Three Essays on Testing for Cross-Sectional Dependence and Specification in Large Panel Data Models

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1 Syracuse University SURFACE Dissertations - ALL SURFACE July 06 Three Essays on Testing for Cross-Sectional Dependence and Specification in Large Panel Data Models Bin Peng Syracuse University Follow this and additional works at: Part of the Social and Behavioral Sciences Commons Recommended Citation Peng, Bin, "Three Essays on Testing for Cross-Sectional Dependence and Specification in Large Panel Data Models" (06. Dissertations - ALL. Paper 5. This Dissertation is brought to you for free and open access by the SURFACE at SURFACE. It has been accepted for inclusion in Dissertations - ALL by an authorized administrator of SURFACE. For more information, please contact surface@syr.edu.

2 Three Essays on Testing for Cross-Sectional Dependence and Specification in Large Panel Data Models Bin Peng B.A., Huazhong University of Science & Technology, 007 M.A., Huazhong University of Science & Technology, 00 DISSERTATION Submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Economics in the Graduate School of Syracuse University. July 06

3 Copyright 06 Bin Peng All Rights Reserved

4 Acknowledgments I am using this opportunity to express my gratitude to everyone who helped me complete this dissertation at Syracuse University. I am thankful for their aspiring guidance, invaluable constructive criticism, and friendly advice. I am sincerely grateful to them for sharing their honest and illuminating views. Particularly, I would like to express my sincere gratitude to my advisors Prof. Chihwa Kao, Prof. Badi H. Baltagi and Prof. Yoonseok Lee for the continuous support, their patience, motivation, and immense knowledge. Their guidance helped me in all the time of writing this dissertation. I would also like to thank the rest of my thesis committee: Prof. William Horrace and Prof. Jan Ondrich, for their insightful comments and encouragement. Besides, I am especially grateful to Prof. Sarah Hamersma for chairing my oral defense. My sincere thanks also go to all the other faculty members, staffs of the economics department, for their input, guidance, accessibility, and helpfulness. I also would like to thank my classmates for the stimulating discussions, for the sleepless nights we were working together, and for all the fun we have had in the last five years. Last but not the least, I would like to thank my wife, my parents, my parents in law and my sister for supporting me spiritually throughout writing this thesis and my life in general. Their unconditional support, continuous encouragement, quiet patience, and unwavering love are the wealth of all my life. v

5 Table of Contents Page Essay I: On Testing for Sphericity with Non-normality in a Fixed Effects Panel Data Model.... Introduction.... The Model and Assumptions J u Test Asymptotics of the J u Test Monte Carlo Simulations Experiment Design Results Conclusion... References... 3 Tables... 5 Appendix... 9 Essay II: Testing Cross-Sectional Dependence in Large Panel Data Models with Serial Correlation Introduction Model and Tests LM and CD Tests Assumptions and Modified CD Test Statistic Asymptotics Monte Carlo Simulations Experimental Design Simulation Results Conclusions References Tables... Appendix... 7 Essay III: Tests of Specification for Large Dynamic Panel Data Models Introduction The Model and the Estimators Existing Tests of Specification Tests for Serial Correlation... 0 vi

6 3.. Tests of Overidentifying Restrictions Issues Raised by Large T Test Statistics and Asymptotics Asymptotics of the Tests for Serial Correlation Corrected Tests of Overidentifying Restrictions and Their Asymptotics Power Properties Power Analysis under the Local AR(q and MA(q Errors Slope Heterogeneity and Error Cross-Sectional Dependence Choice of p and κ Monte Carlo Simulations Experimental Design Simulation Results Concluding Remarks References... 6 Tables Appendix vii

7 List of Tables Page Essay I: On Testing for Sphericity with Non-normality in a Fixed Effects Panel Data Model Table. Size of Tests... 5 Table. Size Adjusted Power of Tests: Factor Model... 6 Table 3. Size Adjusted Power of Tests: SAR( Model... 7 Essay II: Testing Cross-Sectional Dependence in Large Panel Data Models with Serial Correlation Table. Size of Tests with IID Errors over Time... Table. Size of Tests with MA( Errors... Table 3. Size of Tests with AR( Errors... 3 Table. Size of Tests with ARMA(, Errors... Table 5. Size Adjusted Power of CD R : Factor Model... 5 Table 6. Size Adjusted Power of CD R : SAR( Model... 6 Essay III: Tests of Specification for Large Dynamic Panel Data Models Table. Size of Tests Table. Power (Size Adjusted Power of Tests: MA( Table 3. Power (Size Adjusted Power of Tests: MA( Table. Power (Size Adjusted Power of Tests: MA( Table 5. Power (Size Adjusted Power of Tests: MA( Table 6. Power (Size Adjusted Power of Tests: AR( Table 7. Power (Size Adjusted Power of Tests: AR(... 7 Table 8. Power (Size Adjusted Power of Tests: AR(... 7 Table 9. Power (Size Adjusted Power of Tests: AR( Table 0. Power (Size Adjusted Power of Tests: Heterogeneous Slopes... 7 Table. Power (Size Adjusted Power of Tests: Heterogeneous Slopes Table. Power (Size Adjusted Power of Tests: Cross-Sectional Dependence (Factor.. 76 Table 3. Power (Size Adjusted Power of Tests: Cross-Sectional Dependence (Factor.. 77 viii

8 Essay I: On Testing for Sphericity with Non-normality in a Fixed Effects Panel Data Model

9 Introduction This paper proposes testing the null of sphericity of the variance-covariance matrix in a fixed effects panel data model which does not require the normality assumption on the disturbances. This builds on the paper by Chen et al.(00 who use U-statistics to test for sphericity of the variance-covariance matrix in statistics. The null of sphericity means that the variance-covariance matrix is proportional to the identity matrix. Rejecting the null means having cross-sectional dependence among the individual units of observation or heteroskedasticity or both. In empirical economic studies, individuals are affected by common shocks. For example, investors decisions may be influenced by the way they interact with each other and also by common macro-economic shocks or public policies. These potentially cause cross-sectional dependence among the units. In statistics, the n n sample covariance matrix S n is widely used for tests of sphericity since it is a consistent estimator for the variance-covariance matrix Σ n. One could either use the likelihood ratio test, see Anderson (003, or test the Frobenius norm of the difference between S n and Σ n, see John (97,97. However, with panel data sets where n the number of individuals is larger than the time series dimension of the data T, the sample covariance matrix becomes singular. This causes problems for the likelihood ratio test which is based on the inverse of S n. Even when n is smaller than T, the sample covariance matrix S n is ill-conditioned as shown in the Random Matrix Theory (RMT literature. In fact, the eigenvalues of the sample covariance matrix S n are no longer consistent for their population counterpart, see Johnstone (00. Ledoit and Wolf (00 show that the scaled Frobenius norm of S n does not converge to that of Σ n with n/t c (0,. As a result, John s test, see John (97,97, is no longer applicable. Hence, Ledoit and Wolf (00 propose a new test for the null of sphericity which could be applied even when n is relatively as large as T. However, these statistical tests for raw data are not directly applicable to testing sphericity in panel data regressions since the disturbances are unobservable. Baltagi et al. (0 extend the Ledoit and Wolf (00 s John test to the fixed effects panel data model and correct for

