Amsler and Lee (Econometric Theory, 1995) LM t-stat

-----------------------------------------

LM test statistic = -3.8193
Selected lag = 7.0000
Given break point = 50.0000

Estimated coeff. of dummy var. = -0.5138
Its t-stat = -0.6425
Standard error .. = 0.7626
Standardized dummy coeff -0.6737

LS Min LM t-stat
*********************************************************************************
-----------------------------------------

Model (1=A, 2=C) = 1.0000

Min. test statistic = -3.4737
Estimated break point = 54.0000
Selected lag = 4.0000

Estimated coeff. of dummy var. = 2.1530
Its t-stat = 1.9255
Standard error .. = 1.0459
Standardized dummy coeff = 2.0584

Coeff and t-stat
Z(t) = [S(t-1), (lags..omitted), 1, B(t)]

-0.2386 -3.4737
0.2585 2.1288
2.1530 1.9255

*********************************************************************************
-----------------------------------------

Model (1=A, 2=C) = 2.0000

Min. test statistic = -3.3989
Estimated break point = 29.0000
Selected lag = 4.0000

Estimated coeff. of dummy var. = -0.2737
Its t-stat = -0.9946
Standard error .. = 1.0470
Standardized dummy coeff = -0.2614

Coeff and t-stat
Z(t) = [S(t-1), (lags..omitted), 1, B(t), D(t)]

-0.2233 -3.3989
0.5250 1.9474
1.7854 1.6781
-0.2737 -0.9946

xuehe你有没有这本书？

请zhaomn200145和xuelida二位看一下:前短时间似乎可以run这个m-break的,现在一搞就出这个,不知道为什么?

:\gauss7.0\src\indices2.src(67) : error G0474 : ' (4) = indices2(dataset,var1,var2)' : Illegal creation of global
f:\gauss7.0\src\indices2.src() : error G0017 : 'local t1,i,nvec,kd,indx1,indx2,err,errmsg,f1,flag' : WARNING: LOCAL outside of procedure
f:\gauss7.0\src\indices2.src(70) : error G0474 : 'flag' : Illegal creation of global

还有:

f:\gauss7.0\src\ols.src(264) : error G0474 : ' (11) = ols(dataset, depvar, indvars)' : Illegal creation of global
f:\gauss7.0\src\ols.src(266) : error G0474 : 'dataset' : Illegal creation of global
f:\gauss7.0\src\ols.src(266) : error G0474 : 'dataset' : Illegal creation of global

[此贴子已经被作者于2007-8-11 17:56:05编辑过]

\gauss7.0\src\indices2.src(67) ，这个是你的indices2.src第67行出现了问题，运行的程序与这个不匹备。

f:\gauss7.0\src\ols.src(264) : ' (11) = ols(dataset, depvar, indvars)' : Illegal creation of global

这个可能是你的数据集与ols.src不匹备。

我不知道你的数据不对，还是你修改了运行程序了，gauss里调用的程序是容易出现这些问题的。

[此贴子已经被作者于2007-8-11 18:20:12编辑过]

m-break可以run一些,出来一下结果后就有上面的警告,不知道可否用了?还有没有程序没用?

