请问：如何检验一个方程中两个变量是否有内生性？

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<p>大家好。</p><p>伍德里奇的书里只提到了如何检验一个变量是否具有内生性（hausman检验），</p><p>如果想检验一个方程两个变量是否有内生性呢？</p><p>如方程：y=x1+x2+z+e</p><p>需要一一判断x1和x2分别是否具有内生性。</p><p>下列几种检验方式，哪种正确呢？</p><p>1.一一检验。检验X1时，采用hausman检验（此时把x2视同外生）；然后再检验x2，其方式如同x1。</p><p>2.同时检验。ivreg2把两个都视同内生—>ivendog x1—>ivendog x2。对于这一种，虽然能得到检验统计量及p值，可我不知道stata在检验x1时，是如何处理x2的，像第1种方式那样把x2看成外生的吗？反之亦然。</p><p>此外，我们检验内生性常用的hausman检验，在stata中对应的是Wu－Hausman统计量吗？</p><p>困惑很久了，望高人解答。非常感谢！</p><p></p>

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关键词：内生性 Hausman检验 hausman ivendog ausman 我不知道 如何 统计

Actually what I thought is to combine 2 and 1 together to determine whether X1 or X2 or both are endogenous.

i) Perform 2 first by treating both two variables to be endogenous and using Ivreg2 in STATA which gives you a joint F test of residuals obtained from runing both 1st stage regressions. If the joint F statistic is greater than the critical F value (or you can simply see the p value), we reject the null and conclude at least one suspected variable, X1 or X2, or both are endogenous

ii) Then from there, you can go ahead to run separately endogeneity test from 1

As to the Hausman test or Wu-Hausman test or even Durbin-Wu-Hausman test, my understanding is that they are asyptotically equivalent, but numerically might be different, especially in finite samples.

Hopefully I will be right.

[此贴子已经被作者于2009-5-19 22:13:55编辑过]

Thank you very much!

So excited when I saw a reply.

I understood the first stage you suggest. Just as what you said,if p value is significant(less than 10%),we can conclude that at least one of the two is endogenous.

With regard to the second stage, does it mean that both of X1 and X2 are endogenous if the p values of seperate hausman tests which are executed treating the other suspected regressor as exogenous are significant ? If the p value of hausman test for X1 is significant, but not significant for X2, we can conclude it is X1 that is endogneous, and vice versa.

Is my understanding right?

此外，如果ivreg2……(x1 x2=……)—>ivendog结果显示p值较大（大于0.1），是不是意味着X1和X2都不是内生的？

望高手解答。谢谢！

我觉得，你再去看看woodridge的书

我记得上面对进行内生性检验提出了要求（假设）： 必须有绝对外生的变量（不需要进行检验的），否则不能进行。