10 the bias due to substituting within residuals for the actual disturbances. However, their test relies on the normality assumption and their simulation results show that the test has size distortion under non-normality of the disturbances. To account for the possible non-normality of the disturbances as well as the large n, small T issues in testing the null of sphericity, Chen et al.(00 propose a modified John test by constructing U-statistics of observable samples for estimating trσ n and trσ n. Based on their work, this paper proposes a new test for the null of sphericity of the disturbances in a fixed effects regression panel data model. This test does not require the assumption of normality of the disturbances, and can be applied to the case where n is larger than T. The limiting distribution of this test statistic under the null is derived. Also, its finite sample properties are studied using Monte Carlo simulations. The paper is organized as follows. Section specifies the fixed effects panel data regression model and the assumptions required. Section 3 introduces the test statistic. Section derives the limiting distribution of this test statistic under the null and discusses its power properties. Section 5 reports the results of Monte Carlo simulations, while Section 6 concludes. All the proofs and technical details can be found in an Appendix available upon request from the authors. Notation: B (tr(b B / is the Frobenius norm of a matrix B or the Euclidean norm of a vector B, and tr(b is the trace of B. d denotes convergence in distribution and p denotes convergence in probability. For two matrices B (b ij and C (c ij, we define B C (b ij c ij. The Model and Assumptions Consider the following fixed effects panel data regression model y it α + x itβ + µ i + v it, for i,,..., n; t,,..., T, (. 3

11 where i indexes the cross-sectional dimension and t indexes the time series dimension. y it is the dependent variable, x it denotes the k vector of exogenous regressors, and β is the corresponding k vector of parameters. µ i denotes the time-invariant individual effects which can be fixed or random and could be correlated with the regressors. Define the vector of disturbances v t (v t,..., v nt and its corresponding variance-covariance matrix Σ n. The null hypothesis of interest is sphericity: H 0 : Σ n σ vi n vs H : Σ n σ vi n. (. The alternative hypothesis allows cross-sectional dependence or heteroskedasticity or both. For the panel data regression model, v it is unobserved, and the test statistic is based upon consistent estimates of variance-covariance matrix, denoted by S n or its correlation coefficients matrix counterpart, see Breusch and Pagan (980. Baltagi et al. (0 extend the Ledoit and Wolf (00 test to a fixed effects panel data model with large n and large T. They show that the noise resulting from using within residuals rather than the actual disturbances accumulates and causes bias for the proposed test statistic. However, their simulations show that their test is oversized under non-normality of the disturbances. This paper extends Chen et al. (00 to test the null of sphericity of the variance-covariance matrix of the disturbances in a fixed effects panel data regression model without assuming normality of the disturbances. We use the within residuals which are given by ˆv it ỹ it x it β v it v i. x it ( β β, (.3 where x it x it x i. and x i. T t x it. Similarly, ỹ it y it ȳ i., ȳ i. T t y it, ( T and v i. T T t v it. The within estimator of β is given by β n t i x it x it ( T n i x itỹ it. Let ỹ t (ỹ t,..., ỹ nt, ˆv t (ˆv t,..., ˆv nt, v. ( v.,..., v n., and t x t ( x t,..., x nt. The within residuals can be rewritten in matrix form as ˆv t v t v. ( β β. To facilitate our analysis, we require the following assumptions: x t

12 Assumption The n vectors v, v,..., v T are independent and identically distributed (i.i.d. with mean vector 0 and covariance matrix Σ n ΓΓ, where Γ is an n m (m matrix, v t can be written as v t ΓZ t, where Z t (z t,..., z tm are i.i.d.random vectors with mean vector 0 and covariance matrix I m. We also assume that each v it, for i,..., n has uniformly bounded 8th moment and there exists a finite constant such that E(zl 3 +, for l,..., m. Assumption The regressors x it, i,..., n, t,..., T are independent of the idiosyncratic disturbances v it, i,..., n, t,..., T. The regressors x it have finite fourth moments: E[ x it ] K <, where K is a positive constant. Assumption 3 As (n, T, tr(σ n, tr(σ n/tr (Σ n 0. The asymptotics follow the framework employed by Chen et al. (00. Assumption 3 requires tr(σ n to grow at a slower rate than tr (Σ n. This assumption is flexible. In fact, if all the eigenvalues of Σ n are bounded away from zero and infinity, tr(σ n/tr (Σ n 0, is always true for any n as n. Moreover, this assumption allows n to be much larger than T, which is more suitable for micro-panel data. 3 J u Test For testing the null hypothesis (., the test statistic is based on the scaled distance measure between σ v Σ n and I n : U 0 n tr [S n ( n trs n I n ] ( ( n trs n n tr(s n, (3. where S n is the n n sample covariance matrix and I n is an n n identity matrix. With the normality assumption, John (97 shows that for fixed n, and as T : nt U d 0 χ n(n+/. (3. 5

13 But when n goes to infinity, the test statistic diverges. Ledoit and Wolf (00 propose a modified test statistic under the null, as (n, T and n/t c (0, : T U 0 n d N(,. (3.3 Define J 0 T U 0 n, then under the null J 0 d N(0,. However, this test cannot be used directly in a fixed effects panel data regression model. The raw data sample covariance matrix S n is replaced by its counterpart Ŝn T T t ˆv tˆv t, where ˆv t is the within residual ( given by (.3. The residual-based Û0 is defined as Û0 (Ŝ n trŝn tr n n and the corresponding residual-based J 0 test is Ĵ0 T Û0 n. Baltagi et al. (0 propose a bias correction: J BF K Ĵ0 n (T. (3. They show that in a fixed effects panel data regression, as (n, T and n/t c, J BF K d N(0, under the null. However, their result relies on the normality assumption of v t. Without the normality assumption, the bias-corrected John test is not robust, see the simulations in Baltagi et al. (0. Chen et al. (00 propose a new test statistic for the sphericity of the variance-covariance matrix of the disturbances without the normality assumption and under much relaxed conditions where n could be much larger than T. They construct the U-statistics for estimating trσ n and trσ n. Following their framework, we propose a residual-based test statistic for testing the null of sphericity described in (. in a fixed effects panel data model. Define ˆM,T T t ˆv tˆv t ; ˆM,T C T s ˆv tˆv s ; ˆM3,T C T (ˆv tˆv s ; s ˆM,T C 3 T τ τ ˆv tˆv sˆv sˆv τ ; ˆM5,T C T τ η η τ η η ˆv tˆv sˆv τ ˆv η, where C i T T!/(T i!. Also, let ˆR ˆM,T ˆM,T and ˆR ˆM 3,T ˆM,T + ˆM 5,T. 6

