identification of simultaneous equation model 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!
 A: 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 $\mathbb{E}\left(xv^{2}\right)\neq0$.
