The impact of a Hausman pretest on the size of a hypothesis test - PowerPoint PPT Presentation

The impact of a Hausman pretest on the size of a hypothesis test Patrik Guggenberger Department of Economics UCLA WORKSHOP ON CURRENT TRENDS AND CHALLENGES IN MODEL SELECTION AND RELATED AREAS Vienna 2008 Supported by NSF grant SES-0748922

Outline of talk � Applications and “justi…cation” of Hausman (1978, Ecta) pretests � Summary of main results � Finite sample results � Asymptotic size formula, Andrews and Guggenberger (2005)

Hausman (1978) (pre)tests: applications General setup: Goal of interest � Under H 0 , b � 1 & b � 2 consistent and asy normal and b � 1 has some “favorable” properties When H 0 fails, b � 1 is inconsistent, b � 2 remains consistent ! Test H 0 based on quadratic form of b � 1 � b � 2 ! Use b � 1 or b � 2

Hausman (pre)tests: applications � Among the most often used pretests in Economics; � 500 Jstor–citations � About 75 papers in the AER (25 in 2000–2004 alone) use a Hausman test � Discussed in most Econometric textbooks � On syllabus of most graduate Econometric courses � YET: no discussion anywhere of impact

Hausman (pre)tests: applications (cont’) 1) linear instrumental variables (IV) model: y 1 = y 2 � + X� + u Focus of interest: inference on � 2 R Problem: y 2 ;i and u i may be correlated X i 2 R k 1 assumed “exogenous” , project out X; M X = I � X ( X 0 X ) � 1 X 0 y 1 = y 2 � + u

Hausman (pre)tests: applications (cont’) 1) linear instrumental variables (IV) model: y 1 = y 2 � + u; y 2 = Z� + v: Z i 2 R k 2 are “instrumental variables” EZ i u i = 0 ; “exogenous” EZ i y 2 ;i = ( EZ i Z 0 i ) � 6 = 0 ; “relevant”

Hausman (pre)tests: applications (cont’) 1) linear IV model: y 1 = y 2 � + u; y 2 = Z� + v: 2SLS estimation: b � 2 SLS = ( y 0 2 P Z y 2 ) � 1 y 0 2 P Z y 1 y 2 = P Z y 2 for P Z = Z ( Z 0 Z ) � 1 Z 0 1. Regress y 2 on Z to get b 2. Regress y 1 on b y 2

Hausman (pre)tests: applications (cont’) 1) linear IV model: y 1 = y 2 � + u; y 2 = Z� + v: If y 2 is exogenous i.e. corr ( u i ; v i ) = 0 ; would like to use t-test based on OLS estimator Problem: y 2 may be endogenous. In the latter case, can use t-test based on 2SLS estimator

Hausman (pre)tests: applications (cont’) Use the Hausman pretest based on H n = n ( b � 2 SLS � b � OLS ) 2 V 2 SLS � b b V OLS to test pretest null hypothesis: H 0 : y 2 is exogenous. Pretest not rejected: continue with t-test based on OLS estimator Pretest rejected: continue with t-test based on 2SLS estimator

Hausman (pre)tests: applications (cont’) 2) panel data model: y it = x it � + c i + u it Under E ( c i j x i ) = 0 where x i = ( x i 1 ; :::; x iT ) , would like to use random e¤ects based inference; problem if individual e¤ect c i and regressor x it are “related”. Use a Hausman pretest to test the latter. 3) test of overidentifying restrictions: Eg ( z i ; � ) = 0 ; g = ( g 0 1 ; g 0 2 ) 0

This talk: linear IV model y 1 = y 2 � + u; y 2 = Z� + v: Test f H 0 : � = � 0 using two-stage procedure: Stage 1: Hausman pretest based on H n Stage 2: Use T 2 SLS or T OLS depending on outcome of Stage 1.

“ Justi…cation ” of Hausman pretests: linear IV case Two stage test: reject f H 0 : � = � 0 when 1( H n > � 2 1 � � & T 2 SLS > z 1 � � ) + 1( H n � � 2 1 � � & T OLS > z 1 � � ) = 1 � Under a false pretest hypothesis, the Hausman statistic H n diverges to in…nity � Under a true pretest hypothesis, the Hausman statistic H n converges to a � 2 limit distribution (under certain assumptions...) � Even a full justi…cation for the control of the pointwise null rejection prob- ability of the two stage test requires some work; let alone of its size!

Summary of results Interested in the asymptotic size properties of a two stage test : Stage 1: Hausman (1978) pretest (nominal size � ) Stage 2: simple hypothesis test f H 0 : � = � 0 (nominal size � ) � The asymptotic size of the two stage test is 1 � Conditional on the Hausman pretest not rejecting the null, the size is 1 � Theoretical results well re‡ected in …nite sample simulations

Finite sample evidence: Linear IV case � Use grid for � = Corr ( u i ; v i ) and � 2 = n� 0 EZ i Z 0 i �=Ev 2 i “concentration parameter” � Hansen, Hausman, and Newey (2004) Five years of AER, JPE, and QJE # of papers Q 10 Q 25 Q 50 Q 75 Q 90 � 2 28 8 : 95 12 : 7 23 : 6 105 588 � 22 : 022 : 0735 : 279 : 466 : 555

� ( u i ; v i ; Z i ) i.i.d. normal, zero mean, unit variances, EZ i Z 0 i = I k 2 ; Z i independent of u i and v i � = � 0 (1 ; :::; 1) 0 2 R k 2 for � 0 2 R n = 1000 ; k 2 = 5 instruments nominal sizes of pretest & second stage test: � = � = : 05