14 If we observe the true v t, then R, R and M j,t, for j,, 3,, 5 are obtained similarly by replacing (ˆv t, ˆv s, ˆv τ, ˆv η with (v t, v s, v τ, v η. R and R are unbiased estimators for trσ n and trσ n, respectively. The scaled distance measure between σv Σ n and I n is [ ( ] given by U T nr. Define A Γ Γ and ψ + 8 tr Σ n Σn R T T tr(σ n tr(σ n + [( ( ] tr A A A T tr(σ n tr(σ n A tr(σ n tr(σ n. Chen et al. (00 show that as (n, T : ψ [( UT + n ( ] tr (Σ n tr (Σ n d N(0,. (3.5 Let J CZZ T U T, then under the null J CZZ propose the following test statistic: d N(0,. Following this framework, we J u T ÛT T ( n ˆR ˆR. (3.6 J u is the residual-based statistic corresponding to J CZZ. There are two important issues to be considered. First, whether the residual-based ˆR and ˆR are consistent estimates for trσ n and trσ n under the null, respectively. Second, the asymptotics of the proposed test need to be derived. Both concerns are tackled in the next Section. Asymptotics of the J u Test In this Section, we prove that, under the null, n ˆR and n ˆR are consistent estimators for trσ n n σv and n trσ n σv, respectively. Next, we show J u converges to N(0, under the null and we discuss its power properties. To examine the asymptotics of J u, we rewrite it as T (ÛT U T J u J CZZ + (J u J CZZ J CZZ +. (. The first term J CZZ is asymptotically standard normal under the null. The second term J u J CZZ is the scaled difference between the residual-based ÛT and the true U T. From 7

15 Section 3, this difference can be rewritten as follows: J u J CZZ T [ ( ( ( ( ] ( ( n ˆR n R n R n ˆR n ˆR n R. (. From equation (., it is clear that this term depends upon the two differences: n ˆR n R and n ˆR n R. Their asymptotic behavior is given in the following propositions: Proposition Under Assumptions - and the null, ( ˆM ( n,t M n,t σ v T + O p ( ˆM ( n,t M n,t σ v T + O p ; (3 ˆR n R ( n O p nt. T n Proposition Under Assumptions - and the null, ( ˆM n 3,T M n 3,T + ( n ( ( O p + O T p T + O T p T ; ( ˆM ( n n,t M n T 5 n,t + σ T v + O p T n ( O p T ; (3 ˆM T n 5,T M n 5,T + n T ( O p. T nt Propositions and show that the differences n ˆR n R (n, T. Therefore, since n R T n ; n T 6 T σ v + + O p ( n T + σ v + O p ( n T ; ( n ˆR n R O p ( T + Op ( nt + p σ v and n R and n ˆR n R vanish as p σ v, we conclude that n ˆR and ˆR n are consistent estimates for σv and σv respectively. The following corollary gives these conclusions: Corollary Under Assumptions - and the null, as (n, T, ( n ˆR n ˆR p σ v. p σ v; ( Note that n ˆR is a consistent estimator of σ v under the null with large n and large T. However, n trŝ n is not consistent, see Baltagi et al. (0. Proposition 3 Under Assumptions - and the null, T(ÛT U T O p ( n +Op ( T +Op ( nt. Propositions, and 3 give the asymptotics of the bias term J u J CZZ. Compared with the statistic based on raw data, the test statistic based on the within residuals, defined ( below equation (.3, can be expressed by ˆv t v t v. x t β β. This has the additional 8

16 ( terms v. and x t β β. These two terms can be regarded as extra noise resulting from the regression. Based on equations (. and (., the extra noise T(ÛT U T depends upon ˆR n R n ( ˆM n,t M,T ( ˆM n,t M,T and ˆR n R n ( ˆM n 3,T M 3,T ( ˆM n,t M,T + ( ˆM n 5,T M 5,T. Hence the magnitude of T(ÛT U T depends upon how v. and x t( β β accumulate in ( n ˆM j,t n M j,t, for j,, 3,, 5. Note that v. is an n dimensional vector, although each element of v. is O p ( T, v. may still accumulate in the above five terms as (n, T, depending upon the relative speed of n and T. x t( β β is ( O p nt which is related to both n and T. We may expect its convergence speed nt to be fast enough so that x t( β β vanishes as (n, T. More specifically, Proposition shows the leading terms of n ˆM j,t n M j,t, for j 3,, 5 will not vanish if n T does not converge to zero. These terms are caused by the accumulation of v.. However, due to the subtraction formulation of the test statistic, the leading terms cancel each other in both ˆR n R n and ˆR ( n ( n R n. Similar cancellations occur for other terms which are O p, O T p T n ( and O p T since their expressions are exactly the same. These cancellations lead us to T ˆR n R ( n O p nt and ˆR n R ( n O ( p T + ( Op nt + Op T, and consequently nt ( T(ÛT U T O p ( + O n p( + O p T p nt. Therefore, J u J CZZ 0 as (n, T and we do not need to correct the bias in the fixed effects panel data regression model. This result is based on our detailed calculation of how v. and x t( β β are accumulating in ( ˆM n j,t M n j,t, for j,, 3,, 5 and the special formulation of J CZZ. As discussed above, the convergence of J u is given by the following theorem: Theorem Under Assumptions -3 and the null, in the fixed effects panel data regression model (., as (n, T J u d N(0,. (.3 Under the alternative, the limiting distribution of J u is the same as (3.5 if T(ÛT U T vanishes as (n, T 0. Similar to Chen et al. (00, we consider an alternative: H : Σ n ( σ i σ j ρ j i n n, where ρ (, and ρ 0. σ l var(v lt, which is uniformly 9

17 bounded away from infinity and zero, for l,..., n. Under this alternative, we can show that J u J CZZ o p (, which in turn implies that J u and J CZZ have the same power [ ( ] properties. Define δ,t tr (Σ n and δ ntr(σ n,t tr Σ n Σn tr(σ n tr(σ n. One can show that T δ,t and δ,t /(T δ,t 0 as (n, T. This satisfies the conditions of Theorem in Chen et al. (00. By using this Theorem, the corresponding power function P (J u z α Σ n (σ i σ j ρ j i n n, as (n, T, where z α is the upper quantile of N(0,. Let us consider a special case under this alternative. More specifically, assume that 0, σ i σ j σ v for any (i, j and T/n 0 as (n, T. It follows that ψ T and ( ρ J u T ρ / d N(0,. 5 Monte Carlo Simulations We conduct Monte Carlo experiments to assess the empirical size and power of the J u test proposed in this paper. We follow the design of Baltagi et al. (0 and assume homoskedasticity on the remainder error term. In this case, the J u test becomes a test for cross-sectional dependence. We also report the performance of J BF K proposed by Baltagi et al. (0 for comparison purposes. 5. Experiment Design Consider the following data-generating process: y it α + βx it + µ i + v it, i,..., n; t,..., T, (5. x it λx i,t + µ i + η it, (5. where µ i is the fixed effects and v it is the idiosyncratic error, η it i.i.d. N(φ η, σ η. The regressor x it is allowed to be correlated with the µ i s. This follows the design by Im et al. (999. 0

