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论坛 计量经济学与统计论坛 五区 计量经济学与统计软件
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2009-03-11
<p>刚学这个东西,看不懂结果,请教下以下结果要怎么看啊??谢谢谢谢</p><p>The options chosen are:<br/>h =  28.0000 <br/>eps1 =  0.1500 <br/>hetdat =  1.0000 <br/>hetvar =  1.0000 <br/>hetomega =  1.0000 <br/>hetq =  1.0000 <br/>robust =  1.0000 (prewhit =  1.0000 )<br/>The maximum number of breaks is:  5.0000 <br/>********************************************************<br/>Output from the global optimization<br/>********************************************************<br/>The model with 1.0000 breaks has SSR :  34.9487 <br/>The dates of the breaks are:   42.0000 <br/>The model with 2.0000 breaks has SSR :  29.4489 <br/>The dates of the breaks are:  <br/> 42.0000 <br/> 84.0000 <br/>The model with 3.0000 breaks has SSR :  29.1108 <br/>The dates of the breaks are:  <br/> 42.0000 <br/> 92.0000 <br/> 120.0000 <br/>The model with 4.0000 breaks has SSR :  28.5707 <br/>The dates of the breaks are:  <br/> 42.0000 <br/> 92.0000 <br/> 120.0000 <br/> 152.0000 <br/>The model with 5.0000 breaks has SSR :  30.1982 <br/>The dates of the breaks are:  <br/> 42.0000 <br/> 70.0000 <br/> 98.0000 <br/> 126.0000 <br/> 154.0000 <br/>********************************************************<br/>Output from the testing procedures<br/>********************************************************<br/>a) supF tests against a fixed number of breaks<br/>--------------------------------------------------------------<br/>The supF test for 0 versus 1.0000 breaks (scaled by q) is: 11.8300 <br/>The supF test for 0 versus 2.0000 breaks (scaled by q) is: 7.1363 <br/>The supF test for 0 versus 3.0000 breaks (scaled by q) is: 6.3308 <br/>The supF test for 0 versus 4.0000 breaks (scaled by q) is: 7.3353 <br/>The supF test for 0 versus 5.0000 breaks (scaled by q) is: 3.6518 <br/>-------------------------<br/>The critical values at the  10.0000 % level are (for k=1 to  5.0000 ):<br/> 7.0400  6.2800  5.2100  4.4100  3.4700 <br/>The critical values at the  5.0000 % level are (for k=1 to  5.0000 ):<br/> 8.5800  7.2200  5.9600  4.9900  3.9100 <br/>The critical values at the  2.5000 % level are (for k=1 to  5.0000 ):<br/> 10.1800  8.1400  6.7200  5.5100  4.3400 <br/>The critical values at the  1.0000 % level are (for k=1 to  5.0000 ):<br/> 12.2900  9.3600  7.6000  6.1900  4.9100 <br/>--------------------------------------------------------------<br/>b) Dmax tests against an unknown number of breaks<br/>--------------------------------------------------------------<br/>The UDmax test is:  11.8300 <br/>(the critical value at the  10.0000 % level is:  7.4600 )<br/>(the critical value at the  5.0000 % level is:  8.8800 )<br/>(the critical value at the  2.5000 % level is:  10.3900 )<br/>(the critical value at the  1.0000 % level is:  12.3700 )<br/>********************************************************<br/>---------------------<br/>The WDmax test at the  10.0000 % level is:  11.8300 <br/>(The critical value is:  8.2000 )<br/>---------------------<br/>The WDmax test at the  5.0000 % level is:  12.6127 <br/>(The critical value is:  9.9100 )<br/>---------------------<br/>The WDmax test at the  2.5000 % level is:  13.5524 <br/>(The critical value is:  11.6700 )<br/>---------------------<br/>The WDmax test at the  1.0000 % level is:  14.5640 <br/>(The critical value is:  13.8300 )<br/>********************************************************<br/>supF(l+1|l) tests using global otimizers under the null<br/>--------------------------------------------------------------<br/>The supF( 2.0000 | 1.0000 ) test is :  6.4337 <br/>It corresponds to a new break at:  84.0000 <br/>The supF( 3.0000 | 2.0000 ) test is :  2.0548 <br/>It corresponds to a new break at:  152.0000 <br/>The supF( 4.0000 | 3.0000 ) test is :  6.7518 <br/>It corresponds to a new break at:  152.0000 <br/>Given the location of the breaks from the global optimization<br/>with  4.0000 breaks there was no more place to insert <br/>an additional breaks that satisfy the minimal length requirement.