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  • heterogeneous treatment effects and ivregress

    I have a question regarding the heterogeneous treatment effects. For example, I want to estimate the heterogeneous treatment effects of informal caregiving on caregivers' health by household income. Let y denote the binary health outcome (1=good; 0=poor), CG is the caregiving status (1=caregivers; 0 otherwise), Z1 is an instrumental variable for CG and is a continuous variable taking values from [0,1], HHinc is a binary indicating whether the HH is rich or poor (1=rich; 0 otherwise), I also have age and edu as additional covariates.

    My primary purpose is to estimate the heterogeneous treatment effects by HH income. CG is a binary variable but I used c.CG instead of i.CG because I have just one instrumental variable. I wanted to confirm if this is the right way to go about it. I have also controlled for HHinc.
    ivregress 2sls y (CG c.CG#i.HHinc = Z1 c.Z1#i.HHinc) age edu HHinc, robust
    I'd appreciate your help. Thanks!

    (similar question was posted by somebody in 2021 but wasnt answered)

  • #2
    Given the interaction, I'd consider a control function.

    Comment


    • #3
      Originally posted by George Ford View Post
      Given the interaction, I'd consider a control function.
      Thank you for your response. This is the first time I have heard about the control function. I went over some material. Would it be possible for you to please confirm if this looks better? Thank you in advance!

      CG: binary
      Z: continuous between [0,1]
      HHinc: binary
      health: binary

      Stage 1:
      logit CG Z1 c.Z1#i.HHinc age edu HHinc

      predict v_hat, residuals

      Stage 2:
      logit health c.CG##i.HHinc v_hat c.v_hat#i.HHinc age edu HHinc

      Comment


      • #4
        Guest: If you’re going to to this, you should use biprobit. I’ll show how later.
        Last edited by sladmin; 19 Nov 2024, 08:39. Reason: anonymize original poster

        Comment


        • #5
          Because your endogenous variable CG is binary, the CF method is at best an approximation. If you're happy with an approximation, then you might as well use the linear method you proposed -- which is perfectly fine. If you want to take the binary nature of health and CG seriously, this is what I would do:

          Code:
          biprobit (health = i.CG c.HHinc i.CG#c.HHinc age edu) (CG = Z c.HHinc age educ)
          margins, at(CG = (0,1)) predict(pmarg1) post
          lincom _b[2._at] - _b[1._at]
          Note how you don't need another IV for i.CG#c.HHinc because this is maximum likelihood estimation. You are free to add c.Z#c.HHinc to the equation for CG if you want to. It can help strengthen the instruments. The margins + lincom commands estimate the effect on the probabilities -- the average partial effects.

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          • #6
            Originally posted by Jeff Wooldridge View Post
            Because your endogenous variable CG is binary, the CF method is at best an approximation. If you're happy with an approximation, then you might as well use the linear method you proposed -- which is perfectly fine. If you want to take the binary nature of health and CG seriously, this is what I would do:
            oh okay, makes sense

            Code:
            biprobit (health = i.CG c.HHinc i.CG#c.HHinc age edu) (CG = Z c.HHinc age educ)
            margins, at(CG = (0,1)) predict(pmarg1) post
            lincom _b[2._at] - _b[1._at]
            Originally posted by Jeff Wooldridge View Post
            Note how you don't need another IV for i.CG#c.HHinc because this is maximum likelihood estimation. You are free to add c.Z#c.HHinc to the equation for CG if you want to. It can help strengthen the instruments. The margins + lincom commands estimate the effect on the probabilities -- the average partial effects.
            Thank you very much Dr. Wooldridge for your detailed instructions. I appreciate your help!

            Comment


            • #7
              I missed the fact the two variables of interest were binary. Sorry, Glad Jeff jumped in.

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