Is LIML consistent under heteroskedastic errors?

Question

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!

Answer 1

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!