Specification Test for Instrumental Variables Regression with Many Instruments

Transcription

1 Specification Test for Instrumental Variables Regression with Many Instruments Yoonseok Lee and Ryo Okui April 009 Preliminary; comments are welcome Abstract This paper considers specification testing for instrumental variables estimation in the presence of many instruments. The test is similar to the overidentifying restrictions test of Sargan 958) but the test statistic asymptotically follows the standard normal distribution under the null hypothesis when the number of instruments is proportional to the sample size. It turns out that the new test is equivalent to the test proposed by Hahn and Hausman 00) for the scalar endogenous regressor case. Keywords and phrases: instrumental variables estimation; many instruments; overidentifying restrictions test; specification test. JEL classification: C; C3. Introduction Instrumental variables estimation is an important tool for economic analysis. However, it has been observed that conventional asymptotic theory often provides a poor approximation to the finite sample distribution of estimators or test statistics for instrumental variables estimation. Many studies document this problem in the presence of weak instruments see, e.g., Staiger and Stock Valuable comments are obtained from Jinyong Hahn, Han Hong, and the participants of the Kansai Econometric Society Kobe meeting. The usual disclaimer applies. Department of Economics, University of Michigan, 6 Tappan Street, 365C Lorch Hall, Ann Arbor, MI , USA. yoolee@umich.edu Department of Economics, Hong Kong University of Science and Technology, Clear Water Bay, Kowloon, Hong Kong. okui@ust.hk

2 997) and Stock and Wright 000)) and in the presence of many instruments see, e.g., Kunitomo 980), Morimune 983) and Bekker 994)). It is also well known that asymptotic approximation based on many instruments provides more accurate approximation in many situations see e.g., Hansen, Hausman and ewey 006)). This paper considers specification tests for instrumental variables estimation when the number of instruments is increasing with the same size viz., Bekker asymptotics). In particular, we consider the Sargan test Sargan, 958) that examines the correlation between the residuals from the instrumental variables estimation and the instruments. We derive the limiting behavior of the Sargan test statistics using many instruments asymptotics, under which the number of instruments is proportional to the sample size. Based on this result, a new test is constructed by modifying the Sargan test so that the test statistic asymptotically follows the standard normal distribution under the null hypothesis even under the many instruments asymptotics. Kunitomo, Morimune and Tsukuda 983) is closely related to the current discussion. They derive higher order asymptotic approximation to the distribution of the overidentifying restrictions test. However, the number of instruments is fixed in their paper. We show that the proposed test is equivalent to the test by Hahn and Hausman 00) when the dimension of the parameter is one. Hahn and Hausman 00) observe that an instrumental variables estimator has the same probability limit as the inverse of the instrumental variable estimator from the reverse regression does, under the standard asymptotics i.e., with fixed number of instruments); these two estimators, however, have different limits when conventional asymptotic is not adequate. The Hahn and Hausman test is based on the difference between these two instrumental variables estimators, and it is claimed that this test can be used to check whether conventional asymptotic results are reliable. We show that this test is in fact a test of overidentifying restrictions. Our new interpretation of the Hahn and Hausman test is useful because it overcomes several limitations of the Hahn and Hausman test and it provides us some guideline on how to extend the Hahn and Hausman test to more general settings. For example, we can easily handle cases with more than two endogenous variables in our framework. Our result also demonstrates that, while the Hahn-Hausman test involves the inverse of the estimate from the reverse regression so it is difficult to interpret the test when the regression parameter is zero, the case with zero parameter value is not a problem because the test is essetially an overidentifying restrictions test. Moreover, In fact, a close inspection of the formula for the Hahn-Hausman test reveals that the test statistics is well defined even if the parameter is zero.

