5楼推导，6楼通俗解释，我明白了，谢谢版主，跪谢！

玩stata时间不长，膜拜一下。

要是谁再能按照上述方式解释下hansen检验就好了，那就比较系统地掌握工具变量的使用了，特别是二者有何区别。。。等待达人回复！

大侠，小弟想请教以下几个问题：
1) I have estimated the system GMM by using the -twostep robust- option, is the value of Difference-in-Hansen exactely the same as Difference-in-Sargan's value?

2) I want to report the statistic and p-value of Difference-in-Sargan/Hansen, but i am not sure how to obtain it.

3) 为什么有时候在屏幕下端出现4行Difference-in-Hansen tests 有的时候只出现2行Difference-in-Hansen tests?  2者有什么区别吗

4）On the bottom 4 lines of results table, it displays as:
Difference-in-Hansen tests of exogeneity of instrument subsets:
  
Hansen test excluding group:     chi2(14)   =  16.92  Prob > chi2 = 0.261
Difference (null H = exogenous): chi2(5)    =   5.76  Prob > chi2 = 0.330
  
Hansen test excluding group:     chi2(4)    =   1.33  Prob > chi2 = 0.857
Difference (null H = exogenous): chi2(15)   =  21.35  Prob > chi2 = 0.126
  
I am not sure whether my following understaning is correct or not:
i) the first line tests the IV relevance of GMM-type instruments in the level equation
ii) the second line tests the IV exogeneity of GMM-type instruments in the level equation
iiii) the third line tests the IV relevance of IV-type instruments in the level equation
iv) the fourth line tests the IV exogeneity of IV-type instruments in the level equation
Does the failure of rejection imply system GMM is better than difference GMM?

5) can i make the IVs for endogenous variables are dated from t-3 ? If i can, is the command -xtabond, gmm(L.Y, lag(2. ))- correct? I understand that the conventional IV is dated from t-2.

热切期待大侠的指教

Pan

我回答不了你啊。。。。。请大人回复。。。。。。。虽然帖子是我建的。。。

热切期待大侠的指教


本文来自: 人大经济论坛 Stata专版 版，详细出处参考： https://bbs.pinggu.org/forum.php? ... &from^^uid=125847

解释的不错。

弱弱地表示没看懂版主的回复

good

一直想问，Sargan是检验工具变量外生还是检验有没有过度约束。。。？多谢。。。！

看原文1958_Sargan_The Estimation of Economic Relationships using IV

感谢五楼六楼大大的解答！

The Sargan-Hansen test is a test of overidentifying restrictions.  The joint null hypothesis is that the instruments are valid instruments, i.e., uncorrelated with the error term, and that the excluded instruments are correctly excluded from the estimated equation.
Sargan检验原假设：工具变量有效，即工具变量外生。要通过这个检验，应该要接受原假设。

是外生性检验？那为何又称为过度识别？

kanbudong

学习一下，还不会用stata,我先顶一下，回头再来看

检验工具变量外生和过度识别是一样的，参见“A crucial assumption for the validity of GMM is that the instruments are exogenous. If the model is exactly identified, detection of invalid instruments is impossible because even when E[zε] 6= 0, the estimator will choose β so that Z 0 ˆ E = 0 exactly. But if the model is overidentified, a test statistic for the joint validity of the moment conditions (identifying restrictions) falls naturally out of the GMM framework. Under the null of joint validity, the vector of empirical moments1N Z0 E is randomly distributed around 0.”
from“How to Do xtabond2”

豪斯曼检验在内生性问题里面设定：如果X没有内生性，IV估计和OLS估计无显著差别。
步骤：
1、用需要检验存在内生性的Xk对其他外生的解释变量X和工具变量IV进行ols回归，用predict uhat 的命令得到残差v
2、建立回归方程.Y=aX+bv+u
3、检验b的显著性（t值）
结论：如果b 不显著，则表明Xk没有内生性

所以Sagan检验针对过度识别，hausman检验针对是否具有内生性，检验出没有内生性就不用再用ivregress等工具变量才需要用的命令了。

更准确的表述应该是不能拒绝原假设