hz1124 发表于 2019-7-29 16:02
解决了那就好。
楼主不用太在意空间自回归系数(rho)的显著性。
一些学者(Lung-fei Lee, 李龙飞)指出, ...
看您的回复好像对空间计量模型比较了解可否问您个问题。就是我在进行空间面板模型的豪斯曼检验时 sqrt(diag(V_b-V_B))是缺失值,chi-2也是负数。如下。
. qui xsmle chzs lncfsp hlw lnczsr lnfmzl lncsfj lnwztzze lnjyzc lnkxjszc lnzhong,emat(ww1) model(sem)
> nolog
. est sto re
. qui xsmle chzs lncfsp hlw lnczsr lnfmzl lncsfj lnwztzze lnjyzc lnkxjszc lnzhong,emat(ww1) model(sem)
> nolog fe
. est sto fe
. hausman fe re
---- Coefficients ----
| (b) (B) (b-B) sqrt(diag(V_b-V_B))
| fe re Difference S.E.
-------------+----------------------------------------------------------------
lncfsp | -19.37703 -14.86388 -4.513152 .
hlw | .5255573 1.431621 -.9060634 .
lnczsr | 12.18053 13.93069 -1.750158 .
lnfmzl | 2.149375 2.745353 -.5959786 .
lncsfj | -4.960823 -4.204073 -.7567499 .
lnwztzze | 2.75794 3.394447 -.6365069 .
lnjyzc | -2.023616 -11.65596 9.632345 .
lnkxjszc | 3.235641 3.937528 -.7018867 .
lnzhong | -12.96071 18.25933 -31.22004 6.335719
------------------------------------------------------------------------------
b = consistent under Ho and Ha; obtained from xsmle
B = inconsistent under Ha, efficient under Ho; obtained from xsmle
Test: Ho: difference in coefficients not systematic
chi2(9) = (b-B)'[(V_b-V_B)^(-1)](b-B)
= -13.05 chi2<0 ==> model fitted on these
data fails to meet the asymptotic
assumptions of the Hausman test;
see suest for a generalized test
根据这个帖子的评论
https://bbs.pinggu.org/thread-337115-1-1.html,我重新调整了命令:
qui xsmle chzs lncfsp hlw lnczsr lnfmzl lncsfj lnwztzze lnjyzc lnkxjszc lnzhong schzs ,emat(ww1) model(sem) nolog re
est sto re
qui xsmle chzs lncfsp hlw lnczsr lnfmzl lncsfj lnwztzze lnjyzc lnkxjszc lnzhong schzs ,emat(ww1) model(sem) nolog fe
est sto fe
hausman fe
hausman fe,sigmaless
hausman fe,sigmamore
结果显示这样的错误
the two models need to be different
目前还没找到解决办法。