The options chosen are:
h = 15.0000
eps1 = 0.1500
hetdat = 1.0000
hetvar = 1.0000
hetomega = 1.0000
hetq = 1.0000
robust = 1.0000 (prewhit = 1.0000 )
The maximum number of breaks is: 5.0000
********************************************************
Output from the global optimization
********************************************************
The model with 1.0000 breaks has SSR : 644.9955
The dates of the breaks are: 79.0000
The model with 2.0000 breaks has SSR : 455.9502
The dates of the breaks are:
47.0000
79.0000
The model with 3.0000 breaks has SSR : 445.1819
The dates of the breaks are:
24.0000
47.0000
79.0000
The model with 4.0000 breaks has SSR : 444.8797
The dates of the breaks are:
24.0000
47.0000
64.0000
79.0000
The model with 5.0000 breaks has SSR : 449.6395
The dates of the breaks are:
16.0000
31.0000
47.0000
64.0000
79.0000
********************************************************
Output from the testing procedures
********************************************************
a) supF tests against a fixed number of breaks
--------------------------------------------------------------
The supF test for 0 versus 1.0000 breaks (scaled by q) is: 57.9058
The supF test for 0 versus 2.0000 breaks (scaled by q) is: 43.0143
The supF test for 0 versus 3.0000 breaks (scaled by q) is: 33.3228
The supF test for 0 versus 4.0000 breaks (scaled by q) is: 24.7706
The supF test for 0 versus 5.0000 breaks (scaled by q) is: 18.3259
-------------------------
The critical values at the 10.0000 % level are (for k=1 to 5.0000 ):
7.0400 6.2800 5.2100 4.4100 3.4700
The critical values at the 5.0000 % level are (for k=1 to 5.0000 ):
8.5800 7.2200 5.9600 4.9900 3.9100
The critical values at the 2.5000 % level are (for k=1 to 5.0000 ):
10.1800 8.1400 6.7200 5.5100 4.3400
The critical values at the 1.0000 % level are (for k=1 to 5.0000 ):
12.2900 9.3600 7.6000 6.1900 4.9100
--------------------------------------------------------------
b) Dmax tests against an unknown number of breaks
--------------------------------------------------------------
The UDmax test is: 57.9058
(the critical value at the 10.0000 % level is: 7.4600 )
(the critical value at the 5.0000 % level is: 8.8800 )
(the critical value at the 2.5000 % level is: 10.3900 )
(the critical value at the 1.0000 % level is: 12.3700 )
********************************************************
---------------------
The WDmax test at the 10.0000 % level is: 57.9058
(The critical value is: 8.2000 )
---------------------
The WDmax test at the 5.0000 % level is: 57.9058
(The critical value is: 9.9100 )
---------------------
The WDmax test at the 2.5000 % level is: 57.9058
(The critical value is: 11.6700 )
---------------------
The WDmax test at the 1.0000 % level is: 57.9058
(The critical value is: 13.8300 )
********************************************************
supF(l+1|l) tests using global otimizers under the null
--------------------------------------------------------------
The supF( 2.0000 | 1.0000 ) test is : 33.9275
It corresponds to a new break at: 47.0000
The supF( 3.0000 | 2.0000 ) test is : 14.7246
It corresponds to a new break at: 24.0000
The supF( 4.0000 | 3.0000 ) test is : 0.0330
It corresponds to a new break at: 64.0000
Given the location of the breaks from the global optimization
with 4.0000 breaks there was no more place to insert
an additional breaks that satisfy the minimal length requirement.
The supF( 5.0000 | 4.0000 ) test is : 0.0000
It corresponds to a new break at: 0.0000
********************************************************
The critical values of supF(i+1|i) at the 10.0000 % level are (for i=1 to 5.0000 ) are:
7.0400 8.5100 9.4100 10.0400 10.5800
The critical values of supF(i+1|i) at the 5.0000 % level are (for i=1 to 5.0000 ) are:
8.5800 10.1300 11.1400 11.8300 12.2500
The critical values of supF(i+1|i) at the 2.5000 % level are (for i=1 to 5.0000 ) are:
10.1800 11.8600 12.6600 13.4000 13.8900
The critical values of supF(i+1|i) at the 1.0000 % level are (for i=1 to 5.0000 ) are:
12.2900 13.8900 14.8000 15.2800 15.7600
********************************************************
Output from the application of Information criteria
--------------------------------------------------------------
Values of BIC and lwz with 0.0000 breaks: 2.4677 2.4775
Values of BIC and lwz with 1.0000 breaks: 1.9245 2.0095
Values of BIC and lwz with 2.0000 breaks: 1.6676 1.8282
Values of BIC and lwz with 3.0000 breaks: 1.7337 1.9703
Values of BIC and lwz with 4.0000 breaks: 1.8231 2.1360
Values of BIC and lwz with 5.0000 breaks: 1.9237 2.3136
The number of breaks chosen by BIC is : 2.0000
The number of breaks chosen by LWZ is : 2.0000
********************************************************
Output from the sequential procedure at significance level 10.0000 %
--------------------------------------------------------------
The first break found is at: 79.0000
The next break found is at: 47.0000
The next break found is at: 24.0000
----------------------------------------------------
The sequential procedure estimated the number of breaks at: 3.0000
********************************************************
Output from the sequential procedure at significance level 5.0000 %
--------------------------------------------------------------
The first break found is at: 79.0000
The next break found is at: 47.0000
The next break found is at: 24.0000
----------------------------------------------------
The sequential procedure estimated the number of breaks at: 3.0000
********************************************************
Output from the sequential procedure at significance level 2.5000 %
--------------------------------------------------------------
The first break found is at: 79.0000
The next break found is at: 47.0000
The next break found is at: 24.0000
----------------------------------------------------
The sequential procedure estimated the number of breaks at: 3.0000
********************************************************
Output from the sequential procedure at significance level 1.0000 %
--------------------------------------------------------------
The first break found is at: 79.0000
The next break found is at: 47.0000
----------------------------------------------------
The sequential procedure estimated the number of breaks at: 2.0000
********************************************************
Output from the repartition procedure for the 10.0000 % significance level
----------------------------------------
The updated break dates are :
24.0000
47.0000
79.0000
********************************************************
Output from the repartition procedure for the 5.0000 % significance level
----------------------------------------
The updated break dates are :
24.0000
47.0000
79.0000
********************************************************
Output from the repartition procedure for the 2.5000 % significance level
----------------------------------------
The updated break dates are :
24.0000
47.0000
79.0000
********************************************************
Output from the repartition procedure for the 1.0000 % significance level
----------------------------------------
The updated break dates are :
47.0000
79.0000
********************************************************
Output from the estimation of the model selected by BIC
--------------------------------------------------------------
Valid cases: 103 Dependent variable: Y
Missing cases: 0 Deletion method: None
Total SS: 1214.922 Degrees of freedom: 100
R-squared: 0.625 Rbar-squared: 0.617
Residual SS: 455.950 Std error of est: 2.135
F(3,100): 55.486 Probability of F: 0.000
Durbin-Watson: 1.942

