Let the model $y=x+x^2-1$ be exactly correct, where $x\sim N(0,1)$, then $x^2\sim \chi^2_{(1)}$. Say we want to estimate the model $y=\beta x-1+\epsilon$ by least squares. Let $(y_i,x_i)_{i=1}^n\overset{iid}{\sim}$. So we should have an omitted variable bias. Solving for $\beta$, you get $$\overset{\wedge}{\beta}\rightarrow\frac{cov(x,y)+E(x)}{var(x)}=\frac{cov(x,x+x^2)+E(x)}{var(x)}=\beta+\frac{cov(x,x^2)}{var(x)}+\frac{E(x)}{var(x)}=\beta$$ as $cov(x,x^2)=0$, and $E(x)=0$. So it seems like we identified $x$ correctly by $\beta$, but the marginal effect of $x$ is $\frac{dy}{dx}=\beta+2x$, so we haven't identified the effect of $x$. So why don't we need the error term and the regressors to be independent, instead of just uncorrelated? This question is really a precursor for why in GMM, specifically IV, we need our instruments to only be uncorrelated with the error term, and not independent of it, as without that, we don't identify the complete effect of $x$. It seems like especially for non-linear regressions, we would need independence to identify the effect of our independent variable.

  • $\begingroup$ There seems to be a great deal of overloading of concepts and notation in this question. For instance, the relevant sense of expectation and covariance is that of a sample statistic--but then there is no basis to conclude the expectations of $x^3$ and $x$ are zero because these are "odd functions." Integrability of these functions is irrelevant, making one wonder what you have in mind. If you could clarify the meanings of your notation and lay out your assumptions explicitly it would help focus this question--and maybe even lead you to a good answer yourself. $\endgroup$
    – whuber
    Dec 18, 2018 at 21:25
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    $\begingroup$ Sorry, I definitely started to ramble a bit in the post. As I was writing it, I realized that I wasn't being rigorous at all, and people were probably going to be annoyed by that. I'll clean it up a bit. $\endgroup$ Dec 18, 2018 at 21:37
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    $\begingroup$ related: stats.stackexchange.com/questions/190703/non-linear-endogeneity/… $\endgroup$ Dec 19, 2018 at 8:29


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