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multiequationgmmslides - Multiple Equation Linear GMM Eric...

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Multiple Equation Linear GMM Eric Zivot November 23, 2009 Multiple Equation Linear GMM
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Notation y iM , i = individual; M = equation There are M linear equations, y im = z 0 im (1 × L m ) δ m ( L m × 1) + ε im , m = 1 , . . . , M ; i = 1 , . . . , n Remarks: 1. No a priori assumptions about cross equation error correlation 2. No cross equation parameter restrictions
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Giant Regression Representation y 1 n × 1 . . . y M n × 1 = Z 1 n × L 1 . . . Z M n × L M δ 1 L 1 × 1 . . . δ M L M × 1 + ε 1 n × 1 . . . ε M n × 1 or y ¯ nM × 1 = Z ¯ nM × L δ L × 1 + e ¯ nM × 1 L = M X m =1 L m
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Main Issues and Questions: 1. Why not just estimate each equation separately? (a) Joint estimation may improve e ciency (b) Joint estimation is sensitive to misspeci fi cation of individual equations 2. Theory may provide cross equation restrictions (a) Improve e ciency (b) test restrictions
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Example: 2 equation wage equation LW i = φ 1 + β 1 S i + γ 1 IQ i + πEXPR i + ε i 1 , L 1 = 4 KWW i = φ 2 + β 2 S i + γ 2 IQ i + ε i 2 , L 2 = 3 z i 1 = (1 , S i , IQ i , EXPR i ) 0 z i 1 = (1 , S i , IQ i ) 0 Note, ε i 1 and ε i 2 may be correlated (eg. due to common omitted variable ability)
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Example: Panel data for wage equation LW 69 i = φ 1 + β 1 S i + γ 1 IQ i + π 1 EXPR i + ε i 1 , LW 80 i = φ 2 + β 2 S i + γ 2 IQ i + π 2 EXPR i + ε i 2 , If all coe cients do not change over time then φ 1 = φ 2 , β 1 = β 2 , γ 1 = γ 2 , π 1 = π 2 ε im = α i + η im α i = unobserved individual fi xed e ff ect
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Instruments x im ( K m × 1) = instruments for m th equation E [ x im ε im ] = 0 , m = 1 , . . . , M K = M X m =1 K m orthogonality conditions Note: We are not assuming cross-equation orthogonality conditions. That is, we may have E [ x im ε ik ] 6 = 0 unless x im and x ik have variables in common.
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Example: 2 equation wage equation Assume 1. IQ is endogenous in both equations 2. MED is exogenous in both equations E [ MED i ε i 1 ] = 0 , E [ MED i ε i 2 ] = 0 x i 1 = x i 2 = (1 , S i , EXPR i , MED i ) 0
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GMM Moment Conditions and Identi fi cation De fi ne δ ( L × 1) = ( δ 0 1 , . . . , δ 0 M ) 0 , L = M X m =1 L m g i ( δ ) K × 1 = g i 1 ( δ 1 ) . . . g iM ( δ M ) = x i 1 ε i 1 . . . x iM ε iM = x i 1 ( y i 1 z 0 i 1 δ 1 ) . . . x iM ( y iM z 0 iM δ M ) Then there are K linear moment equations such that E [ g i ( δ )] = 0 E [ g i ( ˜ δ )] 6 = 0 for ˜ δ 6 = δ
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Now, E [ g i ( δ )] = E [ x i 1 y i 1 ] . . . E [ x iM y iM ] E [ x i 1 z 0 i 1 ] δ 1 . . . E [ x iM z 0 iM ] δ M = E [ x i 1 y i 1 ] . . . E [ x iM y iM ] E [ x i 1 z 0 i 1 ] · · · 0 . . . . . . . . . 0 · · · E [ x iM z 0 iM ] δ 1 . . . δ M = σ xy ( K × 1) Σ xz ( K × L ) δ ( L × 1) Note: The above moment conditions for each equation are the same as the moment conditions we derived for the single equation linear GMM model.
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If K m = L m for m = 1 , . . . , M then δ is exactly identi fi ed and δ 1 .
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