协整检验的这个结果什么意思〉？

tracy771

13685

收藏 2015-01-08

我有5个变量，都是一阶单整的，现在结果是这样的。谁能告诉我这是什么意思？接下去该怎么做呀？

Date: 01/08/15 Time: 19:54

Sample (adjusted): 1993 2012

Included observations: 20 after adjustments

Trend assumption: Linear deterministic trend

Series: LOGX1 LOGX2 LOGX4 X3 Y

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 * 0.876485 101.0433 69.81889 0.0000

At most 1 * 0.731947 59.21551 47.85613 0.0030

At most 2 * 0.592410 32.88412 29.79707 0.0214

At most 3 0.479993 14.93425 15.49471 0.0606

At most 4 0.088623 1.855969 3.841466 0.1731

Trace test indicates 3 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 * 0.876485 41.82784 33.87687 0.0046

At most 1 0.731947 26.33138 27.58434 0.0716

At most 2 0.592410 17.94987 21.13162 0.1317

At most 3 0.479993 13.07828 14.26460 0.0763

At most 4 0.088623 1.855969 3.841466 0.1731

Max-eigenvalue 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

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tracy771

2015-1-8 20:19:54

Unrestricted Cointegrating Coefficients (normalized by b'*S11*b=I):

LOGX1 LOGX2 LOGX4 X3 Y
4.555150    3.531539 -14.43942 -3.731899 -24.97873
-3.926265 -3.336182    9.026019 -16.10135 -96.69235
-10.41815 -2.038886    13.71172 -4.974609 -106.0608
-1.950990 -2.831214    7.134506    4.098511    20.04481
-1.637222 -2.396634    6.625832 -0.391537 -75.28070

Unrestricted Adjustment Coefficients (alpha):

D(LOGX1)    0.010415 -0.020709    0.067667 -0.009469 -0.022247
D(LOGX2) -0.107561    0.079388    0.005397 -0.065421    0.003834
D(LOGX4)    0.023705 -0.009015    0.011244 -0.012998 -7.25E-06
D(X3)    0.093855    0.072165    0.027063 -0.010486 -0.023812
D(Y) -0.000154 -0.000400    0.002937 -0.000774    0.003456

1 Cointegrating Equation(s): Log likelihood    161.0236

Normalized cointegrating coefficients (standard error in parentheses)
LOGX1 LOGX2 LOGX4 X3 Y
1.000000    0.775285 -3.169911 -0.819270 -5.483624
   (0.09645)    (0.22251)    (0.36889)    (2.63493)

Adjustment coefficients (standard error in parentheses)
D(LOGX1)    0.047442
   (0.15003)
D(LOGX2) -0.489954
   (0.17584)
D(LOGX4)    0.107980
   (0.03473)
D(X3)    0.427522
   (0.16090)
D(Y) -0.000703
   (0.01551)

2 Cointegrating Equation(s): Log likelihood    174.1893

Normalized cointegrating coefficients (standard error in parentheses)
LOGX1 LOGX2 LOGX4 X3 Y
1.000000    0.000000 -12.24357 -52.07384 -319.1515
   (1.69291)    (8.31515)    (59.5010)
0.000000    1.000000    11.70364    66.11063    404.5840
   (2.13634)    (10.4932)    (75.0862)

Adjustment coefficients (standard error in parentheses)
D(LOGX1)    0.128752    0.105871
   (0.19503)    (0.15755)
D(LOGX2) -0.801651 -0.644706
   (0.19068)    (0.15404)
D(LOGX4)    0.143374    0.113790
   (0.04331)    (0.03499)
D(X3)    0.144182    0.090695
   (0.17502)    (0.14139)
D(Y)    0.000866    0.000788
   (0.02047)    (0.01654)

3 Cointegrating Equation(s): Log likelihood    183.1642

Normalized cointegrating coefficients (standard error in parentheses)
LOGX1 LOGX2 LOGX4 X3 Y
1.000000    0.000000    0.000000    4.080851    35.45928
   (0.54368)    (3.96165)
0.000000    1.000000    0.000000    12.43229    65.61112
   (1.50739)    (10.9839)
0.000000    0.000000    1.000000    4.586464    28.96302
   (0.51651)    (3.76369)

Adjustment coefficients (standard error in parentheses)
D(LOGX1) -0.576214 -0.032095    0.590524
   (0.31816)    (0.13935)    (0.57824)
D(LOGX2) -0.857874 -0.655709    2.343663
   (0.38100)    (0.16687)    (0.69245)
D(LOGX4)    0.026227    0.090863 -0.269472
   (0.07809)    (0.03420)    (0.14193)
D(X3) -0.137761    0.035517 -0.332765
   (0.33826)    (0.14815)    (0.61477)
D(Y) -0.029730 -0.005200    0.038890
   (0.03976)    (0.01741)    (0.07225)

4 Cointegrating Equation(s): Log likelihood    189.7034

Normalized cointegrating coefficients (standard error in parentheses)
LOGX1 LOGX2 LOGX4 X3 Y
1.000000    0.000000    0.000000    0.000000    16.27194
   (6.80115)
0.000000    1.000000    0.000000    0.000000    7.157017
   (17.7722)
0.000000    0.000000    1.000000    0.000000    7.398393
   (6.82274)
0.000000    0.000000    0.000000    1.000000    4.701798
   (1.53878)

Adjustment coefficients (standard error in parentheses)
D(LOGX1) -0.557739 -0.005285    0.522964 -0.080848
   (0.32072)    (0.15741)    (0.60524)    (0.46689)
D(LOGX2) -0.730237 -0.470487    1.876914 -1.171819
   (0.31636)    (0.15527)    (0.59701)    (0.46055)
D(LOGX4)    0.051585    0.127662 -0.362203 -0.052523
   (0.06580)    (0.03229)    (0.12417)    (0.09578)
D(X3) -0.117303    0.065206 -0.407580 -1.689820
   (0.34084)    (0.16729)    (0.64321)    (0.49618)
D(Y) -0.028221 -0.003009    0.033370 -0.010771
   (0.04019)    (0.01973)    (0.07584)    (0.05851)

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祝贺人大

2015-1-8 22:16:47

第一个说明存在2个协整关系，第二个不存在协整关系

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tracy771

2015-1-11 13:34:13

祝贺人大发表于 2015-1-8 22:16
第一个说明存在2个协整关系，第二个不存在协整关系

能具体解释下吗？或者告诉我哪里可以找到解释结果的书。谢谢！

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冰凉凉

2019-1-9 15:37:42

楼上错了欢迎反驳先给结论第一个存在三个协整关系第二个存在一个协整关系
首先 None \ At most1\这些都是原假设第一个 none p值小于0.05 说明拒绝不存在协整关系的原假设 at most1p值也小于0.05 说明拒绝最多存在一个协整关系的原假设同理 at most2 也是所以最终拒绝了不存在、最多存在一个、最多存在两个的原假设，结论是存在三个协整关系。
再者实在不懂下面还有一行字给你总结好了 Trace test indicates 3 cointegrating eqn(s) at the 0.05 level
即在5%的置信水平下，存在三个协整关系！！！

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水船纵横

2019-2-2 16:20:47

祝贺人大发表于 2015-1-8 22:16
第一个说明存在2个协整关系，第二个不存在协整关系

别误人子弟了，第一个trace检验说明5%水平下有三个协整方程，第二个max-eigen检验说明有一个协整方程，说明至少有一个协整方程。

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