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

CC BY-SA 4.0
2022-04-22T16:00:58.323Z
Thanks a lot, Michael and Oda. Could you also suggest a textbook/article where the non-linear and interaction-based endogenous variables are discussed. Will help in citing these in my paper.
PhD Student
CC BY-SA 4.0
2022-04-25T08:43:22.293Z
https://www.tandfonline.com/doi/full/10.1080/07474938.2018.1427486
Michael Gmeiner
CC BY-SA 4.0
2022-05-05T07:27:14.483Z
Hi, Suppose instead of IV 2SLS, if we use the control function approach, would we then also need to estimate separate error terms for the endogenous variable, its quadratic term and interaction term and use these three error terms in the final model?
PhD Student
CC BY-SA 4.0
2022-04-21T07:32:08.753Z
2022-04-21T07:32:08.753Z
Michael Gmeiner
<p>I have a model of the form <span class="math-container">$$y_i = ax_{1i} + bx_{1i}^2 + cx_{3i} + \varepsilon_i$$</span> Where <span class="math-container">$x_1$</span> is an endogenous variable. Would I need instruments for both <span class="math-container">$x_1$</span> and <span class="math-container">$x_1^2$</span> or do I only need to implement the first stage of 2SLS for <span class="math-container">$x_1$</span>?</p>
<p>Thanks,
Nitin</p>

CC BY-SA 4.0
2022-04-19T18:59:28.763Z
You should do both. Sometimes a natural extra one is the instrument squared.
oda
CC BY-SA 4.0
2022-04-20T22:03:20.893Z
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?
PhD Student
CC BY-SA 4.0
2022-04-21T08:58:33.120Z
Yes. See @MichaelGmeiner's answer.
oda
<econometrics><endogeneity>
Instrumental variable analysis for non-linear endogenous variable
PhD Student
# Instrumental variable analysis for non-linear endogenous variable

(Asked by: username_0 on 2022-04-19)

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

**Tags:** econometrics, endogeneity
> **Comments:**
> **username_1**: You should do both. Sometimes a natural extra one is the instrument squared.  
> **username_0**: 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?  
> **username_1**: Yes. See @MichaelGmeiner's answer.  

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## 1 answers

### Answer by username_2 on 2022-04-21. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: Thanks a lot, Michael and Oda. Could you also suggest a textbook/article where the non-linear and interaction-based endogenous variables are discussed. Will help in citing these in my paper.  
> **username_2**: https://www.tandfonline.com/doi/full/10.1080/07474938.2018.1427486  
> **username_0**: Hi, Suppose instead of IV 2SLS, if we use the control function approach, would we then also need to estimate separate error terms for the endogenous variable, its quadratic term and interaction term and use these three error terms in the final model?  

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