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lect18_06apr10 - Imbens Lecture Notes 18 ARE213 Spring'06...

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Imbens, Lecture Notes 18, ARE213 Spring ’06 1 ARE213 Econometrics Spring 2006 UC Berkeley Department of Agricultural and Resource Economics Endogeneity II: Two-Stage-Least-Squares, Control Function, and Limited-Information-Maximum-Likelihood Estimation 1. Two-Stage-Least-Squares A more systematic way to combine the multiple instruments is through two-stage-least- squares estimation. Let us do this in more generality. The equation of interest is Y i = X i β + ε i = X i 1 β 1 + X i 2 β 2 + ε i . Let σ 2 be the variance of ε i . The vector of covariates X i can be split into two parts, a possibly endogenous part X i 1 and an exogenous part X i 2 . The vector of instruments is Z i . It can be split into the excluded instruments Z i 1 and the exogenous covariates X i 2 , or Z i = ( Z i 1 , X i 2 ). Typically the common part X i 2 of the vectors Z i and X i will at least contain the intercept. The TSLS estimation method consists of two stages. In the first stage all the endogenous regressors are regressed on all the instruments and exogenous variables. That is, we estimate X i 1 = Z i Π + η i = Z i 1 Π 1 + X i 2 Π 2 + η i . Note that X i 1 is a K - vector, so that with Z i an L - vector, Π is a L × K matrix of parameters. Estimating this by least squares leads to ˆ Π = ( Z Z ) - 1 Z X 1 . We then calculate the predicted values for X based on this regression: ˆ X 1 = Z ˆ Π .
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Imbens, Lecture Notes 18, ARE213 Spring ’06 2 Note that if we have a similar equation for X i 2 , X i 2 = Z i Π + η i = Z i 1 Π 1 + X i 2 Π 2 + η i , the result would be Π 2 = I and Π 1 = 0, so that the predicted value is ˆ X 2 = X 2 . Hence in the end we could treat all regressors symmetrically and just regress X on Z to get ˆ X = Z ( Z Z ) - 1 Z X . In the second stage the outcome is regressed on the predicted regressors: Y i = ˆ X i β + ν = ( Z i ˆ Π) β + ν i . We can write the estimator for β as: ˆ β = ( ( X Z ) · ( Z Z ) - 1 · ( Z X ) ) - 1 · ( X Z ) · ( Z Z ) - 1 · ( Z Y ) . In large samples N · ( ˆ β - β ) ∼ N 0 , σ 2 · ( ( X Z ) · ( Z Z ) - 1 · ( Z X ) ) - 1 . The error variance is E [( Y - X β ) 2 ], estimated as i ( Y i - X i ˆ β ) 2 /N . Note that this variance is not the variance you would get as the standard ols variance from regressing Y i on ˆ X i . Let us see what we get from this for the Angrist-Krueger data. The first stage is the same regression we did before: educ i = 12 . 6881 + 0 . 0566 · qob 2 i + 0 . 1173 · qob 3 i + 0 . 1514 · qob 4 i (0 . 0115) (0 . 0163) (0 . 0160) (0 . 0163)
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