Geek Answers Handbook

Why can't I use the dependent variable in the selection equation as an independent variable in the outcome equation?

I use the example from the sampleSelection package, where I switch the dependent variable for the equation of the result to family income, because I want the woman’s participation in the workforce, lfp, to be a logical determinant.

library("sampleSelection")
data( Mroz87 )
Mroz87$kids <- ( Mroz87$kids5 + Mroz87$kids618 > 0 )
selection2 <- selection(selection = lfp ~ age + I(age^2) + faminc + kids + educ, 
                    outcome = faminc ~ lfp + wage + age + exper, 
                    data = Mroz87, method = "2step")

My problem is that adding lfp as an independent variable to the result equation prevents the solution. I get:

Error in solve.default(crossprod(xMat)) : 
Lapack routine dgesv: system is exactly singular: U[6,6] = 0

Removing lfp from the result allows you to decide, but not the one I'm looking for. I seem to read that the problem is the matrix math that works under it (a feature prevents inversion, maybe?). Is there a way to get the solution I want, maybe work? Please, help.

+4
1

tobit-2, , , , faminc , lfp = 1. , lfp 1 . , ( ).

sampleSelection, ( r-forge /treatmentReg ). .

0

More articles:


All Articles