Normal Selection Model Results from Heckman and Honor (1990) James - - PowerPoint PPT Presentation

Normal Selection Model Results from Heckman and Honoré (1990)

James J. Heckman University of Chicago This draft, March 28, 2006

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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

normal random variable and let ()

def

[ | ]. For all ( ), we prove the following results: (N-1) () =

1

• 2 exp

n 2

2

• ()

max{0 } (N-2) 0 ()

• = 0() = ()(() ) 1

(N-3) 2() 2 (N-4) 0 [ | ] = 1 + () 2() 1

2

(N-5) [ | ]

• (N-6)

[( ())3 | ] = () ¡ 22() 3() + 2 1 ¢ = 2() 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.

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1 Proofs of Results (N-1) to (N-10)

The moment generating function for a truncated normal dis- tribution with truncation point is: () = 2 R

• 1
• 2 exp

µ 1 22 ¶

• R
• 1
• 2 exp

µ 1 22 ¶

• The equality in (N-1) follows from:

() = [ | ] =

• ¯

¯ ¯ ¯

=0

The inequality is obvious.

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By direct calculation, 0() = ()(() ). Now note that: [2 |

• ] = 2

2 |=0 = 1 + () [3 |

• ] = 3

2 |=0 = ()(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() 2 = Var[ | ]

• 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).

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The Truncated Normal Analysis

Truncated Standard Normal Density Function

• 2
• 1

1 2 3 4 5 6 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Z Conditional p.d.f. of Truncated Z ~ N(0,1) p.d.f. of Z | Z > -1 p.d.f. of Z | Z > -0.5 p.d.f. of Z | Z > 0 p.d.f. of Z | Z > 0.5 p.d.f. of Z | Z > 1

Z ∼ N0,1

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The Truncated Standard Normal Expectation

Truncated Standard Normal Expectation

• 4
• 3
• 2
• 1

1 2 3 4 0.5 1 1.5 2 2.5 3 3.5 4 E(Z|Z>c) c

EZ|Z > c; Z ∼ N0,1

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The Truncated Standard Normal Variance

Truncated Standard Normal Variance

• 4
• 3
• 2
• 1

1 2 3 4 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 var(Z|Z>c) c var(Z|Z>c) = 1-(E(Z-c|Z>c)*(E(Z|Z>c)

varZ|Z > c; Z ∼ N0,1

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Truncated Standard Normal Expectations

• 4
• 3
• 2
• 1

1 2 3 4 0.5 1 1.5 2 2.5 3 3.5 4 E(Z|Z>c) c E(Z|Z>c) E(Z-c|Z>c)

EZ|Z > c and EZ − c|Z > c; Z ∼ N0,1

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Truncated Standard Normal Expectations

• 4
• 3
• 2
• 1

1 2 3 4 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 (E(Z-c|Z>c)*(E(Z|Z>c) c E(Z|Z>c) E(Z-c|Z>c) E(Z|Z>c)*E(Z-c|Z>c)

EZ|Z > c and EZ − c|Z > c; Z ∼ N0,1

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Truncated Standard Normal Density Function

• 4
• 3
• 2
• 1

1 2 3 4 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 X p.d.f. of X conditional on Y>0; [X,Y] ~ Standard Normal X | Y > 0, ρ = -0.9 X | Y > 0, ρ = -0.75 X | Y > 0, ρ = -0.5 X | Y > 0, ρ = -0.25 X | Y > 0, ρ = 0

X Y ∼ N , 1 ρ ρ 1

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Truncated Standard Normal Density Function

• 4
• 3
• 2
• 1

1 2 3 4 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 X p.d.f. of X conditional on Y>0; [X,Y] ~ Standard Normal X | Y > 0, ρ = 0 X | Y > 0, ρ = 0.25 X | Y > 0, ρ = 0.5 X | Y > 0, ρ = 0.75 X | Y > 0, ρ = 0.9

X Y ∼ N , 1 ρ ρ 1

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Truncated Standard Normal Density Function

• 3
• 2
• 1

1 2 3 4 5 6 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 X p.d.f. of X conditional on Y>0; [X,Y] ~ Standard Normal X|Y>0, ρ=0.5, μX=0, μY=0 X|Y>0, ρ=0.5, μX=0.5, μY=0 X|Y>0, ρ=0.5, μX=1, μY=0 X|Y>0, ρ=0.5, μX=1.5, μY=0 X|Y>0, ρ=0.5, μX=2, μY=0

X Y ∼ N μX , 1 0.5 0.5 1

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Truncated Standard Normal Density Function

• 4
• 3
• 2
• 1

1 2 3 4 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 X p.d.f. of X conditional on Y>0; [X,Y] ~ Standard Normal X|Y>0, ρ=0.5, μY=0, μX=0 X|Y>0, ρ=0.5, μY=0.5, μX=0 X|Y>0, ρ=0.5, μY=1, μX=0 X|Y>0, ρ=0.5, μY=1.5, μX=0 X|Y>0, ρ=0.5, μY=2, μX=0

X Y ∼ N μY , 1 0.5 0.5 1

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Extended Roy Model

Treatment Effects and Weights for the Extended Roy Model

0.2 0.4 0.6 0.8 1 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 UD ATE, MTE, TT, TUT ATE(UD) MTE(U D) TT(U D) TUT(U D) 0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 UD MTE(U D), IVθ1(UD), IVθ2(UD), IVpropensity Score (UD),OLSweights(UD) MTE(U D) IVθ1(UD) IVθ2(UD) IVpropensity Score (UD) OLSweights(UD)

Treatment Values conditional on Propensity Score Weights conditional on Propensity Score Model and Parameters Y = D ⋅ Y1 + 1 − D ⋅ Y0; D = 1γZ − V > 0; V ∼ N0,1 γZ = 0.2 + 0.3 ⋅ Z1 + 0.1 ⋅ Z2; U1 = −0.012 ⋅ V; U0 = 0.05 ⋅ V; Y1 = 0.04 + 0.8 ⋅ X1 + 0.4 ⋅ X2 + U1; Y0 = 0.22 + 0.5 ⋅ X1 + 0.1 ⋅ X2 + U0; X1 X2 ∼ N −2 2 , 4 0 0 4 ; Z1 Z2 ∼ N −1 1 , 9 0 0 9

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