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Bài giảng Phân tích số liệu mảng - Chương 4: Random effect model (REM)

Chia sẻ: Cao Ngữ Lam | Ngày: | Loại File: PDF | Số trang:18

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Bài giảng Phân tích số liệu mảng - Chương 4: Random effect model (REM). Chương này cung cấp cho sinh viên những nội dung gồm: giới thiệu về Mô hình tác động ngẫu nhiên; ước tính các tham số độ dốc trong FEM bằng công cụ ước tính bên trong, công cụ ước tính giữa; ước tính FEM bằng phương pháp biến giả bình phương tối thiểu (LSDV);... Mời các bạn cùng tham khảo!

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Nội dung Text: Bài giảng Phân tích số liệu mảng - Chương 4: Random effect model (REM)

  1. 6/6/2022 Chapter 4 Random effect model (REM) Mr U_KHOA TOÁN KINH TẾ 57
  2. Objectives 58 (1) Introduce about Random Effect Model (2) Estimates the slope paramaters in FEM by Within Estimator, Between Estimator (3) Estimates FEM by Least Square Dummy Variables (LSDV) method Mr U_KHOA TOÁN KINH TẾ 6/6/2022
  3. Notes. There are too many parameters in the fixed effects model and the 59 loss of degrees of freedom can be avoided if the α*i can be assumed random 4.1 Introduce Random effect model yit = x itb + a * + u it i i = 1,N; t = 1,T (4.1) Here, α*i is assumed to be random If the individual effects α*i are supposed to have non zero mean, with E (α*i)= α0 Then we can define cross section units effects α*i= α0 + αi Pre Eq. (4.1) Mr U_KHOA TOÁN KINH TẾ 6/6/2022
  4. 4.1.1 The assumptions on the components of errors 60 About ai , ( ) ( ) ( ) ( ) ( ) E a i = 0, ,V a i = E a 2 = s m , ,E a i x it = 0, ,E a ia j = 0 i 2 About u it , ( ) ( ) ( ) ( ) E u it = 0, ,V u it = E u 2 = s 2 , ,E u it u js = 0 for i ¹ j and t ¹ s it u The components of the error are not correlated E (αiuit) =0 Remark. The αi are independent of the error term uit and the regressors xit, for all i and t 4.1.2 Mean and variance of errors The mean and variance of the component errors are ( ) U_KHOA 0, ,V ( ) ( ) EMr e it =TOÁN KINH TẾ e it = V yit = s a + s 2 2 u 6/6/2022
  5. The covariance of the composite error, 61 Cov (εit, εjs ) = E(εitεjs)= E(αi+ uit) (αj+ ujs) = E (αiαj + uit αj + αi ujs + uitujs) Or Case 1. Cov (εit, εjs ) = σ2α + σ2u i = j, t= s Case 2. Cov (εit, εjs ) = σ2α i = j, t ≠ s Case 3. Cov (εit, εjs ) = 0 i ≠ j, t ≠ s Mr U_KHOA TOÁN KINH TẾ 6/6/2022
  6. For cross section unit i, Eq. (4.1) can be written as 62 The variance- covariance matrix of εi (for individual i) is ææ e i1 ö ö çç ÷ ÷ ( ) ' çç E e ie i = E ç ç çç e i2 ÷ ... ÷ ÷ (e i1 e i2 ... e iT ) ÷ ÷ ÷ çç e iT ÷ ÷ Mr U_KHOA TOÁN KINH TẾ èè ø ø 6/6/2022
  7. æ e2 e i1e i2 ... e i1e iT ö 63 ç i1 ÷ ç e i2 e i1 e2 ... e i2 e iT ÷ = Eç i2 ÷ ç ... ... ... ... ÷ ç e e e e ... e 2 ÷ è iT i1 iT i2 iT ø æ s2 + s2 s 2 ... s 2 ö ç a u a a ÷ ç sa 2 sa + s2 2 ... sa 2 ÷ =ç u ÷ ç ... ... ... ÷ ç sa 2 sa2 ... s a + s 2 2 ÷ è u ø æ 1 'ö 2 1 = U = s I + s ee = s ç Q + ee ÷ + Ts a ee' 2 2 a ' 2 u T è T ø u T u ( = Mr U_KHOA TOÁN KINH TẾTs a P s 2Q + s 2 + 2 u ) (4.2) 6/6/2022
  8. 1 ( ) -1 here P = ee' = e e'e e' = I T - Q 64 T 1 -1 (4.2) Þ U = 2 Q + qP su ( ) (4.3) s2 where q = u (4.4) (s 2 u + Ts 2 a ) -1/2 1 1 Therefore, U = Q+ P (4.5) su ( s 2 + Ts a u 2 ) 1 æ 1 ö Or, U -1/2 = çQ+ P ÷ (4.6) su ç è ( s 2 + Ts a u 2 ) ÷ ø And U = s 2(T-1) s 2 + Ts a u u 2 ( ) (4.7) Mr U_KHOA TOÁN KINH TẾ 6/6/2022
