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⛳ Stata 系列推文:
作者:肖蕊 (上海交通大学)
邮箱:1327536202@qq.com
编者按:本文主要摘译自下文,特此致谢!
Source:OLS vs WLS: Dealing with heteroskedasticity -Link-
目录
众所周知,处理线性回归模型 (LRM) 中异方差问题的一种常用策略是使用加权最小二乘法 (WLS),或者更准确地说,可行最小二乘法。
其思路在于,如果你认为残差项存在异方差问题,那么在估计模型系数和标准误时,将这个因素考虑在内就可以得到更好 (更有效) 的结果。这种有效性是通过给予数据中精准的部分更多比重,给予相对分散的部分较小比重得到的。
大多数情况下,你会发现 OLS 和 WLS 得到的系数是类似的,WLS 方法得到的标准误可能更小一些。然而问题在于,当 (两种方法) 估计系数不同的时候会发生什么?
Wooldridge 教授最近的推特提醒我们,如果模型设定是正确的,那么 OLS 和 WLS 方法都是一致的,但是 WLS 是有效的;然而如果模型设定错误,OLS 和 WLS 就可能呈现不同的结果,并且两者的估计结果都是错误的。
于是一个简单的问题出现了:我们如何检验 OLS 和 WLS 两者的估计系数是否等价呢?
首先我并不知道问题的答案。但是为了解决这个问题我有一些猜想,接下来我会向你展示这些想法。当然,我们需要一些数据来做这些事情。
我使用的数据是可以从网上获得的,与此同时剔除了缺少工资数据的观测值。
. use "http://fmwww.bc.edu/RePEc/bocode/o/oaxaca.dta", clear
. keep if lnwage!=.
我们首先从一个简单的线性估计开始,来检验异方差问题。为了增加趣味性,我将使用手动和 “现成的” 方法:
. reg lnwage educ exper tenure female age agesq
Source | SS df MS Number of obs = 1,434
-------------+---------------------------------- F(6, 1427) = 123.27
Model | 137.953858 6 22.9923096 Prob > F = 0.0000
Residual | 266.165946 1,427 .186521336 R-squared = 0.3414
-------------+---------------------------------- Adj R-squared = 0.3386
Total | 404.119804 1,433 .282009633 Root MSE = .43188
------------------------------------------------------------------------------
lnwage | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
educ | 0.063 0.005 12.56 0.000 0.053 0.073
exper | -0.000 0.002 -0.14 0.890 -0.004 0.003
tenure | 0.006 0.002 3.32 0.001 0.003 0.010
female | -0.151 0.024 -6.27 0.000 -0.198 -0.104
age | 0.112 0.008 14.63 0.000 0.097 0.127
agesq | -0.001 0.000 -13.17 0.000 -0.001 -0.001
_cons | 0.333 0.143 2.33 0.020 0.052 0.614
------------------------------------------------------------------------------
. estat hett, iid rhs
Breusch-Pagan / Cook-Weisberg test for heteroskedasticity
Ho: Constant variance
Variables: educ exper tenure female age agesq
chi2(6) = 78.98
Prob > chi2 = 0.0000
. predict resid, resid
. gen resid2=resid^2
. regress resid2 educ exper tenure female age agesq
Source | SS df MS Number of obs = 1,434
-------------+---------------------------------- F(6, 1427) = 13.86
Model | 20.9491235 6 3.49152058 Prob > F = 0.0000
Residual | 359.403977 1,427 .25185983 R-squared = 0.0551
-------------+---------------------------------- Adj R-squared = 0.0511
Total | 380.353101 1,433 .265424355 Root MSE = .50186
------------------------------------------------------------------------------
resid2 | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
educ | -0.010 0.006 -1.67 0.095 -0.021 0.002
exper | -0.007 0.002 -3.34 0.001 -0.011 -0.003
tenure | -0.003 0.002 -1.25 0.211 -0.007 0.002
female | 0.116 0.028 4.16 0.000 0.061 0.171
age | -0.047 0.009 -5.26 0.000 -0.064 -0.029
agesq | 0.001 0.000 6.03 0.000 0.000 0.001
_cons | 1.094 0.167 6.57 0.000 0.767 1.421
------------------------------------------------------------------------------
. display "Chi2: `=e(N)*e(r2)'"
Chi2: 78.9819853933182
