手动选择R模型中的Heckman

Question

我想在R中手动计算一个Heckman选择模型.我的问题是标准误差是有偏差的.有没有办法手动纠正这些？

在sampleSelection模型的我的(样本)代码下面(正确的SE)和手动代码(正确的估计,错误的SE)

 require(sampleSelection)

data( Mroz87 )
Mroz87$kids <- ( Mroz87$kids5 + Mroz87$kids618 > 0 )

使用sampleSelection

heckman <- selection(selection = lfp ~ age + I(age^2) + faminc + kids + educ, outcome = wage ~ exper + I(exper^2) + educ + city, 
                                data = Mroz87, method = "2step")
summary(heckman)

手动

seleqn1 <- glm(lfp ~ age + I(age^2) + faminc + kids + educ, family=binomial(link="probit"), data=Mroz87)
summary(seleqn1)

# Calculate inverse Mills ratio by hand ##
Mroz87$IMR <- dnorm(seleqn1$linear.predictors)/pnorm(seleqn1$linear.predictors)

# Outcome equation correcting for selection ## ==> correct estimates, wrong SEs
outeqn1 <- lm(wage ~ exper + I(exper^2) + educ + city + IMR, data=Mroz87, subset=(lfp==1))
summary(outeqn1)

Answer 1

myprobit    <- probit(lfp ~ age + I(age^2) + faminc + kids + educ - 1, x = TRUE, 
                           iterlim = 30, data=Mroz87)

imrData     <- invMillsRatio(myprobit) # same as yours in this particular case
Mroz87$IMR1 <- imrData$IMR1

outeqn1     <- lm(wage ~ -1 + exper + I(exper^2) + educ + city + IMR1, 
                  data=Mroz87, subset=(lfp==1))

最主要的是你使用拦截模型而不是无拦截。