1
$\begingroup$

I'm approaching for the first time GLS estimators.

Suppose that $\operatorname{Var}(u|x)=\sigma^2 h(x)$, where $h(x)$ is some function of the explanatory variables that determines the heteroscedasticity. Since variances must be positive, $h(x)>0$ for all possible values of the independent variables. Let's say, we don't know $h(x)$.

The book (Wooldridge, the introductory one) proposes the following way for modelling heteroscedasticity:

$$\operatorname{Var}(u|x)=\sigma^2 {\rm e}^{\delta_0 +\delta_1 x_1 + \cdots + \delta_k x_k}\tag 1$$

where $\delta_j$ are unknown parameters. Thus, $h(x)={\rm e}^{\delta_0 +\delta_1 x_1 + ... + \delta_k x_k}$

Then, the book says, under assumption $[1]$ I can write

$$u^2 =\sigma^2 ({\rm e}^{\delta_0 +\delta_1 x_1 + ... + \delta_k x_k}) \nu\tag 2$$

enter image description here

Where do equations $[2]$ and $[8.31]$ come from?

$\endgroup$

1 Answer 1

3
$\begingroup$

The assumptions of the generalized linear regression model is $\mathbb E[\mathbf u\mid \mathbf X]=\mathbf 0$ and $\mathbb E[\mathbf u\mathbf u^\top\mid\mathbf X]=\sigma^2\mathbf\Omega.$

If $\bf\Omega$ is unknown owing to its dependence on unknown parameter vector $\boldsymbol\gamma, $ then we seek a consistent estimator of $\boldsymbol\gamma$ eventually to get $\boldsymbol\Omega\left(\hat{\boldsymbol\gamma}\right). $

This is the essence of Feasible GLS.

Consider $$y_t=\mathbf X^\top\boldsymbol\beta+u_t, ~~~~\mathbb Eu^2=\exp(\mathbf Z_t^\top\boldsymbol\gamma),$$ where $\mathbf Z_t$ can be a function of $\mathbf X_t$ but more importantly it's based on all the exogenous variables of the information set on which conditioning is being done.

For consistent estimator of $\boldsymbol\gamma, $ we would calculate the OLS residuals $\hat{u}_t$ for $\hat{\boldsymbol\beta}$ and then we would run the auxiliary linear regression $$\ln{\hat{u}^2_t}=\mathbf Z_t^\top\boldsymbol\gamma+v_t.$$

Wooldridge is also following the same line of attack: take expectation of both sides of $[2]$ conditional on $\mathbf x$ and you would reach $(8.30) $ as $\mathbb E[\nu\mid\mathbf x]=1.$ And if $\nu$ is independent of $\mathbf x, $ the regression of $\ln u^2$ on $x_i$ follows suit from $[2].$


Reference:

$\rm [I]$ Econometric Theory and Methods, Russell Davidson, James G. MacKinnon, Oxford University Press, $2021, $ sec. $7.4.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.