工具变量法地Stata命令及实例

1、实用标准工具变量法的Stata命令及实例本实例使用数据集“ grilic.dta先看一下数据集的统计特征：.sumVariableOpsMeanStd. Dev.MinMaxrns7582691293.443800101rns80758.292876.455382501mrt758.5145119.500119401mrt80758.8984169.302298801smsa758.7044855.45657501smsa80758.7124011.45294201med75810.910292.74112018iq758103.8562 13.6186754145kww75836.57388

2、7.3022471256year75869.031662.6317946673age75821.835092.9817561630age8075833.011873.0855042838s75813.40501 2.231828918s8075813.707122.214693918expr7581.7354292.105542011.444expr8075811.394264.210745.69222.045tenure7581.8311351.67363010tenure807587.3627975.05024022lw7585.686739 .42894944.6057.051lw807

3、5816.826555.40992684.7498.032考察智商与受教育年限的相关关系：.corr iq s (obs=758)iqsiq 1.0000s 0.51311.0000上表显示，智商（在一定程度上可以视为能力的代理变量）与受教育年限具有强烈的正相关关系（相关系数为 0.51 ）。作为一个参考系，先进行 OLS回归，并使用稳健标准差：文案大全实用标准.reg Iw s iq expr tenure rns smsa,rLinear regression758Number of obs =F(5, 752)=84.05Prob > F =0.0000R-squared= 0.3

4、521Root MSE=.34641Robustlw Coef.Std. Err. t P>|t|95% Conf. Intervals .102643.0062099 16.53 0.000.0904523 .1148338expr .0381189 .00661445.76 0.000.025134 .0511038tenure .0356146 .00799884.45 0.000.0199118 .0513173rns -.08407$币.029533 -2.85 0.005 -.1420566 -.0261029smsa .1396i666 .0280564.98 0.000.

5、0845893.194744_cons4.103i675 .0876665 46.81 0.0003.931575 4.275775其中expr, tenure, rns, smsa均为控制变量，而我们主要感兴趣的是变量受教育年限（s）o回归的结果显示，教育投资的年回报率为10.26%,这个似乎太高 了。可能的原因是，由于遗漏变量“能力”与受教育正相关，故“能力”对工资的贡献也被纳入教育的贡献，因此高估了教育的 回报率。引入智商iq作为能力的代理变量，再进行 OLS回归：文案大全Linear regression758Number of obs = F(6, 751)=71.89Prob &g

6、t; F =0.0000R-squared= 0.3600Root MSE=.34454Robustlw Coef.Std. Err. t P>|t|95% Conf. Intervals .0927874 .0069763 13.30 0.000.0790921 .1064826iq .0032792 .00113212.90 0.004.0010567 .0055016expr .0393443 .00666035.91 0.000.0262692 .0524193tenure .034209 .0078957 4.33 0.000.0187088 .0497092rns -.074

7、532>5 .0299772 -2.49 0.013 -.1333815 -.0156834smsa .1367：369 .02777124.92 0.000.0822186 .1912553_cons 3.895172 .1159286 33.60 0.0003.667589 4.122754虽然教育的投资回报率有所下降，但是依然很高。由于用iq作为能力的代理变量有测量误差，故 iq是内生变量， 考虑使用变量（med（母亲的受教育年限）、kww（在“knowledge of the World of Work ”中的成绩）、mrt （婚姻虚拟变量，已婚 =1） age （年龄）作为iq

8、的工具变量，进行2SLS回归，并使用 稳健的标准差：实用标准.ivregress 2sls lw s expr tenure rns smsa (iq=med kww mrt age),rInstrumental variables (2SLS) regression Number of obs =758Wald chi2(6) =355.73Prob > chi2 =0.0000R-squared= 0.2002Root MSE=.38336Robustlw Coef.Std. Err. z P>|z|95% Conf. Intervaliq -.0115468 .0056376

9、 -2.05 0.041 -.0225962 -.0004974s .1373477 .01749897.85 0.000.1030506 .1716449expr .0338041 .00748444.52 0.000.019135 .0484732tenure .040564 .0095848 4.23 0.000.0217781.05935rns -.1176984 .0359582 -3.27 0.001 -.1881751 -.0472216smsa .149983 .03222764.65 0.000.0868182 .2131479_cons4.8371875 .3799432

10、12.73 0.0004.09325.58255Instrumented: iqInstruments: s expr tenure rns smsa med kww mrt age在此2SLS回归中，教育回报率反而上升到13.73%,而iq对工资 的贡献居然为负值。使用工具变量的前提是工具变量的有效性。 为此，进行过度识别检验，考察是否所有的工具变量均外生，即 与扰动项不相关： .estat overidTest of overidentifying restrictions:Score chi2(3)= 51.5449 (p = 0.0000)结果强烈拒绝所有工具变量均外生的原假设。考虑仅

