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I'm working on a a bivariate logistic regression. But I have an endogeneity problem and I want solve it through 2sls with 2 instrumental variables.

My thought was to regress (OLS) in first stage and later a bilogit in second stage. Is this a suitable analysis? Is it a "forbidden regression"? What would be the best option that helps to solve it?

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    $\begingroup$ If your only question is about (the existence of R) code, it is off topic here. If you have a statistical question about 2sls when 1 stage is a binary variable, please edit to clarify. $\endgroup$ – gung - Reinstate Monica♦ Aug 11 '17 at 15:36
  • $\begingroup$ I see what you mean. Ok. On other hand, to carry out a 2sls with ols in 1st stage and logit in 2nd is a "forbbiden regression", isn't it? thanks for you answer, gung. $\endgroup$ – SMD Aug 11 '17 at 15:49
  • $\begingroup$ If you have a statistical question about 2sls when 1 stage is a binary variable, please edit your question to focus on that. Otherwise, this will end up being closed. $\endgroup$ – gung - Reinstate Monica♦ Aug 11 '17 at 16:00
  • $\begingroup$ I've edited question, gung $\endgroup$ – SMD Aug 11 '17 at 16:17
  • $\begingroup$ This question is now on topic here. I'm voting to leave open. $\endgroup$ – gung - Reinstate Monica♦ Aug 11 '17 at 17:14
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Short answer : No, you can't. That is indeed forbidden regression what you're trying to do.

Possible solutions : some brief discussion, google the keywords to learn more.

  1. Use LPM(linear probability model) - just use regular 2SLS with your data. sounds crazy, but a lot of people do this in economics literature. Make sure that you get the standard errors right.
  2. MLE - specify the distribution of your first stage equation, jointly with the second stage binary outcome equation. (i.e. use probit model with bi-variate normal errors)
  3. Control function approach - basically add the residuals from the first stage equation to covariates in the second stage.

for details, see Blundell & Powell(2004) or Rivers & Vuong(1988).

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  • $\begingroup$ If I could, I would rate your answer as useful. Thank you so much $\endgroup$ – SMD Nov 17 '17 at 20:46

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