3 the Sargan test is a version of the J-test by Hansen 98). Hence, it would be easy to consider an extension of our new test statistics to other models such as moment-condition-based nonlinear models, whereas the idea of using reverse regression in Hahn and Hausman 00) seems to be limited to the linear regression models and thus difficult to be further generalized. The remainder of this paper is organized as follows. Section describes the basic framework. Section 3 proposes new specification tests for instrumental variables estimation based on the Sargan test under many instruments asymptotics. Section 4 establishes the equivalence between our new test and the test of Hahn and Hausman 00). Section 5 concludes the article with several important remarks. All the mathematical proofs are provided in the Appendix. Model We consider a linear instrumental variable regression model given by y i = X iδ + u i for i =,,,, where y i is the scalar outcome variable and X i is the m vector of regressors that is possibly correlated with an unobserved error u i. We assume a K vector of instruments, Z i, which we treat as deterministic. The results hold when Z i is random provided that all the assumptions given below are stated conditional on Z = Z,, Z ). We also let P = Z Z Z) Z. Throughout the paper, we consider the asymptotic sequence under which both the sample size,, and the number of instruments, K, tend to infinity i.e.,, K ) with satisfying α K α for some 0 α <. We further assume that X i = Π Z i + V i, ) where Π is the K m matrix of nuisance parameters whose value might depend on as well as K. The unobserved random variable V i is assumed to be uncorrelated with Z. Under the null hypothesis of instrument validity, it is also assumed that Eu i Z i ) = 0 for all i. For the independently and identically distributed vector of unobservables ε i = u i, V i ), we define V ar ε i ) Σ = σ u σ V u σ V u Σ V ) 3

4 conformably as u i, V i ), where σ V u 0 so that X i is correlated with u i through the correlation between u i and V i. We precisely make the following assumptions. Assumption. i) K/ = α = α + o /) for some 0 < α <, where, K. ii) Z and Π are of full column rank. iii) ε i = u i, V i ) are independently and identically distributed for i =,,,, with mean zero and positive definite variance matrix Σ given in ). The fourth moment of ε i exists. iv) Π Z ZΠ/ Θ as, K, where Θ is positive definite. v) sup i Z i π j < for all j =,, m, where π j is the jth column of Π. vi) sup sup j i= P ij <, where P ij is the i, j)th element of P. vii) i= P ii α )/α ) converges as, K. Assumption is similar to that imposed in van Hasselt 008, Assumptions, and 4). assumption guarantees that the Sargan-type test statistics in the next section has a well-defined asymptotic distribution under many instruments asymptotics. We note that the normality of the unobservables is not assumed here but we implicitly assume homoskedasticity. The full-rankness of Π rules out under-identification. Assumptions iv) and v) imply that the information accumulation from the new instruments addition is limited and thus bounded even with K. The following conditions are needed to show the consistency of the asymptotic variance estimator. Assumption. i) X i and u i have finite eighth moments. ii) sup sup i P ii <. This 3 Sargan-type Specification Tests We let ˆδ sls be the two stage least squares SLS) estimator given by ˆδ sls = X P X ) X P y, where X = X,, X ) and y = y,, y ). The Sargan test statistic Sargan, 958) is then defined as S ˆδsls ) = û P û ˆσ u, 3) where û = y X ˆδ sls and ˆσ u = û û/. It is well known that the standard asymptotic theory i.e., when K is fixed) gives S ˆδsls ) d χ K m as. 4) Conventional asymptotics, however, may not provide an accurate approximation to the finite sample distribution, particularly when the number of instruments is large. In this section, we instead 4

5 consider higher order approximation based on the many instruments asymptotics, which provides more accurate finite sample results e.g., Bekker, 994). In particular, we develop specification tests similar to the Sargan test, which is suitable under, K, and we consider their asymptotic behavior. Based on Bekker 994), since û = I XX P X) X P )u for the -dimensional identity matrix I, we can show that û P û = u P u u P X X P X ) X P u p ασ u α σ V uθ + ασ V ) σ V u, 5) as, K. ote that the probability limit 5) is simply zero when α = 0, which includes the traditional asymptotic result with but K fixed. Apparently, the probability limit 5) can be consistently estimated by where ˆb = α y X ˆδ bsls ) y X ˆδ bsls ) ) y X ˆδ bsls ) P X X P X X P y X ˆδ bsls ), ˆδ bsls = X P α I) X ) X P α I) y 6) is the bias corrected SLS estimator e.g., agar, 959; Donald and ewey, 00; Hahn and Hausman, 00). Using similar techniques as Bekker 994), Hahn and Hausman 00) and van Hasselt 008), we can derive that ) α û P û ˆb d 0, w) as, K, 7) where w = α)σ 4 u + lim ) Pii α)/α Eu 4 i ) 3σu). 4 i= Based on 7), we define the t-test statistic given by t = ˆd ŵ, 8) where ˆd = ) α y X ˆδ sls ) P y X ˆδ sls ) ˆb, ) ŵ = α ) y X ˆδ bsls ) y X ˆδ bsls ) 9) + ) Pii α)/α i= ) ) y i X iˆδ bsls ) 4 3 y X ˆδ bsls ) y X ˆδ bsls ). i= 5

