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# singleequationgmmslides - Single Equation Linear GMM...

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Single Equation Linear GMM Consider the linear regression model y t = z 0 t δ 0 + ε t , t = 1 , . . . , n z t = L × 1 vector of explanatory variables δ 0 = L × 1 vector of unknown coe cients ε t = random error term Engodeneity The model allows for the possibility that some or all of the elements of z t may be correlated with the error term ε t (i.e., E [ z tk ε t ] 6 = 0 for some k ) . If E [ z tk ε i ] 6 = 0 , then z tk is called an endogenous variable . If z t contains endogenous variables, then the least squares estimator of δ 0 in is biased and inconsistent.

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Instruments It is assumed that there exists a K × 1 vector of instrumental variables x t that may contain some or all of the elements of z t . Let w t represent the vector of unique and nonconstant elements of { y t , z t , x t } . It is assumed that { w t } is a stationary and ergodic stochastic process.
Moment Conditions for General Model De fi ne g t ( w t , δ 0 ) = x t ε t = x t ( y t z 0 t δ 0 ) It is assumed that the instrumental variables x t satisfy the set of K orthogo- nality conditions E [ g t ( w t , δ 0 )] = E [ x t ε t ] = E [ x t ( y t z 0 t δ 0 )] = 0 Expanding gives the relation Σ xy ( K × 1) = Σ xz ( K × L ) δ 0 ( L × 1) where Σ xy = E [ x t y t ] and Σ xz = E [ x t z 0 t ] .

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Example: Demand-Supply model with supply shifter demand: q d i = α 0 + α 1 p i + u i supply: q s i = β 0 + β 1 p i + β 2 temp i + v i equilibrium: q d i = q s i = q i E [ temp i u i ] = E [ temp i v i ] = 0 Goal: Estimate demand equation; α 1 = demand elasticity if data are in logs Remark Simultaneity causes p i to be endogenous in demand equation: E [ p i u i ] 6 = 0 Why? Use equilibrium condition, solve for p i and compute E [ p i u i ]
Demand/Supply model in Hayashi notation y i = z 0 i δ + ε i y i = q i , z i = (1 , p i ) 0 , δ = ( α 0 , α 1 ) 0 x i = (1 , temp i ) 0 , w i = ( q i , p i , temp i ) 0 L = 2 , K = 2 Note: 1 is common to both z i and x i .

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Example: Wage Equation ln W i = δ 1 + δ 2 S i + δ 3 EXPR i + δ 4 IQ i + ε i W i = wages S i = years of schooling EXPR i = years of experience IQ i = score on IQ test δ 2 = rate of return to schooling Assume E [ S i ε i ] = E [ EXPR i ε i ] = 0 Endogeneity: IQ i is a proxy for ability but is measured with error IQ i = π 0 + π 1 ABILITY i + error i E [ IQ i ε i ] 6 = 0
Instruments: AGE i = age in years MED i = years of mother’s education E [ AGE i ε i ] = E [ MED i ε i ] = 0

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Wage Equation in Hayashi notation y i = z 0 i δ + ε i y i = ln W i , z i = (1 , S i , EXPR i , IQ i ) 0 x i = (1 , S i , EXPR i , AGE i , MED i ) 0 w i = (ln W i , S i , EXPR i , IQ i , AGE i , MED i ) 0 L = 4 , K = 5 Remarks 1. 1 , S i , EXPR i are often called included exogenous variables. That is, they are included in the behavioral wage equation. 2. AGE i , MED i are often called excluded exogenous variables. That is, they are excluded from the behavioral wage equation.
Identi fi cation Identi fi cation means that δ 0 is the unique solution to the moment equations E [ g t ( w t , δ 0 )] = 0 That is, δ 0 is identi fi ed provided E [ g t ( w t , δ 0 )] = 0 and E [ g t ( w t , δ )] 6 = 0 for δ 6 = δ 0 For the linear GMM model, this is equivalent to Σ xy ( K × 1) = Σ xz ( K × L ) δ 0 ( L × 1) Σ xy ( K × 1) 6 = Σ xz ( K × L ) δ ( L × 1) for δ 6 = δ 0

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Rank Condition For identi fi cation of δ 0 , it is required that the K × L matrix E [ x t z 0 t ] = Σ xz be of full rank L . This
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singleequationgmmslides - Single Equation Linear GMM...

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