楼主: bigdog_1
2101 0

一篇关于 HAUSMAN TEST 为负的原因 [推广有奖]

  • 0关注
  • 0粉丝

大专生

41%

还不是VIP/贵宾

-

威望
0
论坛币
8757 个
通用积分
0.1200
学术水平
2 点
热心指数
3 点
信用等级
2 点
经验
1012 点
帖子
61
精华
0
在线时间
22 小时
注册时间
2006-3-29
最后登录
2019-11-12

+2 论坛币
k人 参与回答

经管之家送您一份

应届毕业生专属福利!

求职就业群
赵安豆老师微信:zhaoandou666

经管之家联合CDA

送您一个全额奖学金名额~ !

感谢您参与论坛问题回答

经管之家送您两个论坛币!

+2 论坛币
It is hard to discuss the Hausman test without being specific about how the
test is performed. Let B be the parameter estimates from a fully efficient
estimator (random-effects regression in this case) and b be the estimates from
a less efficient estimator (fixed-effects regression), but one that is
consistent in the face of one or more violated assumptions, in this case that
the effects are correlated with one or more of the regressors. If the
assumption is violated then we expect that the estimates from the two
estimators will not be the same, b~=B.

The Hausman test is essentially a Wald test that (b-B)==0 for all coefficients
where the covariance matrix for b-B is taken as the difference of the
covariance matrices (VCEs) for b and B. What is amazing about the test is
that we can just subtract these two covariance matrices to get an estimate of
the covariance matrix of (b-B) without even considering that the VCEs of the
two estimators might be correlated -- they are after all estimated on the same
data. We can just subtract, but only because the the VCE of the fully
efficient estimator is uncorrelated with the VCEs of all other estimators, see
Hausman and Taylor (1981), "panel data and unobservable individual effects",
econometrica, 49, 1337-1398). The VCE of the efficient estimator will also be
smaller than the less efficient estimator. Taken together, these results
imply that the subtraction of the two VCE (V_b-V_B) will be positive definite
(PD) and that we need not consider the covariance between the two VCEs.

These results, however, hold only asymtotically. For any given finite sample
we have no reason to believe that (V_b-V_B) will be PD. So, it is amazing
that we can just subtract these two matrices, but the price we pay is that we
can only do so safely if we have an infinite amount of data. The Hausman
test, unlike most tests, relies on asymptotic arguments not only for its
distribution, but for its ability to be computed! Let's discuss what we do
what we do when (V_b-V_B) in not PD in the context of Eric's results.

Aside: If anyone is interested in a Hausman-like test that drops the
assumption that either estimator is fully efficient, actually estimates the
covariance between the VCEs, and can always be computed, see Weesie (2000)
"Seemingly unrelated est. and cluster-adjusted sandwich estimator", STB
Reprints Vol 9, pp 231-248. The test unfortunately requires the scores from
the estimator, and -xtreg, fe- does not directly produce these.


二维码

扫码加我 拉你入群

请注明:姓名-公司-职位

以便审核进群资格,未注明则拒绝

关键词:hausman ausman test Man Aus test hausman

您需要登录后才可以回帖 登录 | 我要注册

本版微信群
加好友,备注jltj
拉您入交流群

京ICP备16021002-2号 京B2-20170662号 京公网安备 11010802022788号 论坛法律顾问:王进律师 知识产权保护声明   免责及隐私声明

GMT+8, 2024-4-28 20:16