Standard Prob Standardized Cor with
Variable Estimate Error t-value >|t| Estimate Dep Var
-------------------------------------------------------------------------------
X1 1.355037 0.311465 4.350523 0.000 0.247421 0.247421
X2 -1.796138 0.377471 -4.758347 0.000 -0.270615 -0.270615
X3 5.642890 0.435866 12.946384 0.000 0.736282 0.736282
--------------------------------------------------------------
Corrected standard errors for the coefficients
--------------------------------------------------------------
The corrected standard error for coefficient 1.0000 is: 0.1555
The corrected standard error for coefficient 2.0000 is: 0.5110
The corrected standard error for coefficient 3.0000 is: 0.6029
--------------------------------------------------------------
Confidence intervals for the break dates
--------------------------------------------------------------
The 95% C.I. for the 1.0000 th break is: 36.0000 48.0000
The 90% C.I. for the 1.0000 th break is: 39.0000 48.0000
The 95% C.I. for the 2.0000 th break is: 77.0000 81.0000
The 90% C.I. for the 2.0000 th break is: 77.0000 81.0000
********************************************************
********************************************************
Output from the estimation of the model selected by the
sequential method at significance level 10.0000 %
--------------------------------------------------------------
Valid cases: 103 Dependent variable: Y
Missing cases: 0 Deletion method: None
Total SS: 1214.922 Degrees of freedom: 99
R-squared: 0.634 Rbar-squared: 0.622
Residual SS: 445.182 Std error of est: 2.121
F(4,99): 42.794 Probability of F: 0.000
Durbin-Watson: 1.982

Standard Prob Standardized Cor with
Variable Estimate Error t-value >|t| Estimate Dep Var
-------------------------------------------------------------------------------
X1 1.823617 0.432858 4.212967 0.000 0.237945 0.237945
X2 0.866085 0.442168 1.958724 0.053 0.110627 0.110627
X3 -1.796138 0.374866 -4.791414 0.000 -0.270615 -0.270615
X4 5.642890 0.432858 13.036351 0.000 0.736282 0.736282
--------------------------------------------------------------
Corrected standard errors for the coefficients
--------------------------------------------------------------
The corrected standard error for coefficient 1.0000 is: 0.1899
The corrected standard error for coefficient 2.0000 is: 0.1535
The corrected standard error for coefficient 3.0000 is: 0.5110
The corrected standard error for coefficient 4.0000 is: 0.6029
--------------------------------------------------------------
Confidence intervals for the break dates
--------------------------------------------------------------
The 95% C.I. for the 1.0000 th break is: 17.0000 35.0000
The 90% C.I. for the 1.0000 th break is: 19.0000 32.0000
The 95% C.I. for the 2.0000 th break is: 32.0000 48.0000
The 90% C.I. for the 2.0000 th break is: 36.0000 48.0000
The 95% C.I. for the 3.0000 th break is: 77.0000 81.0000
The 90% C.I. for the 3.0000 th break is: 77.0000 81.0000
********************************************************
for the 5.0000 % level, the model is the same as for the 10.0000 % level.
The estimation is not repeated.
----------------------------------------------------------------
for the 2.5000 % level, the model is the same as for the 5.0000 % level.
The estimation is not repeated.
----------------------------------------------------------------
Output from the estimation of the model selected by the
sequential method at significance level 1.0000 %
--------------------------------------------------------------
Valid cases: 103 Dependent variable: Y
Missing cases: 0 Deletion method: None
Total SS: 1214.922 Degrees of freedom: 100
R-squared: 0.625 Rbar-squared: 0.617
Residual SS: 455.950 Std error of est: 2.135
F(3,100): 55.486 Probability of F: 0.000
Durbin-Watson: 1.942

