工具变量法的Stata命令和实例

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：.sumVariableObsMeanStd.Dev.MinMaxrns758.2691293.443800101rns80758.292876.455382501mrt758.5145119.500119401mrt80758.8984169.302298801smsa758.7044855.45657501smsa80758.7124011.45294201med75810.910292.74112018iq758103.856213.6186754145kww75836.573887.3022471256year75869.031662.6317946673age75821.835092.9817561630age8075833.011873.0855042838s75813.405012.231828918s8075813.707122.214693918expr7581.7354292.105542011.444expr8075811.394264.210745.69222.045tenure7581.8311351.67363010tenure807587.3627975.05024022Iw7585.686739.42894944.6057.051lw807586.826555.40992684.7498.032•考察智商与受教育年限的相关关系：.corriqs(obs=758)iqsiq1.0000s0.51311.0000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0LS回归•并使用稳健标准差:Linearregression75884.05Linearregression75884.050.0000F(5,752)Prob>FR-squared = 0.3521RootMSE = .34641lwCoef.RobustStd.Err.tP>lt|[95%Conf.Interval]s.102643.006209916.530.000.0904523.1148338expr.0381189.00661445.760.000.025134.0511038tenure.0356146.00799884.450.000.0199118.0513173rns-.0840797.029533-2.850.005-.1420566-.0261029smsa.1396666.0280564.980.000.0845893.194744_cons4.103675.087666546.810.0003.9315754.275775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）。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:Linearregression75871.89Linearregression75871.890.0000F(6,751)Prob>FR-squared = 0.3600RootMSE = .34454lwCoef.RobustStd.Err.tP>|t|[95%Conf.Interval]s.0927874.006976313.300.000.0790921.1064826iq.0032792.00113212.900.004.0010567.0055016expr.0393443.00666035.910.000.0262692.0524193tenure.034209.00789574.330.000.0187088.0497092rns-.0745325.0299772-2.490.013-.1333815-.0156834smsa.1367369.02777124.920.000.0822186.1912553_cons3.895172.115928633.600.0003.6675894.122754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“knowledgeoftheWorldofWork”中的成绩）、mrt（婚姻虚拟变量.已婚=1）age（年龄））作为iq的工具变量•进行2SLS回归•并使用稳健的标准差：Instrumentalvariables(2SLS)regressionNumberofobs=758Waldchi2(6) = 355.73Prob>chi2 = 0.0000R-squared = 0.2002RootMSE = .38336lwCoef.RobustStd.Err.zP>|z|[95%Conf.Interval]iq-.0115468.0056376-2.050.041-.0225962-.0004974s.1373477.01749897.850.000.1030506.1716449expr.0338041.00748444.520.000.019135.0484732tenure.040564.00958484.230.000.0217781.05935rns-.1176984.0359582-3.270.001-.1881751-.0472216smsa.149983.03222764.650.000.0868182.2131479_cons4.837875.379943212.730.0004.09325.58255Instrumented:iqInstruments:sexprtenurernssmsamedkwwmrtage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：.estatoveridTestofoveridentifyingrestrictions:Scorechi2(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：First-stageregressionsNumberofobsF(7,750)Prob>FR-squaredAdjR-squaredRootMSE75847.740.00000.3066NumberofobsF(7,750)Prob>FR-squaredAdjR-squaredRootMSE75847.740.00000.30660.300111.3931iqCoef.RobustStd.Err.tP>|t|[95%Conf.Interval]s2.467021.232775510.600.0002.0100522.92399expr-.4501353.2391647-1.880.060-.9196471.0193766tenure.2059531.2695620.760.445-.3232327.7351388rns-2.689831.8921335-3.020.003-4.441207-.938455smsa.2627416.94653090.280.781-1.5954242.120907med.3470133.16813562.060.039.0169409.6770857kww.3081811.06467944.760.000.1812068.4351553_cons56.671223.07695518.420.00050.6307562.71169Instrumentalvariables(2SLS)regressionNumberofobsWaldchi2(6)Prob>chi2R-squared758370.040.00000.2775RootMSE=.36436lwCoef.RobustStd.Err.zP>|z|[95%Conf.Interval]iq.0139284.00603932.310.021.0020916.0257653s.0607803.01895053.210.001.023638.0979227expr.0433237.00741185.850.000.0287968.0578505tenure.0296442.0083173.560.000.0133432.0459452rns-.0435271.0344779-1.260.207-.1111026.0240483smsa.1272224.02974144.280.000.0689303.1855146_cons3.218043.39836838.080.0002.4372563.998831Instrumented:iqInstruments: sexprtenurernssmsamedkww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：.estatoveridTestofoveridentifyingrestrictions:Scorechi2(1)=.151451(p=0.6972)接受原假设•认为(med・kww)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：.estatfirststage,allforcenonrobustFirst-stageregressionsummarystatisticsVariableR-sq.AdjustedR-sq.PartialR-sq.RobustF(2,750)Prob>Fiq0.30660.30010.038213.4028 0.0000Shea'spartialR-squaredShea'sShea'sVariablePartialR-sq. Adj.PartialR-sq.iq0.03820.0305Minimumeigenvaluestatistic=14.9058CriticalValuesHo:Instrumentsareweak2SLSrelativebias#ofendogenousregressors: CriticalValuesHo:Instrumentsareweak2SLSrelativebias#ofexcludedinstruments: 25% 10% 20% 30%(notavailable)2SLSSizeofnominal5%WaldtestLIMLSizeofnominal2SLSSizeofnominal5%WaldtestLIMLSizeofnominal5%Waldtest10% 15% 20%19.93 11.59 8.758.68 5.33 4.4225%7.253.92从以上结果可以看出•虽然Shea'spartialR"2不到0.04•但是F统计量为13.40>10。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”(sizedistortion)・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”(nominalsize)为5%的沃尔德检验•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):.ivregresslimllwsexprtenurernssmsa(iq=medkww),rInstrumentalvariables(LIML)regressionNumberofobs=758Waldchi2(6) = 369.62Prob>chi2 = 0.0000R-squared = 0.2768RootMSE = .36454lwCoef.RobustStd.Err.zP>|z|[95%Conf.Interval]iq.0139764.00606812.300.021.0020831.0258697s.0606362.0190343.190.001.0233303.0979421expr.0433416.00741855.840.000.0288016.0578816tenure.0296237.0083233.560.000.0133109.0459364rns-.0433875.034529-1.260.209-.1110631.0242881smsa.1271796.02975994.270.000.0688512.185508_cons3.214994.40014928.030.0002.4307163.999272Instrumented:iqInstruments: sexprtenurernssmsamedkww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”

