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2013-07-15

Pool unit root test: Summary

Date: 07/15/13   Time: 19:39

Sample: 2006 2011

Series: Y600078, Y600141, Y600143, Y600146, Y600160, Y600227,

        Y600230, Y600389, Y600423, Y600426, Y600469, Y600470,

        Y600480, Y600486, Y600527, Y600589, Y600596, Y600618,

        Y600636, Y600725, Y600731, Y600746, Y600810, Y600889,

        Y600985, Y000059, Y000407, Y000422, Y000510, Y000523,

        Y000525, Y000553, Y000565, Y000589, Y000599, Y000615,

        Y000619, Y000627, Y000635, Y000659, Y000687, Y000698,

        Y000731, Y000755, Y000782, Y000819, Y000830, Y002018,

        Y002054, Y002061

Exogenous variables: Individual effects

Automatic selection of maximum lags

Automatic selection of lags based on SIC: 0

Newey-West bandwidth selection using Bartlett kernel

Balanced observations for each test

Cross-

Method

Statistic

Prob.**

sections

Obs

Null: Unit root (assumes common unit root process)

Levin, Lin & Chu t*

-9.14764

0.0000

50

250

Breitung t-stat

-1.39469

0.0816

50

200

Null: Unit root (assumes individual unit root process)

Im, Pesaran and Shin W-stat

-1.16947

0.1211

50

250

ADF - Fisher Chi-square

120.375

0.0808

50

250

PP - Fisher Chi-square

157.099

0.0002

50

250

Null: No unit root (assumes common unit root process)

Hadri Z-stat

7.73405

0.0000

50

300

** Probabilities for Fisher tests are computed using an asympotic Chi

        -square distribution. All other tests assume asymptotic normality.




就是这样~~不知道这么多种检验结果不一样的话~~应该看哪一种~~算不算数据平稳呀~
我问同学说~~一般只用ADF一种来检验是这样吗?
求大声指点下下~~
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全部回复
2014-11-3 15:27:34
不要一组变量一起进行平稳性检验,需要一个一个进行检验,原假设是被检验的序列含有一个单位根
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