题目

如何从EViews里面的Johansen检验结果看出协整方程?
Date:04/30/14 Time:11:21
Sample (adjusted):1983 2012
Included observations:30 after adjustments
Trend assumption:Linear deterministic trend
Series:M N P
Lags interval (in first differences):1 to 1
Unrestricted Cointegration Rank Test (Trace)
Hypothesized Trace 0.05
No.of CE(s) Eigenvalue Statistic Critical Value Prob.**
None * 0.467116 30.53590 29.79707 0.0410
At most 1 0.314106 11.65237 15.49471 0.1744
At most 2 0.011316 0.341410 3.841466 0.5590
Trace test indicates 1 cointegrating eqn(s) at the 0.05 level
* denotes rejection of the hypothesis at the 0.05 level
**MacKinnon-Haug-Michelis (1999) p-values
Unrestricted Cointegration Rank Test (Maximum Eigenvalue)
Hypothesized Max-Eigen 0.05
No.of CE(s) Eigenvalue Statistic Critical Value Prob.**
None 0.467116 18.88352 21.13162 0.1003
At most 1 0.314106 11.31096 14.26460 0.1393
At most 2 0.011316 0.341410 3.841466 0.5590
Max-eigenvalue test indicates no cointegration at the 0.05 level
* denotes rejection of the hypothesis at the 0.05 level
**MacKinnon-Haug-Michelis (1999) p-values
Unrestricted Cointegrating Coefficients (normalized by b'*S11*b=I):
M N P
1.912314 7.897272 7.298349
0.892072 0.329896 -10.51537
4.812426 -4.492027 -26.96774
Unrestricted Adjustment Coefficients (alpha):
D(M) -0.045222 -0.019310 0.008725
D(N) -0.054281 -0.028006 0.001551
D(P) -0.008705 0.010221 0.001037
1 Cointegrating Equation(s):Log likelihood 147.6442
Normalized cointegrating coefficients (standard error in parentheses)
M N P
1.000000 4.129693 3.816501
(1.00692) (2.00876)
Adjustment coefficients (standard error in parentheses)
D(M) -0.086479
(0.03872)
D(N) -0.103802
(0.02980)
D(P) -0.016647
(0.00867)
2 Cointegrating Equation(s):Log likelihood 153.2997
Normalized cointegrating coefficients (standard error in parentheses)
M N P
1.000000 0.000000 -13.32235
(2.75020)
0.000000 1.000000 4.150150
(0.70247)
Adjustment coefficients (standard error in parentheses)
D(M) -0.103704 -0.363500
(0.04194) (0.15708)
D(N) -0.128786 -0.437911
(0.03069) (0.11494)
D(P) -0.007529 -0.065374
(0.00854) (0.03200)

提问时间：2021-03-15

答案

关键看这几行：
None * 0.467116 30.53590 29.79707 0.0410
At most 1 0.314106 11.65237 15.49471 0.1744
At most 2 0.011316 0.341410 3.841466 0.5590

尤其是最后的P值,None后的P是0.04,小于0.05,拒绝原假设,而原假设是None,就是没有一个协整方程,拒绝以了,就表明有.但要知道几个,往下看.
第二行后面的P是0.17,大于0.05,接受原假设,而原假设是最多一个协整方程,因此,分析结束.结果是：存在一个协整方程.
若有帮助,请及时采纳,谢谢!
统计人刘得意