18 To study the power of the tests, we consider two different types of cross-sectional dependence models: a factor model and a spatial model. For the factor model, see Pesaran (00, Pesaran and Tosetti (0, Baltagi et al. (0, we assume: v it γ i f t + ɛ it, (5.3 where f t (t,..., T are the factors and γ i (i,..., n are the loadings. For the spatial model, we consider a first-order spatial autocorrelation model SAR(, see Anselin and Bera (998 and Baltagi et al. (003, given by: v it δ(0.5v i,t + 0.5v i+,t + ɛ it. (5. The ɛ it in (5.3 and (5. are assumed to be i.i.d.(0, σɛ across individuals and over time. Under the null, we have γ i 0 and δ 0. Under the null, the v it comes from some i.i.d. distribution across individuals and over time with mean zero and variance σv. These are not necessarily normally distributed. For models (5. and (5., we set α and β ; µ i is drawn from i.i.d. N(φ µ, σµ with φ µ 0 and σµ 0.5. We also set λ 0.7, φ η 0 and ση. For models (5.3 and (5., γ i i.i.d. U( 0.5, 0.55; f t is set to be i.i.d. N(0, and δ 0.. Various distributions are considered in generating the model errors, v it in (5. and ɛ it in (5.3 and (5. are assumed be normal, lognormal, gamma, chi-squared with mean zero and variance 0.5. The Monte Carlo experiments are conducted for n 0, 0, 60, 80, 00, 00, 00 and T 0, 0, 60, 80. We perform, 000 replications to compute the J u and J BF K test statistics. We conduct the tests at the positive one-sided 5% nominal significance level to obtain the empirical size.

19 5. Results Table gives the empirical size of the J u and J BF K tests allowing v it to be generated from different distributions. When the disturbances are normally distributed, the size of J u and J BF K are both close to 5%, which is consistent with the theoretical results. The rest of Table shows the results with v it coming from alternative non-normal distributions. The size of J u is close to 5% when n and T are large; for small n or small T, it is slightly oversized. However, J BF K is no longer robust to non-normality and suffers from size distortions. Table presents the size adjusted power of the tests under the alternative specification of a factor model. Both tests have size adjusted power that is almost when n and T are large with v it normally distributed. For small n and small T, the size adjusted power of J u works as well as J BF K. Note that the size adjusted power of J BF K is quite good even when n is a lot larger than T for the normal distribution scenario. However, for non-normal distributions, the size adjusted power of J u is as n and T become large; and it is larger than the size adjusted power of J BF K for all (n, T combinations. Table 3 reports the size adjusted power of both tests under the alternative specification of SAR(. The results are similar to the factor model. J u works as well as J BF K for the normal distribution scenario, but better for all combinations of n and T for non-normal distribution scenarios. 6 Conclusion Though the John test proposed by Baltagi et al. (0 has been shown to perform well for a large panel data regression model with fixed effects, it relies heavily on the normality assumption. This paper proposes a new test, J u, for the null of sphericity of the disturbances which does not rely on the normality assumption. Instead of n/t c, we allow n to be a larger order of T which is consistent with micro-panel data sets with large n and small T.

20 Acknowledgments Essay I is based on the paper of Baltagi, Kao and Peng (05. References [] Anderson, T. W., 003. An Introduction to Multivariate Statistical Analysis, Wiley. [] Anselin, L., and Bera, A. K., 998. Spatial Dependence in Linear Regression Models with an Introduction to Spatial Econometrics. Handbook of Applied Economic Statistics, Amman Ullah and David E. A. Giles, eds., Marcel Dekker, New York. [3] Bai, Z.D., and Saranadasa, H., 996. Effect of High Dimension: By an Example of a Two Sample Problem. Statistical Sinica. 6, [] Baltagi, B. H., Song, S. H., and Koh, W., 003. Testing Panel Data Regression Models with Spatial Error Correlation. Journal of Econometrics. 7, [5] Baltagi, B. H., Feng, Q., and Kao, C., 0. Testing for Sphericity in a Fixed Effects Panel Data Model. The Econometrics Journal., 5-7. [6] Baltagi, B. H., Kao, C., and Peng, B., 05. On Testing for Sphericity with Nonnormality in a Fixed Effects Panel Data Model. Statistics & Probability Letters. 98, [7] Breusch, T. S., and Pagan, A. R., 980. The Lagrange Multiplier Test and Its Application to Model Specification in Econometrics. Review of Economic Studies. 7, [8] Chen, S., Zhang, L., and Zhong, P., 00. Tests for High-dimensional Covariance Matrices. Journal of the American Statistical Association. 05, [9] Im, K. S., Ahn, S. C., Schmidt, P., and Wooldridge, J. M., 999. Efficient Estimation of Panel Data Models with Strictly Exogenous Explanatory Variables. Journal of Econometrics. 93,

21 [0] John, S., 97. Some Optimal Multivariate Test. Biometrika. 58, 3-7. [] John, S., 97. The Distribution of a Statistic Used for Testing Sphericity of Normal Distributions. Biometrika. 59, [] Johnstone, I., 00. On the Distribution of the Largest Principal Component. Annals of Statistics. 9, [3] Ledoit, O., and Wolf, M., 00. Some Hypothesis Tests for the Covariance Matrix When the Dimension is Large Compared to the Sample Size. Annals of Statistics. 30, [] Ledoit, O., and Wolf, M., 00. A Well-conditioned Estimator for Large-Dimensional Covariance Matrices. Journal of Multivariate Analysis. 88, [5] Pesaran, M. H., 00. General Diagnostic Test for Cross Section Dependence in Panels. CESifo Working Paper, No.9, University of Cambridge. [6] Pesaran, M. H., and Tosetti, E., 0. Large Panels with Common Factors and Spatial Correlations. Journal of Econometrics. 6, 8-0.

