Panel Data Analysis Part I Classical Methods: Background Material - PowerPoint PPT Presentation

Panel Data Analysis Part I – Classical Methods: Background Material James J. Heckman University of Chicago Econ 312, Spring 2019 Heckman Part I

Review of the Early Econometric Literature on the Problem Heckman Part I

• Y it = X it β + ε it i = 1 , . . . , I , and t = 1 , . . . , T • Y i · = X i · β + ε i · i = 1 , . . . , I • Y i · = � T X i · = � T Y it X it T , t =1 t =1 T • Y i = ( Y i 1 , . . . , Y iT ) , X i = ( X i 1 , . . . , X iT ) • ε it = f i + U it Heckman Part I

• E ( f i ) = E ( U it ) = 0 • E ( f 2 i ) = σ 2 E ( U 2 it ) = σ 2 f U • E ( f i f i ′ ) = 0 i � = i ′ • U it is iid for all i , t . • Assume that X it is strictly exogenous: • E ( U it + f i | X i 1 , . . . , X iT ) = 0 all t . • Also distribution of the X i = ( X i 1 , . . . , X iT ) ′ does not depend on β . Cov ( ε i , t , ε i , t ′ ) = σ 2 f σ 2 f ρ = (intraclass correlation coefficient) σ 2 f + σ 2 u Heckman Part I

• Look at covariance matrix for ε i = ( ε i 1 , . . . , ε iT ) ′   1 ρ ρ ...   i ε i ) = ( σ 2 f + σ 2 E ( ε ′ u )  = A . ρ ρ  ρ ρ 1 Heckman Part I

• Now stack all of the disturbances (in groups of T ) • ε = ( ε 1 , ε 2 . . . ε I )   0 0 0 A 0 A 0 0   E ( εε ′ ) = Ω � = I =   0 0 0 A   0 0 0 A Heckman Part I

• T × T blocks • stack X i into a supervector X • stack Y i into a supervector Y • Then, we have that GLS estimator is ˆ β GLS = ( X ′ Ω − 1 X ) − 1 ( X ′ Ω − 1 Y ) • OLS applied to the data yields unbiased but inefficient estimators of β (because of exogeneity of X it with respect to ε it ). • But computer program produces the wrong standard errors. • Correct standard errors are: σ 2 ( X ′ X ) − 1 ( X ′ Ω X )( X ′ X ) − 1 • OLS standard errors to assume to be σ 2 ( X ′ X ) − 1 . • ∴ inferences based on OLS models are incorrect. Heckman Part I

• Write A = (1 − ρ ) I + ριι ′ • ι is a T × 1 vector of 1 ’ s   . .   1- ρ 0 0 . ρ ρ ρ ρ .   . ρ ρ ρ ρ  0 1- ρ 0 .    A = +     . ρ ρ ρ ρ   .   0 0 1- ρ .   · · · · · · · · · · · · · · · · · · · · · · · · A − 1 = λ 1 ιι ′ + λ 2 I Heckman Part I

• Where − ρ λ 1 = (1 − ρ )(1 − ρ + T ρ ) 1 λ 2 = 1 − ρ . Heckman Part I

• Proof: AA − 1 = I (direct multiplication). What is the GLS estimator doing? ˆ β GLS = ( X ′ Ω − 1 X ) − 1 ( X ′ Ω − 1 Y )   . . A − 1 0 0 0 .   . .   A − 1 0 0 0 . Ω − 1 =   .   . A − 1 0 0 0 .     . . A − 1 0 0 0 .     X 1 Y 1 X 2 Y 2         X = Y =  . . .  .   .  . .    X I Y I Heckman Part I

• Thus, we conclude that � � − 1 � � I I � � ˆ i A − 1 X i i A − 1 Y i β GLS = X ′ X ′ i =1 i =1 • Use the expression for A − 1 given above: � � I I I X ′ i ιι ′ X i � � � i A − 1 X i = T λ 1 X ′ X ′ + λ 2 i X i T i =1 i =1 i =1 Heckman Part I

• Look at ι ′ X i = X i · T a K vector of means. • Now look at the GLS estimator it is a function of within and between variation. Heckman Part I

• Total sum of squares I � X ′ T XX = i X i i =1 Within Deviations ( T × K ) X i − ιι ′ X i − ι X ′ = T X i i · [ I − ιι ′ = T ] X i Heckman Part I

• Observe: � � � � I − ιι ′ I − ιι ′ = I − ιι ′ (idempotent) T T T X i is T × K , X ′ i is 1 × K . • Thus, we have that within variation is given by I i ( I − ιι ′ � X ′ T ) X i = W XX i =1 I I X ′ i ιι ′ X i � � B XX = T X ′ i · X i · = T i =1 i =1 T XX = W XX + B XX ���� ���� within between Heckman Part I

• GLS estimator is given by taking � � � � I I X ′ i ιι ′ X i X ′ i ιι ′ X i � � + λ 2 + W XX T λ 1 T T i =1 i =1 � � I X ′ i ιι ′ X i � = λ 2 W XX + ( λ 2 + T λ 1 ) T i =1 = λ 2 W XX + ( λ 2 + T λ 1 ) B XX . Heckman Part I

• There is a similar decomposition for other term and we get that ˆ [ λ 2 W XX + ( λ 2 + T λ 1 ) B XX ] − 1 β GLS = · [ λ 2 W XY + ( λ 1 + T λ 2 ) B XY ] Heckman Part I