结果清晰地表明我们遇到了异方差问题。因此,为了说明这个问题,我们使用 FGLS 方法再次进行估计。换句话说,我们首先建立模型刻画异方差问题,然后得到条件异方差的预测值。
. gen logresid2=log(resid2)
. regress logresid2 educ exper tenure female age agesq
Source | SS df MS Number of obs = 1,434
-------------+---------------------------------- F(6, 1427) = 17.75
Model | 557.714323 6 92.9523871 Prob > F = 0.0000
Residual | 7473.81328 1,427 5.23743047 R-squared = 0.0694
-------------+---------------------------------- Adj R-squared = 0.0655
Total | 8031.5276 1,433 5.60469477 Root MSE = 2.2885
------------------------------------------------------------------------------
logresid2 | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
educ | -0.010 0.027 -0.37 0.710 -0.062 0.042
exper | -0.033 0.010 -3.50 0.000 -0.052 -0.015
tenure | -0.008 0.010 -0.81 0.416 -0.028 0.012
female | 0.648 0.128 5.09 0.000 0.398 0.899
age | -0.266 0.040 -6.57 0.000 -0.345 -0.187
agesq | 0.004 0.000 7.28 0.000 0.003 0.005
_cons | 1.286 0.759 1.69 0.091 -0.204 2.775
------------------------------------------------------------------------------
. predictnl h_x=exp(xb())
接着使用 WLS ,或者对数据进行转换处理,重新估计模型:
. qui:reg lnwage educ exper tenure female age agesq
. est sto mols
. qui:reg lnwage educ exper tenure female age agesq [aw=1/h_x]
. est sto mwls
. foreach i in lnwage educ exper tenure female age agesq {
1. gen `i'w=`i'*sqrt(1/h_x)
2.}
. gen one =sqrt(1/h_x)
. qui:reg lnwagew educw experw tenurew femalew agew agesqw one, nocons
. est sto mtls
. esttab mols mwls mtls,se nogaps b(4) mtitle(ols wls tls)
------------------------------------------------------------
(1) (2) (3)
ols wls tls
------------------------------------------------------------
educ 0.0633*** 0.0558***
(0.0050) (0.0044)
exper -0.0002 -0.0015
(0.0018) (0.0017)
tenure 0.0063*** 0.0033*
(0.0019) (0.0015)
female -0.1508*** -0.1537***
(0.0241) (0.0218)
age 0.1118*** 0.0976***
(0.0076) (0.0082)
agesq -0.0012*** -0.0011***
(0.0001) (0.0001)
educw 0.0558***
(0.0044)
experw -0.0015
(0.0017)
tenurew 0.0033*
(0.0015)
femalew -0.1537***
(0.0218)
agew 0.0976***
(0.0082)
agesqw -0.0011***
(0.0001)
one 0.7189***
(0.1615)
_cons 0.3332* 0.7189***
(0.1433) (0.1615)
------------------------------------------------------------
N 1434 1434 1434
------------------------------------------------------------
Standard errors in parentheses
* p<0.05, ** p<0.01, *** p<0.001
正如我们所知道的,WLS 方法和使用转换后的数据会得到完全一致的结果。现在,如果观察上表中第 1 列和第 2 列的结果,你会发现两者系数上存在些许差异。这些差异看上去无足轻重,但是它们在统计上是否存在显著区别的?换句话说:我们如何比较 WLS 和 OLS 的系数呢?
一种受到推崇的选择是使用 Hausman 检验,当我们将 OLS的估计结果视为一致的,WLS 结果视为有效时:
. hausman mols mwls
---- Coefficients ----
| (b) (B) (b-B) sqrt(diag(V_b-V_B))
| mols mwls Difference Std. err.
-------------+----------------------------------------------------------------
educ | .0632923 .0557789 .0075134 .0024064
exper | -.0002496 -.0014931 .0012435 .0006326
tenure | .0062977 .0032773 .0030204 .0011022
female | -.1508285 -.1536601 .0028316 .0101524
age | .1117635 .0975919 .0141716 .
agesq | -.0012397 -.0010556 -.0001841 .
