Q:

Is LIML consistent under heteroskedastic errors?

Please let the answer be yes.
Suppose we have a model
\begin{eqnarray}
y= X \beta + \epsilon \\
X = Z \Pi + V
\end{eqnarray}
and we compute the LIML estimator
\begin{eqnarray}
\hat{\beta}_{LIML} = (X'(I - \lambda M_Z) X)^{-1} (X'(I-\lambda M_Z) y),
\end{eqnarray}
do you guys know whether this one is consistent even under heteroskedastic errors? That would be sweet!

A:

Usually consistency does not depend on the variance of the error term! So I would guess it is still consistent. Here is a half proof:
\begin{eqnarray}
\hat{\beta}_{LIML} &=& (X'(I - \lambda M_Z) X)^{-1} (X'(I-\lambda M_Z) y)\\
&=& (X'(I - \lambda M_Z) X)^{-1} (X'(I-\lambda M_Z)) (X\beta +u)\\
&=& \beta+(X'(I - \lambda M_Z) X)^{-1} (X'(I-\lambda M_Z)) u
\end{eqnarray}
Now assuming the probability limit of $\lambda$ is finite (to verify!), that $(X'(I-\lambda M_Z)X)$ converges to some Q and using an assumption similar to E[u]=0, you would get something like:
$$(X'(I - \lambda M_Z) X)^{-1} (X'(I-\lambda M_Z)) u\to (QX)^ {-1}Qu\to 0$$
and hence: $\hat{\beta}_{LIML} \to\beta +0\to\beta$
The proof might not be 100% rigorous, but is shows that heteroskedasticity does not come into play at any point regarding consistency!