11、使用变量(med, kww作为iq的工具变量，再次进行2SLS 回归，同时显示第一阶段的回归结果：.ivregress 2sls lw s expr tenure rns smsa (iq=med kww),r firstFirst-stage regressionsNumber of obs =758F( 7, 750) =47.74Prob > F=0.0000R-squared=0.3066Adj R-squared =0.3001Root MSE= 11.3931iqR Coef.obustStd. Err. tP>|t|95% Conf. Intervals2.46702

12、1 .232775510.600.0002.0100522.92399expr-.4501353.2391647-1.880.060-.9196471.0193766tenure.2059(；31.2695620.760.445-.3232327.7351388rns-2.6898：，18921335-3.020.003-4.441207-.938455smsa.2627,416.94653090.280.781-1.5954242.120907med.3470133.16813562.060.039.0169409.6770857kww.30818111.06467944.760.000.1

13、812068.4351553_cons56.671223.07695518.42 0.00050.6307562.71169Instrumental variables (2SLS) regression Number of obs =758Wald chi2(6) =370.04Prob > chi2 =0.0000R-squared=0.2775Root MSE=.36436Robustlw Coef. Std. Err. z P>|z|95% Conf. Intervaliq.0139284 .00603932.31 0.021s.0607803 .01895053.21 0

14、.001expr.0433237 .00741185.850.000tenure.0296L42 .0083173.560.000rns-.0435271 .0344779-1.260.207smsa.1272：224 .02974144.280.000_cons3.218043 .39836838.080.000.0020916 .0257653.023638 .0979227.0287968 .0578505.0133432 .0459452-.1111026 .0240483.0689303 .18551462.437256 3.998831Instrumented: iqInstrum

15、ents: s expr tenure rns smsa med kww上表显示，教育的回报率为 6.08%,较为合理，再次进行过度识别检验:.estat overidTest of overidentifying restrictions:Score chi2(1)= .151451 (p = 0.6972)接受原假设，认为(meet kwv外生，与扰动项不相关。进一步考察有效工具变量的第二个条件，即工具变量与内生变量 的相关性。从第一阶段的回归结果可以看出，工具变量对内生变 量具有较好的解释力。更正式的检验如下：.estat firststage,all forcenonrobustFir

16、st-stage regression summary statisticsAdjuVariable R-sq.>tedR-sqPartial RobustR-sq. F(2,750)Prob > Fiq 0.30660.30010.038213.40280.0000Shea's partial R-squaredVariableShea'sPartial F匕sq.Shea'sAdj. Partial R-sq.iq0.03820.0305Minimum eigenvalue statistic = 14.9058Critical Values# of e

17、ndogenous regressors: 1Ho: Instruments are weak # of excluded instruments: 25%10%20%30%2SLS relative bias(not available)10%15%20%25%2SLS Size of nominal 5% Wald test 19.93 11.59 LIML Size of nominal 5% Wald test 8.68 5.338.75 7.254.42 3.92从以上结果可以看出，虽然 Shea' s partial R2不到0.04 ,但是F统计量为13.40>10

18、。我们知道，虽然2SLS是一致的，但却是有偏的，故使用 2SLS会 带来“显著性水平扭曲”(size distortion),而且这种扭曲随着弱工具变量而增大。上表的最后部分显示，如果在结构方程中 对内生解释变量的显著性进行“名义显著性水平” (nominal size ) 为5%勺沃尔德检验，加入可以接受的“真实显著性水平”(truesize )不超过15%则可以拒绝“弱工具变量”的原假设，因为 最小特征值统计量为14.91,大于临界值11.59。总之我们有理由认为不存在弱工具变量。但为了稳健起见，下面 使用对弱工具变量更不敏感的有限信息最大似然法(LIML): .ivregress lim

19、l lw s expr tenure rns smsa (iq=med kww),rNumber of obs =369.62=0.0000=0.2768=.36454758Instrumental variables (LIML) regression Wald chi2(6) Prob > chi2 R-squared Root MSERobustlwCoef.Std. Err. zP>|z|95% Conf. Intervaliq.0139764 .00606812.30 0.021.0020831.0258697s.0606362 .0190343.19 0.0010233

20、303 .0979421expr.0433416 .00741855.84 0.000.0288016.0578816tenure.02962>37 .0083233.56 0.000.0133109.0459364rns-.04338775 .034529-1.26 0.209-.1110631.0242881smsa.1271796 .02975994.27 0.000.0688512.185508_cons3.2141994 .40014928.03 0.0002.4307163.999272Instrumented: iqInstruments: s expr tenure rn