------------------------------------------------------------------------------
b = Consistent under H0 and Ha; obtained from regress.
B = Inconsistent under Ha, efficient under H0; obtained from regress.
Test of H0: Difference in coefficients not systematic
chi2(6) = (b-B)'[(V_b-V_B)^(-1)](b-B)
= -10.54
Warning: chi2 < 0 ==> model fitted on these data
fails to meet the asymptotic assumptions
of the Hausman test; see suest for a
generalized test.
但这样做似乎违背了检验背后的假设,导致了我们得到了一个出乎意料的 chi2 统计量。按照输出结果的建议,我们可以尝试 suest
命令获得相应系数,但是由于加权方法不同,运行时会出现报错情况。
. suest mols mwls
inconsistent weighting types
r(322);
我们当然可以同时估计标准 OLS,以及使用转换后数据的 WLS 结果:
. suest mols mtls
Simultaneous results for mols, mtls
Number of obs = 1,434
------------------------------------------------------------------------------
| Robust
| Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
mols_mean |
educ | 0.063 0.006 11.43 0.000 0.052 0.074
exper | -0.000 0.002 -0.14 0.891 -0.004 0.003
tenure | 0.006 0.002 3.51 0.000 0.003 0.010
female | -0.151 0.023 -6.52 0.000 -0.196 -0.106
age | 0.112 0.010 11.15 0.000 0.092 0.131
agesq | -0.001 0.000 -10.01 0.000 -0.001 -0.001
_cons | 0.333 0.199 1.68 0.093 -0.056 0.722
-------------+----------------------------------------------------------------
mols_lnvar |
_cons | -1.679 0.073 -23.02 0.000 -1.822 -1.536
-------------+----------------------------------------------------------------
mtls_mean |
educw | 0.056 0.005 10.91 0.000 0.046 0.066
experw | -0.001 0.002 -0.92 0.355 -0.005 0.002
tenurew | 0.003 0.002 2.02 0.043 0.000 0.006
femalew | -0.154 0.022 -7.10 0.000 -0.196 -0.111
agew | 0.098 0.008 11.75 0.000 0.081 0.114
agesqw | -0.001 0.000 -10.50 0.000 -0.001 -0.001
one | 0.719 0.161 4.46 0.000 0.403 1.035
-------------+----------------------------------------------------------------
mtls_lnvar |
_cons | 1.581 0.066 24.00 0.000 1.451 1.710
------------------------------------------------------------------------------
. test ([mols_mean]educ=[mtls_mean]educw) ([mols_mean]exper=[mtls_mean]experw) ///
> ([mols_mean]tenure=[mtls_mean]tenurew) ([mols_mean]female=[mtls_mean]femalew) ///
> ([mols_mean]age=[mtls_mean]agew) ([mols_mean]agesq=[mtls_mean]agesqw) ///
> ([mols_mean]_cons=[mtls_mean]one)
( 1) [mols_mean]educ - [mtls_mean]educw = 0
( 2) [mols_mean]exper - [mtls_mean]experw = 0
( 3) [mols_mean]tenure - [mtls_mean]tenurew = 0
( 4) [mols_mean]female - [mtls_mean]femalew = 0
( 5) [mols_mean]age - [mtls_mean]agew = 0
( 6) [mols_mean]agesq - [mtls_mean]agesqw = 0
( 7) [mols_mean]_cons - [mtls_mean]one = 0
chi2( 7) = 32.11
Prob > chi2 = 0.0000
这种方法暗示 OLS 和 WLS 估计是存在差异的,但是我们的分析还没有结束。然后我们使用由 Alan 推荐的策略,这种方法又指向了由 Justin McCrary 提出的方法,后者可以从以下「网址」找到。
这种方法建议同时使用原始变量和转换后的变量进行辅助回归,并检验转换后的变量对应的估计系数是否显著区别于 0。在这里,转换后数据的估计结果和之前略有不同:
. foreach i in educ exper tenure female age agesq {
1. gen `i'z=`i'*(1/h_x)
2. }
. gen onez =(1/h_x)
. reg lnwage educ exper tenure female age agesq educz experz tenurez femalez ///