6 All technical details are provided in the Appendix. We note that t in 8) is nothing but a properly standardized quadratic form û P û and thus has very a similar structure with the standard Sargan test statistic 3). The modification is based on the nonstandard asymptotics with many instruments. The following theorem shows that the asymptotic null distribution of the test statistic t, which can be understood as a Sargan-type or a modified Sargan) test statistic, is standard normal. The proof is available in the Appendix. Theorem. Suppose that Assumptions and hold. Under the null hypothesis of the correct specification, t d 0, ) as, K. Under the many instruments asymptotics, this result shows that the properly standardized quadratic form û P û follows asymptotic normal distribution. Therefore, we can expect that the standard Chi-square approximation as 4) does poor job for the Sargan-type) model specification tests particularly when K is large relative to. Alternatively, we can consider the Sargan-type test based on the bias corrected SLS estimator ˆδ bsls. By construction, we have y X ˆδ bsls ) P y X ˆδ bsls ) p ασ u as, K. Therefore, similarly as 7), we can show that ) α y X ˆδ bsls ) P α I)y X ˆδ bsls ) d 0, w) 0) as, K, based on which we can obtain the t-test statistics as t = ˆd ŵ, ) where ˆd = ) α y X ˆδ bsls ) P α I)y X ˆδ bsls ). The following lemma states that these two test statistics, t and t, are numerically equivalent, in fact. Lemma. t = t. A problem may be that we do not know when we should use the Chi-square approximation and when we should use the standard normal approximation. Calhoun 008) considers a similar problem in the context of the F test in linear regressions with many regressors. 6

7 The proof is available in Appendix. Basically, because the SLS estimator is biased in the presence of many instruments, bias correction is necessary to construct the overidentifying restrictions test statistics. Roughly speaking, Lemma demonstrates that, for the SLS estimators and the overidentifying restrictions test statistics based on them, bias correction in the stage of estimation is equivalent to bias correction when constructing the test statistics. Using Lemma and Theorem, we can also conclude that t d 0, ) as, K under Assumptions and. We finally note that the asymptotic variance can be estimated using simpler formulas than 9) if we are willing to make an additional assumption on the distribution of u i. For example, if we further assume that u i is normally distributed and thus Eu 4 i ) = 3σ4 u), then the asymptotic variance w can be estimated by ) w = α ) y X ˆδ bsls ) y X ˆδ bsls ). ) 4 Comparing with the Hahn-Hausman Test In this section, we consider the case that the dimension of X i and δ) is one and the error term u i is normally distributed. Under this particular case, we can show that the modified Sargan test statistics, t and t, are numerically equivalent to the test statistic suggested by Hahn and Hausman 00). The Hahn-Hausman test is based on the difference between the instrumental variables estimator of δ and the inverse of the estimator that uses the same set of instruments but for which the roles of the dependent variable and the regressor are reversed. The basic idea is that the SLS estimator of X on y i.e., the reverse regression) using Z as instruments is asymptotically equivalent to /ˆδ sls when the standard asymptotics with K fixed) is adequate. When the conventional asymptotics does not provide a good approximation e.g., K ), however, these two estimators converge to two different limits. Using the bias-corrected SLS estimator of δ in 6), we can define the difference as X P α I)y X P α I)X y P α I)y X P α I)y. 3) Since y X ˆδ bsls ) P α I)X = 0 by construction, however, the difference 3) can be rewritten 7

