# Instrumental variable analysis for non-linear endogenous variable

I have a model of the form $$y_i = ax_{1i} + bx_{1i}^2 + cx_{3i} + \varepsilon_i$$ Where $$x_1$$ is an endogenous variable. Would I need instruments for both $$x_1$$ and $$x_1^2$$ or do I only need to implement the first stage of 2SLS for $$x_1$$?

Thanks, Nitin

• You should do both. Sometimes a natural extra one is the instrument squared.
– oda
Commented Apr 19, 2022 at 18:59
• Thanks, Oda. Suppose if we have an interaction term involving an endogenous variable with an exogenous one, would then also we would require an instrument for the interaction term? Commented Apr 20, 2022 at 22:03
– oda
Commented Apr 21, 2022 at 8:58

If $$Cov(x_1, \varepsilon_i)\ne 0$$, it is probably the case that $$Cov(x_1^2, \varepsilon)\ne 0$$, and you would need an instrument for $$x_1^2$$.
If $$z$$ is your instrument for $$x_1$$, then as Oda said, it is natural for $$z^2$$ to be the instrument for $$x_1^2$$.
If you have an interaction term, $$x_1\cdot w$$, for which $$Cov(w,\varepsilon)= 0$$, then becasuse $$Cov(x_1, \varepsilon)\ne 0$$ it is probably the case that $$Cov(x_1\cdot w,\varepsilon)\ne 0$$ and you would need an instrument. The natural choice of an instrument is $$z\cdot w$$.
Note that $$Cov(z,\varepsilon)=0$$ and $$Cov(w,\varepsilon)= 0$$ does not imply that $$Cov(z\cdot w,\varepsilon)=0$$. Thus, instrument validity would require a separate assumption.