> agez agesqz onez, robust
Linear regression Number of obs = 1,434
F(13, 1420) = 50.98
Prob > F = 0.0000
R-squared = 0.3554
Root MSE = .42829
------------------------------------------------------------------------------
| Robust
lnwage | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
educ | 0.084 0.014 6.24 0.000 0.058 0.111
exper | 0.006 0.014 0.45 0.651 -0.020 0.033
tenure | 0.015 0.006 2.55 0.011 0.004 0.027
female | -0.116 0.232 -0.50 0.617 -0.572 0.339
age | 0.148 0.091 1.63 0.103 -0.030 0.326
agesq | -0.002 0.001 -1.38 0.167 -0.004 0.001
educz | -0.001 0.000 -1.81 0.070 -0.001 0.000
experz | -0.000 0.000 -0.11 0.912 -0.001 0.001
tenurez | -0.000 0.000 -2.20 0.028 -0.000 -0.000
femalez | -0.003 0.007 -0.47 0.641 -0.018 0.011
agez | -0.001 0.003 -0.26 0.795 -0.006 0.005
agesqz | 0.000 0.000 0.21 0.831 -0.000 0.000
onez | 0.023 0.074 0.31 0.755 -0.122 0.168
_cons | -0.564 0.918 -0.61 0.539 -2.364 1.236
------------------------------------------------------------------------------
. est sto mcry
. test educz experz tenurez femalez agez agesqz onez
( 1) educz = 0
( 2) experz = 0
( 3) tenurez = 0
( 4) femalez = 0
( 5) agez = 0
( 6) agesqz = 0
( 7) onez = 0
F( 7, 1420) = 5.05
Prob > F = 0.0000
这种方法也证明 WLS 和 OLS 估计系数可能是不同的。然而,得到的统计量比使用 suest
方法得到的统计量小得多。然后,我们有一种需要用到“克隆”数据的方法,大体步骤如下:
. gen id=_n
. expand 2, gen(clone)
. gen wgt=1 if clone==0
. replace wgt=(1/h_x) if clone==1
. reg lnwage c.(educ exper tenure female age agesq)##i.clone [w=wgt], cluster(id)
Linear regression Number of obs = 2,868
F(13, 1433) = 48.47
Prob > F = 0.0000
R-squared = 0.2875
Root MSE = .36353
(Std. err. adjusted for 1,434 clusters in id)
-----------------------------------------------------------------------------------
| Robust
lnwage | Coefficient std. err. t P>|t| [95% conf. interval]
------------------------+----------------------------------------------------------
educ | 0.063 0.006 11.40 0.000 0.052 0.074
exper | -0.000 0.002 -0.14 0.891 -0.004 0.003
tenure | 0.006 0.002 3.50 0.000 0.003 0.010
female | -0.151 0.023 -6.51 0.000 -0.196 -0.105
age | 0.112 0.010 11.13 0.000 0.092 0.131
agesq | -0.001 0.000 -9.99 0.000 -0.001 -0.001
clone | 0.386 0.103 3.75 0.000 0.184 0.588
clone#c.educ | -0.008 0.003 -2.51 0.012 -0.013 -0.002
clone#c.exper | -0.001 0.001 -1.18 0.239 -0.003 0.001
clone#c.tenure | -0.003 0.001 -2.79 0.005 -0.005 -0.001
clone#c.female | -0.003 0.011 -0.25 0.803 -0.025 0.019
clone#c.age | -0.014 0.005 -2.67 0.008 -0.025 -0.004
clone#c.agesq | 0.000 0.000 2.66 0.008 0.000 0.000
_cons | 0.333 0.199 1.67 0.094 -0.057 0.724
-----------------------------------------------------------------------------------
. test 1.clone#c.educ 1.clone#c.exper 1.clone#c.tenure 1.clone#c.female 1.clone#c.age ///
> 1.clone#c.agesq 1.clone
( 1) 1.clone#c.educ = 0
( 2) 1.clone#c.exper = 0
( 3) 1.clone#c.tenure = 0
( 4) 1.clone#c.female = 0
( 5) 1.clone#c.age = 0
( 6) 1.clone#c.agesq = 0
( 7) 1.clone = 0