Finite Sample Null Rejection Probabilities of Symmetric Two-stage Test and 2SLS Based t-Test � 2 n � 0 .05 .1 .2 .3 .4 .5 . 0 5.1;0.0 34.9;0.0 88.5;0.1 99.9;0.4 99.9;2.4 99.9;8.5 99.9;22.2 99.9;4 13 6.7;0.7 35.2;0.8 86.8;1.3 95.4;3.1 91.0;5.9 83.8;10.0 71.6;14.9 53.8;2 50 7.8;3.4 34.5;3.5 81.4;3.6 77.1;4.2 50.0;5.3 21.2;6.7 8.0;8.4 7.7;10 113 7.8;4.4 32.3;4.4 74.0;4.4 51.6;4.7 15.3;5.1 5.5;5.7 5.7;6.6 6.3;7 200 7.4;4.8 29.7;4.7 65.1;4.8 29.2;4.9 5.8;5.0 5.2;5.4 5.3;5.8 5.7;6 313 7.1;4.9 27.1;4.9 55.4;4.9 15.1;4.9 5.1;5.1 5.1;5.3 5.3;5.5 5.4;5 450 6.8;5.0 24.6;5.0 46.3;4.9 8.8;5.0 5.1;5.1 5.1;5.2 5.2;5.4 5.3;5 613 6.5;5.0 22.2;5.0 38.4;5.0 6.5;5.0 5.1;5.1 5.1;5.2 5.2;5.3 5.2;5

Finite Sample Null Rejection Probabilities of Two-stage Test and Pretest � = � = : 05 ; � 2 = 23 : 6 ; � = : 279 n k 2 Sym HPre CondlSym 100 1 62.4 15.4 73.0 1000 1 79.4 21.1 100 10000 1 79.1 21.5 100 100 5 63.2 14.0 73.0 1000 5 81.0 19.3 100 10000 5 80.7 19.7 100 100 20 66.2 10.3 73.4 1000 20 85.4 14.8 100 10000 20 84.3 15.9 100

Findings of simulation study � Find extreme size distortion of two-stage test in cases where correlation is nonzero and small, especially for small values of the concentration para- meter � On the other hand, t-test based on 2SLS has good size properties (except when � large and � 2 small) � Hausman pretest seems unable to distinguish between zero correlation and small correlation � BUT: small correlations get picked up by t-test based on OLS � The null rejection probability is not controlled uniformly in �

Asymptotic size � Asymptotic size of test: f H 0 : � = � 0 : AsySz ( � 0 ) = lim sup n !1 sup P � 0 ;� ( T n ( � 0 ) > c 1 � � ) � 2 � � Andrews and Guggenberger (2005) “Asymptotic size and a problem with subsampling and the m out of n bootstrap” Discontinuous limit distributions � f � n;h : n � 1 g denotes sequence in � s.t. n r � n;h ! h as n ! 1 for a given localization parameter; often r = 1 = 2

� H is the set of all h � Assumption: For some r > 0 ; all h 2 H; all sequences f � n;h : n � 1 g ; T n ( � 0 ) ! d J h under f � n;h : n � 1 g ; for some proper distribution J h � at discontinuity point 0: it is not enough to know that � n ! 0 to deter- mine the asymptotic distribution of the test statistic � AG (2005): if critical value c Fix (1 � � ) is used, then: AsySz ( � 0 ) = sup [1 � J h ( c Fix (1 � � ))] h 2 H

General setup � nuisance parameter � has 3 components: � = ( � 1 ; � 2 ; � 3 ) where ( � 1 ; � 2 ) 2 � 1 � � 2 � limit distribution of test stat is discontinuous in � 1 2 R p � limit dist’n is continuous in � 2 2 R q but � 2 a¤ects limit distribution � limit distribution does not depend on � 3 2 T 3 ; arbitrary space – e. g., � 3 could be an error distribution—inf dim’l � General formula: AsySz ( � 0 ) = sup [1 � J h ( c Fix (1 � � ))] ; h 2 H where H = H 1 � H 2 and H 2 = cl (� 2 )

Examples � Inference on AR parameter in AR(1) model � Inference under moment inequalities � Inference under boundary restrictions � Inference under weak identi…cation � Inference based on post-model selection (conservative and consistent)...

Two stage test with Hausman pretest: speci…cation of � y 1 = y 2 � + u; y 2 = Z� + v: � Two stage test statistic: T n ( � 0 ) = 1( H n > � 2 1 � � ) T 2 SLS + 1( H n � � 2 1 � � ) T OLS � Speci…cation of � : i ) 1 = 2 �=� v jj ; where � 1 = �; � 2 = jj ( E F Z i Z 0 � 2 v = E F v 2 i ; � = Corr F ( u i ; v i ) � Parameter spaces: � 1 = [ � 1 ; 1] and � 2 = [ �; � ] for 0 < � < � < 1 (excludes weak IV)

� point of discontinuity: � = 0

Technical step � Want to evaluate AsySz ( � 0 ) = sup [1 � J h ( c Fix (1 � � ))] h 2 H � Need to derive J h ; the limit of T n ( � 0 ) under the sequences f � n;h : n � 1 g ; T n ( � 0 ) ! d J h � Here f � n;h g has components � n;h; 1 = Corr F n ( u i ; v i ) ; � n;h; 2 = jj ( E F n Z i Z 0 i ) 1 = 2 � n = ( E F n v 2 i ) 1 = 2 jj ; n 1 = 2 � n;h; 1 ! h 1 ; � n;h; 2 ! h 2 ; and � n;h; 3 = :::

� Will not state formula; but some details: Under f � n;h g ; H n ! d � 2 1 ( h 2 1 h 2 2 ( h 2 2 + 1) � 1 ) : Therefore, H n ! d � 2 1 if h 1 = 0 :