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Consider the following SEM can be identified: $$ y_i = x_i \alpha + z_i \beta + u_i\\ z_i = x_i \delta + v_i\\ $$ where we have $$ E[u_i] = E[v_i] = E[u_i x_i] = E[v_i x_i] = 0\\ cov(x_i, u_i v_i) = 0\\ cov(x_i, v_i^2) \neq 0\\ E[v_i u_i] \neq 0 $$ Here $x_i$, $y_i$ and $z_i$ are observable. I am wondering is there a way to identify $(\alpha, \beta)'$? To do so we need at least two moment conditions, and one is obviously $E[x_i (y_i - x_i \alpha - z_i \beta)] = 0$. But can we find more moment conditions? Thanks all!

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For notational simplicity, I drop the subindex $i$ in the sequel. Observe that $\mathbb{E}\left(vx\right)=0$ implies that $$ \mathbb{E}\left\{ \left(z-x\delta\right)x\right\} =0. $$ Hence, we have $\delta=\mathbb{E}\left(zx\right)/\mathbb{E}\left(x^{2}\right)$ is identifiable. Then $v=z-x\delta$ is identifiable. In addition to $\mathbb{E}\left(ux\right)=0$, we now can use $\mathrm{cov}\left(x,uv\right)=0$ to identify $\left(\alpha,\beta\right)'$. For simplicity, I assume $\mathbb{E}\left(x\right)=0$ which together with $\mathrm{cov}\left(x,uv\right)=0$ implies $$ \mathbb{E}\left(xuv\right)=0. $$ Otherwise, we need to use $$ \mathbb{E}\left(xuv\right)-\mathbb{E}\left(x\right)\mathbb{E}\left(uv\right)=0. $$ This will create cumbersome notations, though the ideas won't change.

Define $w=xv$, which is observable given the identification of $v$. We have $\mathbb{E}\left(uw\right)=0$ and then $$ \mathbb{E}\left\{ \left(y-x\alpha-z\beta\right)w\right\} =\mathbb{E}\left\{ \left(yw\right)-\left(xw\right)\alpha-\left(zw\right)\beta\right\} =0. $$ The above display and the condition mentioned in your question $$ \mathbb{E}\left\{ \left(yx\right)-x^{2}\alpha-\left(zx\right)\beta\right\} =0 $$ give rise to the following equation $$ \begin{bmatrix}\mathbb{E}\left(x^{2}\right) & \mathbb{E}\left(zx\right)\\ \mathbb{E}\left(xw\right) & \mathbb{E}\left(zw\right) \end{bmatrix}\begin{bmatrix}\alpha\\ \beta \end{bmatrix}=\begin{bmatrix}\mathbb{E}\left(yx\right)\\ \mathbb{E}\left(yw\right) \end{bmatrix}. $$ So $\left(\alpha,\beta\right)'$ is identifiable if the matrix $$ A\equiv\begin{bmatrix}\mathbb{E}\left(x^{2}\right) & \mathbb{E}\left(zx\right)\\ \mathbb{E}\left(xw\right) & \mathbb{E}\left(zw\right) \end{bmatrix} $$ is nonsingular (which is testable from data).

We indeed can show that this matrix cannot be singular. Plugging $w=xv$ into the above matrix and using the assumption $\mathbb{E}\left(uw\right)=0$, we have $$ A=\begin{bmatrix}\mathbb{E}\left(x^{2}\right) & \mathbb{E}\left(x^{2}\right)\delta\\ \mathbb{E}\left(x^{2}v\right) & \mathbb{E}\left(x^{2}v\right)\delta+\mathbb{E}\left(xv^{2}\right) \end{bmatrix}. $$ Note that $\mathbb{E}\left(xv^{2}\right)=\mathrm{cov}\left(x,v^{2}\right)\neq0$. The rows (or columns) of $A$ cannot be linearly dependent. Take columns for example. Suppose the two columns are dependent. If $\mathbb{E}\left(x^{2}v\right)\neq0$, then we must have $$ \frac{\mathbb{E}\left(x^{2}\right)\delta}{\mathbb{E}\left(x^{2}\right)}=\frac{\mathbb{E}\left(x^{2}v\right)\delta+\mathbb{E}\left(xv^{2}\right)}{\mathbb{E}\left(x^{2}v\right)}, $$ or $\mathbb{E}\left(xv^{2}\right)/\mathbb{E}\left(x^{2}v\right)=0$, which contradicts to $\mathbb{E}\left(xv^{2}\right)\neq0$. If $\mathbb{E}\left(x^{2}v\right)=0$, $A$ becomes $$ A=\begin{bmatrix}\mathbb{E}\left(x^{2}\right) & \mathbb{E}\left(x^{2}\right)\delta\\ 0 & \mathbb{E}\left(xv^{2}\right) \end{bmatrix}. $$ And again $A$ cannot be singular for $\mathbb{E}\left(xv^{2}\right)\neq0$.

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