8 as ˆδ bsls y X ˆδ bsls + X ˆδ bsls ) P α I)y X P α I)y = y X ˆδ bsls ) P α I)y X P α I)y = y X ˆδ bsls ) P α I)y X ˆδ bsls ) X. 4) P α I)y ote that y X ˆδ bsls ) P y X ˆδ bsls ) is a quadratic form of the sample covariance between the regression residual and the instruments, on which Sargan s overidentifying restrictions test 3) is based. As shown in the previous section, we can understand α y X ˆδ bsls ) y X ˆδ bsls ) as the term that demeans the Sargan statistic. This basic result shows that both the Hahn-Hausman test and the Sargan test are based on the asymptotic behavior of the same quantity. More precisely, the Hahn-Hausman test statistic based on the bias-corrected SLS estimator is defined as 3 m = K y X ˆδ bsls ) y X ˆδ ) bsls ) ) K ˆδ bsls X P K K I P ))X X ) P α I)y X P α I)X y P α I)y X P α I)y y X ˆδ bsls ) y X ˆδ ) bsls ) = α ) ˆδ bsls X P α I)X) α X ) P α I)y X P α I)X y P α I)y X, P α I)y where the term in the square root is the standard error of the difference. / / In comparison, the modified Sargan test t in ) with w in ), which reflects the many instruments asymptotics and normality of the error, is given by t = { } S ˆδ bsls ) α α α ) 5) = α ) y X ˆδ bsls ) y X ˆδ ) ) / bsls ) α y X ˆδ ) bsls P α I) y X ˆδ ) bsls, 3 In addition to m, Hahn and Hausman 00) also discuss a different statistic m which is based on the LIML estimator. Theorem 4.3 of Hahn and Hausman 00) shows that, however, m and m are asymptotically equivalent. Therefore, we can focus our attention to m in this discussion. 8

9 where S ˆδ bsls ) = {y X ˆδ bsls ) P y X ˆδ bsls )}/{y X ˆδ bsls ) y X ˆδ bsls )/} is the standard Sargan statistic based on the bias-corrected SLS estimator. The following theorem shows that the m test of Hahn and Hausman 00) is equivalent to the test based on t in 5). Theorem. It follows that m argument. = t sgn [ X P α I)y], where sgn[ ] gives the sign of its This result shows that the test of Hahn and Hausman 00) can be understood as a modification of the Sargan s overidentifying restrictions test reflecting many instruments asymptotics up to the scalar multiplication of sgn [ X P α I)y]). Hahn and Hausman 00) document good finite sample properties of their tests compared with the Sargan test. Our result shows that this good performance comes from that many instruments asymptotics provides better approximation, not from that the test statistics are different from Sargan s. Another implication is that the possibility of the coefficient being zero is no longer a problem in using the Hahn-Hausman type test because we can reformulate it as the modified Sargan test. See, for example, Theorem 4. of Hahn and Hausman 00). The coefficient appears in the denominator of the asymptotic variance, which makes the test difficult to interpret when the coefficient is indeed zero.) 5 Discussions This paper develops a new specification test for instrumental variables estimation. We examine the asymptotic distribution of the Sargan test statistics under many instruments asymptotic and modify the Sargan test such that it asymptotically follows the standard normal distribution under the null hypothesis when the number of instruments is proportional to the sample size. We also show that this new test is equivalent to the test developed by Hahn and Hausman 00) for the scalar endogenous regressor case. Our equivalence result is useful when we consider an extension of the Hahn-Hausman test to more general settings. For example, Section 5 of Hahn and Hausman 00) considers the case in which there are two endogenous regressors in the model. Unfortunately, the test statistics given there is extremely complicated. On the other hand, it is straightforward to consider models with multiple endogenous regressors in our framework. Furthermore, we can also consider extensions to more general nonlinear moment restriction models, whereas it is difficult to extend the idea of using reverse equation to general moment restriction models. The Sargan test is a version of the J-test by Hansen 98) and it suggests that we may obtain a better test by examining the properties of the J-test in the presence of many moment 9

