连老师:
下面是我碰到的具体问题,请赐教!谢谢!
采用系统GMM命令:
xtdpdsys cs ls Lag_rent2 Lag_si Lag_pro Lag_cov Lag_mb smg nfpg
得到的结果:
System dynamic panel-data estimation Number of obs = 669
Group variable: company Number of groups = 67
Time variable: year
Obs per group: min = 2
avg = 9.985075
max = 15
Number of instruments = 128 Wald chi2(9) = 248.50
Prob > chi2 = 0.0000
One-step results
------------------------------------------------------------------------------
cs | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
cs |
L1. | .6405133 .0489364 13.09 0.000 .5445998 .7364268
ls | .1089056 .0429461 2.54 0.011 .0247327 .1930785
Lag_rent2 | -.0008363 .0066601 -0.13 0.900 -.0138898 .0122172
Lag_si | -.0884034 .0299857 -2.95 0.003 -.1471744 -.0296325
Lag_pro | -.2270473 .0908122 -2.50 0.012 -.405036 -.0490586
Lag_cov | .0799174 .0395652 2.02 0.043 .0023711 .1574637
Lag_mb | .0009354 .0048314 0.19 0.846 -.008534 .0104049
smg | -.0102183 .00719 -1.42 0.155 -.0243104 .0038738
nfpg | .0026641 .0028123 0.95 0.343 -.0028479 .0081761
_cons | -.5397587 .4232628 -1.28 0.202 -1.369339 .2898211
------------------------------------------------------------------------------
Instruments for differenced equation
GMM-type: L(2/.).cs
Standard: D.ls D.Lag_rent2 D.Lag_si D.Lag_pro D.Lag_cov D.Lag_mb D.smg
D.nfpg
Instruments for level equation
GMM-type: LD.cs
Standard: _cons
此时,ls(我主要关注的变量)前的系数是显著的,然后采用Sargan检验,得到:
estat sargan
Sargan test of overidentifying restrictions
H0: overidentifying restrictions are valid
chi2(118) = 286.3493
Prob > chi2 = 0.0000
说明工具变量有效。
然后进行steptwo:
命令:
xtdpdsys cs logsloan Lag_rent2 Lag_si Lag_pro Lag_cov Lag_mb smg nfpg, twostep vce(robust)
System dynamic panel-data estimation Number of obs = 669
Group variable: company Number of groups = 67
Time variable: year
Obs per group: min = 2
avg = 9.985075
max = 15
Number of instruments = 128 Wald chi2(9) = 51.43
Prob > chi2 = 0.0000
Two-step results
------------------------------------------------------------------------------
| WC-Robust
cs | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
cs |
L1. | .6112183 .2212093 2.76 0.006 .177656 1.044781
logsloan | .106656 .180621 0.59 0.555 -.2473546 .4606667
Lag_rent2 | -.0023285 .0149491 -0.16 0.876 -.0316281 .0269712
Lag_si | -.0560543 .1457963 -0.38 0.701 -.3418097 .2297012
Lag_pro | -.218882 .2758682 -0.79 0.428 -.7595737 .3218097
Lag_cov | .072324 .1341872 0.54 0.590 -.1906782 .3353261
Lag_mb | .0011677 .0141426 0.08 0.934 -.0265513 .0288868
smg | -.0059688 .0149468 -0.40 0.690 -.0352639 .0233263
nfpg | .0032441 .0042373 0.77 0.444 -.0050608 .0115491
_cons | -.7525732 1.779216 -0.42 0.672 -4.239773 2.734626
------------------------------------------------------------------------------
Instruments for differenced equation
GMM-type: L(2/.).cs
Standard: D.logsloan D.Lag_rent2 D.Lag_si D.Lag_pro D.Lag_cov D.Lag_mb
D.smg D.nfpg
Instruments for level equation
GMM-type: LD.cs
Standard: _cons
在采用steptwo后,ls(我主要关注的变量)前的系数变得不显著了。
在采用二步时,我最初得到的结果是显著的,但出现了一个warning:
Warning: gmm two-step standard errors are biased; robust standard errors are recommended.
于是我就采用了twostep vce(robust)。
再进行Sargan检验,得到:
estat sargan
Sargan test of overidentifying restrictions
H0: overidentifying restrictions are valid
cannot calculate Sargan test with vce(robust)
chi2(118) = .
Prob > chi2 = .
我的问题:
1.“cannot calculate Sargan test with vce(robust)”这说明什么?
2.我该如何判断我该采取一步结果还是两步结果呢?
3.Lag_rent2等变量,我是否可以采用一阶滞后的形式?
非常感谢!
|