I have the following outcome (second-stage) equation:

$$y = \beta_0 + \beta_1w + \beta_2x + \beta_3w x + \cdots$$

$y$, $w$ and $x$ are all binary. Both $w$ and $x$ are endogenous, but I have an excellent instrument for $x$ ($z$ say). As such, I also can write a selection (first-stage) equation:

$$x = \alpha_0 + \alpha_1w + \alpha_2z + \cdots$$

The typical approach to estimating the above equations is to use joint likelihood methods (e.g. a recursive bivariate probit model, which assumes that the errors of the treatment and outcome equations are jointly distributed following a standard bivariate normal distribution with correlation $\rho$).

Let's say that I do this. Now my question is: can I interpret the coefficient $\beta_3$ in the outcome equation, even though $w$ is also endogenous and I do not have a valid instrument for this?

Why might this not be possible?

  • Even if $x$ can now effectively be treated as exogenous, the interaction $w*x$ is still endogenous, because $w$ is endogenous.

Why might this be possible?

  • I found a paper (https://www.tandfonline.com/doi/pdf/10.1080/07474938.2018.1427486) which has the same model as in the outcome equation above, but where $w$ is endogenous and $x$ is exogenous.

  • In this case, they show that the OLS estimate of $\beta_3$ is asymptotically valid even if you do not use IV techniques. So the interpretation of the coefficient of an interaction using OLS is valid, even though the variable is itself endogenous.

Two differences between their set-up and ours:

  1. In theirs, $x$ is exogenous. In ours, it is endogenous, but we instrument is (so can we treat it as "exogenous" as that point)?
  2. They show this holds for OLS, we are using probit link functions and bivariate error dependencies. It is not clear to me that this result would extent to our case?

Any thoughts, advice, or references on this would be greatly appreciated!


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