Normal Selection Model Results from Heckman and Honoré (1990) James J. Heckman University of Chicago This draft, March 28, 2006 1
�� � The properties of the normal selection model are generated by the properties of a truncated normal model which we now es- tablish. See Heckman and Honoré (1990). Let � be a standard def normal random variable and let � ( � ) � � [ � | � � � ] . For all � � ( �� � � ) , we prove the following results: n o � � 2 1 2 � exp 2 (N-1) � ( � ) = � max { 0 � � } � � ( � � ) (N-2) 0 � �� ( � ) = � 0 ( � ) = � ( � )( � ( � ) � � ) � 1 � (N-3) � 2 � ( � ) � 0 � �� 2 (N-4) 0 � � �� [ � | � � � ] = 1 + � ( � ) � � � 2 ( � ) � 1 � 2
� ��� �� � �� �� �� � (N-5) �� �� [ � | � � � ] � 0 � � [( � � � ( � )) 3 | � � � ] (N-6) = � 2 � ( � ) ¡ ¢ 2 � 2 ( � ) � 3 �� ( � ) + � 2 � 1 = � ( � ) �� 2 (N-7) � [ � | � � � ] � ���� [ � | � � � ] � (N-8) lim � ��� � ( � ) = 0 � lim � �� � ( � ) = � � �� ( � ) �� ( � ) (N-9) lim = 0 � lim = 1 � (N-10) lim � ��� � �� [ � | � � � ] = 1 � lim � �� � �� [ � | � � � ] = 0 � 3
Results (N-2), (N-4) and (N-5) are implications of log concav- ity. (N-7) is an implication of symmetry and log concavity. (N-1) and (N-3) are consequences of normality. The left hand side limits of (N-8) and (N-10) are true for any distribution. So is the right hand limit of (N-8) provided that the support of � is not bounded on the right. The right hand limits of (N-9) and (N-10) are consequences of normality. 4
�� � �� � � �� � � � 1 Proofs of Results (N-1) to (N-10) The moment generating function for a truncated normal dis- tribution with truncation point � is: µ ¶ R � 1 � 1 2 � 2 2 � exp ��� ( � ) = � �� 2 µ ¶ R � 1 � 1 2 � exp 2 � 2 The equality in (N-1) follows from: ¯ ¯ ¯ � ( � ) = � [ � | � � � ] = ���� ¯ � =0 The inequality is obvious. 5
�� � � By direct calculation, � 0 ( � ) = � ( � )( � ( � ) � � ) . Now note that: � ] = � 2 ��� � [ � 2 | � | � =0 = 1 + � ( � ) � �� 2 � ] = � 3 ��� � [ � 3 | � | � =0 = � ( � )(2 + � 2 ) � �� 2 Therefore: � �� [ � | � � � ] = 1 + � 2 ( � ) = 1 � �� ( � ) �� � As Var [ � | � � � ] � 0 and � ( � )( � ( � ) � � ) � 0 by (N-1), this proves (N-2) and (N-4). To prove (N-3) notice that Var [ � | � � � ] = 1 � �� ( � ) �� , and therefore: � 2 � ( � ) = � � Var [ � | � � � ] � 0 � �� 2 6
� �� �� �� � �� �� where the inequality follows from Proposition 1 in Heckman and Honore (1990). (N-5) also follows from Proposition 1, whereas (N-6) follows by direct calculation from the expression for � [( � � � ( � )) 3 | � � � ] . (N-7) is trivial. (N-8) is obvious. The first part of (N-9) follows directly from � 0 Hôpital’s rule. (N-2) and (N-3) imply that �� ( � ) is increasing �� ( � ) and bounded by 1. Therefore ��� exists and does not �� ( � ) exceed 1. If ��� � 1 then � ( � ) would eventually be less than � , contradicting (N-1). This proves the second part of (N-9). (N-9) and (N-4) imply (N-10). 7
The Truncated Normal Analysis Truncated Standard Normal Density Function 1.6 p. d .f. o f Z | Z > -1 p. d .f. o f Z | Z > -0.5 1.4 p. d .f. o f Z | Z > 0 C onditional p. d .f. o f Truncated Z ~ N (0,1) p. d .f. o f Z | Z > 0.5 p. d .f. o f Z | Z > 1 1.2 1 0.8 0.6 0.4 0.2 0 -2 -1 0 1 2 3 4 5 6 Z Z ∼ N 0,1 8