22 Table : Size of Tests n Normal Errors T J u J BF K Gamma Errors J u J BF K Lognormal Errors J u J BF K Chi-squared Errors J u J BF K Notes: This table reports the size of J u and J BF K with different error distribution specification in a fixed effects panel data model without cross-sectional dependence among the errors. The tests are one-sided and are conducted at the 5% nominal significance level. We conduct the simulation with four distributions: normal, gamma, lognormal and chi-squared with mean 0, and variance

23 Table : Size adjusted power of tests: factor model n Normal Errors T Ju J BF K Gamma Errors J u J BF K Lognormal Errors Ju J BF K Chi-squared Errors J u J BF K Notes: This table computes the size adjusted power for a factor structure model that allows for crosssectional dependence in the error. We conduct the simulation with four distributions: normal, gamma, lognormal and chi-squared with mean 0, and variance

24 Table 3: Size adjusted power of tests: SAR( model n Normal Errors T J u J BF K Gamma Errors J u J BF K Lognormal Errors J u J BF K Chi-squared Errors J u J BF K Notes: This table computes the size adjusted power for a SAR( structure model that allows for crosssectional dependence in the error. We conduct the simulation with four distributions: normal, gamma, lognormal and chi-squared with mean 0, and variance

25 Appendix This appendix includes all the proofs of the Propositions and Theorems in the text, and it also presents some useful Lemmas, which are frequently used in the proofs of the Propositions and Theorems. In the fixed effects panel data regression model: y it x itβ + µ i + v it (i,..., n, t,..., T. β is the within estimator and the within residual is ˆv it ỹ it x β, it where ỹ it y it ȳ i. and x it x it x i.. Define ṽ it v it v i., then the residuals ˆv it ṽ it x it( β β and in vector form we have ˆv t ṽ t x t ( β β. Before we go to the proofs, we introduce the following lemmas which will be used frequently later. Lemma For a random sequence Z n, if E(Z n O(n v, where v is a constant, then Z n O p (n v/. It is Lemma in Baltagi, Feng and Kao(0, we put it here since we use it frequently. Lemma Under Assumption, and the null, (a T nt t i v it σ v T + O p ( T ; nt (b T nt 3 s i v itvis T σ T v + O p ( T ; n (c T nt 3 t j v itvjt n σ T v + O p ( T ; T (d n( T nt t i v it n σ T v + O p ( n T ; T (e T nt 3 (T s j i i v itvjs n σ T v + O p (. T 3 8

26 Proof. Consider (a, nt t i v it σ v T + nt by using the fact nt (v it σ v O p (. For (b t (vit σv σ v T + O p( T nt. i nt 3 s i v itv is T T σ v + nt 3 s (vitv is σv T σ T v + O p ( T n. i by using the fact T nt s i (v itvis σv O p (. For (c nt 3 t j v itv jt n T σ v + nt 3 t (vitv jt σv n σ T v + O p ( T T. j using T n T t j (v itvjt σv O p (. For (d n( nt t i n vit [ v T it] σv [ T n + T n t i n [ T σ v + O p ( T T ] n T σ v + O p ( t (vit σv] i n T T. For (e nt 3 (T s j i i vitv js n σ T v + nt 3 (T n T σ v + O p ( T 3, s j i (vitv js σv i by using the fact nt s j i i (v itv js σ v O p (. The following lemma verifies the consistency propositions of n R and n R. 9

27 Lemma 3 Under Assumption,, and the null, (a n M,T σ v + O p ( nt ; (b n M,T O p ( T n ; (c n M 3,T σ v + O p ( T ; (d n M,T O p ( T n + O p( T T ; (e n M 5,T O p ( T ; (f n R σ v + O p ( nt ; (g n R σ v + O p ( T. Proof. Before the proofs, to simplify the notation, let s define T T (T (T (T 3, t,s,τ,η τ η η τ η η First consider (a, n M,T n T σ v + nt t v tv t nt t t (vit σv i σv + O p (. nt i v it 0

28 Here we get the results by part (a of Lemma. For (b, n M,T n T (T nt (T O p ( T n. s v tv s s v it v is i Since E[( T nt (T s i v itv is ] E[ T n T (T s i v itvis] O p (, nt we prove it. For (c, n M 3,T n T (T nt (T nt (T s s (v tv s s i i v it v is v jt v js j σv + O p ( T n + O p( T. vitv is + nt (T s v it v is v jt v js j using part (b of Lemma and T nt (T s j v itv is v jt v js O p (, hence T n M 3,T σv + O p ( T n + O p( T σ v + O p ( T.

29 For (d, n M,T n T (T (T nt (T (T nt (T (T + nt (T (T τ τ v tv s v sv τ τ τ i j O p ( T n + O p( T T. τ τ i v it v is v js v jτ v it visv iτ τ τ j v it v is v js v jτ For (e, n M 5,T n T (T (T (T 3 nt t,s,τ,η i v it v is v iτ v iη + nt t,s,τ,η O p ( T n + O p( T O p( T. v tv s v τv η t,s,τ,η j v it v is v jτ v jη For (f and (g, we can use the results of (a, (b, (c, (d, (e and show that n R n M,T n M,T σv + O p ( O p ( nt T n σ v + O p (. nt n R n M 3,T n M,T + n M 5,T σv + O p ( T (O p( T n + O p( T T + O p( T σv + O p ( T.

30 A Proof of Proposition A. Proof of part ( Proof. Recall in a panel data model ỹ it x itβ + ṽ it, we have β β ( x it x it ( x itṽ it. t i t i It is easy to show that β β O p (. nt Now ˆv it ỹ it x it β x itβ + ṽ it x it β ṽ it x it(β β. In vector form ˆv t ṽ t x t ( β β, where ṽ t v t v. Consider (a, n ˆM,T n T nt 3 k ˆv tˆv t n t (ṽ t x t ( β β (ṽ t x t ( β β t (ṽ tṽ t ṽ t x( β β ( β β x tṽ t + ( β β x t x t ( β β t A k, where A T nt t ṽ tṽ t ; A T nt t ṽ t x t ( β β; and A 3 T nt t ( β β x t x t ( β β. Lemma Under Assumption, and the null, (a A n M,T σ v T + O p ( T n ; (a A O p ( nt ; (a 3 A 3 O p ( T. 3

31 Proof. Proof of a A nt nt t t ṽ tṽ t nt v tv t nt n M,T nt n M,T nt (v t v (v t v t v t v + n v v t v t( T t t n M,T nt ( t s v s + n ( T s v tv s + nt vit + i s t v t ( T t v tv s s v it v is. i v s s With part (a of Lemma 3, we can easily get A n M,T σ v T + O p( T n. Proof of a Now we consider A. A nt nt nt nt (v t v x t ( β β t t t t v t x t ( β β nt v it x it ( β β nt v x t ( β β t t v it x it ( β β nt ( t ( T i v is x it ( β β s v it x it ( β β + O p ( nt + O p( nt (O p( nt + O p (T no p ( nt O p ( nt + O p( nt T O p( nt. i s v is x it ( β β i

32 Here we use the fact that β β O p ( nt. 3 Proof of a 3 A nt ( β β t x it( β β i O p ( nt O p( nt O p(nt O p ( nt. Hence with Lemma, we have n ˆM,T n M,T σ v T + O p( T n. A. Proof of part ( n ˆM,T n T (T nt (T 3 B k, k ˆv tˆv s s (ṽ t x t ( β β (ṽ s x s ( β β s where B T nt (T s ṽ tṽ s ; B T nt (T s ( β β x tṽ s and B 3 T nt (T s ( β β x t x s ( β β. Lemma 5 Under Assumption, and the null, (b B n M,T σ v T + O p ( T n ; (b B O p ( nt ; (b 3 B 3 O p ( nt. 5