21、s smsa med kww结果发现，LIML的系数估计值与2SLS非常接近，这也从侧面印 证了 “不存在弱工具变量”。使用工具变量法的前提是存在内生解释变量，为此须进行豪斯曼 检验，其原假设是“所有的解释变量均为外生”：.qui reg lw iq s expr tenure rns smsa.estimates store ols.qui ivregress 2sls lw s expr tenure rns smsa (iq=med kww).estimates store iv .hausman iv ols, constant sigmamoreNote: the rank of t

22、he differenced variance matrix (1) does not equal the number of coefficients being tested (7); be sure this is what you expect, or there may be problems computing the test. Examine the output of your estimators for anything unexpected and possibly consider scaling your variables so that the coeffici

23、ents are on a similar scale.Coefficients(b) iv(B)ols(b-B)Differentsqrt(diag(V_ ceS.E._b-V_B)iq.0139284.0032792.0106493.0054318s.06078()3.0927874-.032007.0163254expr.04332>37.0393443.0039794.0020297tenure.0296442.034209-.0045648.0023283rns-.0435271-.0745325.0310054.0158145smsa.1272224.1367369-.009

24、5145.0048529_cons3.2180433.895172-.6771285.3453751b = consistent under Ho and Ha; obtained from ivregressB = inconsistent under Ha, efficient under Ho; obtained from regressTest: Ho: difference in coefficients not systematicchi2(1) = (b-B)'(V_b-V_B)A(-1)(b-B) =3.84Prob>chi2 =0.0499(V_b-V_B is

25、 not positive definite)上表显示，可以在5%勺显著性水平下拒绝“所有解释变量均外生 的原假设”，即认为存在内生解释变量iq。由于传统的豪斯曼检 验建立在同方差的前提下，故在上述回归中均没有使用稳健标准 差。由于传统的豪斯曼检验在异方差的情形下不成立，下面使用异方 差稳健的DWK验：.estat endogenousTests of endogeneityHo: variables are exogenousDurbin (score) chi2(1)= 3.87962 (p = 0.0489)Wu-Hausman F(1,750)= 3.85842 (p = 0.0499

26、)据此可认为iq为内生解释变量。如果存在异方差，则GMM匕2SLS更有效。为此进行如下的最优GMMfr计：.ivregress gmm lw s expr tenure rns smsa (iq=med kww)Instrumental variables (GMM) regression Number of obs =758Wald chi2(6) =372.75Prob > chi2 =0.0000R-squared =0.2750GMM weight matrix: RobustRoot MSE =.36499lwR Coef.obustStd. Err. zP>|z|95%

27、 Conf. Intervaliq.0140888 .00603572.330.020.0022591.0259185s.0603672 .01895453.180.001.0232171.0975174expr.0431117 .00741125.820.000.0285861.0576373tenure.02997'64 .00827283.620.000.013762.0461908rns-.044516 .0344404-1.290.196-.1120179.0229859smsa.1267：368 .02976334.260.000.0684018.1850718_cons3

28、.207298 .3980838.060.0002.4270693.987526Instrumented: iqInstruments: s expr tenure rns smsa med kww进行过度识别检验:.estat overidTest of overidentifying restriction:Hansen's J chi2(1) = .151451 (p = 0.6972)由于p值为0.70,故认为所有的工具变量均为外生。考虑迭代GMM.ivregress gmm lwsexprtenure rns smsa (iq=medkww),igmmIteration 1:

29、 changeinbeta=1.753e-05changeinW =1.100e-02Iteration 2: changeinbeta=4.872e-08changeinW =7.880e-05Iteration 3: changeinbeta=2.495e-10changeinW =2.304e-07Instrumental variables (GMM) regression Number of obs =758Wald chi2(6) =372.73Prob > chi2 =0.0000R-squared =0.2750GMM weight matrix: RobustRoot

30、MSE =.36499lwR Coef.obustStd. Err. zP>|z|95% Conf. Intervaliq.0140901 .00603572.33 0.020.0022603.02592s.0603629 .01895483.18 0.001.0232122.0975135expr.0431101.00741135.820.000.0285841.057636tenure.02997，52.00827293.620.000.0137606.0461898rns-.0445114.0344408-1.290.196-.1120142.0229913smsa.1267：399.02976374.260.000.0684041.1850757_cons3.207224.39808788.060.0002.4269863.987462Instrumented: iqInstruments: s expr tenure rns smsa med kww如果希望将以上各种估计法的系数估计值及其标准差列在同一 张表中，可使用如下命令：文案大全实用标准.