How to show that GLS estimator is consistent in regression model?

Question

$$\mathbf Y=\mathbf X\beta + \mathbf U$$ $$\mathbf E[\mathbf U\mathbf U']=\Omega $$

$$\hat \beta - \beta=(\mathbf X'\Omega^{-1}\mathbf X)^{-1}\mathbf X'\Omega^{-1}\mathbf U$$

let $\mathbf X^{*}=\Omega^{-0.5}\mathbf X$, and $\mathbf U^{*}=\Omega^{-0.5}\mathbf U$. it is obviously that $\mathbf u^{*}_i$,which is the i-th element of $\mathbf U^{*}$, is iid follows the standard normal distribution, if we assume that $\mathbf U$ follows normal distribution. But $x_i^{*'}$, the i-th row of $\mathbf X^{*}$ is not necessarily independent, for it is the linear combination of all rows in $\mathbf X$ (which is assumed iid for they stand for observations from the same distribution).

Thus the followings are not necessarily true under independent Law of Large number and Central limit theorem:

1. $plim\frac{\mathbf X'\Omega^{-1}\mathbf U}{n}=plim\frac{\mathbf X^{*'}\mathbf U^{*}}{n}= plim\frac{1}{n}\Sigma x_i^{*}u_i^{*} =0$

2. $\frac{\mathbf X'\Omega^{-1}\mathbf U}{\sqrt n}=\frac{\mathbf X^{*'}\mathbf U^{*}}{\sqrt n}= \frac{1}{\sqrt n}\Sigma x_i^{*}u_i^{*} $, converges to normal distribution

I understand that when $\mathbf \Omega $ is a diagonal matrix then $x_i^{*}$ is independent and it is not a problem, what if when $\mathbf \Omega $ is not a diagonal matrix? What kind of dependent LLN and CLT can be applied here? should we make other assumptions?

Answer 1

If the regressors are strictly exogenous to the error term (which is the initial standard assumption made), meaning, if the error vector is mean-independent of the regressor matrix

$$E(\mathbf u \mid \mathbf X) = \mathbf 0$$

then consistency follows immediately. Set for convenience $\mathbf \Omega ^{-1/2} \equiv \mathbf C$. Then

$$\text{plim}\left(n^{-1} (\mathbf C\mathbf X)'\mathbf u \right)= \text{plim}\left(n^{-1}\mathbf X'\mathbf C'\mathbf u\right)$$

This is a $k \times 1$ matrix. What would a typical element will be, say the first?

The matrix $\mathbf C'\mathbf u$ is a $n\times 1$ vector,

$$\mathbf C'\mathbf u = \left(\sum_{i=1}^nc_{1i}u_i,..., \sum_{i=1}^nc_{ni}u_i\right)'$$

Mutliplied by the first row of $\mathbf X' = (x_{11}\;...\;x_{1n})$ gives

$$\left [\mathbf X'\mathbf C'\mathbf u\right]_1 =x_{11}\cdot \sum_{i=1}^nc_{1i}u_i\;+...\;+ x_{1n}\cdot\sum_{i=1}^nc_{ni}u_i$$

Now strict exogeneity comes into play: it is stronger and so implies that

$$E(\mathbf u \mid \mathbf X) = \mathbf 0 \implies E(x_{1i}u_j)=0\;\;\forall i,j$$

So the above sum of sums, scaled by $n^{-1}$ will tend to zero asymptotically, since all products in each sum will tend to a zero expected value. This proves consistency.

Note that strict exogeneity is really needed, for the case where the matrix $\Omega$ is a function of the regressors. In such a case, we would be looking at products of the form

$$x_{1i}\cdot h(\mathbf x) \cdot u_j$$

which would still go to zero in expected value terms, because mean-independence implies orthogonality between the error term and any function of the regressors.

If $\Omega$ does not depend on the regressors, then we could weaken strict exogeneity to the RHS condition, which is stronger than "contemporaneous uncorrelatedness", the latter needed for OLS.

I leave the case for asymptotic normality to anyone interested.