33 Proof. Proof of b B nt (T nt (T (v t v (v s v s s v tv s nt t n M,T n v v n M,T nt ( n M,T σ v T + O p( T n. v t v nt t v v s + n v v s vit + i s v it v is i Proof of b B nt (T nt (T nt (T nt (T ( β β x t(v s v s s ( β β x tv s + nt s s i i ( β β x t v t ( β β x itv is + nt ( β β x itv is + nt O p ( nt T + O p( nt O p( nt. t t ( β β x it( T i ( β β x itv it i v is s 3 Proof of b 3 B 3 nt (T ( β β s x it x is ( β β i O p ( nt O p( nt O p(nt O p ( nt. 6

34 Then Lemma 5 is proved and we could prove part ( of Proposition, which is n ˆM,T n M,T σ v T + O p( T n. A.3 Proof of part (3 With the results of part ( and (, we obtain n ˆR n ˆM,T n ˆM,T ( n M,T σ v T + O p( T n ( n M,T σ v T + O p( T n n M,T n M,T + O p ( T n n R + O p ( T n. B Proof of Proposition Proof. Due to space limit, we put the proofs for part (, (, (3 in supplementary appendix. With part (, (, (3 of Proposition, we prove part ( as the following n ˆR n ˆM 3,T n ˆM,T + n ˆM 5,T n M 3,T n M,T + n n M 5,T + O p ( T + O p( T T + O p( T n n n R + O p ( T + O p( T T + O p( T n. 7

35 C Proof of Corollary Proof. With part (3 of Proposition and Lemma, as (n, T, we have n ˆR n ˆM,T n ˆM,T n R + O p ( T n σv p + O p ( σv. nt D Proof of Corollary Proof. With part ( of Proposition and part (g of Lemma, as (n, T and n/t 0, we have n ˆR n ˆM 3,T n ˆM,T + n ˆM 5,T n n R + O p ( T + O p( T n + O p( T T p σv. E Proof of Proposition 3 Proof. With Proposition and, we have J u J czz T (Ûn U n T ( ˆR n ( R n ( R n ( ˆR n ( ˆR n ( R n T [( R n + O p ( n + O T p ( T + O n p( T ( R T n ( R n ( R n + O p ( T n ] ( R n ( R n + O p (. T n 8

36 Let s first consider the numerator, since n R O p ( and n R O p (, as (n, T 0 and n/t p 0, the numerator becomes: T [( n n R + O p ( T + O p( T n + O p( T T ( n R ( n R ( n R + O p ( T n ] T [( n R ( n n R + (O p ( T + O p( T n + O p( T T ( n R ( n R ( n R O p ( T n ( n R (O p ( T n ] n T [O p ( T + O p( T n + O p( T T + O p( nt ] n O p ( T + O p( + O p ( p 0. n T Next we consider the denominator, we expand it as the follows: ( n R ( + O p ( R T n ( n R + ( 3 O p ( R T n + ( n R O p ( nt O p ( + O p ( T n + O p( nt O p (. From the results above, we can easily show that T (Ûn U n o p ( p 0. 9

37 F Proof of Theorem Proof. Since J u J czz + T (Ûn Un, J czz d N(0, and with Proposition 3, we obtain J u d N(0,. 30

38 The following Sections prove part (, (, (3 of Proposition. G Proof of part ( Proof. Since n ˆM 3,T is a scaler, we could expand it as the follows: n ˆM 3,T nt (T nt (T tr{ nt (T 7 C i. i (ˆv tˆv s s ((ṽ t x t ( β β (ṽ s x s ( β β s (ṽ tṽ s ( β β x tṽ s ṽ t x s ( β β + ( β β x t x s ( β β } s where C T nt (T s (ṽ tṽ s ṽ tṽ s ; C ( β β T nt (T ( β β T nt (T C 5 nt (T ( β β s ṽ tṽ s x t x s ( β β; C nt (T ( β β s (ṽ tṽ s x tṽ s ; C 3 s ( x tṽ s ṽ s x t ( β β; s ( x tṽ s ṽ t x s ( β β; C 6 ( β β ( nt (T s x tṽ s ( β β x t x s ( β β; C 7 nt (T ( β β ( s x t x s ( β β( β β x t x s ( β β. Lemma 6 Under Assumption,, 3 and the null, (c C n M 3,T σ v T T + ( T (c C O p ( nt + O p( T + O p ( T nt ; (c 3 C 3 O p ( T ; (c C O p ( nt ; (c 5 C 5 O p ( nt ; (c 6 C 6 O p ( nt ; (c 7 C 7 O p ( nt. + n σ T v 3n σ T v + O p ( n + O T p ( T + O T p( T ; n 3

39 G. Proof of part (c C nt (T nt (T nt (T n M 3,T + (ṽ tṽ s ṽ tṽ s s (v t v (v s v (v t v (v s v s (v tv s v v s v t v + v v s 5 C k. k Where C T nt (T s v tv s v t v ; C T nt (T s v tv s v v ; C 3 T nt t v t v v t v ; C nt (T s v t v v v s and C 5 3 n v v v v. Lemma 7 Under Assumption,, 3 and the null, ( C σ v T + O p( T n + O p( n T ; ( C O p ( n T ; (3 C 3 ( T T ( C n σ T v + O p ( n ; T (5 C 5 3n T σ v + O p ( n T. + n σ T v + O p ( n + O T p ( T + O T p( T ; n 3

40 Proof. Proof of part ( C nt (T nt (T nt (T nt (T v tv s v t( T s s τ v τ τ v tv s v tv τ s τ i j s τ i v it v is v jt v jτ vitv is v iτ nt (T s τ j v it v is v jt v jτ. Here we need to discuss three cases of (s, t, τ: ( t s; t τ; ( t s; s τ; (3 t s τ. Then the first term becomes the following: nt (T nt (T nt (T s τ i s vitv is v iτ vitv is i,τ τ i vitv is v iτ nt (T σ v T + O p( T n + O P ( T n + O p( T n σ v T + O p( T n. s i v 3 itv is 33

41 Next we consider the second term nt (T nt (T s τ j s v it v is v jt v jτ v it v is vjt j v nt it v is v jt v jτ (T,τ τ j n O p ( T + O p( T + O n p( T T O p( T. nt (T s v it v is v jt v js j Therefore, we obtain that C σ v T + O n p( T n + O p( T. Proof of part ( C + nt (T nt 3 (T nt 3 (T nt 3 (T nt 3 (T v tv s ( T s s τ η v τ ( T τ v tv s v τv η s τ η i j s τ η i v η η v it v is v jτ v jη v it v is v iτ v iη s τ η j v it v is v jτ v jη. To consider the order of the two terms of C. First, we distinct the cases of the first term, for (t s, τ, η, we have ten cases: I Two s: ( (s τ η t; ( (t τ η s; (3 (s τ (t η; ( (s η (t τ; 3