10 conditions. We note that ewey and Windmeijer 008) provide an asymptotic result for the J-test under many weak moments asymptotics. It is also interesting to consider the properties of the Sargan test under alternative asymptotic sequences. Hausman, Stock and Yogo 005) examines the performance of the Hahn-Hausman test in the presence of weak instruments. They find that the Hahn-Hausman test does not have a strong power in detecting the presence of weak or irrelevant instruments. This result also applies to our test because of the equivalence result Theorem ). On the other hand, this finding is natural from our point of view. The Hahn-Hausman test is essentially testing overidentifying restrictions and does not examine the strength of instruments. Another interesting extension is to derive the asymptotic distribution of the Sargan test under many weak instruments asymptotics see, e.g., Chao and Swanson 005)). A Appendix: Mathematical Proofs A. Proof of Lemma We show the equivalence between t and t. Since the denominators are the same, it is sufficient to show that d = d. We note that y X ˆδ bsls ) P α I)y X ˆδ bsls ) = y X ˆδ bsls ) P y X ˆδ bsls ) α y X ˆδ bsls ) y X ˆδ bsls ) = y X ˆδ sls X ˆδ bsls + X ˆδ) P y X ˆδ sls X ˆδ bsls + X ˆδ sls ) α y X ˆδ bsls ) y X ˆδ bsls ) = y X ˆδ sls ) P y X ˆδ sls ) α y X ˆδ bsls ) y X ˆδ bsls ) +ˆδ sls ˆδ bsls ) X P Xˆδ sls ˆδ bsls ), where the last equality follows because y X ˆδ) P X = 0. ow we have Therefore we have ˆδ sls ˆδ bsls = X P X) X P y ˆδ bsls = X P X) X P y X ˆδ bsls ). y X ˆδ bsls ) P α I)y X ˆδ bsls ) = y X ˆδ sls ) P y X ˆδ sls ) α y X ˆδ bsls ) y X ˆδ bsls ) + y X ˆδ bsls ) P XX P X) X P y X ˆδ bsls ) = y X ˆδ sls ) P y X ˆδ sls ) ˆB. It follows therefore that d = d. A. Proof of Theorem We first present two technical lemmas that are used to show the theorem. 0

11 Lemma A.. Under Assumption, we have u P α I)u d 0, w), A.6) α u P α I)X = O p ), A.7) X P α I)X p α)θ. A.8) Proof of Lemma A. We use Theorem of van Hasselt 008) to show A.6). The matrices U, M, V and C in Theorem of van Hasselt 008) are u, 0, u, P α I)/ α in our case, respectively. The conditions for Theorem of of van Hasselt 008) is summarized in Assumption 3 in van Hasselt 008). Assumption.ii) implies Assumption 3i) in van Hasselt 008), Assumptions vi) and vii) imply Assumption 3iii) in van Hasselt 008). Assumptions 3ii)and 3iv) in van Hasselt 008) are automatically satisfied, Therefore, under Assumption, the conditions for Theorem of van Hasselt 008) are satisfied. It follows that u P αi)u/ α d 0, w). We also use Theorem of van Hasselt 008) to show A.7) and A.8) Let U = u, X ). We consider the asymptotic distribution of U P α I)U. Theorem of van Hasselt 008) is used to derive the results. The matrices U, M, V and C in Theorem of van Hasselt 008) are U, 0, π Z ), u, V ), P α I) in our case, respectively. Assumption ii) implies Assumption 3i) in van Hasselt 008), Assumption iv) implies Assumption 3ii) in van Hasselt 008), Assumptions v) and vi) imply Assumption 3iii) in van Hasselt 008) and Assumptions iii) and vii) imply Assumption 3iv) in van Hasselt 008). Therefore, under Assumption, the conditions for Theorem of van Hasselt 008) are satisfied. It follows that U P α I)U EU P α I)U) ) = O p ). Since Eu P α I)X) = 0, A.7) holds. Since EX P α I)X) = α )π Z Zπ, A.8) holds. Lemma A.3. Under Assumptions and, we have p ) α α) y X ˆδ bsls ) y X ˆδ bsls ) +4 α) Z iˆπp ii α) y i X iˆδ bsls ) 3 i= i= ) + Pii/ α ) ) y i X iˆδ bsls ) 4 3 y X ˆδ bsls ) y X ˆδ bsls ) i= i= α α)σϵ α) lim + lim i= Z iπp ii α)eϵ 3 i ) i= P ii α ) Eϵ 4 i ) 3σ 4 ϵ ). Proof of Lemma A.3 We need to show y X ˆδ bsls ) y X ˆδ bsls ) p σ u,

