Weak IV. 1. What are Weak Instruments?

1 1. What are Weak instruments ? Time Series Analysis, Fall 2007 Professor Anna MikushevaPaul Schrimpf, scribeNovemeber 13, 2007corrected September 2012 LectureWeak lecture extensively uses lectures given by Jim Stock as a part of mini-course at the NBER What are Weak Instruments? Consider the simplest classical homoskedastic IV model:yt= xt+utxt=Zt +vt,whereytare one-dimensional,xtisn 1,Ztisk 1 andEutZt= 0, one observes data{yt,xt,Zt}.Assume thatn k. In generalutandvtare correlated, and thus,xtis an endogenous (since we assumed thatEutZt= 0), if it also relevant (EZ xhas rankn), then it can serveas instrument and the model is identified. The usual TSLS will be TSLS= (x PZx) 1x PZy,wherePZ=Z(Z Z) 1Z .The problem of weak identification arises when moment conditions are not very informative about theparameter of interest, that is, when the rank of the matrix g0( 0)| = R 0=( 0) =EZ xisn, but atthe same time itis very close to a reduced rank matrix (for example, the smallest eigenvalue ofn nmatrixx Z(Z Z) 1Z xis very closeto zero).

2 In the case of 1 instrument and 1 endogenous regressor weakidentification corresponds to a weak correlation between the instrument and the explain the essence of weak IV problem we start with a toy example of totally irrelevant relevant a situation when one has 1 endogenous regressor and 1 instrumentwhich is independent of everything (totally irrelevant, = 0). That is, the instrument is not valid and isnot identified. The question is how TSLS behaves? This should explain what we see in Bounder, Jaeger,Baker s (1995) random quarter of birth exercise: TSLS 0= 1 Ztut=TZtv1t T 2u uv Ztut tvt u,Z vwhere ( u, v) N(0, ), =(2). Let = uv/ 2v, then u= v+ , and uv TSLSv 0 + . vConclusions: TSLSis inconsistent (as expected, since is not identified). TSLSis centered around 0+ (since has symmetric distribution ), which is the limit of Asymptotically TSLShas heavy tails (since has Cauchy distribution )vCite as: Anna Mikusheva, course materials for Time Series Analysis, Fall 2007.

3 MIT OpenCourseWare ( ),Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].7-8 Concentration parameter2 Non-uniform asymptoticsIf the instrument is relevantEZtxt= 0, then as the sample size (T )increasesTSLSis consistent and asymptotically normal. IfEZtxt= 0, as shown before, the asymptoticsbreaks down. So, the correlation equals zero is a point of discontinuity of asymptotics. That is, the limitof T( TSLS 0) depends on the value ofEZtxt, which is anuisance parameter(parameter that wedo not care about per se, but which affects the distribution) in this case. It means that the convergenceof T( TSLS 0) to normal distribution is not uniform with respect to the nuisance parameter. Thatis, ifEZtxt= 0 but is very small, the convergence is slow and it requires a larger sample to allow fornormal approximation to be accurate. One may hope that another asymptotic embedding will providebetter asymptotic parameterConsider the same IV model as beforeyt= xt+utxt=Zt +vt,but now assume one endogenous regressor (n= 1) and several instrumentsk 1.

4 Introduce a concentrationparameter 2= Z Z / 2v. Then x PZu(Z +v) P Zu+v P TSLS ZuZu 0===.x PZx(Z +v) Z(Z Z) 1Z(Z +v) 2 2v+2 Zv+v PZvLet s assume that instruments are fixed and errors are normals. Let us introduce= Zuu , Z v=Z u Zv,S=v PZv, andS=u PZv. Then and are standard normal, and distribution ofS Z vv2uvuvvvZ v u vandSuvdoes not depvend on sample size (chi-squared). Finally, we have: ( ) =u u+Svu/ TSLS 0. v1 + 2 v/ +Svv/ 2 Notice, that in this expression plays the role of the sample size! If is large, ( TSLS 0) will beapproximately normal, if is small, then the distribution is non-standard. That is, is an effective nuisanceparameter here, and it measures the amount of information data have about the parameter .Weak instruments asymptoticsWeak instrument asymptotics is the name for asymptotic embedding modeling correlation as converging tozero at speed is the same as modeling being constant. So, assume that =C/ T.

5 Then( + zTSLS 0 v) zu,( +zv) ( +zv) 2 where (zu,zv) N(0, ), =(uuvQ 2). =C 1/2 QZZ,ZZ=EZtZt .uvvWhat is also important here is that under this nesting (weak instrument asymptotics) the first-stageF-statistic (for testing all coefficients on instruments are zeros) converges in distribution to a non-central 2kwith non-centrality parameter 2 Detecting Weak InstrumentsThere are several approaches to detect whether one has a weak instrument problem:Cite as: Anna Mikusheva, course materials for Time Series Analysis, Fall 2007. MIT OpenCourseWare ( ),MassachusettsInstitute of Technology. Downloaded on [DD Month YYYY].663. Inference methods robust towards weak instruments3(1) Compare the first-stage F statistic with a cut-off (Stock, Yogo). Assume we have a homoskedastic IVmodel with one endogenous variablextand some exogenous variablesWtyt= xt+ Wt+utwhereytandxtare 0. The data you observe is LetZtbe ak 1instrument, in particular,EutZt= 0.

