0
$\begingroup$

$$y_i = \beta_1 + \beta_2 x_{2i} + \beta_3 x_{3i} + \beta_4 x_{4i} + e_i$$

So you usually i'd take this and compure for $$U_i^2 = \beta_1 + \beta_2 x_{2i} + \beta_3 x_{3i} + \beta_4 x_{4i} + \beta_2^2 x_{2i}^2 + \beta_3 x_{3i}^2 + \beta_4 x_{4i}^2 + \beta_2\beta_3+\beta_2\beta_4 +\beta_3\beta_4 + \nu_i$$

Which is the full white test, but i don't know how this statement changes things

Test the null hypothesis that the error term is homoskedastic against the alternative hypothesis that the variance of the error term is a function of $x_{2i}$, and none of the other explanatory variables.

Does it just mean i eliminate everything and only add a square of $\beta_2x_2i$

$$U_i^2 = \beta_1 + \beta_2 x_{2i} + \beta_3 x_{3i} + \beta_4 x_{4i} + \beta_2^2 +\nu_i$$

As a bonus question, how do i interpret the effect of adding both a square products of variables together in the auxiliary equation? - Oh the product just adds data on linearity

$\endgroup$
0
$\begingroup$

It means that in the projection equation for $U_i^2$ you keep the constant term and just various powers of $x_{2i}$ (but exclude all the interactions of $x_{2i}$ with other regressors and functions of purely other regressors).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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