12 and y i X iˆδ bsls ) 4 p Eu 4 i ). i= We have y X ˆδ bsls ) y X ˆδ bsls ) = δ ˆδ bsls ) X Xδ ˆδ bsls ) + δ ˆδ bsls ) X u + u Xδ ˆδ bsls ) + u u. Theorem 3 of van Hasselt 008) says that δ ˆδ bsls p 0. Under Assumption, we have X X/ = O p ) and X u/ = O p ) and u u/ p σ u. Thus, it holds that y X ˆδ bsls ) y X ˆδ bsls ) p σ u. We now consider the estimation of the fourth moment of u i. y i X iˆδ bsls ) 4 = i= i= X iδ ˆδ ) 4 4 bsls ) + i= i= X iδ ˆδ bsls )) u i i= X iδ ˆδ ) bsls ) u 3 i + X iδ ˆδ bsls )) 3 ui u 4 i. Under Assumption, we have δ ˆδ bsls p 0 by Theorem 3 of van Hasselt 008). Therefore, assuming Assumption. in addition, we have i= y i X iˆδ bsls ) 4 = i= i= u 4 i + o p ) p Eu 4 i ). The equality follows because X iδ ˆδ ) 4 bsls ) X i 4 δ ˆδ bsls 4 = O p )o p ) = o p ) i= by the existence of the eighth order moment of X i Assumption.) and δ ˆδ bsls p 0, where is the Euclidean norm, and a similar argument can show that i= X i δ ˆδ bsls )) 3 u i / = o p ), i= X i δ ˆδ bsls )) u i / = o p) and i= X i δ ˆδ bsls ))u 3 i / = o p). Proof of Theorem i= oting that t = t, we consider t only. We observe that y X ˆδ bsls ) P α I)y X ˆδ bsls ) = ˆδ bsls δ) X P α I)Xˆδ bsls δ) ˆδ bsls δ) X P α I)u u P α I)Xˆδ bsls δ) + u P α I)u = u P α I)X X P α I)X ) X P α I)u u P α I)X X P α I)X ) X P α I)u u P α I)X X P α I)X ) X P α I)u + u P α I)u = u P α I)X X P α I)X ) X P α I)u + u P α I)u.

13 Therefore we have ) α y X ˆδ bsls ) P α I)y X ˆδ bsls ) = ) u P α I)X K X P α I)X X P α I)u + u P α I)u. α Lemma A. implies that ) α y X ˆδ bsls ) P α I)y X ˆδ bsls ) = u P α I)u + o p ) d 0, w). α Lemma A.3 states that It follows that ŵ p w. t d 0, ). A.3 Proof of Theorem We observe that = = m α α ) y X ˆδ bsls ) y X ˆδ / bsls )) X P α I)y ˆδ bsls X P α I)X) X ) X P α I)y) P α I)y α w X P α I)X y P α I)y X. P α I)y ow, we have X P α I)y X P α I)X y P α I)y X P α I)y X P α I)X y P α I)y X P α I)y = ˆδ bsls y X ˆδ bsls + X ˆδ bsls ) P α I)y X P α I)y = y X ˆδ bsls ) P α I)y X. P α I)y Since y X ˆδ bsls ) P α I)X = 0 by the definition of ˆδ, it follows that Thus, we have m = y X ˆδ bsls ) P α I)y = y X ˆδ bsls ) P α I)y X ˆδ bsls + X ˆδ bsls ) = y X ˆδ bsls ) P α I)y X ˆδ bsls ). sgn[ X P α I)y]y X ˆδ bsls ) P α I)y X ˆδ bsls ) α w ) ˆd = sgn[ X P α I)y]. w 3

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