6 The first-stage regression in this case isxt=Zt + Wt+vt,and the relevance condition means that = 0. The weak instrument problem arises when 0. Stockand Yogo showed that the first-stageF-statistic is distributed as a non-central 2with a non-centralityparameter directly related to the concentration parameter . As a result, the first-stageF-statisticcan serveas an indicator of the value of .Idea: lookat the first-stage F-statistics since it is an indicator of 2and choose a cut-off that wouldguarantee either relative bias less than 10% (for estimation) or the size of 5% test being less than 10%(for testing and confidence sets). By relative bias we mean the following: the maximum bias of TSLSis no more than 10% of the bias of : then the first stage F-statistic is the statistic for testing = 0 in the first stage regressionxt= Zt+ Wt+vt. We know thatEF= 1 + 2/k, so we can estimate 2/kasF 1. Compare the obtained 2/kwith cut-off (tables in Stock, Wright and Yogo).

7 Note that the cut-offs are far higher than the critical values for the F-test (weak instruments are more often thanwhat would be detected by pre-test = 0) Rule of thumbF <10 indicates weak procedure described above works only for a single endogenous variablext. If the regression hasmore than one endogenous (instrumented) regressor, then the analog of the F-test will be the first-stagematrix and a test for rank of this matrix. See Cragg and Donald (1993) for more ! This test for weak IV assumes a homoskedastic setting! What to do in the heteroskedasticcase or when one has autocorrelation isan open question.(2) The Hahn-Hausman test of the null of strong instruments . The idea is that if instruments are strongthen the regression and the reverse regression should give estimates of and 1/ , which are consistentwith each other. Think about the Elasticity of inter-temporal substitution example from the lastlecture. The test is based on comparing them.

8 Problem: it tests the null of strong identification anddoes not control for the probability of a type-II error (mistake of not-detecting weak IV when it ispresent). The test may experience power problems as well.(3) Do not test for weak-strong instruments , but rather use methods robust towards weak Inference methods robust towards weak instrumentsInferences include tests and confidence sets. We concentrate mainly on tests. Tests robust towards weakinstruments are supposed to maintain the correct size no matter whether instruments are weak or can be achieved in two ways: using statistics whose distribution do not depend on or using con-ditioning on sufficient statistics for . The problem of robust inferences is fully solved for the case of oneendogenous is still an open question for the case of more than one endogenous as: Anna Mikusheva, course materials for Time Series Analysis, Fall 2007. MIT OpenCourseWare ( ),MassachusettsInstitute of Technology.

9 Downloaded on [DD Month YYYY]. Case of one endogenous Case of one endogenous are two widely known statistics whose distributions do not depend on : Anderson-Rubin (AR) andLagrange Multiplier (LM).AR testConsider our modely=X +u,X=Z +v,whereXis one-dimensional and test for hypothesisH0: = 0. Under the null, vectory X is equal tothe errorutand is uncorrelated withZ(due to exogeneity of instruments ). The suggested statistics is(y AR( 0= X) P)Z(y X ).(y X ) MZ(y X )/(T k)herePZ=Z(Z Z) 1Z ,MZ=I Pz. The distribution of AR does not depend on asymptoticallyAR 2 k/k. The formula may remind you of the J-test for over-identifying restrictions. It would be a J-testif one were to plugs in a more general situation of more than one endogenous variable and/or included exogenous regressorsAR statistic is F-statistic testing that all coefficients onZare zero in the regression ofy 0 XonZandW. Note, that one tests all coefficients simultaneously (as a set) in a case of more than one confidence setOne can construct a confidence set robust towards weak instruments based on theAR test by inverting it.

10 That is, by finding all which are not rejected by the data. In this case, it is theset :Conf. set ={ 0:AR( 0)< 2k,1 }.The nice thingabout this procedure is that solving for the confidence set is equivalent to solving a quadraticinequality. This confidence set can be empty with positive probability (caution!).LM testThe LM test formula can be found in Kleibergen (2002). It has a 21distribution irrespective ofthe strength of the instruments . The problem with this test, though, is that it has non-monotonic powerand tends to produce wider confidence sets than the CLR test described testsThe idea comes from Moreira (2003). He suggested that one consider any test statisticconditional on a sufficient statistic for be calledQT. By definition of sufficient statistic, the conditional(onQT) distribution of any variable does not depend on . So, instead of using fixed critical values, onewould usecritical values depending on realization ofQT(that is, random)q1 (QT).