Condition for nonlinear IV Consider the linear model $$y=x\beta+\epsilon.$$ A necessary condition for $z$ to be a valid instrument for $x$ is that $E[x'z]\neq0$, right? What if we have a nonlinear regression of the form $y=g(x,\beta)+\epsilon$, should we then have $E[(\dfrac{\partial g }{\partial \beta})'z]\neq0$ or still $E[x'z]\neq0$. I have read somewhere the former is true, but I am not sure. Could someone please help me out? Thanks in advance!
 A: If $E(\epsilon|z)=0$ one can identify the function $g$ through the following estimating equation: $$E(y-g(x,\beta)|z)=0.$$ Newey and Powell (2003) explain why in this case the identifying assumption is: for all $\delta(x,\beta)$ with finite expectation, $$E(\delta(x,\beta)|z)=0\Longrightarrow \delta(x,\beta)=0.$$ The intuition behind this result is that $(y,x,z)$ is observable, and so relevant joint and marginal distribution are identified. If the estimating equation has a solution, we have identified $g$ uniquely if the identifying assumption holds: Suppose there are two functions solving the estimating equation, say $\delta_0$ and $\delta_1$. Then since both satisfy the estimating equation, $E(\delta_0(x,\beta)-\delta_1(x,\beta)|z)=0$. But this implies that $\delta_0(x,\beta)=\delta_1(x,\beta)$; since $\delta_0$ and $\delta_1$ were arbitrary, the identifying assumption ensures that any solution to the estimating equation is unique. 
For the 1-variable case $g(x,\beta)\equiv x\beta$, this translates into the condition that if $E(x|z)\beta=0$, then $x\beta=0$. 
The condition you refer to is intuitively plausible if consider a linear approximation by a local Taylor expansion; but I believe that the condition is not sufficient.


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*Newey, W. K., & Powell, J. L. (2003). Instrumental variable estimation of nonparametric models. Econometrica, 71(5), 1565-1578.