  9. By taking all cross section units in the sample, the variance - covariance 65 matrix of the error term (ε) will be of order NT x NT æ U 0 ... 0 ö ç ÷ ( ) E ee ' = E ç ç 0 U ... 0 ... ... ... ... ÷ ÷ ç è 0 0 0 U ÷ ø u ( 2 ) ( ) = U Ä I N = s 2 I T Ä I N + s a J T Ä I N = W (4.8) where J T = ee' Mr U_KHOA TOÁN KINH TẾ 6/6/2022
  10. 4.2 GLS estimation 66 Idea. The generalised least squares (GLS) is used in estimating a random effects model when U is known. Suppose that the variance – covariance matrix (U) is known Pre multiply Eq. (4.1) by U-1/2 to get Homework. Proving why reason Eq. (4.10) equal with IT Mr U_KHOA TOÁN KINH TẾ 6/6/2022
  11. With Eq. (4.9) we can apply OLS to estimate parameter (γ) 67 The GLS estimators of γ are Eq. (4.12) can be written in expanded form as ( )( ) -1 Eq. (4.12) = S W XX + qS B XX SW + qSB Xy Xy (4.13) Homework.KINH TẾ Mr U_KHOA TOÁN Expanding detail (4.12) to finding why (4.12) can similar (4.13) 6/6/2022
  12. Remark. 68 - If θ = 1, then GLS estimator is equivalent to OLS pooled estimator. - If θ = 0, then GLS estimator will be equal to LSDV - The parameter θ measures the weight given to between-group variation. - If U is unknown, we can use a two-step GLS estimation known with name is called FGLS (Feasible Generalized Least Squares) 4.3 FGLS estimator Note. When U is unknown as means as σ2α & σ2u are unidentifed. We can use two-step GLS estimation known as FGLS Step 1. We estimate the “within” estimation and “between” estimation model to find out Mr U_KHOA TOÁN KINH TẾ 6/6/2022
  13. 69 The σ2α & σ2u are obtained from the ”between ” effect estimation, the “within” effect estimation, etc. Then, we have to caculate Mr U_KHOA TOÁN KINH TẾ 6/6/2022
  14. Step 2. We have to estimate the following model: 70 4.4 Testing of Hypotheses Introduction. In a panel regression model, either fixed or random effect is an issue of unobserved variables measuring heterogeneity across the entities which renders the bias in pooled regression estimation. 4.4.1 Measuring of Goodness Fit Panel data can be utilities to calculate within-entity variation (R2W) , between-entity variation (R2B ) and overall variation (R2). Option 1. Testing for Pooled Regression yi =α+ Xiβ+ εi. (i = 1, .., n) Mr U_KHOA TOÁN KINH TẾ 6/6/2022
  15. Pre Example (3.1) with model 71 ROAA = f(HHI, L_A, SIZE, ASSET_GRO, GDP, INF) + ε Option 2. Testing for Fix Effects yi = eαi + Xiβ+ εi. (i = 1, .., n) Method 1. Fix effects model is only valid when we could test the joint significance of the dummies by: H0: α1 = α2 =…= αN = 0 H1: αi ≠ 0 The F test is calculated by the following formular: Mr U_KHOA TOÁN KINH TẾ 6/6/2022
  16. Pre Example (3.1) With model 72 ROAA = f(HHI, L_A, SIZE, ASSET_GRO, GDP, INF) + ε Method 2. F test can alse check by xtreg with option fe in Stata Option 3. Testing of Random Effects H0: σ2α = 0 H1: σ2α > 0 To test this hypothesis, we can use Lagrange Multiplier (LM) test developed by Bresuch and Pagan (1980) Mr U_KHOA TOÁN KINH TẾ 6/6/2022
  17. Pre Example (3.1) With model 73 ROAA = f(HHI, L_A, SIZE, ASSET_GRO, GDP, INF) + ε 4.4.2 Fix or Random effect: Hausman Test yi = eαi + Xiβ+ εi. (i = 1, .., n) (FE) H0: E (εit|Xit) = 0 H1: E (εit|Xit) ≠ 0 Mr U_KHOA TOÁN KINH TẾ 6/6/2022
  18. Using this fact, we have 74 Therefore, With Then The test statistic is Mr U_KHOA TOÁN KINH TẾ Hausman test (H) = q’ (Var(q))-1q 6/6/2022
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