各位坛友，面板同阶单整，但是却不协整，为什么？

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Date: 03/08/14   Time: 10:28   
我在做4个变量的面板数据回归前，做了面板单位根检验，是I(1)，但是他们不协整，为什么？
水平数据不稳定：
Sample: 1980 2012   
Exogenous variables: Individual effects   
Automatic selection of maximum lags   
Automatic selection of lags based on SIC: 0 to 4   
Newey-West bandwidth selection using Bartlett kernel   
   
   Cross-
Method Statistic Prob.** sections Obs
Null: Unit root (assumes common unit root process)     
Levin, Lin & Chu t*  2.16016  0.9846  12  368
   
Null: Unit root (assumes individual unit root process)     
Im, Pesaran and Shin W-stat   4.24724  1.0000  12  368
ADF - Fisher Chi-square  12.0741  0.9791  12  368
PP - Fisher Chi-square  17.3726  0.8323  12  384
   
** Probabilities for Fisher tests are computed using an asymptotic Chi   
        -square distribution. All other tests assume asymptotic normality.   
差分数据却是稳定：
Date: 03/08/14   Time: 10:31   
Sample: 1980 2012   
Exogenous variables: Individual effects   
Automatic selection of maximum lags   
Automatic selection of lags based on SIC: 0 to 3   
Newey-West bandwidth selection using Bartlett kernel   
   
   Cross-
Method Statistic Prob.** sections Obs
Null: Unit root (assumes common unit root process)     
Levin, Lin & Chu t* -6.60803  0.0000  12  365
   
Null: Unit root (assumes individual unit root process)     
Im, Pesaran and Shin W-stat  -9.57945  0.0000  12  365
ADF - Fisher Chi-square  132.852  0.0000  12  365
PP - Fisher Chi-square  125.631  0.0000  12  372
   
** Probabilities for Fisher tests are computed using an asymptotic Chi   
        -square distribution. All other tests assume asymptotic normality.   
但是协整检验却没通过：
Pedroni Residual Cointegration Test     
Series: LGDP? LFX? LEC? LL?      
Date: 03/08/14   Time: 10:32     
Sample: 1980 2012     
Included observations: 33     
Cross-sections included: 3     
Null Hypothesis: No cointegration     
Trend assumption: No deterministic trend     
Lag selection: fixed at 1     
Newey-West bandwidth selection with Bartlett kernel     
     
Alternative hypothesis: common AR coefs. (within-dimension)     
    Weighted
  Statistic Prob. Statistic Prob.
Panel v-Statistic  -0.159765  0.5635 -0.131826  0.5524
Panel rho-Statistic   0.840191  0.7996  0.686589  0.7538
Panel PP-Statistic   0.552616  0.7097  0.351230  0.6373
Panel ADF-Statistic   0.191005  0.5757  0.137149  0.5545
     
Alternative hypothesis: individual AR coefs. (between-dimension)     
     
  Statistic Prob.  
Group rho-Statistic   1.385234  0.9170  
Group PP-Statistic   0.863504  0.8061  
Group ADF-Statistic   0.570304  0.7158  
     
     
Cross section specific results     
     
Phillips-Peron results (non-parametric)     
     
Cross ID AR(1) Variance HAC   Bandwidth Obs
_LN 0.801 0.005429 0.006691 2.00 32
_JL 0.610 0.003365 0.004182 1.00 32
_HLJ 0.763 0.009380 0.009380 0.00 32
     
Augmented Dickey-Fuller results (parametric)     
     
Cross ID AR(1) Variance Lag Max lag Obs
_LN 0.759 0.005197 1 -- 31
_JL 0.583 0.002891 1 -- 31
_HLJ 0.719 0.008299 1 -- 31

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关键词：同阶单整 distribution observations Alternative Integration process common

平稳性检验是做协整的前提，但是没有说两个检验互相替代啊

我的理解是 几个I(1)变量，应该是协整的？对不

这个问题 大家碰见过没有

大家多帮忙

同阶单整未必就协整 否则只需要验证是否同阶单整即可何须协整呢
另外就像平稳或者协整未必有格兰杰因果一样 这几个检验之间并没有必然的关系