendLag Length: 0 (Automatic based on SIC, MAXLAG=11)t-StatisticProb.*Augmented Dickey-Fuller test statistic-7.7181000.0000Test critical values:1% level-

14、4.0682905% level-3.46291210% level-3.157836*MacKinnon (1996) one-sided p-values.可以看出ADF值小于所有临界值，且P值为0,由此可以判定经过一阶差分 的利润是平稳的。然后对红利进行一阶差分，选择带时间趋势和截距项Null Hypothesis: D(BNU) has a unit rootExogenous: Constant, Linear TrendLag Length: 2 (Automatic based on SIC, MAXLAG=11)t-StatisticProb.*Augmented Dickey

15、-Fuller test statistic-6.2335670.0000Test critical values:1% level-4.0710065% level-3.46419810% level-3.158586*MacKinnon (1996) one-sided p-values.可以看出ADF值小于所有临界值，且P值为0,由此可以判定经过一阶差分 的红利也是平稳的。由此可以看出利润和红利都是一阶单整。10.5答：(1)首先判断lnY和ln X的平稳性，先看图LNYLNX12 11 从图中我们可以初步判断，ln Y和ln X可能是非平稳的。之后进行ADF检验, 检验结果如下表1 l

16、n Y的AD检验Null Hypothesis: LNY has a unit rootExogenous: Constant, Linear TrendLag Length: 0 (Automatic based on SIC, MAXLAG=7)t-StatisticProb.*Augmented Dickey-Fuller test statistic-1.6599900.7439Test critical values:1% level-4.2967295% level-3.56837910% level-3.218382可以看到ADF值大于所有临界值且P值很大，所以不能拒绝原假设，即

17、lnY 是非平稳的。表2 ln X的AD检验Null Hypothesis: LNX has a unit rootExogenous: Constant, Linear TrendLag Length: 7 (Automatic based on SIC, MAXLAG=7)t-StatisticProb.*Augmented Dickey-Fuller test statistic1.2596500.9999Test critical values:1% level-4.4163455% level-3.62203310% level-3.248592可以看到ADF值大于所有临界值且P值很

18、大，所以不能拒绝原假设，即ln X 是非平稳的。(2)确定ln 丫和ln X的单整数，对ln 丫和ln X进行一阶差分，并进行ADF检验, 检验结果如下表3心lnY的AD检验Null Hypothesis: D(LNY) has a unit rootExogenous: Constant, Linear TrendLag Length: 0 (Automatic based on SIC, MAXLAG=7)t-StatisticProb.*Augmented Dickey-Fuller test statistic-3.4010950.0708Test critical values:1%

19、 level-4.3098245% level-3.57424410% level-3.221728表4 .Mn X的AD检验Null Hypothesis: D(LNX) has a unit rootExogenous: Constant, Linear TrendLag Length: 6 (Automatic based on SIC, MAXLAG=7)t-StatisticProb.*Augmented Dickey-Fuller test statistic-8.6837070.0000Test critical values:1% level-4.4163455% level-

20、3.62203310% level-3.248592从表中我们可以看出® nY和厶ln X不是同阶单整，因此不存在协整关系10.6答：(1)首先对In C进行平稳性检验，选择带时间趋势和截距项Null Hypothesis: X has a unit rootExogenous: Constant, Linear TrendLag Length: 0 (Automatic based on SIC, MAXLAG=11)t-StatisticProb.*Augmented Dickey-Fuller test statistic1.4552181.0000Test critical

21、values:1% level-4.2050045% level-3.52660910% level-3.194611*MacKinnon (1996) one-sided p-values.可以看到ADF值大于所有临界值且P值很大，所以不能拒绝原假设，即InC 是非平稳的。然后对ln 丫进行平稳性检验，选择带时间趋势和截距项Null Hypothesis: 丫 has a unit rootExogenous: Constant, Linear TrendLag Length: 0 (Automatic based on SIC, MAXLAG=9)t-StatisticProb.*Augm

22、ented Dickey-Fuller test statistic0.8663970.9997Test critical values:1% level-4.2050045% level-3.52660910% level-3.194611*MacKinnon (1996) one-sided p-values.可以看到AD值大于所有临界值且P值很大，所以不能拒绝原假设，即ln Y也 是非平稳的。对ln C和ln Y分别进行一阶差分 首先对ln C进行一阶差分，选择带时间趋势和截距项Null Hypothesis: D(X) has a unit rootExogenous: Constan

23、t, Linear TrendLag Length: 0 (Automatic based on SIC, MAXLAG=9)t-StatisticProb.*Augmented Dickey-Fuller test statistic-5.3935450.0004Test critical values:1% level-4.2118685% level-3.52975810% level-3.196411*MacKinnon (1996) one-sided p-values.可以看出adFS小于所有临界值，且p值几乎为0,由此可以判定经过一阶差分的In C是平稳的。然后对In 丫进行一阶

24、差分，选择只带截距项Null Hypothesis: D(Y) has a unit rootExogenous: ConstantLag Length: 0 (Automatic based on SIC, MAXLAG=9)t-StatisticProb.*Augmented Dickey-Fuller test statistic-4.9391430.0002Test critical values:1% level-3.6104535% level-2.93898710% level-2.607932*MacKinnon (1996) one-sided p-values.可以看出ad

25、F值小于所有临界值，且p值几乎为o,由此可以判定经过一阶差 分的ln Y也是平稳的。由此可以看出ln C和lnY都是一阶单整。EG两步：第一步得协整回归方程：Dependent Variable: XMethod: Least SquaresDate: 06/03/10 Time: 20:54Sample: 1950 1990Included observations: 41CoefficientStd. Errort-StatisticProb.Y0.9418490.01079387.261900.0000C-0.2240000.063217-3.5433550.0010R-squared0

26、.994904Mean dependent var5.263551Adjusted R-squared0.994774S.D. dependent var0.572245S.E. of regression0.041369Akaike info criterion-3.485005Sum squared resid0.066745Schwarz criterion-3.401416Log likelihood73.44261Hannan-Quinn criter.-3.454567F-statistic7614.640Durbin-Watson stat1.434445Prob(F-stati

27、stic)0.000000In C 二-0.2240941849ln Y2R 二 0.994904 DW 二 1.434445第二步：得到残差序列et，对残差序列进行平稳性检验Null Hypothesis: E has a unit rootExogenous: NoneLag Length: 0 (Automatic based on SIC, MAXLAG=9)t-StatisticProb.*Augmented Dickey-Fuller test statistic-4.6577030.0000Test critical values:1% level-2.6240575% level-1.94931910% level-1.611711*M