42 II One : (5 (t τ s η; (6 (t η s τ; (7 (s τ t η; (8 (s η t τ; (9 (τ η t s; III No : (0 t s τ η. Then the first term of C can be calculated respectively as the following: ( (s τ η t : T nt 3 (T s i v itvis 3 O p ( T 3 ; n ( (t τ η s : T nt 3 (T s i v3 itv is O p ( T 3 ; n (3 (s τ (t η : T nt 3 (T s i v itvis O p ( ; T ( (s η (t τ : T nt 3 (T s i v itvis O p ( ; T (5 (t τ s η : T nt 3 (T,η s η η i v itv is v iη O p ( (6 (t η s τ : T nt 3 (T,τ τ i v itv is v iτ O p ( (7 (s τ t η : T nt 3 (T,η s η η i v itvisv iη O p ( (8 (s η s τ : T nt 3 (T,τ τ i v itvisv iτ O p ( (9 (τ η t s : T nt 3 (T,η s η η i v itv is viη O p ( (0t s τ η : nt 3 (T,τ,η,η τ η η T n ; T n ; T n ; T n ; T n ; i v itv is v iτ v iη O p ( T n. Similarly, we can also expand the second term as the following ten cases: ( (s τ η t : T nt 3 (T s j v itv is vjs O p ( n ; T 3 ( (t τ η s : T nt 3 (T s j v itv is vjt O p ( n ; T 3 (3 (s τ (t η : T nt 3 (T s j v itv is v jt v js O p ( ; T 3 ( (s η (t τ : T nt 3 (T s j v itv is v js v jt O p ( ; T 3 (5 (t τ s η : T nt 3 (T,η s η η j v itv is v jt v jη O p ( (6 (t η s τ : T nt 3 (T,η s η η j v itv is v jτ v jt O p ( (7 (s τ t η : T nt 3 (T,η s η η j v itv is v js v jη O p ( (8 (s η s τ : T nt 3 (T,η s η η j v itv is v jτ v js O p ( (9 (τ η t s : T nt 3 (T,η s η η j v itv is vjτ O p ( n ; T (0t s τ η : T nt 3 (T,τ,η,η τ η η i v itv is v jτ v jη O p (. T Hence n C O p ( T. T T ; T T ; T T ; T T ; 35

43 3 Proof of part (3 C 3 nt nt 3 nt 3 nt 3 v t( T t t v s v t( T s s τ v tv s v tv τ t s τ i j t s τ i v τ τ v it v is v jt v jτ v itv is v iτ + nt 3 t s τ j v it v is v jt v jτ. There are five cases for the first term, they are t s τ; t s τ, t τ s, s τ t, t s τ. Then the first term can be expressed by the sum of the following five cases: ( t s τ : T nt 3 t i v it O p ( ; T ( t s τ : T nt 3 t τ τ i v3 itv iτ O p ( T ; n (3 t τ s : T nt 3 s i v3 itv is O p ( ( s τ t : nt 3 s (5 t s τ : nt 3,τ τ T n ; i v itv is T T σ v + O p ( T n ; i v itv is v iτ O p ( T n. Similarly, we can also get the five cases of the second term: ( t s τ : T nt 3 t j v itvjt n σ T v + O p ( ( t s τ : T nt 3 t τ τ j v itv jt v jτ O p ( n ; T (3 t τ s : nt 3 s ( s τ t : nt 3 s (5 t s τ : nt 3,τ τ Add all the terms up, we can easily get T T ; j v itv is vjt O p ( n ; T j v itv is v jt v js O p ( ; T j v itv is v jt v jτ O p ( T T. C 3 ( T + n n σ T T v + O p ( T + O p( T T + O p( T n. 36

44 Proof of part ( C nt (T nt (T nt 3 (T nt 3 (T + nt 3 (T v t v v v s s v t( T s v τ ( T τ s τ η v tv τ v ηv s s τ η i v η v s η v it v iτ v iη v is s τ η j To distinguish the two terms, there are 0 cases to discuss: v it v iτ v jη v js. I Two s: ( t τ η s; ( t s τ η; (3 t τ s η; ( t η s τ; II One : (5 (t τ s η; (6 (t η s τ; (7 t (s τ η; (8 t (s η τ; (9 t s (τ η; III No : (0 t s τ η. Then the first term of C can be expanded to 0 cases as the following: ( t τ η s : T nt 3 (T s i v3 itv is O p ( T 3 ; n ( t s τ η : T nt 3 (T s i v itvis 3 O p ( T 3 ; n (3 t τ s η : T nt 3 (T s i v itvis O p ( ; T ( t η s τ : T nt 3 (T s i v itvis O p ( ; T (5 (t τ s η : T nt 3 (T η s η η i v itv iη v is O p ( (6 (t η s τ : T nt 3 (T τ τ i v itv iτ v is O p ( (7 t (s τ η : T nt 3 (T η s η η i v itv iη vis O p ( (8 t (s η τ : T nt 3 (T τ τ i v itv iτ vis O p ( (9 t s (τ η : T nt 3 (T τ τ i v itviτv is O p ( (0 t s τ η : nt 3 (T τ η η τ η η T n ; T n ; T n ; T n ; T n ; i v itv iτ v iη v is O p ( T n. 37

45 Similarly, we can also get 0 terms for the second term s expansion: ( t τ η s : T nt 3 (T s j v itv jt v js O p ( n ; T 3 ( t s τ η : T nt 3 (T s j v itv is vjs O p ( n ; T 3 (3 t τ s η : T nt 3 (T s j v itvjs n σ T v + O p ( ; T 3 ( t η s τ : T nt 3 (T s j v itv is v jt v js O p ( ; T 3 (5 (t τ s η : nt 3 (T η s η η (6 (t η s τ : nt 3 (T τ τ (7 t (s τ η : nt 3 (T η s η η (8 t (s η τ : nt 3 (T τ τ (9 t s (τ η : nt 3 (T τ τ (0 t s τ η : nt 3 (T τ η η τ η η Hence, j v itv jη v js O p ( n ; T j v itv iτ v jt v js O p ( j v itv is v jη v js O p ( T T ; T T ; j v itv iτ vjs O p ( n ; T j v itv iτ v jτ v js O p ( T T ; j v itv iτ v jη v js O p ( T. 5 Proof of part (5 C n n σ T v + O p ( T. C 5 3 n v v v v 3n( n v v. 3n( nt 3n(( nt + ( nt t t s i i v it + nt v it + ( nt v it v is i 3n(( v nt it + ( nt t i 3n n T σ v + O p ( T T + O p( s t t v it v is i i i v it( nt v it( nt n T 3n T σ v + O p ( s s n T. v it v is i v it v is + o p i 38