<br/>The supF( 5.0000 | 4.0000 ) test is :  0.0000 <br/>It corresponds to a new break at:  0.0000 <br/>********************************************************<br/>The critical values of supF(i+1|i) at the  10.0000 % level are (for i=1 to  5.0000 ) are: <br/> 7.0400  8.5100  9.4100  10.0400  10.5800 <br/>The critical values of supF(i+1|i) at the  5.0000 % level are (for i=1 to  5.0000 ) are: <br/> 8.5800  10.1300  11.1400  11.8300  12.2500 <br/>The critical values of supF(i+1|i) at the  2.5000 % level are (for i=1 to  5.0000 ) are: <br/> 10.1800  11.8600  12.6600  13.4000  13.8900 <br/>The critical values of supF(i+1|i) at the  1.0000 % level are (for i=1 to  5.0000 ) are: <br/> 12.2900  13.8900  14.8000  15.2800  15.7600 <br/>********************************************************<br/>Output from the application of Information criteria<br/>--------------------------------------------------------------<br/>Values of BIC and lwz with  0.0000  breaks: -1.5965 -1.5911 <br/>Values of BIC and lwz with  1.0000  breaks: -1.6324 -1.5693 <br/>Values of BIC and lwz with  2.0000  breaks: -1.7482 -1.6271 <br/>Values of BIC and lwz with  3.0000  breaks: -1.7042 -1.5251 <br/>Values of BIC and lwz with  4.0000  breaks: -1.6675 -1.4302 <br/>Values of BIC and lwz with  5.0000  breaks: -1.5566 -1.2610 <br/>The number of breaks chosen by BIC is : 2.0000 <br/>The number of breaks chosen by LWZ is : 2.0000 <br/>********************************************************<br/>Output from the sequential procedure at significance level  10.0000 %<br/>--------------------------------------------------------------<br/>The first break found is at:  42.0000 <br/>----------------------------------------------------<br/>The sequential procedure estimated the number of breaks at: 1.0000 <br/>********************************************************<br/>Output from the sequential procedure at significance level  5.0000 %<br/>--------------------------------------------------------------<br/>The first break found is at:  42.0000 <br/>----------------------------------------------------<br/>The sequential procedure estimated the number of breaks at: 1.0000 <br/>********************************************************<br/>Output from the sequential procedure at significance level  2.5000 %<br/>--------------------------------------------------------------<br/>The first break found is at:  42.0000 <br/>----------------------------------------------------<br/>The sequential procedure estimated the number of breaks at: 1.0000 <br/>********************************************************<br/>Output from the sequential procedure at significance level  1.0000 %<br/>--------------------------------------------------------------<br/>----------------------------------------------------<br/>The sequential procedure estimated the number of breaks at: 0.0000 <br/>********************************************************<br/>Output from the repartition procedure for the  10.0000 % significance level<br/>----------------------------------------<br/>The updated break dates are : 42.0000 <br/>********************************************************<br/>Output from the repartition procedure for the  5.0000 % significance level<br/>----------------------------------------<br/>The updated break dates are : 42.0000 <br/>********************************************************<br/>Output from the repartition procedure for the  2.5000 % significance level<br/>----------------------------------------<br/>The updated break dates are : 42.0000 <br/>********************************************************<br/>Output from the repartition procedure for the  1.0000 % significance level<br/>********************************************************<br/>The sequential procedure found no break and <br/>the repartition procedure is skipped.