Standard Prob Standardized Cor with
Variable Estimate Error t-value >|t| Estimate Dep Var
-------------------------------------------------------------------------------
X1 1.355037 0.311465 4.350523 0.000 0.247421 0.247421
X2 -1.796138 0.377471 -4.758347 0.000 -0.270615 -0.270615
X3 5.642890 0.435866 12.946384 0.000 0.736282 0.736282
--------------------------------------------------------------
Corrected standard errors for the coefficients
--------------------------------------------------------------
The corrected standard error for coefficient 1.0000 is: 0.1555
The corrected standard error for coefficient 2.0000 is: 0.5110
The corrected standard error for coefficient 3.0000 is: 0.6029
--------------------------------------------------------------
Confidence intervals for the break dates
--------------------------------------------------------------
The 95% C.I. for the 1.0000 th break is: 36.0000 48.0000
The 90% C.I. for the 1.0000 th break is: 39.0000 48.0000
The 95% C.I. for the 2.0000 th break is: 77.0000 81.0000
The 90% C.I. for the 2.0000 th break is: 77.0000 81.0000
********************************************************

Computation and Analysis of Multiple Structural Change Models

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本附件包括：

• Computation and Analysis of Multiple Structural Change Models.pdf

你这个计算结果已经出来了

如果用自己的数据可能会出现些问题

[此贴子已经被作者于2007-8-11 20:39:03编辑过]

原程序是这样的：

new;
T = 100 + 50; /* 100 observations + 50 presample values */
N= 10000; /* the number of replications is 10000 */
seed1=124564; /* fix the seed */
rstat=0;

bstat=0;

size=0; /* initialize statistics */
i=1; /* I start the loop */
do while i<=n;
e=rndns(t,2,seed1); /*generating 2 pseudo normal random data*/
/* generate 2 stationary autoregressive processes,
** rho=0.25 with drifts.
/* Replace 0.25 by 1 for random walks
rho=0.25;
y1= recserar(0.3 + e[.,1],0, rho);
y2= recserar(0.5 + e[.,2],0, rho);
y1= y1[51:t,.]; /*discard the first 50 observations */

y2= y2[51:t,.];
x = ones(t-50,1)~y2;
xxi=invpd(x’x); b=xxi*(x’y1); e=y1-x*b;
s2=e’*e/(rows(x)-cols(x));
sd=sqrt(diag(s2*xxi));
tstud=b./sd;
t2= tstud[2,1]; /* take the t-stat for the slope coeff */
bols2 = b[2,1]; /* idem */
r2= 1 - sumc(e^2) / sumc((y1-meanc(y1))^2);
size = sumc(abs(t2).> 1.96) + size ;
rstat =rstat|r2; /* stack r-squared */
bstat = bstat|bols2; /* stack coefficients */
i=i+1;
endo;
output file = es3.res on; /* or reset;*/
print; format /rz 10, 5;

但是我用gauss运行的时候它就不认rstat=0;这是怎么回事？是我的gauss少什么插件吗？

我的可以运行呀，但是我建议不用这样的格式写，比较繁琐。像y1的生成，可以直接用公式就可以赋值。

seed命令就是产生随机数时用的，没什么特别之处，这是因为计算机调用的都是产生伪随机数。

我还是初学者啊，连入门还谈不上哩，主要研究门限同积，但是这方面已经非常完善（也或许我没有找到突破口吧，惭愧），所以哩，目前想把同积与变点结合起来研究一下！希望姐姐们给点建议和意见！

　　　在此谢过！

学校图书馆有吧，我们学校的图书馆我借过。。。。。。。。。

xuelida你在啊，太好了。

我现在学习用gauss编程，象模特卡罗和bootstrap也都可以做，但有个问题一直搞不明白，就是如何用象模特卡罗这样仿真去计算某一个具体统计分布的size和power，如何通过gauss程序来实现呢？比如用仿真试验得出单位根的ADF检验的统计分布。你对这个问题的研究比我要深，能简单介绍一下吗？