Instrumental variables with 2 datasets

[Olga Shanks]

Hello,

This is not a Stata-specific question, but I think this may still be the best forum to ask about IV. I have 2 datasets: 1 small where I have outcome, treatment, and instrument variables, and 1 large dataset where I only have instrument and outcome variables. Is it "legal" to do the first-stage regression using dataset 1, then use the first-stage estimated model to generate predicted values of the treatment in dataset 2, and then run the second stage in dataset 2? And if it is legal, what are the implications for standard errors and any other pitfalls?

Thank you for any help.
Olga

[Dario Maimone Ansaldo Patti]

I would say no. I do not understand why you do not match the two datasets. If you want to use fitted values from one dataset to the other, i guess you can match them and using ivreg or ivreg2. However, without having more information about your datasets, it is difficult to give you an answer. On top of this, if you want to make an IV estimation manually, using the fitted values, you also need to correct standard errors in the second stage.

[Olga Shanks]

Thank you for responding, Dario. The "small" dataset has only 420 observations, while the "large" one has 20,000. The IV works out using just the small dataset. But I wanted to be more persuasive, I guess. If I can't use the 20,000 observations, I feel like it's a waste. It's such a rich dataset but does not have the instrument.
But I agree with you. I shouldn't use the method incorrectly just because of my data.

Olga

[Dario Maimone Ansaldo Patti]

If you do not have instruments (or they are difficult to find), you can still estimate your model using an IV estimator following the procedure described here. The user written routine ivreg2h performs the estimation explained in the paper, while this presentation describes stata routine.

I hope it helps.

[David Radwin]

See https://www.statalist.org/forums/for...65#post1594965

[Olga Shanks]

David and Dario, thank you very much! These are helpful.