• Define θ = 1 + T λ 1 ρ = 1 − T (1 − ρ + T ρ ) λ 2 θ = 1 − ρ + T ρ − T ρ 1 − ρ = 1 − ρ + T ρ 1 − ρ + T ρ ˆ β GLS = [ W XX + θ B XX ] − 1 [ W XY + θ B XY ] . • 2 estimators are averaged together: Take first estimator the within estimator. Heckman Part I

• This is simply given by taking derivations from mean: Y it − Y i · = ( X it − X ι · ) β + U it − U i · X i · β + f i + ¯ Y i · = U i · • ∴ subtracting Y i · produces an estimator free of f i . • Doing that eliminates the fixed effects from the model. Thus, we have that ˆ ( W XX ) − 1 W XY = β W ˆ ( B XX ) − 1 B XY β B = Heckman Part I

• Simply average over the groups and we are done. ( W XX )ˆ = β W W XY ( B XX )ˆ β B = B XY [ W XX + θ B XX ]ˆ W XX ˆ β W + θ ( B XX )ˆ = β GLS β B • ∴ we have that β GLS = [ W XX + θ B XX ] − 1 [ W XX ˆ ˆ β W + θ B XX ˆ β B ] . • For a scalar regressor ˆ β GLS lies between ˆ β W and ˆ β B . • (But not, necessarily so, for the general regressor case). Heckman Part I

• Now suppose ρ = 0 = ⇒ λ 1 = 0 = ⇒ θ = 1 . • Then ˆ β GLS is simply OLS. Suppose that ρ = 1. • A is singular, A − 1 doesn’t exist. • If we have that regressors are fixed over the spell for the case W XX = 0 and GLS is between estimator. Heckman Part I

• Suppose that T → ∞ , ρ � = 0 ( T λ 1 ) − ρ T lim = lim (1 − ρ + T ρ ) λ 2 T →∞ T →∞ − ρ = lim → − 1 1 − ρ T →∞ + ρ T • ∴ θ = 0 • ∴ [ˆ β GLS = ˆ β W ]. Heckman Part I

• In this case, the within estimator is the efficient estimator. • Now A − 1 matrix itself can be written in an interesting fashion and provides an example of another interpretation of the estimator. • A − 1 = λ 2 ( I − k ιι ′ ) where k is given by − λ 1 λ 2 [ I − c ιι ′ ][ I − c ιι ′ ] I − c ιι ′ − c ιι ′ + Tc 2 ιι ′ = I − (2 c − Tc 2 ) ιι ′ = where 2 c − Tc 2 = − λ 1 F = I − c ιι ′ . , λ 2 Heckman Part I

• Solve to get � � � c = 1 1 − ρ 1 − 1 − ρ + ρ T T • ∴ GLS estimator is of the form � � − 1 � � I I � � ˆ i A − 1 X i ) i A − 1 Y i β GLS = ( X ′ X ′ . i =1 i =1 Heckman Part I

• But this is ⇔ to transforming the data in the following way Y i = X i β + ε i = FX i β + F ε i FY i FY i = Y i − ( cT ι ) Y i · ι ′ Y i Y i · = T • The mean value of Y i for person i over sample period T . FX i = X i − ( cT ) ι X ′ i . • Now, suppose that we have ρ = 0 , c = 0 , GLS is OLS applied to data. Heckman Part I

Standard Errors for Fixed Effect; Estimator IS Produced by OLS Formula; GLS is OLS Heckman Part I

• Proof:         0 0 Y 1 ι Y 2 0 ι 0         =  =  f i +  f 2 + Y  f N         · · · · · · · · · · · ·     Y N 0 0 ι     X 1 U 1  β + + · · · · · ·    X N U N     Y i 1 X i 1 Y i = · · · X i = · · ·     Y iT X iT i ) = σ 2 I T E ( U i U ′ E ( U i U ′ j ) = 0 , i � = j . Heckman Part I

• Define F = I − ιι ′ T . • Partition Y i = X i β + if i + U i Heckman Part I

• Using ( F ) + ιι ′ T = I � ιι ′ � F = 0 T FY i = FX i β + FU i ιι ′ T Y i = ιι ′ T X i β + ιι ′ T U i � � − 1 � � I I � � ˆ X ′ X ′ β W = i FX i i FY i . i =1 i =1 Heckman Part I

• Let � � I � B = X ′ i FX i . i =1 �� � � I N � � Var ˆ = ( B ) ′ E X ′ U ′ β W i FU i i FX i B i =1 i =1 � � I � σ 2 = B ′ ( X ′ i F ) FX ′ B U i i =1 � σ 2 i FX i ) − 1 = U ( X ′ where � ιι ′ � � ιι ′ � = ιι ′ T . T T • This is OLS variance covariance. Heckman Part I

• Between Estimator: � N �� − 1 � N � � ιι ′ ιι ′ � � ˆ = X i T X ′ X i T Y i β B i i =1 i =1 N N ιι ′ ii ′ � � = β + X i T U i + X i T f i i =1 i =1 • Now ˆ β B uncorrelated with ˆ β W because f i ⊥ ⊥ U j all i , j ; Heckman Part I

�� � − 1 � σ 2 � � � ιι ′ ˆ T + σ 2 U = T X ′ Var β B X i i f �� X i ιι ′ X ′ � �� � − 1 ιι ′ i · T X ′ X i i T � � �� � − 1 f + σ 2 ιι ′ σ 2 U = X i T X ′ . i T Heckman Part I