.estimatesstoreols.quiivregress2slslwsexprtenurernssmsa(iq=medkww).estimatesstoreiv.hausmanivols,constantsigmamore―fficients⑹―fficients⑹iv(B)olsiq.0139284.0032792s.0607803.0927874expr.0433237.0393443tenure.0296442.034209rns-.0435271-.0745325smsa.1272224.1367369_cons3.2180433.895172Coefficients(b-B)Differencesqrt(diag(V_b-V_B))S.E..0106493.0054318-.032007.0163254.0039794.0020297-.0045648.0023283.0310054.0158145-.0095145.0048529-.6771285.3453751Note:therankofthedifferencedvariancematrix(1)doesnotequalthenumberofcoefficientsbeingtested(7);besurethisiswhatyouexpect,ortheremaybeproblemscomputingthetest.Examinetheoutputofyourestimatorsforanythingunexpectedandpossiblyconsiderscalingyourvariablessothatthecoefficientsareonasimilarscale.b=consistentunderHoandHa;obtainedfromivregressB=inconsistentunderHa,efficientunderHo;obtainedfromregressTest:Ho:differenceincoefficientsnotsystematicchi2(1)(b-B)'[(V_b-V_B厂(-1)](b-B)3.84Prob>chi20.0499(V_b-V_Bisnotpositivedefinite)上表显示.可以在5%的显著性水平下拒绝“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DWH检验：.estatendogenousTestsofendogeneityHo:variablesareexogenousDurbin(score)chi2(1) =3.87962(p=0.0489)Wu-HausmanF(1,750) =3.85842(p=0.0499)据此可认为iq为内生解释变量。•如果存在异方差•则GMM比2SLS更有效。为此进行如下的最优GMM估计：.ivregressgmmlwsexprtenurernssmsa(iq=medkww)Instrumentalvariables(GMM)regressionGMMweightmatrix:RobustNumberofobsInstrumentalvariables(GMM)regressionGMMweightmatrix:RobustNumberofobs=758Waldchi2(6)=372.75Prob>chi2=0.0000R-squared=0.2750RootMSE=.36499lwCoef.RobustStd.Err.zP>|z|[95%Conf.Interval]iq.0140888.00603572.330.020.0022591.0259185s.0603672.01895453.180.001.0232171.0975174expr.0431117.00741125.820.000.0285861.0576373tenure.0299764.00827283.620.000.013762.0461908rns-.044516.0344404-1.290.196-.1120179.0229859smsa.1267368.02976334.260.000.0684018.1850718_cons3.207298.3980838.060.0002.4270693.987526Instrumented:iqInstruments: sexprtenurernssmsamedkww•进行过度识别检验：.estatoveridTestofoveridentifyingrestriction:Hansen'sJchi2(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：

.ivregressgmmlwsexprtenurernssmsa(iq=medkww),igmmIteration1:changeinbeta=1.753e-05changeinW= 1.100e-02Iteration2:changeinbeta=4.872e-08changeinW=7.880e-05Iteration3:changeinbeta=2.495e-10changeinW=2.304e-07Instrumentalvariables(GMM)regressionNumberofobs758Waldchi2(6)372.73Prob>chi20.0000R-squared0.2750GMMweightmatrix:RobustRootMSE.36499RobustIwCoef.Std.Err.zP>|z|[95%Conf.Interval]iq.0140901.00603572.330.020.0022603.02592s.0603629.01895483.180.001.0232122.0975135expr.0431101.00741135.820.000.0285841.057636tenure.0299752.00827293.620.000.0137606.0461898rns-.0445114.0344408 --1.290.196-.1120142.0229913smsa.1267399.02976374.260.000.0684041.1850757cons3.207224.39808788.060.0002.4269863.987462Instrumented:iqInstruments:sexprtenurernssmsamedkww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：.