The Truncated Standard Normal Ex p ectation Truncated Standard Normal Expectation 4 3.5 3 2.5 E (Z|Z> c ) 2 1.5 1 0.5 0 -4 -3 -2 -1 0 1 2 3 4 c E Z|Z > c ; Z ∼ N 0,1 9
The Truncated Standard Normal Variance Truncated Standard Normal Variance 1 v ar (Z|Z> c ) = 1-( E (Z- c |Z> c )*( E (Z|Z> c ) 0.9 0.8 0.7 0.6 v ar (Z|Z> c ) 0.5 0.4 0.3 0.2 0.1 0 -4 -3 -2 -1 0 1 2 3 4 c v ar Z|Z > c ; Z ∼ N 0,1 10
Truncated Standard Normal Expectations 4 E (Z|Z> c ) E (Z- c |Z> c ) 3.5 3 2.5 E (Z|Z> c ) 2 1.5 1 0.5 0 -4 -3 -2 -1 0 1 2 3 4 c E Z|Z > c and E Z − c |Z > c ; Z ∼ N 0,1 11
Truncated Standard Normal Expectations 2 E (Z|Z> c ) 1.8 E (Z- c |Z> c ) E (Z|Z> c )* E (Z- c |Z> c ) 1.6 1.4 ( E (Z- c |Z> c )*( E (Z|Z> c ) 1.2 1 0.8 0.6 0.4 0.2 0 -4 -3 -2 -1 0 1 2 3 4 c E Z|Z > c and E Z − c |Z > c ; Z ∼ N 0,1 12
Truncated Standard Normal Density Function 0.8 X | Y > 0, ρ = -0.9 p. d .f. o f X conditional on Y>0; [X,Y] ~ Standard Normal X | Y > 0, ρ = -0.75 0.7 X | Y > 0, ρ = -0.5 X | Y > 0, ρ = -0.25 0.6 X | Y > 0, ρ = 0 0.5 0.4 0.3 0.2 0.1 0 -4 -3 -2 -1 0 1 2 3 4 X ρ X 0 1 ∼ N , ρ Y 0 1 13
Truncated Standard Normal Density Function 0.8 X | Y > 0, ρ = 0 p. d .f. o f X conditional on Y>0; [X,Y] ~ Standard Normal 0.7 X | Y > 0, ρ = 0.25 X | Y > 0, ρ = 0.5 X | Y > 0, ρ = 0.75 0.6 X | Y > 0, ρ = 0.9 0.5 0.4 0.3 0.2 0.1 0 -4 -3 -2 -1 0 1 2 3 4 X ρ X 0 1 ∼ N , ρ Y 0 1 14
Truncated Standard Normal Density Function 0.8 X|Y>0, ρ =0.5, μ X =0, μ Y =0 p. d .f. o f X conditional on Y>0; [X,Y] ~ Standard Normal X|Y>0, ρ =0.5, μ X =0.5, μ Y =0 0.7 X|Y>0, ρ =0.5, μ X =1, μ Y =0 0.6 X|Y>0, ρ =0.5, μ X =1.5, μ Y =0 X|Y>0, ρ =0.5, μ X =2, μ Y =0 0.5 0.4 0.3 0.2 0.1 0 -3 -2 -1 0 1 2 3 4 5 6 X μ X X 1 0.5 ∼ N , Y 0 0.5 1 15
Truncated Standard Normal Density Function 0.8 X|Y>0, ρ =0.5, μ Y =0, μ X =0 p. d .f. o f X conditional on Y>0; [X,Y] ~ Standard Normal X|Y>0, ρ =0.5, μ Y =0.5, μ X =0 0.7 X|Y>0, ρ =0.5, μ Y =1, μ X =0 0.6 X|Y>0, ρ =0.5, μ Y =1.5, μ X =0 X|Y>0, ρ =0.5, μ Y =2, μ X =0 0.5 0.4 0.3 0.2 0.1 0 -4 -3 -2 -1 0 1 2 3 4 X X 0 1 0.5 ∼ N , μ Y Y 0.5 1 16
Extended Roy Model Treatment Effects and Weights for the Extended Roy Model MTE (U D ), I V θ 1 (U D ), I V θ 2 (U D ), I V p ro p ensity Score (U D ),OL S w ei g hts (U D ) 0.4 ATE (U D ) MTE (U D ) MTE (U D ) I V θ 1 (U D ) 0.35 0.6 TT (U D ) I V θ 2 (U D ) T U T (U D ) I V p ro p ensity Score (U D ) 0.3 0.5 OL S w ei g hts (U D ) ATE , MTE , TT , T U T 0.25 0.4 0.2 0.3 0.15 0.2 0.1 0.1 0.05 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 U D U D Treatment Values conditional on Propensity Score Weights conditional on Propensity Score Model and Parameters Y = D ⋅ Y 1 + 1 − D ⋅ Y 0 ; D = 1 γ Z − V > 0 ; V ∼ N 0,1 γ Z = 0.2 + 0.3 ⋅ Z 1 + 0.1 ⋅ Z 2 ; U 1 = − 0.012 ⋅ V ; U 0 = 0.05 ⋅ V ; Y 1 = 0.04 + 0.8 ⋅ X 1 + 0.4 ⋅ X 2 + U 1 ; Y 0 = 0.22 + 0.5 ⋅ X 1 + 0.1 ⋅ X 2 + U 0 ; − 2 − 1 X 1 4 0 Z 1 9 0 ∼ N ∼ N , ; , X 2 2 0 4 Z 2 1 0 9 17
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