46 Hence, with the results above we have C n M 3,T σ v T + (T + n T T σv 3n n T σ v + O p ( T + O p( T T + O p( T n. G. Proof of part c C nt (T ( β β nt (T ( β β 8 C k. k (ṽ tṽ s x tṽ s s ((v t v (v s v x t(v s v Where C ( β β T nt (T s v tv s x tv s ; C ( β β T nt (T C 3 nt (T ( β β ( β β T nt (T β s s v t v x tv s ; C nt (T ( β β s v v s x tv s ; C 6 nt (T ( β β s v v x tv s ; C 8 nt ( β β t v v x t v. Lemma 8 Under Assumption,, 3 and the null, s v tv s x t v ; s v t v x t v ; C 5 s v v s x t v ; C 7 ( β nt (T ( C O p ( nt + O p( T nt ; ( C O p ( T n ; (3 C 3 O p ( T ; ( C O p ( T ; (5 C 5 O p ( T + O p ( nt + O p( T nt ; (6 C 6 O p ( T ; 39

47 (7 C 7 O p ( T ; (8 C 8 O p ( T. Proof. Proof of part ( C nt (T ( β β nt (T ( β β O p ( nt + O p( T nt. s s i i v it v is x jt v js j v it vis x it nt (T ( β β s v it v is x jt v js j Proof of part ( C nt (T ( β β nt (T ( β β nt (T ( β β + nt (T ( β β v tv s x t v s s i v it v is x jt ( T j s τ i v it v is x it v iτ s τ j v jτ τ v it v is x jt v jτ. To expand the two terms of C, we consider 3 cases : ( t τ s; ( t τ s; (3 t s τ. The first term of can be expanded as the following terms: ( t τ s : ( β β T nt (T s i v itv is x it O p ( ; nt ( t s τ : ( β β T nt (T,τ τ i v itv is x it v iτ O p ( (3 t s τ : nt (T ( β β,τ s The second term can be expanded as: ( t τ s : nt (T ( β β s ( t s τ : nt (T ( β β,τ τ i v itv is x it O p ( nt. j v itv is x jt v jt O p ( nt ; T nt ; j v itv is x jt v jτ O p ( T n ; 0

48 (3 t s τ : ( β β T nt (T,τ s j v itv is x jt v js O p ( T. nt Then we have Proof of part (3 C O p ( T n. C 3 nt (T ( β β nt (T ( β β nt (T ( β β + nt (T ( β β v t v x tv s s v t( T s v τ x tv s η s τ i v it v iτ x itv is s τ j v it v iτ x jtv js. As C, the two terms can be expanded as the following three cases, let s first consider the first term: ( t τ s : nt (T ( β β s ( t s τ : nt (T ( β β s (3 t s τ : nt (T ( β β η s η i v it x it v is O p ( i v itv is x it O p ( τ nt ; nt ; i v itv iτ x it v is O p ( nt. The second term can be expanded as the following: ( t τ s : ( β β T nt (T s j v it x jt v js O p ( ; T ( t s τ : ( β β T nt (T s j v itv is x jt v js O p ( (3 t s τ : nt (T ( β β τ Then we have τ C 3 O p ( T. T nt ; j v itv iτ x jt v js O p ( T n.

49 Proof of part ( C nt ( β β nt ( β β nt 3 ( β β nt 3 ( β β v t v x t v t v t( T t v s x t( T s t s τ i v τ τ v it v is x itv iτ t s τ j v it v is x jtv jτ. We distinguish 5 cases for the expansion of the two terms, consider the first term: ( t s τ : ( β β T nt 3 t i v3 it x it O p ( ; nt 3 ( t s τ : ( β β T nt 3 t τ τ i v it x it v iτ O p ( ; nt (3 t τ s : ( β β T nt 3 s i v itv is x it O p ( ; nt ( t s τ : ( β β T nt 3 s i v itvis x it O p ( ; nt (5 t s τ : nt 3 ( β β τ i v itv is x it v iη O p ( nt. Similarly the second term also can be expanded as the following terms: ( t s τ : ( β β T nt 3 t j v it x jt v jt O p ( ; T 3 ( t s τ : ( β β T nt 3 t τ η j v it x jt v jτ O p ( ; T (3 t τ s : ( β β T nt 3 s j v itv is x jt v jt O p ( ( t s τ : ( β β T nt 3 s j v itv is x jt v js O p ( (5 t s τ : nt 3 ( β β τ With the results above, we get C O p ( T. T nt ; T nt ; j v itv is x jt v jτ O p ( T n.

50 Proof of part (5 C 5 nt (T ( β β nt (T ( β β nt (T ( β β + nt (T ( β β v v s x tv s s ( T s v τ v s x tv s τ s τ i v iτ vis x it s τ i The first term then can be expressed as the following 3 terms: ( t τ s : ( β β T nt (T s i v itvis x it O p ( ( t s τ : ( β β T nt (T s i v3 is x it O p ( (3 t s τ : nt (T ( β β τ τ The second term of C 5 can be expressed as: ( t τ s : nt (T ( β β s ( t s τ : nt (T ( β β s (3 t s τ : nt (T ( β β τ τ Then, we have v iτ v is x jt v js. nt ; nt ; i v iτv is x it O p ( nt. i v itv is x jt v js O p ( i v is x jt v js O p ( ; T C 5 O p ( T + O p( nt + O p( T nt. T nt ; i v iτv is x jt v js O p ( T nt. 3

51 Proof of part (6 C 6 nt (T ( β β nt (T ( β β nt 3 (T ( β β nt 3 (T ( β β There are 0 cases for expanding: v v s x t v s ( T s v τ v s x t( T τ s τ η i v η η v iτ v is x it v iη s τ η j v iτ v is x jt v jη. I Two s: ( t τ η s; ( t s τ η; (3 t τ s η; ( t η s τ; II One : (5 t τ s η; (6 t η s τ;(7 t τ s η; (8 t η s τ; (9 t s τ η; III No : (0 t s τ η. The first term of C 6 can be expanded as the following 0 terms: ( t τ η s : ( β β T nt 3 (T s i v itv is x it O p ( ; nt 3 ( t s τ η : ( β β T nt 3 (T s i v3 is x it O p ( ; nt 3 (3 t τ s η : ( β β T nt 3 (T s i v itvis x it O p ( ; nt 3 ( t η s τ : ( β β T nt 3 (T s i v is x it v it O p ( ; nt 3 (5 t τ s η : ( β β T nt 3 (T η s η η i v itv is x it v iη O p ( (6 t η s τ : ( β β T nt 3 (T τ τ i v iτv is x it v it O p ( (7 t τ s η : ( β β T nt 3 (T τ τ i v iτvis x it O p ( (8 t η s τ : ( β β T nt 3 (T η s η η i v is x it v iη O p ( ; nt (9 t s τ η : nt 3 (T ( β β τ τ nt 3 ; nt 3 ; nt ; i v iτv is x it O p ( nt ; (0 t s τ η : nt 3 (T ( β β τ η η τ η η i v iτv is x it v iη O p ( nt. Similarly, there are also 0 terms for expanding the second term of C 6 :