F( 7, 1433) = 4.57
Prob > F = 0.0000
这种方法和之前描述的 suest
方法几乎完全一致。并且它也暗示 WLS 和 OLS 估计系数可能不同。其他不那么常见的方法,举例来说,涉及使用模拟的方法,特别是自抽样。但是在这里,我们还有一处可以探讨。
所以,让我们都试一下:
. keep if clone==0
. program bs_wls_ols, eclass
1. reg lnwage educ exper tenure female age agesq
2. matrix b1=e(b)
3. capture drop lres
4. predictnl lres=log((lnwage-xb())^2)
5. reg lres educ exper tenure female age agesq
6. capture drop nwgt
7. predictnl nwgt=1/exp(xb())
8. ** two steps assuming Weights change
. reg lnwage educ exper tenure female age agesq [w=nwgt]
9. matrix b2=e(b)
10. ** two steps assuming Weights do not change
. reg lnwage educ exper tenure female age agesq [w=1/h_x]
11. matrix b3=e(b)
12. ** Finally the differences
. matrix db2=b1-b2
13. matrix db3=b1-b3
14. ** putting all together:
. matrix coleq b2= wols_dw
15. matrix coleq b3= wols_fw
16. matrix coleq db2= dwols_dw
17. matrix coleq db3= dwols_fw
18. matrix b=b2,b3,db2,db3
19. ereturn post b
20. end
. bootstrap, reps(500) seed(10) nodots: bs_wls_ols
Bootstrap results Number of obs = 1,434
Replications = 500
------------------------------------------------------------------------------
| Observed Bootstrap Normal-based
| coefficient std. err. z P>|z| [95% conf. interval]
-------------+----------------------------------------------------------------
wols_dw |
educ | 0.056 0.005 10.63 0.000 0.045 0.066
exper | -0.001 0.002 -0.88 0.379 -0.005 0.002
tenure | 0.003 0.002 2.05 0.040 0.000 0.006
female | -0.154 0.025 -6.16 0.000 -0.203 -0.105
age | 0.098 0.008 12.69 0.000 0.083 0.113
agesq | -0.001 0.000 -11.12 0.000 -0.001 -0.001
_cons | 0.719 0.142 5.07 0.000 0.441 0.997
-------------+----------------------------------------------------------------
wols_fw |
educ | 0.056 0.005 11.24 0.000 0.046 0.066
exper | -0.001 0.002 -0.94 0.346 -0.005 0.002
tenure | 0.003 0.002 2.06 0.040 0.000 0.006
female | -0.154 0.022 -6.97 0.000 -0.197 -0.110
age | 0.098 0.008 11.58 0.000 0.081 0.114
agesq | -0.001 0.000 -10.44 0.000 -0.001 -0.001
_cons | 0.719 0.163 4.42 0.000 0.400 1.037
-------------+----------------------------------------------------------------
dwols_dw |
educ | 0.008 0.004 2.02 0.044 0.000 0.015
exper | 0.001 0.001 1.07 0.286 -0.001 0.004
tenure | 0.003 0.001 2.51 0.012 0.001 0.005
female | 0.003 0.013 0.22 0.828 -0.023 0.028
age | 0.014 0.008 1.79 0.074 -0.001 0.030
agesq | -0.000 0.000 -1.89 0.059 -0.000 0.000
_cons | -0.386 0.163 -2.36 0.018 -0.706 -0.065
-------------+----------------------------------------------------------------
dwols_fw |
educ | 0.008 0.003 2.48 0.013 0.002 0.013
exper | 0.001 0.001 1.14 0.254 -0.001 0.003
tenure | 0.003 0.001 2.92 0.003 0.001 0.005
female | 0.003 0.011 0.25 0.799 -0.019 0.025
age | 0.014 0.005 2.69 0.007 0.004 0.025
agesq | -0.000 0.000 -2.75 0.006 -0.000 -0.000
_cons | -0.386 0.103 -3.73 0.000 -0.588 -0.183
------------------------------------------------------------------------------
在这里有几点需要注意:
