连老师:您好!
根据您的指导,我先对原四个非平稳的序列进行差分,然后再使用sys-GMM 进行估计.具体步骤是:
第一步.先在原数据库中用生成四序列的一阶差分序列。
gen dgdzb=d.gdzb
gen ddeploa=d.deoloa
gen ddeploa=d.deploa
gen dsrload=d.srload
第二步.对原序列gdzb(各省固定资本形成额/GDB,取对数)、deploa(各省固定资本形成额/GDB,取对数
)、deploa(各省固定资本形成额/GDB,取对数)、srload(各省固定资本形成额/GDB,取对数
)生成四个一阶差分序列即dgdzb、ddeploa、ddeploa、 dsrload进行sys-GMM 进行估计。
语句是:
xtabond2 dgdzb l.dgdzb l.(0/1).(ddeploa dloadep dsrload ),gmm( l.dgdzb,lag(2 5)) iv ( l.(0/1).(ddeploa dloadep dsrload ))
运行的结果是:
. xtabond2 dgdzb l.dgdzb l.(0/1).(ddeploa dloadep dsrload ) ,gmm( l.dgdzb,lag(2 5
> )) iv ( l.(0/1).(ddeploa dloadep dsrload ))
Favoring space over speed. To switch, type or click on mata: mata set matafavor speed, pe
> rm.
Warning: Number of instruments may be large relative to number of observations.
Dynamic panel-data estimation, one-step system GMM
------------------------------------------------------------------------------
Group variable: code Number of obs = 870
Time variable : year Number of groups = 30
Number of instruments = 136 Obs per group: min = 29
Wald chi2(7) = 36.67 avg = 29.00
Prob > chi2 = 0.000 max = 29
------------------------------------------------------------------------------
dgdzb | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
dgdzb |
L1. | .2936256 .0613945 4.78 0.000 .1732946 .4139565
ddeploa |
--. | .1420245 .0500944 2.84 0.005 .0438412 .2402078
L1. | .0427101 .0481251 0.89 0.375 -.0516134 .1370335
dloadep |
--. | .1244038 .0502566 2.48 0.013 .0259028 .2229048
L1. | -.0361744 .0490998 -0.74 0.461 -.1324083 .0600594
dsrload |
--. | -.0281362 .0162513 -1.73 0.083 -.0599882 .0037158
L1. | .0221481 .0155019 1.43 0.153 -.0082351 .0525312
_cons | .0258935 .0058311 4.44 0.000 .0144648 .0373223
------------------------------------------------------------------------------
Arellano-Bond test for AR(1) in first differences: z = -9.61 Pr > z = 0.000
Arellano-Bond test for AR(2) in first differences: z = 0.49 Pr > z = 0.622
Sargan test of overid. restrictions: chi2(128) = 227.08 Prob > chi2 = 0.000
(Not robust, but not weakened by many instruments.)
Difference-in-Sargan tests of exogeneity of instrument subsets:
GMM instruments for levels
Sargan test excluding group: chi2(101) = 189.41 Prob > chi2 = 0.000
Difference (null H = exogenous): chi2(27) = 37.67 Prob > chi2 = 0.083
ivstyle(l.(0/1).(ddeploa dloadep dsrload ))
Sargan test excluding group: chi2(122) = 207.42 Prob > chi2 = 0.000
Difference (null H = exogenous): chi2(6) = 19.67 Prob > chi2 = 0.003
从运行的结果看,比较理想,特别是被解释变量dgdzb的滞后一阶 L1.dgdzb的估计系数为
0.2936256,小于1。我担心的问题是:
一是我的上述具体做法正确吗??
二是我在原序列中直接生成四个一阶差分序列dgdzb、ddeploa、ddeploa、 dsrload,与原序列相比都缺了第一年的项,这样直接进行运算行吗?
三是变量含义的解释。差分后序列dgdzb、ddeploa、ddeploa、 dsrload
得出的sys-GMM
回归结果表达式,能否把差分变量直接写成原变量呢,如把差分后被解释变量dgdzb直接写成原变量gdzb
呢?差分后的经济含义如何解释呢?
四是:减少工具变量数。我在运用FD-GMM分析中部地区8省1978-2008金融发展与资本形成关系时,表达式如下:
xtabond2 dgdzb l.(0/1).(ddeploa dloadep dsrload ) if code>11 & code <20
运算结果是所有变量系数不显著,工具变量数203个,而观察数才为224,我怀疑工具变量数是否太多,怎样才能减少呢?
急等连老师的回复!!