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  • Multinomial logit with sample selection

    Dear everyone,


    I am looking for something similar to Heckman selection model/svysemlog with a modification.

    I have a selectiion variable with two values (0 and 1) in the first step, and a mulitnomial non-ordinal categorical variable (with six categories) in the second step.
    I am interested only in positive (1) values in the first step (around 30% of the total sample).

    What I did at the first place was a. logit analysis for the first step b. multinomial logit for the second step. However, I was advised to use the Heckman selection model for multiple reasons.
    However, if I am not mistaken, Heckman (and svysemlog) cannot be used if the outcome variable is a non-ordinal variable.


    I have two questions:


    a. Is there any Stata package that adresses my problem?
    b. Do you have any advice how to proceed, in case there is no ready-made solution in Stata?


    Thanks in advance!




  • #2
    Hi Mile,
    I do not think there is something already made for this type of model in Stata. However, i can see two avenues for proceeding.
    I would take a look into GSEM and CMP (user written command). Not sure about gsem, but for CMP you will have to change your model slightly and go from a mlogit to an mprobit.
    HTH
    Fernando

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    • #3
      Hi Fernando,



      Thanks a lot! So, if I understand you correctly, the CMP allows for logit (or probit) (1st step) and mprobit (2nd step)?



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      • #4
        Hi Mile
        Yes, from what i learned about -cmp-, it allows you to estimate simultaneous models under the assumption of normality. I think aspart of the examples, -cmp- also has a section that replicates ordered probts, as well as heckman selection model (MLE) . I imagine it should be relatively easy to modify those examples to estimate a probit in your first stage, and mprobit in the second step.
        if you hit a wall, the author of the command is pretty active in the forum, and may be able to give you a hint if something goes wrong.
        Best
        Fernando

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        • #5
          Hi @Fernando,

          Thank you once again!

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