<br/>********************************************************<br/>********************************************************<br/>Output from the estimation of the model selected by BIC<br/>--------------------------------------------------------------<br/>Valid cases:                   189      Dependent variable:                   Y<br/>Missing cases:                   0      Deletion method:                   None<br/>Total SS:                   38.294      Degrees of freedom:                 186<br/>R-squared:                   0.231      Rbar-squared:                     0.223<br/>Residual SS:                29.449      Std error of est:                 0.398<br/>F(3,186):                   18.622      Probability of F:                 0.000<br/>Durbin-Watson:               1.337</p><p>                         Standard                 Prob   Standardized  Cor with<br/>Variable     Estimate      Error      t-value     >|t|     Estimate    Dep Var<br/>-------------------------------------------------------------------------------<br/>X1           0.096598    0.061398    1.573319     0.117    0.080250    0.080250<br/>X2           0.722450    0.061398   11.766690     0.000    0.600182    0.600182<br/>X3           0.294283    0.038831    7.578473     0.000    0.386554    0.386554<br/>--------------------------------------------------------------<br/>Corrected standard errors for the coefficients<br/>--------------------------------------------------------------<br/>The corrected standard error for coefficient 1.0000 is: 0.0688 <br/>The corrected standard error for coefficient 2.0000 is: 0.1651 <br/>The corrected standard error for coefficient 3.0000 is: 0.0293 <br/>--------------------------------------------------------------<br/>Confidence intervals for the break dates<br/>--------------------------------------------------------------<br/>The 95% C.I. for the 1.0000 th break is:  7.0000  46.0000 <br/>The 90% C.I. for the 1.0000 th break is:  17.0000  44.0000 <br/>The 95% C.I. for the 2.0000 th break is:  82.0000  159.0000 <br/>The 90% C.I. for the 2.0000 th break is:  83.0000  138.0000 <br/>********************************************************<br/>********************************************************<br/>Output from the estimation of the model selected by the<br/> sequential method at significance level  10.0000 %<br/>--------------------------------------------------------------<br/>Valid cases:                   189      Dependent variable:                   Y<br/>Missing cases:                   0      Deletion method:                   None<br/>Total SS:                   38.294      Degrees of freedom:                 187<br/>R-squared:                   0.087      Rbar-squared:                     0.082<br/>Residual SS:                34.949      Std error of est:                 0.432<br/>F(2,187):                    8.950      Probability of F:                 0.000<br/>Durbin-Watson:               1.139</p><p>                         Standard                 Prob   Standardized  Cor with<br/>Variable     Estimate      Error      t-value     >|t|     Estimate    Dep Var<br/>-------------------------------------------------------------------------------<br/>X1           0.096598    0.066707    1.448105     0.149    0.080250    0.080250<br/>X2           0.416616    0.035656   11.684234     0.000    0.647509    0.647509<br/>--------------------------------------------------------------<br/>Corrected standard errors for the coefficients<br/>--------------------------------------------------------------<br/>The corrected standard error for coefficient 1.0000 is: 0.0688 <br/>The corrected standard error for coefficient 2.0000 is: 0.0619 <br/>--------------------------------------------------------------<br/>Confidence intervals for the break dates<br/>--------------------------------------------------------------<br/>The 95% C.I. for the 1.0000 th break is: -23.0000  60.0000 <br/>The 90% C.I. for the 1.0000 th break is: -5.0000  54.0000 <br/>********************************************************<br/>for the  5.0000 % level, the model is the same as for the  10.0000 % level.<br/>The estimation is not repeated.<br/>----------------------------------------------------------------<br/>for the  2.5000 % level, the model is the same as for the  5.0000 % level.<br/>The estimation is not repeated.<br/>----------------------------------------------------------------</p>
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2009-6-25 03:15:03
偶正想学介个。。。恩。。。求CODE。。。
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2012-9-9 09:51:55
请问这是Gauss,staa还是r的结构?可以讨论下
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