. test [dwols_fw]educ [dwols_fw]exper [dwols_fw]tenure [dwols_fw]female [dwols_fw]age [dwols_fw]agesq
( 1) [dwols_fw]educ = 0
( 2) [dwols_fw]exper = 0
( 3) [dwols_fw]tenure = 0
( 4) [dwols_fw]female = 0
( 5) [dwols_fw]age = 0
( 6) [dwols_fw]agesq = 0
chi2( 6) = 29.33
Prob > chi2 = 0.0001
. test [dwols_dw]educ [dwols_dw]exper [dwols_dw]tenure [dwols_dw]female [dwols_dw]age [dwols_dw]agesq
( 1) [dwols_dw]educ = 0
( 2) [dwols_dw]exper = 0
( 3) [dwols_dw]tenure = 0
( 4) [dwols_dw]female = 0
( 5) [dwols_dw]age = 0
( 6) [dwols_dw]agesq = 0
chi2( 6) = 14.46
Prob > chi2 = 0.0249
最后,我们也可以使用 ml
手动为目标函数添加权重,尝试同时估计两个方程:
. program myolswls1
1. args lnf xb1 xb2
2. qui:{
3. * OLS regression:
. replace `lnf' = -($ML_y1-`xb1')^2
4. * WLS where weights are added "manually"
. replace `lnf' = `lnf'-(1/h_x)*($ML_y2-`xb2')^2
5. }
6. end
. ml model lf myolswls1 (ols:lnwage = educ exper tenure female age agesq) ///
> (wls:lnwage = educ exper tenure female age agesq), robust maximize
. ml display
Number of obs = 1,434
Wald chi2(6) = 506.14
Log pseudolikelihood = -7197.7961 Prob > chi2 = 0.0000
------------------------------------------------------------------------------
| Robust
| Coefficient std. err. z P>|z| [95% conf. interval]
-------------+----------------------------------------------------------------
ols |
educ | 0.063 0.006 11.43 0.000 0.052 0.074
exper | -0.000 0.002 -0.14 0.891 -0.004 0.003
tenure | 0.006 0.002 3.51 0.000 0.003 0.010
female | -0.151 0.023 -6.52 0.000 -0.196 -0.106
age | 0.112 0.010 11.15 0.000 0.092 0.131
agesq | -0.001 0.000 -10.01 0.000 -0.001 -0.001
_cons | 0.333 0.199 1.68 0.093 -0.056 0.722
-------------+----------------------------------------------------------------
wls |
educ | 0.056 0.005 10.91 0.000 0.046 0.066
exper | -0.001 0.002 -0.92 0.355 -0.005 0.002
tenure | 0.003 0.002 2.02 0.043 0.000 0.006
female | -0.154 0.022 -7.10 0.000 -0.196 -0.111
age | 0.098 0.008 11.75 0.000 0.081 0.114
agesq | -0.001 0.000 -10.50 0.000 -0.001 -0.001
_cons | 0.719 0.161 4.46 0.000 0.403 1.035
------------------------------------------------------------------------------
因为这些结果和 suest
方法相似,这样做相比于前文的检验并没有新的发现。
. test [ols=wls]
( 1) [ols]educ - [wls]educ = 0
( 2) [ols]exper - [wls]exper = 0
( 3) [ols]tenure - [wls]tenure = 0
( 4) [ols]female - [wls]female = 0
( 5) [ols]age - [wls]age = 0
( 6) [ols]agesq - [wls]agesq = 0
chi2( 6) = 30.39
Prob > chi2 = 0.0000
所以如何检验 WLS 和 OLS 的估计系数在统计上是否存在差异呢?我的结论是,我并不知道。
而且不幸的是,我在手头的任何一本书中都没有找到可以解决这个问题的检验步骤和建议。尽管如此,在尝试若干可能的步骤后,就目前的这个例子来看,结果显示 OLS 和 WLS 的系数是不同的。
事实上,所有依赖于 “复制” 数据,或者实际上同时估计 WLS 和 OLS 的检验都指向了相同的结论。然而我想知道是否存在其他 (更好的) 方法来实施这个简单的检验。
Note:产生如下推文列表的 Stata 命令为:
lianxh suest MLE 豪斯曼, m
安装最新版lianxh
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ssc install lianxh, replace
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lianxh
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随时搜索连享会推文、Stata 资源,安装命令如下:
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