stata作hausman检验得出负值？

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关键词：Hausman检验 hausman ausman Stata tata

I do not know the detail of Stata doing the Hausman test. There are chi-square, t and F forms of this test.

Different estimates of the asymptotic variances will not necessarily garantee a positive definite variance in the chi-suqare form, in this case a negative test statistic is possile.

Wooldridge cross-sectional textbook chapter 10.

把你的数据和程序贴出来，让大家给你看看。

2楼说的“Wooldridge cross-sectional textbook chapter 10”我还没去看呢，要看看才行。

不过，楼主也可以参考一下这个贴子：

http://jinhe.xjtu.edu.cn/bbs/dispbbs.asp?boardID=5&ID=357

-- 作者：demo
-- 发布时间：2006-4-12 7:38:27

-- 困惑的问题，请解惑！
各位好！我初学，在尝试作Hausman检验的时候，得到
chi2(6) = (b-B)'[(V_b-V_B)^(-1)](b-B)

= -31.50 chi2<0 ==> model fitted on these

data fails to meet the asymptotic

assumptions of the Hausman test;

请问这样该如何处理啊，看过一篇帖子，说这样的结果还是不能随便就判断用固定效应模型

那么接下来该怎么进行呢？？？谢谢了！！

-- 作者：arlion
-- 发布时间：2006-4-12 19:10:34

--
先要看看模型设定有没有问题 。

-- 作者：从零开始
-- 发布时间：2006-4-21 0:30:14

--
我听人说，hausman为负，则失效了，因此他选择固定效应。

但好象在Greene的书上说，此时应选随机效应，因为负值可能是由于样本较小造成的（记得好象在书中见到过，不知记错没有）

-- 作者：从零开始
-- 发布时间：2006-4-21 0:35:31

--
找到了，Greene（2000），chapter 9, p386

“.......The failure of the matrix in the Wald statistic to be positive definite, however, is sometimes a finite sample problem that is not part of the model structure. In such a case, forcing a solution by using a generalized inverse may be misleading. Hausman suggests that in this instance, the appropriate conclusion might be simply to take the result as zero and , by implication, not reject the null hypothesis.

不拒绝原假设，就是使用随机效应吧。

-- 作者：从零开始
-- 发布时间：2006-4-21 0:36:55

--
我的毕业论文中就出现hausman检验为负，我直接说此时检验失效，为安全起见，应采用固定效应模型，不知这样说妥不妥当？

[em10][em10]

[em09][em09]

找到了，Greene（2000），chapter 9, p386

“.......The failure of the matrix in the Wald statistic to be positive definite, however, is sometimes a finite sample problem that is not part of the model structure. In such a case, forcing a solution by using a generalized inverse may be misleading. Hausman suggests that in this instance, the appropriate conclusion might be simply to take the result as zero and , by implication, not reject the null hypothesis.

不拒绝原假设，就是使用随机效应吧。

-- 作者：从零开始
-- 多謝多謝^^

 sigmamore and sigmaless specify that the two covariance matrices used in
        the test be based on a common estimate of disturbance variance
        (sigma2).

        sigmamore specifies that the covariance matrices be based on the
            estimated disturbance variance from the efficient estimator.
            This option provides a proper estimate of the contrast variance
            for so-called tests of exogeneity and overidentification in
            instrumental variables regression.

        sigmaless specifies that the covariance matrices be based on the
            estimated disturbance variance from the consistent estimator.

        These options can only be specified when both estimators save
        e(sigma) or e(rmse), or with command xtreg.  e(sigma_e) is saved
        after command xtreg with options fe or mle.  e(rmse) is saved after
        command xtreg with option re.

        sigmamore or sigmaless are recommended when comparing fixed-effects
        and random-effects linear regression because they are much less
        likely to produce a nonpositive-definite differenced covariance
        matrix (although the tests are asymptotically equivalent whether or
        not one of the options is specified).

太感谢了~~~

good